Physics (Holt McDougal Physics) [1 ed.] 9780547586694, 0547586698

Find the Answers to Any Textbook! 📚
Find the Answers to Any Textbook! 📚

3,577 241 210MB

English Pages [926] Year 2012

This manual describes the laboratory work in molecular physics at the rate of physical workshop. The manual is intended

599 59 3MB Read more

Пособие составлено в соответствии с требованиями СКФУ ВО, в нем приведен краткий обзор теоретического материала, необход

775 63 45MB Read more

This set of notes covers a 3 quarter advanced undergraduate quantum physics course at UCSD, from the beginning of Quantu

159 14 9MB Read more

AUTHORS Raymond A. Serwav, Ph.D. Professor Emeritus James Madison University
On the cover: A soap bubble sprays droplets as it bursts. Cover Photo Credits: Bubble ©Don Farrall/Photodisc/Getty Images; luger ©Rolf Kosecki/ Corbis; laser beam ©Hank Morgan/UMass Amherst/Photo Researchers, Inc.; crash test dummies ©Corbis Wire/Corbis; carnival ride ©Corbis; cyclists ©David Madison/Corbis; plasma ball ©Brand X Pictures/Getty Images Copyright © 2012 by Houghton Mifflin Harcourt Publishing Company All rights reserved. No part of this work may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopying or recording, or by any information storage and retrieval system, without the prior written permission of the copyright owner unless such copying is expressly permitted by federal copyright law. Requests for permission to make copies of any part of the work should be addressed to Houghton Mifflin Harcourt Publishing Company, Attn: Contracts, Copyrights, and Licensing, 9400 South Park Center Loop, Orlando, Florida 32819. Printed in the U.S.A. ISBN 978-0-547-58669-4 1 2 3 4 5 6 7 8 9 10 XXX 20 19 18 17 16 15 14 13 12 11 4500000000
ABCDEFG
If you have received these materials as examination copies free of charge, Houghton Mifflin Harcourt Publishing Company retains title to the materials and they may not be resold. Resale of examination copies is strictly prohibited. Possession of this publication in print format does not entitle users to convert this publication, or any portion of it, into electronic format.
ii
Jerry S. Faughn, Ph.D. Professor Emeritus Eastern Kentuc ky University
ACKNOWLEDGMENTS Contributing Writers Robert W. Avakian Instructor Trinity School Midland , Texas
David Bethel Science Writer San Lorenzo, New Mexico
David Bradford Science Writer Austin, Texas
Robert Davisson Science Writer Delaware, Ohio
John Jewett Jr., Ph.D. Professor of Physics California State Polytechnic University Pomona, California
Jim Metzner Seth Madej Pulse of the Planet radioseries Jim Metzner Productions, Inc. Yorktown Heights, New York
John M. Stokes Science Writer Socorro, New Mexico
Salvatore Tocci Science Writer East Hampton, New York
Lab Reviewers Christopher Barnett Richard Decoster Elizabeth Ramsayer Joseph Serpico Niles West High School Niles, Illinois
Marv L. Brake, Ph.D. Physics Teacher Mercy High School Farmington Hills, Mic higan
Gregory Puskar Laboratory Manager Physics Department West Virginia University Morgantown.West Virginia
Richard Sorensen Vernier Software & Technology Beaverton, Oregon
Martin Taylor
David s. Hall, Ph.D.
Sargent-WelchNWR
Assistant Professor of Physics
Buffalo Grove, Illinois
Academic Reviewers Marv L. Brake, Ph.D. Physics Teacher Mercy High School Farmington Hills, Michigan
James C. Brown, Jr., Ph.D. Adjunct Assistant Professor of Physics Austin Community College Austin, Texas
Amherst College Amherst, Massachusetts
Roy W. Hann, Jr., Ph.D. Professor of Civil Engineering Texas A&M University College Station, Texas
Sally Hicks, Ph.D. Professor Department of Physics University of Dallas Irving, Texas
Anil RChourasia, Ph.D.
Robert C. Hudson
Associate Professor
Associate Professor Emeritus
Department of Physics Texas A&M University-Commerce Commerce, Texas
Physics Department Roanoke College Salem , Virginia
David s. Coco, Ph.D.
William Ingham, Ph.D.
Senior Research Physicist
Professor of Physics
Applied Researc h Laboratories The University of Texas at Austin Austin, Texas
Thomas Joseph Connolly, Ph.D. Assistant Professor Department of Mechanical Engineering and Biomechanics The University of Texas at San Antonio San Antonio , Texas
Brad de Young Professor Department of Physics and Physical Oceanography Memorial University St. John’s, Newfound land, Canada
Bill Deutschmann, Ph.D. President Oregon Laser Consultants Klamath Falls, Oregon
Arthur A. Few Professor of Space Physics and Environmental Science Rice University Houston , Texas
Scott Fricke, Ph.D. Schlumberger Oilfield Services Sugarland, Texas
Simonetta Fritelli Associate Professor of Physics Duquesne University Pittsburgh, Pennsylvania
James Madison University Harrisonburg, Virginia
Karen B. Kwitter, Ph.D. Professor ofAstronomy Williams College Williamstown, Massachusetts
Phillip LaRoe Professor of Physics Helena College of Technology Helena, Montana
Joseph A. McClure, Ph.D. Associate Professor Emeritus Department of Physics Georgetown University Washington, D.C.
Ralph McGrew Associate Professor Engineering Science Department Broome Community College Binghamton, New York
Clement J. Moses, Ph.D. Associate Professor of Physics Utica College Utica, New York
Alvin M. Saperstein, Ph.D. Professor of Physics; Fellow of Center for Peace and Conflict Studies Department of Physics and Astronomy Wayne State University Detroit, Michigan
Acknowledgments
iii
ACKNOWLEDGMENTS, continued Donald E. Simanek, Ph.D. Emeritus Professor of Physics Lock Haven University Lock Haven, Pennsylvania
H. Michael Sommermann, Ph.D. Professor of Physics Westmont College Santa Barbara, California
Jack B. Swift, Ph.D. Professor Department of Physics The University of Texas at Austin Austin, Texas
Thomas H. Troland, Ph.D. Physics Department University of Kentucky Lexington, Kentucky
Marv L. While Coastal Ecology Institute Louisiana State University Baton Rouge, Louisiana
Jerome Williams, M.S. Professor Emeritus Oceanography Department U.S. Naval Academy Annapolis, Maryland
Jack Cooper Ennis High School Ennis, Texas
Norman A. Mankins
Chairman of Science Department
Science Curriculum Specialist
Butler Senior High School Butler, Pennsylvania
Diego Enciso Troy, Michigan
Ron Esman Plano Senior High School Plano, Texas
Bruce Esser Marian High School Omaha, Nebraska
Curtis Goehring Palm Springs High School Palm Springs, California
Herbert H. Gottlieb Science Education Department City College of New York New York City, New York
David J. Hamilton, Ed.D. Benjamin Franklin High School Portland, Oregon
J. Philip Holden, Ph.D.
Exxon Exploration Company Houston, Texas
Physics Education Consultant
Chairperson of Science Department Fenton High School Bensenville, Illinois
John Ahlquist, M.S. Anoka High School Anoka, Minnesota
Maurice Belanger Science Department Head Nashua High School Nashua, New Hampshire
Larry G. Brown Morgan Park Academy Chicago, Illinois
William K. Conway, Ph.D. Lake Forest High School Lake Forest, Illinois
Michigan Dept. of Education Lansing, Michigan
Joseph Hutchinson W ic hita High School East W ic hita, Kansas
Douglas C. Jenkins Chairman, Science Department Warren Central High School Bowling Green, Kentucky
David S. Jones Miami Sunset Senior High School Miami, Florida
Roger Kassebaum Millard North High School Omaha, Nebraska
Mervin W. Koehlinger, M.S. Concordia Lutheran High School Fort Wayne, Indiana
Phillip LaRoe Central Community College Grand Island, Nebraska
iv
Ac knowledgments
Westwood High School Round Rock, Texas
William D. Ellis
Carol J. Zimmerman, Ph.D.
Teacher Reviewers John Adamowski
William Lash
Canton City Schools Canton, Ohio
John McGehee Palos Verdes Peninsula High School Rolling Hills Estates, California
Debra Schell Austintown Fitch High School Austintown, Ohio
Edward Schweber Solomon Schechter Day School West Orange, New Jersey
Larry Stookey, P.E. Science Antigo High School Antigo, Wisconsin
Joseph A. Taylor Middletown Area High School Middletown, Pennsylvania
Leonard L. Thompson North Allegheny Senior High School Wexford , Pennsylvania
Keith C. Tipton Lubbock, Texas
John T. Vieira Science Department Head B.M.C. Durfee High School Fall River, Massachusetts
Virginia Wood Richmond High School Richmond, M ichigan
Tim Wright Stevens Point Area Senior High School, Stevens Point, Wisconsin
Marv R. Yeomans Hopewell Valley Cent ral High School Pennington, New Jersey
G. Patrick Zuber Science Curriculum Coordinator Yough Senior High School Herminie, Pennsylvania
Patricia J. Zuber Ringgold High School Monongahela, Pennsylvania
Bring physics to life through animations.
PREMIUM CONTENT
JC\
\:::,J
Interactive Demo HMDScience.com
See problem-solving techniques in action and get extra practice.
s b a L k c i u Q
r u o y n i s t p e c n o c y e k r t h g i r e r ‘ Encounte y e h T . s b a l k c i u Q h t i w m o o r class ! k o o b r u o in y
s b a L y r i u Open I n q
e k a m u o y y t i v i t c a b a l d n a Drive t h e h c r a e s e r o t t a h w t u o decisions ab . t i o d o t w ho
s b a L M E T S
g n i r e e n i g n e d n a y g o l o n h c e t e r o l p x E . s t c e j o r p n o s d n a h h g u o thr
s b a L l l i k S Core
. s e u q i n h c e t d n a s l l i k s n o s d n a h e c i t Prac
s b a L e r a Probew
y g o l o n h c e t n o i t c e l l o c a t Integrate da . s b a l r u o y into
s b a L s c i s Foren
f o s n o i t a c i l p p a l a c i t c a r p e t a g i . t s s i e s v y l In a n a e n e c s e m i r c s a h c u s , e c n e i sc
CHAPTER 1 1 2 Why It Matters
3
CHAPTER LABS ONLINE
I THE SCIENCE OF PHYSICS
2 3 Why It Matters Take It Further Physics on the Edge Careers in Physics
10 13 21
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT
26 32
The Circumference-Diameter Ratio of a Circle Metric Prefixes Physics and Measurement Graph Matching
IN ONE DIMENSION
Displacement and Velocity Acceleration Falling Objects Sky Diving Angular Kinematics Special Relativity and Time Dilation Science Writer
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT CHAPTER LABS ONLINE
CHAPTER 3 1 2 3 4 Physics on the Edge Careers in Physics
Motion Acceleration Free-Fall Acceleration Free-Fall
TWO-DIMENSIONAL MOTION AND VECTORS Introduction to Vectors Vector Operations Projectile Motion Relative Motion Special Relativity and Velocities Kinesiologist
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT CHAPTER LABS ONLINE
viii
Holt McDougal Physics
4
What Is Physics? Measurements in Experiments STEM The Mars Climate Orbiter Mission The Language of Physics
CHAPTER 21MOTION 1
2
Vector Treasure Hunt Velocity of a Projectile Projectile Motion
~ HMDScience.com
Go online for t he full complement of labs.
34 36
44
~ 0
56
;f_
60 62 66 68 69
E
76
>,
I:;, Cl
“‘ .E
: 32
Cl C:
‘E
“‘ ::c t
“‘
.c
~ HMDScience.com
Go online for t he full complement of labs.
78 80 84 93
100 104 106
107 114 !t1HMDScience.com
Go online for t he full complement of labs.
~
@
=u
CHAPTER 41FORCES AND THE LAWS OF MOTION 1 Z Why It Matters
3 4 Why It Matters
Timeline CHAPTER LABS ONLINE
CHAPTER s 1
z Why It Matters
3 4 Physics on the Edge Careers in Physics
CHAPTER LABS ONLINE
Changes in Motion Newton’s First Law Astronaut Workouts Newton’s Second and Third Laws Everyday Forces STEM Driving and Friction SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT
118 123 126 128 133 140 142 148
Physics and Its World: 1540-1690
150
Discovering Newton’s Laws Force and Acceleration Static and Kinetic Friction Air Resistance
Why It Matters
3 Careers in Physics
CHAPTER LABS ONLINE
~ HMDScience.com
Go online for the full complement of labs.
I WORK AND ENERGY Work Energy The Energy in Food Conservation of Energy Power The Equivalence of Mass and Energy Roller Coaster Designer SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT
Exploring Work and Energy Conservation of Mechanical Energy Loss of Mechanical Energy Power Programming
CHAPTER 61MOMENTUM AND 1 Z
116
COLLISIONS
Momentum and Impulse Conservation of Momentum STEM Surviving a Collision Elastic and Inelastic Collisions High School Physics Teacher SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT
Impulse and Momentum Conservation of Momentum Collisions
152 154 158 162 167 173
176 178 179
186 ~ HMDScience.com
Go online for the full complement of labs.
188 190 197 199 204 213 214 220 ~ HMDScience.com
Go online for the full complement of labs.
-c-
a
Contents
ix
CHAPTER 71CIRCULAR MOTION AND GRAVITATION 1
z Why It Matters
3 4 Take It Further Take It Further Take It Further Physics on the Edge
CHAPTER LABS ONLINE
CHAPTER a 1 Z 3 Take It Further Take It Further
Timeline CHAPTER LABS ONLINE
Circular Motion Newton’s Law of Universal Gravitation Black Holes Motion in Space Torque and Simple Machines Tangential Speed and Acceleration Rotation and Inertia Rotational Dynamics General Relativity SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT
Circular Motion Torque and Center of Mass Centripetal Acceleration Machines and Efficiency
I FLUID MECHANICS Fluids and Buoyant Force Fluid Pressure Fluids in Motion Properties of Gases Fluid Pressure SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Physics and Its World: 1690-1785
Buoyant Vehicle Buoyancy
222 224 230 233 238 244 252 254 256 258 260 266
~ HMDScience.com
Go onl ine for t he full complement of labs.
268 270 276 280 283 285 287 292 294 ~ HMDScience.com
Go onl ine for t he full complement of labs.
CHAPTER 9 1 Z Why It Matters
3 Why It Matters Careers in Physics
STEM CHAPTER LABS ONLINE
x
Holt McDougal Physics
I HEAT Temperature and Thermal Equilibrium Defining Heat Climate and Clothing Changes in Temperature and Phase STEM Earth-Coupled Heat Pumps HVAC Technician SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Engineering and Technology: Global Warming
Temperature and Internal Energy Thermal Conduction Newton’s Law of Cooling Specific Heat Capacity
296 298 305 312 313 316 320 321 326 328 ~ HMDScience.com
Go onl ine for t he f ull complement of labs.
I
CHAPTER 10 THERMODYNAMICS 1 Z Why It Matters Why It Matters
3 Why It Matters
CHAPTER LABS ONLINE
330 332 338 344 346 348 354 355 360
Relationships Between Heat and Work The First Law of Thermodynamics STEM Gasoline Engines STEM Refrigerators The Second Law of Thermodynamics STEM Deep-Sea Air Conditioning SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT
Relationship Between Heat and Work
~
HMDScience.com
Go online for the full complement of labs.
CHAPTER 11 I VIBRATIONS AND WAVES 1 Why It Matters
z 3 4 Physics on the Edge
Timeline E
a
C.
CHAPTER LABS ONLINE
~ .;
~
“‘E
Simple Harmonic Motion STEM Shock Absorbers Measuring Simple Harmonic Motion Properties of Waves Wave Interactions De Broglie Waves SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Physics and Its World: 1785-1830
Pendulums and Spring Waves Simple Harmonic Motion of a Pendulum Pendulum Periods Pendulum Trials
362 364 368 372 378 385 391 393 398 400 ~ HMDScience.com
Go online for t he full complement of labs.
0
a: :ice
–, “‘
@
,=-
e. ii-
“‘
M “
:E
0.
f”‘
C,
~
“‘ ,
” i:3 C:
@
Why It Matters Physics on the Edge Why It Matters
,=-
“”‘ kf
C,
“‘
§
f
STEM
~E
CHAPTER LABS ONLINE
~ 0
‘-‘ @
~
s
Sound Waves STEM Ultrasound Images Sound Intensity and Resonance Hearing Loss Harmonics Reverberation The Doppler Effect and the Big Bang Song of the Dunes SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Engineering and Technology: Noise Pollution
Resonance and the Nature of Sound Speed of Sound Sound Waves and Beats
402 404 406 410 417 418 425 428 430 431 436 438 ~ HMDScience.com
Go online for the full complement of labs.
Contents
xi
CHAPTER 13 1 Z 3 4
I LIGHT AND REFLECTION
440 442 447 451 465 471 478
Characteristics of Light Flat Mirrors Curved Mirrors Color and Polarization
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT CHAPTER LABS ONLINE
CHAPTER 14 1 Z Why It Matters
3 Why It Matters Careers in Physics
Light and Mirrors Brightness of Light Designing a Device to Trace Drawings Polarization of Light
!lJHMDScience.com
Go onl ine for t he full complement of labs.
I REFRACTION
480 482 488
Refraction Thin Lenses STEM Cameras Optical Phenomena STEM Fiber Optics Optometrist
498 500 502 506 507
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT CHAPTER LABS ONLINE
CHAPTER 15 1 Z 3 Why It Matters Careers in Physics
Refraction and Lenses Converging Lenses Fiber Optics
!t)HMDScience.com
Go onl ine for t he full complement of labs.
I INTERFERENCE AND DIFFRACTION Interference Diffraction Lasers STEM Digital Video Players Laser Surgeon
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT CHAPTER LABS ONLINE
514
Diffraction Double-Slit Interference
516 518 524 533 536 538 539 544
!lJHMDScience.com
Go online for t he full complement of labs.
xii
Holt McDougal Physics
I
CHAPTER 16 ELECTRIC FORCES AND FIELDS 1 Z 3 Why It Matters
546 548 554 562 569 570 576
Electric Charge Electric Force The Electric Field STEM Microwave Ovens
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Charges and Electrostatics Electrostatics Electric Force
CHAPTER LABS ONLINE
!?) HMDScience.com
Go online for the full complement of labs.
I
CHAPTER 17 ELECTRICAL ENERGY AND CURRENT 1 Z 3 Why It Matters
4 Why It Matters Physics on the Edge Physics on the Edge Careers in Physics
STEM
Electric Potential Capacitance Current and Resistance STEM Superconductors Electric Power Household Appliance Power Usage Electron Tunneling Superconductors and BCS Theory Electrician
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Engineering and Technology: Hybrid Electric Vehicles Resistors and Current Capacitors Current and Resistance Electrical Energy
CHAPTER LABS ONLINE
I
Why It Matters Why It Matters
Z 3 Why It Matters Careers in Physics
Schematic Diagrams and Circuits CFLs and LEDs STEM Transistors and Integrated Circuits Resistors in Series or in Parallel Complex Resistor Combinations Decorative Lights and Bulbs Semiconductor Technician
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT CHAPTER LABS ONLINE
Exploring Circuit Elements Resistors in Series and in Parallel Series and Parallel Circuits
580 588 594 603 604 608 610 612 614 615 622 624
~ HMDScience.com
Go online for the full complement of labs.
CHAPTER 18 CIRCUITS AND CIRCUIT ELEMENTS 1
578
626 628 631 634 635 645 650 652 653 660
~ HMDScience.com Go online for the full complement of labs.
Contents
xiii
CHAPTER 19 1 Why It Matters 2 3 Why It Matters
STEM CHAPTER LABS ONLINE
CHAPTER 20 1 Why It Matters 2 Why It Matters 3
4 Why It Matters
Timeline CHAPTER LABS ONLINE
CHAPTER 21 1 Why It Matters 2 3 Physics on the Edge
Timeline CHAPTER LABS ONLINE
I MAGNETISM
662
Magnets and Magnetic Fields STEM Magnetic Resonance Imaging Magnetism from Electricity Magnetic Force Auroras SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Engineering and Technology: Can Cell Phones Cause Cancer?
Magnetism Magnetic Field of a Conducting Wire Magnetic Field Strength Magnetism from Electricity
~ HMDScience.com
Go online for the full complement of labs.
I ELECTROMAGNETIC INDUCTION
690
Electricity from Magnetism STEM Electric Guitar Pickups Generators, Motors, and Mutual Inductance STEM Avoiding Electrocution AC Circuits and Transformers Electromagnetic Waves Radio and 1V Broadcasts SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Physics and Its World: 1830-1890
Electricity and Magnetism Electromagnetic Induction Motors
I ATOMIC PHYSICS Quantization of Energy STEM Solar Cells Models of the Atom Quantum Mechanics Semiconductor Doping SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Physics and Its World: 1890-1950
The Photoelectric Effect
664 669 670 673 674 680 686 688
692 699 700 706 707 715 718 722 728 730
!lJ HMDScience.com
Go online for the full complement of labs.
732 734 743 744 753 760 762 766 768 !lJ HMDScience.com
Go online for the full complement of labs.
xiv
Holt McDougal Physics
CHAPTER 22 j SUBATOMIC PHYSICS 1 Z 3 4 Physics on the Edge Careers in Physics
STEM Timeline CHAPTER LABS ONLINE
770 772 779
The Nucleus Nuclear Decay Nuclear Reactions Particle Physics Antimatter Radiologist
789 793 800 802 803 808 810 812
SUMMARY AND REVIEW STANDARDS-BASED ASSESSMENT Engineering and Technology: Nuclear Waste Physics and Its World: 1950-Present Half-Ute
~ HMDScience.com
Go online for the f ull complement of labs.
REFERENCE APPENDIX A
MATHEMATICAL REVIEW
R2
APPENDIX B
THE SCIENTIFIC PROCESS
R17
APPENDIX C
SYMBOLS
R20
APPENDIX D
EQUATIONS
R26
APPENDIX E
SI UNITS
R38
APPENDIX F
REFERENCE TABLES
R40
APPENDIX G
PERIODIC TABLE OF THE ELEMENTS
R44
APPENDIX H
ABBREVIATED TABLE OF ISOTOPES AND ATOMIC MASSES
R46
APPENDIX I
ADDITIONAL PROBLEMS
R52
SELECTED ANSWERS
R69
GLOSSARY
R79
INDEX
R83
Contents
xv
FEATURES The Mars Climate Orbiter Mission (STEM) Sky Diving Astronaut Workouts Driving and Friction (STEM) The Energy in Food Surviving a Collision (STEM) Black Holes Climate and Clothing Earth-Coupled Heat Pumps (STEM) Gasoline Engines (STEM) Refrigerators (STEM) Deep-Sea Air Conditioning (STEM) Shock Absorbers (STEM) Ultrasound Images (STEM) Hearing Loss Reverberation Song of the Dunes Cameras (STEM) Fiber Optics (STEM) Digital Video Players (STEM) Microwave Ovens (STEM) Superconductors (STEM) Household Appliance Power Usage CFLs and LEDs Transistors and Integrated Circuits (STEM) Decorative Lights and Bulbs Magnetic Resonance Imaging (STEM) Auroras Electric Guitar Pickups (STEM) Avoiding Electrocution (STEM) ~ and TV Broadcasts r Cells (STEM)
Science Writer Kinesiologist Roller Coaster Designer High School Physics Teacher HVAC Technician Optometrist Laser Surgeon Electrician Semiconductor Technician Radiologist
xvi
Holt McDougal Physics
13 60
126 140 162 199 233 312 316 344 346 354 368 406 417 425 430 498 502 536 569 603 608 631 634 650 669 674 699 706 718 743
68 106 178 213 320 506 538 614 652 802
Special Relativity and Time Dilation Special Relativity and Velocities The Equivalence of Mass and Energy General Relativity De Broglie Waves The Doppler Effect and the Big Bang Electron Tunneling Superconductors and BCS Theory Semiconductor Doping Antimatter
66 104 176 258 391 428 610 612 760 800
Angular Kinematics Tangential Speed and Acceleration Rotation and Inertia Rotational Dynamics Properties of Gases Fluid Pressure
62 252 254 256 283 285
ST.E.M Global Warming Noise Pollution Hybrid Electric Vehicles Can Cell Phones Cause Cancer? Nuclear Waste
328 438 624 688 810
Physics and Its World: 1540-1690 Physics and Its World: 1690-1785 Physics and Its World: 1785-1830 Physics and Its World: 1830-1890 Physics and Its World: 1890-1950 Physics and Its World: 1950-Present
150 294 400 730 768 812
SAFETY SYMBOLS Remember that the safety symbols shown here apply to a specific activity, but the numbered rules on the following pages apply to all laboratory work.
~ EYE PROTECTION Wear safety goggles when working around chemicals, acids, bases, flames, or heating devices. Contents under pressure may become projectiles and cause serious injury. o Never look directly at the sun through any optical device or use direct sunlight to illuminate a microscope. o
~ CLOTHING PROTECTION o Secure loose clothing and remove dangling jewelry. Do not wear open-toed shoes or sandals in the lab. o Wear an apron or lab coat to protect your clothing when you are working with chemicals.
~ CHEMICAL SAFETY o
Always wear appropriate protective equipment. Always wear eye goggles, gloves, and a lab apron or lab coat when you are working with any chemical or chemical solution.
o Never taste, touch, or smell chemicals unless your
instructor directs you to do so. o

Do not allow radioactive materials to come into contact with your skin, hair, clothing, or personal belongings. Although the materials used in this lab are not hazardous when used properly, radioactive materials can cause serious illness and may have permanent effects.
ELECTRICAL SAFETY Do not place electrical cords in walking areas or let cords hang over a table edge in a way that could cause equipment to fall if the cord is accidentally pulled. o Do not use equipment that has frayed electrical cords or loose plugs. o
o Be sure that equipment is in the “off” position
~ HEATING SAFETY o
o Whenever possible, use an electric hot plate
instead of an open flame as a heat source. o
When heating materials in a test tube, always angle the test tube away from yourself and others.
o
Glass containers used for heating should be made of heat-resistant glass.
. . .SHARP OBJECT SAFETY
Y o
Use knives and other sharp instruments with extreme care.
~ HAND SAFETY o Perform this experiment in a clear area. Attach masses securely. Falling, dropped, or swinging objects can cause serious injury. o
Use a hot mitt to handle resistors, light sources, and other equipment that may be hot. Allow all equipment to cool before storing it.
o
To avoid burns, wear heat-resistant gloves whenever instructed to do so.
o
Always wear protective gloves when working with an open flame, chemicals, solutions, or wild or unknown plants.
o If you do not know whether an object is hot, do not touch it. o Use tongs when heating test tubes. Never hold a test tube in your hand to heat the test tube.
♦ GLASSWARE SAFETY o Check the condition of glassware before and
after using it. Inform your teacher of any broken, chipped, or cracked glassware, because it should not be used. o
before you plug it in. o Never use an electrical appliance around water or
with wet hands or c lothing. o Be sure to turn off and unplug electrical equipment
when you are fini shed using it. o Never close a circuit until it has been approved by your teacher. Never rewire or adjust any element of a c losed c ircuit. o If the pointer on any kind of meter moves off scale,
open the c irc uit immediately by opening the switch. o Do not work with any batteries, electrical devices, or magnets other than those provided by your teacher.
Avoid wearing hair spray or hair gel on lab days.
Do not pick up broken glass with your bare hands. Place broken glass in a specially designated disposal container.
~ WASTE DISPOSAL o
Clean and decontaminate all work surfaces and personal protective equipment as directed by your instructor.
o
Dispose of all broken glass, contaminated sharp objects, and other contaminated materials (biological and chemical) in special containers as directed by your instructor.
Safety in the Physics Laboratory
xvii
SAFETY IN THE PHYSICS LABORATORY Systematic, careful lab work is an essential part of any science program because lab work is the key to progress in science. In t his class, you w ill practice some of the same fu ndamental laboratory procedures and techniques that experimental physicists use to pursue new knowledge. The equipment and apparatus you w ill use involve various safety hazards, just as they do for working physicists. You must be aware of these hazards. Your teacher w ill guide you in properly using the equipment and carrying out the experiments, but you must also take responsibility for your part in this process. Wit h the active involvement of you and your teacher, t hese risks can be minimized so t hat working in the physics laboratory can be a safe, enjoyable process of discovery.
THESE SAFETY RULES ALWAYS APPLY IN THE LAB: 1. Always wear a lab apron and safety goggles. Wear these safety devices whenever you are in the lab, not just when you are working on an experiment.
2. No contact lenses in the lab. Contact lenses should not be worn during any investigations using chemicals (even if you are wearing goggles). In the event of an accident, chemicals can get behind contact lenses and cause serious damage before the lenses can be removed . If your doctor requires that you wear contact lenses instead of glasses, you should wear eye-cup safety goggles in the lab. Ask your doctor or your teacher how to use this very important and special eye protection.
3. Personal apparel should be appropriate for laboratory work. On lab days avoid wearing long necklaces, dangling bracelets, bulky jewelry, and bulky or loose-fitting clothing. Loose, flopping, or dangling items may get caught in moving parts, accidentally contact electrical connections, or interfere with the investigation in some potentially hazardous manner. In addition, chemical fumes may react with some jewelry, such as pearl jewelry, and ruin them. Cotton clothing is preferable to clothes made of wool, nylon, or polyester.
xviii
Holt McDougal Physics
Tie back long hair. Wear shoes that will protect your feet from chemical spills and falling objects. Do not wear open-toed shoes or sandals or shoes with woven leather straps.
4. NEVER work alone in the laboratory. Work in the lab only while under the supervision of your teacher. Do not leave equipment unattended while it is in operation .
5. Only books and notebooks needed for the experiment should be in the lab. Only the lab notebook and perhaps the textbook should be in the lab. Keep other books, backpacks, purses, and similar items in your desk, locker, or designated storage area.
6. Read the entire experiment before entering the lab. Your teacher will review any applicable safety precautions before the lab. If you are not sure of something, ask your teacher.
7. Heed all safety symbols and cautions written in the experimental investigations and handouts, posted in the room, and given verbally by your teacher. They are provided for a reason: YOUR SAFETY.
8. Know the proper fire-drill procedures and the locations of fire exits and emergency equipment. Make sure you know the procedures to follow in case of a fi re or emergency.
9. If your clothing catches on fire, do not run; WALK to the safety shower, stand under it, and turn it on. Call to your teacher while you do this.
10. Report all accidents to the teacher immediately, no matter how minor. In addition, if you get a headache, feel sick to your stomach, or feel dizzy, tell your teacher immediately.
11. Report all spills to your teacher immediately. Call your teacher rather than trying to clean a spill yourself. Your teacher will tell you if it is safe for you to c lean up the spill; if not, your teacher w ill know how the spill should be cleaned up safely.
lZ. Student-designed inquiry investigations, such as Open Inquiry labs, must be approved by the teacher before being attempted by the student. 13. DO NOT perform unauthorized experiments or use equipment and apparatus in a manner for which they are not intended. Use only materials and equipment listed in the activity equipment list or authorized by your teacher. Steps in a procedure should only be performed as described in the book or lab manual or as approved by your teacher.
14. Stay alert in the lab, and proceed with caution. Be aware of others near you or your equipment when you are about to do something in the lab. If you are not sure of how to proceed, ask your teacher.
16. Food, beverages, chewing gum, and tobacco products are NEVER permitted in the laboratory. 17. NEVER taste chemicals. Do not touch chemicals or allow them to contact areas of bare skin. 18. Use extreme CAUTION when working with hot plates or other heating devices. Keep your head, hands, hair, and c lothing away from the flame or heating area, and t urn the devices off when they are not in use. Remember that metal surfaces connected to the heated area will become hot by conduction. Gas burners should only be lit with a spark lighter. Make sure all heating devices and gas valves are turned off before leaving the laboratory. Never leave a hot plate or other heating device unattended when it is in use. Remember that many metal, ceramic, and glass items do not always look hot when they are hot. Allow all items to cool before storing.
19. Exercise caution when working with electrical equipment. Do not use electrical equipment with frayed o r twisted wires. Be sure your hands are dry before using electrical equipment. Do not let electrical cords dangle from work stations; dangling cords can cause tripping or electrical shocks.
ZO. Keep work areas and apparatus clean and neat. Always c lean up any clutter made during the course of lab work, rearrange apparatus in an orderly manner, and report any damaged or missing items.
Zl. Always thoroughly wash your hands with soap and water at the conclusion of each investigation.
15. Fooling around in the lab is very dangerous. Laboratory equipment and apparatus are not toys; never play in the lab or use lab time or equipment for anything other than their intended purpose.
Safety in t he Physics Laboratory
xix
The runner in this photograph is participating in sports science research at the National Institute of Sport and Physical Education in France. The athlete is being filmed by a video camera. The white reflective patches enable researchers to generate a computer model from the video, similar to the diagram. Researchers use the model to analyze his technique and to help him improve his performance.
SECTION 1 Objectives ►
Identify activities and fields that involve the major areas within physics.

I

Describe the processes of the scientific method. Describe the role of models and diagrams in physics.
What Is Physics? Key Terms model system
hypothesis controlled experiment
The Topics of Physics Many people consider physics to be a difficult science that is far removed from their lives. This may be because many of the world’s most famous physicists study topics such as the structure of the universe or the incredibly small particles within an atom, often using complicated tools to observe and measure what they are studying. But everything around you can be described by using the tools of physics. The goal of physics is to use a small number of basic concepts, equations, and assumptions to describe the physical world. These physics principles can then be used to make predictions about a broad range of phenomena. For example, the same physics principles that are used to describe the interaction between two planets can be used to describe the motion of a soccer ball moving toward a goal.
The Physics of Cars Without knowledge of many of the areas of physics, making cars would be impossible.
Many physicists study the laws of nature simply to satisfy their curiosity about the world we live in. Learning the laws of physics can be rewarding just for its own sake. Also, many of the inventions, appliances, tools, and buildings we live with today are made possible by the application of physics principles. Physics discoveries often turn out to have unexpected practical applications, and advances in technology can in turn lead to new physics discoveries. Figure 1.1 indicates how the areas of physics apply to building and operating a car.
Thermodynamics Efficient engines, use of coolants Electromagnetism Battery, starter,
h\lights
l
Optics Headlights, rearview mirrors
4
Chapter 1
Vibrations and mechanical waves Shock absorbers, radio speakers
Mechanics Spinning motion of the wheels, tires that provide enough friction for traction
Physics is everywhere. We are surrounded by principles of physics in our everyday lives. In fact, most people know much more about physics than they realize. For example, when you buy a carton of ice cream at the store and put it in the freezer at home, you do so because from past experience you know enough about the laws of physics to know that the ice cream will melt if you leave it on the counter.
The Physics of Sailboats Sailboat designers rely on knowledge from many branches of physics.
People who design, build, and operate sailboats, such as the ones shown in Figure 1.2, need a working knowledge of the principles of physics. Designers figure out the best shape for the boat’s hull so that it remains stable and floating yet quick-moving and maneuverable. This design requires knowledge of the physics of fluids. Determining the most efficient shapes for the sails and how to arrange them requires an understanding of the science of motion and its causes. Balancing loads in the construction of a sailboat requires knowledge of mechanics. Some of the same physics principles can also explain how the keel keeps the boat moving in one direction even when the wind is from a slightly different direction. Any problem that deals with temperature, size, motion, position, shape, or color involves physics. Physicists categorize the topics they study in a number of different ways. Figure 1.3 shows some of the major areas of physics that will be described in this book.
€”‘ 0
Name
Subjects
Examples
Mechanics
motion and its causes, interactions between objects
falling objects, friction, weight, spinning objects
Thermodynamics
heat and temperature
melting and freezing processes, engines, refrigerators
Vibrations and wave phenomena
specific types of repetitive motions
springs, pendulums, sound
Optics
light
mirrors, lenses, color, astronomy
Electromagnetism
electricity, magnetism, and light
electrical charge, circuitry, permanent magnets, electromagnets
Relativity
particles moving at any speed, including very high speeds
particle collisions, particle accelerators, nuclear energy
Quantum mechanics
behavior of submicroscopic particles
the atom and its parts
u
.,
‘.::
i,’,
E
~
a: ~
8′
a:
@
The Science of Physics
5
The Scientific Method The Scientific Method Physics, like all other sciences, is based on the scientific method. Make observations and collect data that lead to a question.
Formulate and objectively test hypotheses by experiments.
Interpret results, and revise the hypothesis if necessary.
State conclusions in a form that can be evaluated by others.
model a pattern, plan , representation, or description designed to show the structure or workings of an object, system, or concept
Analyzing Basketball Motion This basketball game involves great complexity.
6
Chapter 1
When scientists look at the world, they see a network of rules and relationships that determine what will happen in a given situation. Everything you will study in this course was learned because someone looked out at the world and asked questions about how things work. There is no single procedure that scientists follow in their work. However, there are certain steps common to all good scientific investigations. These steps, called the scientific method, are summarized in Figure 1.4. This simple chart is easy to understand; but, in reality, most scientific work is not so easily separated. Sometimes, exploratory experiments are performed as a part of the first step in order to generate observations that can lead to a focused question. A revised hypothesis may require more experiments.
Physics uses models that describe phenomena. Although the physical world is very complex, physicists often use models to explain the most fundamental features of various phenomena. Physics has developed powerful models that have been very successful in describing nature. Many of the models currently used in physics are mathematical models. Simple models are usually developed first. It is often easier to study and model parts of a system or phenomenon one at a time. These simple models can then be synthesized into morecomprehensive models. When developing a model, physicists must decide which parts of the phenomenon are relevant and which parts can be disregarded. For example, let’s say you wish to study the motion of the ball shown in Figure 1.5. Many observations can be made about the situation,
including the ball’s surroundings, size, spin, weight, color, time in the air, speed, and sound when hitting the ground. The first step toward simplifying this complicated situation is to decide what to study, that is, to define the system. Typically, a single object and the items that immediately affect it are the focus of attention. For instance, suppose you decide to study the ball’s motion in the air (before it potentially reaches any of the other players), as shown in Figure 1.6. To study this situation, you can eliminate everything except information that affects the ball’s motion.
system a set of particles or interacting components considered to be a distinct physical entity for the purpose of study
Motion of a Basketball To analyze the basketball’s motion, isolate the objects that will affect its motion .

You can disregard characteristics of the ball that have little or no effect on its motion, such as the ball’s color. In some studies of motion, even the ball’s spin and size are disregarded, and the change in the position of the ball will be the only quantity investigated. In effect, the physicist studies the motion of a ball by first creating a simple model of the ball and its motion. Unlike the real ball, the model object is isolated; it has no color, spin, or size, and it makes no noise on impact. Frequently, a model can be summarized with a diagram. Another way to summarize these models is to build a computer simulation or small-scale replica of the situation. Without models to simplify matters, situations such as building a car or sailing a boat would be too complex to study. For instance, analyzing the motion of a sailboat is made easier by imagining that the push on the boat from the wind is steady and consistent. The boat is also treated as an object with a certain mass being pushed through the water. In other words, the color of the boat, the model of the boat, and the details of its shape are left out of the analysis. Furthermore, the water the boat moves through is treated as if it were a perfectly smooth -flowing liquid with no internal friction. In spite of these simplifications, the analysis can still make useful predictions of how the sailboat will move. The Science of Physics
7
Galileo’s Thought Experiment If heavier objects fel l faster than slower ones, would two bricks of different masses tied together fall slower (b) or faster (c) than the heavy brick alone (a)? Because of this contradiction, Galileo hypothesized instead that all objects fall at the same rate, as in (d).
Galileo’s Thought Experiment
-• J (a)
l (b)
I
(c)
Galileo’s Hypothesis
-1 l (d)
Models can help build hypotheses. hypothesis an explanation that is based on p rior scientific researc h or observations and that can be tested
A scientific hypothesis is a reasonable explanation for observations- one that can be tested with additional experiments. The process of simplifying and modeling a situation can h elp you determine the relevant variables and identify a hypothesis for testing. Consider the example of Galileo’s “thought experiment;’ in which he modeled the beh avior of falling objects in order to develop a hypothesis about how objects fell. At the time Galileo published his work on falling objects, in 1638, scientists believed that a heavy object would fall faster than a lighter object. Galileo imagined two objects of different masses tied together and released at the same time from the same height, such as the two bricks of different masses shown in Figure 1.7. Suppose that the heavier brick falls faster than the lighter brick when they are separate, as in (a). When tied together, the heavier brick will speed up the fall of the lighter brick somewhat, and the lighter brick will slow the fall of the heavier brick somewhat. Thus, the tied bricks should fall at a rate in between that of either brick alone, as in (b). However, the two bricks together have a greater mass than the heavier brick alone. For this reason, the tied bricks should fall faster than the heavier brick, as in (c). Galileo used this logical contradiction to refute the idea that different masses fall at different rates. He hypothesized instead that all objects fall at the same rate in the absence of air resistance, as in (d).
Models help guide experimental design. Galileo performed many experiments to test his hypothesis. To be certain he was observing differences due to weight, he kept all other variables the same: the objects he tested had the same size (but different weights) and were measured falling from the same point. The measuring devices at that time were not precise enough to m easure the motion of objects falling in air. So, Galileo used the motion of a ball rolling down a ramp as a model of the motion of a falling ball. 8
Chapter 1

The steeper the ramp, the closer the model came to representing a falling object. These ramp experiments provided data that matched the predictions Galileo made in his hypothesis. Like Galileo’s hypothesis, any hypothesis must be tested in a controlled experiment. In an experiment to test a hypothesis, you must change one variable at a time to determine what influences the phenomenon you are observing. Galileo performed a series of experiments using balls of different weights on one ramp before determining the time they took to roll down a steeper ramp.
controlled experiment an experiment that tests only one factor at a time by using a comparison of a control group with an experimental group
The best physics models can make predictions in new situations. Until the invention of the air pump, it was not possible to perform direct tests of Galileo’s model by observing objects falling in the absence of air resistance. But even though it was not completely testable, Galileo’s model was used to make reasonably accurate predictions about the motion of many objects, from raindrops to boulders (even though they all experience air resistance). Even if some experiments produce results that support a certain model, at any time another experiment may produce results that do not support the model. When this occurs, scientists repeat the experiment until they are sure that the results are not in error. If the unexpected results are confirmed, the model must be abandoned or revised. That is why the last step of the scientific method is so important. A conclusion is valid only if it can be verified by other people.
‘ .Did YOU Know? In addition to conducting experiments to test their hypotheses, scientists also research the work of other scientists. The steps of this type of research include • identifying reliable sources • searching the sources to find references • checking for opposing views • documenting sources • presenting findings to other scientists for review and discussion
SECTION 1 FORMATIVE ASSESSMENT 1. Name the major areas of physics.
2. Identify the area of physics that is most relevant to each of the following situations. Explain your reasoning. a. a high school football game b. food preparation for the prom c. playing in the school band d. lightning in a thunderstorm e. wearing a pair of sunglasses outside in the sun 3. What are the activities involved in the scientific method? 4. Give two examples of ways that physicists model the physical world.
Critical Thinking 5. Identify the area of physics involved in each of the following tests of a lightweight metal alloy proposed for use in sailboat hulls: a. testing the effects of a collision on the alloy b. testing the effects of extreme heat and cold on the alloy c. testing whether the alloy can affect a magnetic compass needle The Science of Physics
9
SECTION 2 Objectives ►
List basic SI units and the quantities they describe.
I

I ► I ►
Convert measurements into scientific notation. Distinguish between accuracy and precision. Use significant figures in measurements and calculations.
Measurements in Experiments Key Terms accuracy precision significant figures
Numbers as Measurements Physicists perform experiments to test hypotheses about how changing one variable in a situation affects another variable. An accurate analysis of such experiments requires numerical measurements. Numerical measurements are different than the numbers used in a mathematics class. In mathematics, a number like 7 can stand alone and be used in equations. In science, measurements are more than just a number. For example, a measurement reported as 7 leads to several questions. What physical quantity is being measured-length, mass, time, or something else? If it is length that is being measured, what units were used for the measurement-meters, feet, inches, miles, or light-years?
Standard Kilogram The kilogram is currently the only SI unit that is defined by a material object. The platinum-iridium cylinder shown here is the primary kilogram standard for the United States.
The description of what kind of physical quantity is represented by a certain measurement is called dimension. You are probably already familiar with three basic dimensions: length, mass, and time. Many other measurements can be expressed in terms of these three dimensions. For example, physical quantities such as force, velocity, energy, volume, and acceleration can all be described as combinations oflength, mass, and time. When we learn about heat and electricity, we will need to add two other dimensions to our list, one for temperature and one for electric current. The description of how much of a physical quantity is represented by a certain numerical measurement and by the unit with which the quantity is measured. Although each dimension is unique, a dimension can be measured using different units. For example, the dimension of time can be measured in seconds, hours, or years.
SI is the standard measurement system for science. When scientists do research, they must communicate the results of their experiments with each other and agree on a system of units for their measurements. In 1960, an international committee agreed on a system of standards, such as the standard shown in Figure 2.1. They also agreed on designations for the fundamental quantities needed for measurements. This system of units is called the Systeme International d ‘Unites (SI). In SI, there are only seven base units. Each base unit describes a single dimension, such as length, mass, or time. 10
Chapter 1
Unit
Original standard
Current standard
meter (length)
10 oo6 000 distance from
equator to North Pole
the distance traveled by light in a vacuum in 3.33564095 x 1o- 9 s
mass of 0.001 cubic meters of water
the mass of a specific platinumiridium alloy cylinder
Uo )(~o ) (2~ ) =
9 192 631 770 times the period of a radio wave emitted from a cesium-133 atom
kilogram (mass) second (time)
0.000 011 574 average solar days
The base units oflength, mass, and time are the meter, kilogram, and second, respectively. In most measurements, these units will be abbreviated as m, kg, and s, respectively. These units are defined by the standards described in Figure 2.2 and are reproduced so that every meterstick, kilogram mass, and clock in the world is calibrated to give consistent results. We will use SI units throughout this book because they are almost universally accepted in science and industry. Not every observation can be described using one of these units, but the units can be combined to form derived units. Derived units are formed by combining the seven base units with multiplication or division. For example, speeds are typically expressed in units of meters per second (mis). In other cases, it may appear that a new unit that is not one of the base units is being introduced, but often these new units merely serve as shorthand ways to refer to combinations of units. For example, forces and weights are typically measured in units of newtons (N), but a newton is defined as being exactly equivalent to one kilogram multiplied by meters per second squared (1 kg•m/s2) . Derived units, such as newton s, will be explained throughout this book as they are introduced.
SI uses prefixes to accommodate extremes.
. Did YOU Know?. ___________ , I
‘ NIST-F1, an atomic clock at the National Institute of Standards and : Technology in Colorado, is one of the ‘ most accurate timing devices in the : world. NIST-F1 is so accurate that it , will not gain or lose a second in nearly ‘ : 20 million years. As a public service, : the institute broadcasts the time ‘ given by NIST-F1 through the Internet, radio stations WWV and WWVB, and ‘ satellite signals.
Units with Prefixes The mass of this mosquito can be expressed several different ways: 1 x 1o- 5 kg, 0.01 g, or 10 mg.
Physics is a science that describes a broad range of topics and requires a wide range of measurements, from very large to very small. For exam ple, distance measurements can range from the distances between stars (about 100 000 000 000 000 000 m) to the distances between atoms in a solid (0.000 000 001 m). Because these numbers can be extremely difficult to read and write, they are often expressed in powers of 10, such as 1 x 1017 m or 1 x 10- 9 m. Another approach commonly used in SI is to combine the units with prefixes that symbolize certain powers of 10, as illustrated in Figure 2.3. The Science of Physics
11
Quick LAB MATERIALS
• balance (0.01 g precision or better) • 50 sheets of loose-leaf paper
METRIC PREFIXES Record the following measurements (with appropriate units and metric prefixes): • the mass of a single sheet of paper • the mass of exactly 10 sheets of paper • the mass of exactly 50 sheets of paper Use each of these measurements to determine the mass of a single sheet of paper. How many different ways can you express each of these measurements? Use your results to estimate the mass of one ream (500 sheets) of paper. How many ways can you express this mass? Which is the most practical approach? Give reasons for your answer.
Power Prefix
Abbreviation
Power
Prefix
Abbreviation
10-18
atto-
a
101
deka-
da
10-15
femto-
103
kilo-
k
10- 12
pico-
p
106
mega-
M
10-9
nano-
n
109
giga-
G
10-6
micro-
µ (Greek letter mu)
1012
tera-
T
10-3
milli-
m
1015
peta-
p
10-2
centi-
C
1018
exa-
E
10-1
deci-
d
The most common prefixes and their symbols are shown in Figure 2.4. For example, the length of a housefly, 5 x 10- 3 m, is equivalent to 5 millimeters (mm), and the distance of a satellite 8.25 x 105 m from Earth’s surface can be expressed as 825 kilometers (km). A year, which is about 3.2 x 107 s, can also be expressed as 32 megaseconds (Ms). Converting a measurement from its prefix form is easy to do. You can build conversion factors from any equivalent relationship, including those in Figure 2.4. Just put the quantity on one side of the equation in the numerator and the quantity on the other side in the denominator, as shown below for the case of the conversion 1 mm = 1 x 10-3 m . Because these two quantities are equal, the following equations are also true: 1 mm = 1 and 10- 3 m = 1 1 mm 10-3 m Thus, any measurement multiplied by either one of these fractions will be multiplied by 1. The number and the unit will change, but the quantity described by the m easurement will stay the same. To convert measurements, use the conversion factor that will cancel with the units you are given to provide the units you need, as shown in the example below. Typically, the units to which you are converting should be placed in the numerator. It is useful to cross out units that cancel to help keep track of them. If you have arranged your terms correctly, the units you are converting from will cancel, leaving you with the unit that you want. If you use the wrong conversion, you will get units that don’t cancel. 1 Units don’t cancel: 37.2 mm x ~m 10- m Units do cancel: 37.2 .mnt x
12
Chapter 1
1
= 3.72 x
~~ = 3.72 x
4
2
10 m~
10- 2 m
S.T.E.M. The Mars Climate Orbite r Mission he Mars Climate Orbiter was a NASA spacecraft designed to take pictures of the Martian surface, generate daily weather maps, and analyze the Martian atmosphere from an orbit about 80 km (50 mi) above Mars. It was also supposed to relay signals from its companion, the Mars Polar Lander, which was scheduled to land near the edge of the southern polar cap of Mars shortly after the orbiter arrived.
lf
The orbiter was launched from Cape Canaveral, Florida, on December 11, 1998. Its thrusters were fired several times along the way to direct it along its path. The orbiter reached Mars nine and a half months later, on September 23, 1999. A signal was sent to the orbiter to fire the thrusters a final time in order to push the spacecraft into orbit around the planet. However, the orbiter did not respond to this final signal. NASA soon determined that the orbiter had passed closer to the planet than intended, as close as 60 km (36 mi). The orbiter most likely overheated because of friction in the Martian atmosphere and then passed beyond the planet into space, fatally damaged.
The $125 million Mars Orbiter mission failed because of a miscommunication about units of measurement. failures. A later Mars mission, the Exploration Rover mission, successfully placed two rovers named Spirit and Opportunity on the surface of Mars, where they collected a wide range of data. Among other things, the rovers found convincing evidence that liquid water once flowed on the surface of Mars. Thus, it is possible that Mars supported life sometime in the past.
The Mars Climate Orbiter was built by Lockheed Martin in Denver, Colorado, while the mission was run by a NASA flight control team at Jet Propulsion Laboratory in Pasadena, California. Review of the failed mission revealed that engineers at Lockheed Martin sent thrust specifications to the flight control team in English units of pounds of force, while the flight control team assumed that the thrust specifications were in newtons, the SI unit for force. Such a problem normally would be caught by others checking and double-checking specifications, but somehow the error escaped notice until it was too late. Unfortunately, communication with the Mars Polar Lander was also lost as the lander entered the Martian atmosphere on December 3, 1999. The failure of these and other space exploration missions reveals the inherent difficulty in sending complex technology into the distant, harsh, and often unknown conditions in space and on other planets. However, NASA has had many more successes than
The Spirit and Opportunity rovers have explored the surface of Mars with a variety of scientific instruments, including cameras, spectrometers, magnets, and a rock-grinding tool. 13
(a)
Choosing Units When determining area by multiplying measurements of length and width, be sure the measurements are expressed in the same units.
203 5c m
~ 1017.5 4 0 70
~ 25437.5
aboui;–

2 .54 X 104crn-rn (b)
20 3 5 rn
~ 1 o. 1 75
4 0.70
~ – 254.375 about _ 2 2 2 .54 X 10 m
-(c)
Both dimension and units must agree. Measurements of physical quantities must be expressed in units that match the dimensions of that quantity. For example, measurements oflength cannot be expressed in units of kilograms because units of kilograms describe the dimension of mass. It is very important to be certain that a measurement is expressed in units that refer to the correct dimension. One good technique for avoiding errors in physics is to check the units in an answer to be certain they are appropriate for the dimension of the physical quantity that is being sought in a problem or calculation. In addition to having the correct dimension, m easurements used in calculations should also have the same units. As an example, consider Figure 2.5(a), which shows two people measuring a room to determine the room’s area. Suppose one person measures the length in meters and the other person measures the width in centimeters. When the numbers are multiplied to find the area, they will give a difficult-to-interpret answer in units of cm•m, as shown in Figure 2.5(b). On the other hand, if both measurements are made using the same units, the calculated area is much easier to interpret because it is expressed in units of m 2 , as shown in Figure 2.5(c). Even if the measurements were made in different units, as in the example above, one unit can be easily converted to the other because centimeters and meters are both units of length. It is also necessary to convert one unit to another when working with units from two different systems, such as meters and feet. In order to avoid confusion, it is better to make the conversion to the same units before doing any more arithmetic. 14
Chapter 1
Sample Problem A A typical bacterium has a mass of about 2.0 fg. Express this measurement in terms of grams and kilograms.
0
ANALYZE
E)
SOLVE
Given:
mass= 2.0fg
Unknown:
mass = ? g mass = ? kg
Build conversion factors from the relationships given in Figure 2.4. Two possibilities are shown below.
1 X 10- 15 g 1 fg —–and—lfg 1 X 10- 15 g Only the first one will cancel the units of femtograms to give units of grams.
Then, take this answer and use a similar process to cancel the units of grams to give units of kilograms.
(2.0 X 10-15.g) ( 1 kg 1
X
103 _g-
)
=
Practice 1. A human hair is approximately 50 µm in diameter. Express this diameter
in meters.
2. If a radio wave has a period of 1 µs, what is the wave’s period in seconds? 3. A hydrogen atom has a diameter of about 10 nm.
a. Express this diameter in meters. b. Express this diameter in millimeters. c. Express this diameter in micrometers.
4. The distance between the sun and Earth is about 1.5 x 10 11 m. Express this distance with an SI prefix and in kilometers. 5. The average mass of an automobile in the United States is about 1.440 x 106 g.
Express this mass in kilograms.
The Science of Physics
15
Accuracy and Precision
accuracy a description of how close a measurement is to the correct or accepted value of the quantity measured
precision the degree of exactness of a measurement
Because theories are based on observation and experiment, careful measurements are very important in physics. But no measurement is perfect. In describing the imperfection of a measurement, one must consider both the accuracy, which describes how close the measurement is to the correct value, and the precision, which describes how exact the measurement is. Although these terms are often used interchangeably in everyday speech, they have specific meanings in a scientific discussion. A numeric measure of confidence in a measurement or result is known as uncertainty. A lower uncertainty indicates greater confidence. Uncertainties are usually expressed by using statistical methods.
Error in experiments must be minimized. Experimental work is never free of error, but it is important to minimize error in order to obtain accurate results. An error can occur, for example, if a mistake is made in reading an instrument or recording the results. One way to minimize error from human oversight or carelessness is to take repeated measurements to be certain they are consistent. If some measurements are taken using one method and some are taken
using a different method, a type of error called method error will result. Method error can be greatly reduced by standardizing the method of taking measurements. For example, when measuring a length with a meterstick, choose a line of sight directly over what is being measured, as shown in Figure 2.G(a). If you are too far to one side, you are likely to overestimate or underestimate the measurement, as shown in Figure 2.G(b) and Figure 2.G(c). Another type of error is instrument error. If a meterstick or balance is not in good working order, this will introduce error into any measurements made with the device. For this reason, it is important to be careful with lab equipment. Rough handling can damage balances. If a wooden meterstick gets wet, it can warp, making accurate measurements difficult.
Line of Sight Affects Measurements If you measure this window by keeping your line of sight directly over the measurement {a), you will find that it is 165.2 cm long. If you do not keep your eye directly above the mark, as in {b) and {c) , you may report a measurement with significant error.
16
Chapter 1
Because the ends of a meterstick can be easily damaged or worn, it is best to minimize instrument error by making measurements with a portion of the scale that is in the middle of the meterstick. Instead of measuring from the end (0 cm), try measuring from the 10 cm line.
Precision describes the limitations of the measuring instrument. Poor accuracy involves errors that can often be corrected. On the other hand, precision describes how exact a measurement can possibly be. For example, a measurement of 1.325 m is more precise than a measurement of 1.3 m. A lack of precision is typically due to limitations of the measuring instrument and is not the result of human error or lack of calibration. For example, if a meterstick is divided only into centimeters, it will be difficult to measure something only a few millimeters thick with it. In many situations, you can improve the precision of a measurement. This can be done by making a reasonable estimation of where the mark on the instrument would have been. Suppose that in a laboratory experiment you are asked to measure the length of a pencil with a meterstick marked in centimeters, as shown in Figure 2.7. The end of the pencil lies somewhere between 18 cm and 18.5 cm. The length you have actually measured is slightly more than 18 cm. You can make a reasonable estimation of how far between the two marks the end of the pencil is and add a digit to the end of the actual measurement. In this case, the end of the pencil seems to be less than halfway between the two marks, so you would report the measurement as 18.2cm.
Estimation in Measurement Even though this ruler is marked in only centimeters and half-centimeters, if you estimate, you can use it to report measurements to a precision of a millimeter.
f
I
I 20
‘1
11 l I I I I I I
Significant figures help keep track of imprecision. It is important to record the precision of your measurements so that other
people can understand and interpret your results. A common convention used in science to indicate precision is known as significant figures. The figures that are significant are the ones that are known for certain, as well as the first digit that is uncertain.
significant figures those digits in a measurement that are known with certainty plus the first digit that is uncertain
In the case of the measurement of the pencil as about 18.2 cm, the measurement has three significant figures. The significant figures of a measurement include all the digits that are actually measured (18 cm), plus one estimated digit. Note that the number of significant figures is determined by the precision of the markings on the measuring scale. The last digit is reported as a 0.2 (for the estimated 0.2 cm past the 18 cm mark). Because this digit is an estimate, the true value for the measurement is actually somewhere between 18.15 cm and 18.25 cm. When the last digit in a recorded measurement is a zero, it is difficult to tell whether the zero is there as a placeholder or as a significant digit. For example, if a length is recorded as 230 mm, it is impossible to tell whether this number has two or three significant digits. In other words, it can be difficult to know whether the measurement of 230 mm means the measurement is known to be between 225 mm and 235 mm or is known more precisely to be between 229.5 mm and 230.5 mm.
The Science of Physics
17
Precision If a mountain’s height is known with an uncertainty of 5 m, the addition of 0.20 m of rocks will not appreciably change the height.
One way to solve such problems is to report all values using scientific notation. In scientific notation, the measurement is recorded to a power of 10, and all of the figures given are significant. For example, if the length of 230 cm has two significant figures, it would be recorded in scientific notation as 2.3 x 102 cm. If it has three significant figures, it would be recorded as 2.30 x 102 cm. Scientific notation is also helpful when the zero in a recorded measurement appears in front of the measured digits. For example, a measurement such as 0.000 15 cm should be expressed in scientific notation as 1.5 x 10- 4 cm if it has two significant figures. The three zeros between the decimal point and the digit 1 are not counted as significant figures because they are present only to locate the decimal point and to indicate the order of magnitude. The rules for determining how many significant figures are in a measurement that includes zeros are shown in Figure 2.9.
Significant figures in calculations require special rules. In calculations, the number of significant figures in your result depends on the number of significant figures in each measurement. For example, if someone reports that the height of a mountaintop, like the one shown in Figure 2.8, is 1710 m, that implies that its actual height is between 1705 and 1715 m. If another person builds a pile of rocks 0.20 m high on top of the mountain, that would not suddenly make the mountain’s new height known accurately enough to be measured as 1710.20 m. The final reported height cannot be more precise than the least precise measurement used to find the answer. Therefore, the reported height should be rounded off to 1710 m even if the pile of rocks is included.
Rule
Examples
1. Zeros between other nonzero digits are significant.
a. 50.3 m has three significant figures. b. 3.0025 s has five significant figures.
2. Zeros in front of nonzero digits are not significant.
a. 0.892 kg has three significant figures. b. 0.0008 ms has one significant figure.
3. Zeros that are at the end of a number and also to the right of the decimal are significant.
a. 57.00 g has four significant figures.
4. Zeros at the end of a number but to the left of a decimal are significant if they have been measured or are the first estimated digit; otherwise, they are not significant. In this book, they will be treated as not significant. (Some books place a bar over a zero at the end of a number to indicate that it is significant. This textbook will use scientific notation for these cases instead.)
a. 1000 m may contain from one to four significant figures, depending on the precision of the measurement, but in this book it will be assumed that measurements like this have one significant figure. b. 20 m may contain one or two significant figures, but in this book it will be assumed to have one significant figure.
18
Chapter 1
b. 2.000 000 kg has seven significant figures.
Similar rules apply to multiplication. Suppose that you calculate the area of a room by multiplying the width and length. If the room’s dimensions are 4.6 m by 6. 7 m, the product of these values would be 2 30.82 m . However, this answer contains four significant figures, which implies that it is more precise than the measurements of the length and width. Because the room could be as small as 4.55 m by 6.65 m or as large as 4.65 m by 6. 75 m, the area of the room is known only to be between 30.26 m 2 and 31.39 m 2 . The area of the room can have only two significant figures because each measurement has only two. So, the area must be rounded off to 31 m 2 • Figure 2.10 summarizes the two basic rules for determining significant figures when you are performing calculations.
Type of calculation
Rule
Examples
addition or subtraction
Given that addition and subtraction take place in columns, round the final answer to the first column from the left containing an estimated digit.
97.3 + 5.85 103.15
round off
103.2
The final answer has the same number of significant figures as the measurement having the smallest number of significant figures.
123 X 5.35 658.05
round off
658
multiplication or division
Calculators do not pay attention to significant figures. When you use a calculator to analyze problems or measurements, you may be able to save time because the calculator can compute faster than you can. However, the calculator does not keep track of significant figures. Calculators often exaggerate the precision of your final results by returning answers with as many digits as the display can show. To reinforce the correct approach, the answers to the sample problems in this book will always show only the number of significant figures that the measurements justify. Providing answers with the correct number of significant figures often requires rounding the results of a calculation. The rules listed in Figure 2.11 on the next page will be used in this book for rounding, and the results of a calculation will be rounded after each type of mathematical operation. For example, the result of a series of multiplications should be rounded using the multiplication/ division rule before it is added to another number. Similarly, the sum of several numbers should be rounded according to the addition/ subtraction rule before the sum is multiplied by another number. Multiple roundings can increase the error in a calculation, but with this method there is no ambiguity about which rule to apply. You should consult your teacher to find out whether to round this way or to delay rounding until the end of all calculations. The Science of Physics
19
What to do
When to do it
Examples
round down
• whenever the digit following the last significant figure is a 0, 1, 2, 3, or 4
30.24 becomes 30.2
• if the last significant figure is an even number and the next digit is a 5, with no other nonzero digits
32.25 becomes 32.2 32.650 00 becomes 32.6
• whenever the digit following the last significant figure is a 6, 7, 8, or 9
22.49 becomes 22.5
• if the digit following the last significant figure is a 5 followed by a nonzero digit
54.7511 becomes 54.8
• if the last significant figure is an odd number and the next digit is a 5, with no other nonzero digits
54.75 becomes 54.8 79.3500 becomes 79.4
round up

SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Which SI units would you use for the following measurements? a. the length of a swimming pool b. the mass of the water in the pool c. the time it takes a swimmer to swim a lap 2. Express the following measurements as indicated. a. 6.20 mg in kilograms b. 3 x 10-9 sin milliseconds c. 88.0 km in meters 3. Perform these calculations, following the rules for significant figures. a. 26 X 0.025 84 = ? b. 15.3 -:- 1.1 = ? c. 782.45 – 3.5328 = ? d. 63.258 + 734.2 = ?
Critical Thinking 4. The following students measure the density of a piece of lead three times. The density oflead is actually 11.34 g/cm3 • Considering all of the results, which person’s results were accurate? Which were precise? Were any both accurate and precise? a. Rachel: 11.32 g/cm3 , 11.35 g/cm3, 11.33 g/cm3 b. Daniel: 11.43 g/cm3, 11.44 g/cm3 , 11.42 g/cm3 3 3 3 c. Leah : 11.55 g/cm , 11.34 g/cm , 11.04 g/cm
20
Chapter 1
The Language ol
Phvsics Mathematics and Physics Just as physicists create simplified models to better understand the real world, they use the tools of mathematics to analyze and summarize their observations. Then they can use the mathematical relationships among physical quantities to help predict what will happen in new situations.
Tables, graphs, and equations can make data easier to understand. There are many ways to organize data. Consider the experiment shown in Figure 3.1, which tests Galileo’s hypothesis that all objects fall at the same rate in the absence of air resistance. In this experiment, a table-tennis ball and a golf ball are dropped in a vacuum. The results are recorded as a set of numbers corresponding to the times of the fall and the distance each ball falls. A convenient way to organize the data is to form a table like Figure 3.2.
Two Balls Falling in a Vacuum This experiment tests Galileo’s hypothesis by having two balls with different masses dropped simultaneously in a vacuum.
Distance golf ball falls (cm)
Distance table-tennis ball falls (cm)
0.067
2.20
2.20
0.133
8.67
8.67
0.200
19.60
19.59
0.267
34.93
34.92
0.333
54.34
54.33
0.400
78.40
78.39
Time (s)
One method for analyzing the data in Figure 3.2 is to construct a graph of the distance the balls have fallen versus the elapsed time since they were released. This graph is shown in Figure 3.3 on the next page. Because the graph shows an obvious pattern, we can draw a smooth curve through the data points to make estimations for times when we have no data. The shape of the graph also provides information about the relationship between time and distance. The Science of Physics
21
Distance Dropped Balls Have Fallen versus Time
Graph of Dropped Ball Data The graph of these data provides a convenient way to summarize the data and indicate the relationship between the time an object has been falling and the distance it has fallen.
100.00 90.00 80.00
e e a,
‘-‘ C
.l9
c”‘
70.00 60.00 50.00 40.00 30.00 20.00 10.00 0.00 0.100
0.200
0.300
0.400
0.500
Time (s)
We can also use the following equation to describe the relationship between the variables in the experiment: (change in position in meters)= 4.9 x (time of fall in seconds)2 This equation allows you to reproduce the graph and make predictions about the change in position for any arbitrary time during the fall.
Physics equations describe relationships. While mathematicians use equations to describe relationships between variables, physicists use the tools of mathematics to describe measured or predicted relationships between physical quantities in a situation. For example, one or more variables may affect the outcome of an experiment. In the case of a prediction, the physical equation is a compact statement based on a model of the situation. It shows how two or more variables are related. Many of the equations in physics represent a simple description of the relationship between physical quantities. To make expressions as simple as possible, physicists often use letters to describe specific quantities in an equation. For example, the letter vis used to denote speed. Sometimes, Greek letters are used to describe mathematical operations. For example, the Greek letter b. (delta) is often used to mean “difference or change in;’ and the Greek letter I: (sigma) is used to mean “sum” or “total:’ With these conventions, the word equation above can be written as follows : b.y = 4.9(b.t) 2
The abbreviation b.y indicates the vertical change in a ball’s position from its starting point, and b.t indicates the time elapsed. The units in which these quantities are measured are also often abbreviated with symbols consisting of a letter or two. Most physics books provide some clues to help you keep track of which letters refer to quantities and variables and which letters are used to indicate units. Typically, variables and other specific quantities are abbreviated with letters that are boldfaced or italicized. (The difference between the two is described 22
Chapter 1
in the chapter “Two-Dimensional Motion and Vectors:’) Units are abbreviated with regular letters (sometimes called roman letters). Some examples of variable symbols and the abbreviations for the units that measure them are shown in Figure 3.4. As you continue to study physics, carefully note the introduction of new variable quantities, and recognize which units go with them. The tables provided in Appendices C-E can help you keep track of these abbreviations.
Quantity
Symbol
Units
Unit abbreviations
change in vertical position
.6.y
meters
m
time interval
.6.t
seconds
s
mass
m
kilograms
kg
Evaluating Physics Equations Although an experiment is the ultimate way to check the validity of a physics equation, several techniques can be used to evaluate whether an equation or result can possibly be valid.
Dimensional analysis can weed out invalid equations. Suppose a car, such as the one in Figure 3.5, is moving at a speed of 88 km/h and you want to know how much time it will take it to travel 725 km. How can you decide a good way to solve the problem? You can use a powerful procedure called dimensional analysis. Dimensional analysis makes use of the fact that dimensions can be treated as algebraic quantities. For example, quantities can be added or subtracted only if they have the same dimensions, and the two sides of any given equation must have the same dimensions. Let us apply this technique to the problem of the car moving at a speed of 88 km/h. This measurement is given in dimensions of length over time. The total distance traveled has the dimension of length. Multiplying these numbers together gives the following dimensions:
Dimensional Analysis in Speed Calculations Dimensional analysis can be a useful check for many types of problems, including those involving how much time it would take for this car to travel 725 km if it moves with a speed of 88 km/h.
Clearly, the result of this calculation does not have the dimensions of time, which is what you are calculating. This equation is not a valid one for this situation.
The Science of Physics
23
To calculate an answer that will have the dimension of time, you should take the distance and divide it by the speed of the car, as follows: length l.eagcti x time . —–=——=time length/time leRgctl.
725 km x 1.0 h -_ 8.2 h 88 km
In a simple example like this one, you might be able to identify the valid equation without dimensional analysis. But with more complicated problems, it is a good idea to check your final equation with dimensional analysis before calculating your answer. This step will prevent you from wasting time computing an invalid equation.
Order-of-magnitude estimations check answers.
, .Did YOU Know? ___________ _ , ,
‘ ‘
The physicist Enrico Fermi made the first nuclear reactor at the University of Chicago in 1942. Fermi was also well known for his ability to make quick order-of-magnitude calculations, such as estimating the number of piano tuners in New York City.
Because the scope of physics is so wide and the numbers may be astronomically large or subatomically small, it is often useful to estimate an answer to a problem before trying to solve the problem exactly. This kind of estimate is called an order-of-magnitude calculation, which means determining the power of 10 that is closest to the actual numerical value of the quantity. Once you have done this, you will be in a position to judge whether the answer you get from a more exacting procedure is correct. For example, consider the car trip described in the discussion of dimensional analysis. We must divide the distance by the speed to find the time. The distance, 725 km, is closer to 103 km (or 1000 km) than to 102 km (or 100 km), so we use 103 km. The speed, 88 km/h, is about 102 km/h (or 100 km/h). 103k,m = 10h
102 k,m/h This estimate indicates that the answer should be closer to 10 than to 1 or to 100 (or 102 ). The correct answer (8.2 h) certainly fits this range. Order-of-magnitude estimates can also be used to estimate numbers in situations in which little information is given. For example, how could you estimate how many gallons of gasoline are used annually by all of the cars in the United States? To find an estimate, you will need to make some assumptions about the average household size, the number of cars per household, the distance traveled, and the average gas mileage. First, consider that the United States has about 300 million people. Assuming that each family of about five people has two cars, an estimate of the number of cars in the country is 120 million. Next, decide the order of magnitude of the average distance each car travels every year. Some cars travel as few as 1000 mi per year, while others travel more than 100 000 mi per year. The appropriate order of magnitude to include in the estimate is 10 000 mi, or 104 mi, per year.
If we assume that cars average 20 mi for every gallon of gas, each car needs about 500 gal per year.
mr)(20lgal_) = 500 gal/year for each car m-r
10/ 00 ( year
24
Chapter 1

Multiplying this by the estimate of the total number of cars in the United States gives an annual consumption of 6 x 1010 gal. 7
(12 x 10
~
)
500 gal ) 10 c;;ar = 6 x 10 gal ( 1
SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Indicate which of the following physics symbols denote units and which denote variables or quantities. a. C b. C c. C d. t e. T f. T
2. Determine the units of the quantity described by each of the following combinations of units: a. kg (m/ s) (1/s) b. (kg/s) (m/s2 ) c. (kg/s) (m/s) 2 d. (kg/s) (m/s) 3. Which of the following is the best order-of-magnitude estimate in meters of the height of a mountain? a. 1 m b. l0m c. l00m d. 1000m
Interpreting Graphics 4. Which graph in Figure 3.6 best matches the data? Volume of air (m3)
Mass of air (kg)
0.50
0.644
1.50
1.936
2.25
2.899
4.00
5.159
5.50
7.096
•aMihlilil 8.000
10.000 cii 8.000 c 6.000 “‘ 4.000 “‘ “‘ 2.000 :ii: 0
cii 6.000 c 4.000
“‘
i
2.00~ 1.00
(a)
3.00
Volume {m3)
5.00
30.000 cii ~
“‘ “‘ “‘
:ii: 10.000 0 1.00
(b)
20.000
3.00
5.00
Volume {m3)
1.00
(C)
3.00
5.00
Volume {m3)
Critical Thinking 5. Which of the following equations best matches the data from item 4? a. (mass)2 = 1.29 (volume) b. (mass)(volume) = 1.29 c. mass = 1.29 (volume) d. mass = 1.29 (volume) 2
The Science of Physics
25
SECTION 1
What Is Physics?
1 1 ,
1 , , ·.,
• Physics is the study of the physical world, from motion and energy to light and electricity.
model system
• Physics uses the scientific method to discover general laws that can be used to make predictions about a variety of situations.
hypothesis
• A common technique in physics for analyzing a complex situation is to disregard irrelevant factors and create a model that describes the essence of a system or situation.
SECTION 2
Measurements in Experiments
controlled experiment
, c · Tc,· 1
• Physics measurements are typically made and expressed in SI , a system that uses a set of base units and prefixes to describe measurements of physical quantities.
accuracy precision significant figures
• Accuracy describes how close a measurement is to reality. Precision results from the limitations of the measuring device used. • The significant figures of a measurement include all of the digits that are actually measured plus one estimated digit. • Significant-figure rules provide a means to ensure that calculations do not report results that are more precise than the data used to make them.
SECTION 3
The Language of Physics
• Physicists make their work easier by summarizing data in tables and graphs and by abbreviating quantities in equations. • Dimensional analysis can help identify whether a physics equation is invalid. • Order-of-magnitude calculations provide a quick way to evaluate the appropriateness of an answer.
.6.y .6.t
m 26
change in vertical position time interval mass Chapter 1
m s kg
meters seconds kilograms
Problem Solving If you need more problem-solving practice, see Appendix I: Additional Problems.
The Science of Physics REVIEWING MAIN IDEAS 1. Refer to Figure 1.3 of this chapter to identify at least two areas of physics involved in the following: a. building a louder stereo system in your car b. bungee jumping c. judging how hot an electric stove burner is by looking at it d. cooling off on a hot day by diving into a swimming pool 2. Which of the following scenarios fit the approach of
the scientific method? a. An auto mechanic listens to how a car runs and comes up with an idea of what might be wrong. The mechanic tests the idea by adjusting the idle speed. Then the mechanic decides his idea was wrong based on this evidence. Finally, the mechanic decides the only other problem could be the fuel pump, and he consults with the shop’s other mechanics about his conclusion. b. Because of a difference of opinions about where to take the class trip, the class president holds an election. The majority of the students decide to go to the amusement park instead of to the shore. c. Your school’s basketball team has advanced to the regional playoffs. A friend from another school says their team will win because their players want to win more than your school’s team does. d. A water fountain does not squirt high enough. The handle on the fountain seems loose, so you try to push the handle in as you turn it. When you do this, the water squirts high enough that you can get a drink. You make sure to tell all your friends how you did it. 3. You have decided to select a new car by using the scientific method. How might you proceed?
4. Consider the phrase, “The quick brown fox jumped over the lazy dog:’ Which details of this situation would a physicist who is modeling the path of a fox ignore?
SI Units REVIEWING MAIN IDEAS 5. List an appropriate SI base unit (with a prefix as
needed) for measuring the following: a. the time it takes to play a CD in your stereo b. the mass of a sports car c. the length of a soccer field d. the diameter of a large pizza e. the mass of a single slice of pepperoni f. a semester at your school g. the distance from your home to your school h. yourmass i. the length of your physics lab room j. your height 6. If you square the speed expressed in meters per second, in what units will the answer be expressed? 7. If you divide a force measured in newtons
(1 newton= 1 kg•m/s2 ) by a speed expressed in meters per second, in what units will the answer be expressed?
CONCEPTUAL QUESTIONS 8. The height of a horse is sometimes given in units of
“hands:’ Why was this a poor standard of length before it was redefined to refer to exactly 4 in.? 9. Explain the advantages in having the meter officially defined in term s of the distance light travels in a given time rather than as the length of a specific metal bar. 10. Einstein’s famous equation indicates that E
= me 2,
where c is the speed of light and mis the object’s m ass. Given this, what is the SI unit for E?
Chapter Review
27
PRACTICE PROBLEMS For problems 11-14, see Sample Problem A. 11. Express each of the following as indicated: a. 2 dm expressed in millimeters b. 2 h 10 min expressed in seconds c. 16 g expressed in micrograms d. 0. 75 km expressed in centimeters e. 0.675 mg expressed in grams f. 462 µm expressed in centimeters g. 35 km/h expressed in meters per second 12. Use the SI prefixes in Figure 2.4 of this chapter to
convert these hypothetical units of measure into appropriate quantities: a. 10 rations b. 2000 mockingbirds c. 10- 6 phones d. 10- 9 goats e. 1018 miners 13. Use the fact that the speed of light in a vacuum is about 3.00 x 108 m/s to determine how many kilometers a pulse from a laser beam travels in exactly one hour.
14. If a metric ton is 1.000 x 103 kg, how many 85 kg people can safely occupy an elevator that can hold a maximum mass of exactly 1 metric ton?
Accuracy, Precision, and Signilicant Figures REVIEWING MAIN IDEAS 15. Can a set of m easurements be precise but not
accurate? Explain. 16. How many significant figures are in the following
measurements? a. 300 000 000 m is b. 3.00 x 108 m/s c. 25.030°C d. 0.006 070°C e. 1.004 J f. 1.305 20 MHz
28
Chapter 1
17. The photographs below show unit conversions on the labels of some grocery-store items. Check the accuracy of these conversions. Are the manufacturers using significant figures correctly? (a)
(b)
(c)
(d)
l=
NETWTSLS
18. The value of the speed of light is now known to be 2.997 924 58 x 108 m /s. Express the speed oflight in the following ways: a. with three significant figures b. with five significant figures c. with seven significant figures 19. How many significant figures are there in the following measurements? a. 78.9 ± 0.2 m b. 3.788 x 109 s C. 2.46 x 106 kg d. 0.0032mm 20. Carry out the following arithmetic operations: a. find the sum of the measurements 756 g, 37.2 g, 0.83 g, and 2.5 g b. find the quotient of3.2 m/3.563 s c. find the product of 5.67 mm x 7r d. find the difference of 27.54 sand 3.8 s 21. A fisherman catches two sturgeons. The smaller of the
two has a measured length of 93.46 cm (two decimal places and four significant figures), and the larger fish has a measured length of 135.3 cm (one decimal place and four significant figures). What is the total length of the two fish? 22. A farmer m easures the distance around a rectangular
field. The length of each long side of the rectangle is found to be 38.44 m, and the length of each short side is found to be 19.5 m. What is the total distance around the field?
Dimensional Analysis and Order-of-Magnitude Estimates Note: In developing answers to order-of-magnitude calculations, you should state your important assumptions, including the numerical values assigned to parameters used in the solution. Since only order-of-magnitude results are expected, do not be surprised if your results differ from those of other students.
REVIEWING MAIN IDEAS 23. Suppose that two quantities, A and B, have different dimensions. Which of the following arithmetic operations could be physically meaningful? a. A+B b. A/B C. AX B
d. A-B 24. Estimate the order of magnitude of the length in meters of each of the following: a. aladybug b. your leg c. your school building d. a giraffe e. a city block 25. If an equation is dimensionally correct, does this mean that the equation is true?
26. The radius of a circle inscribed in any triangle whose sides are a, b, and c is given by the following equation, in which sis an abbreviation for (a+ b + c)..,… 2. Check this formula for dimensional consistency.
r = ~ (s-a)(s~b)(s-c) 27. The period of a simple pendulum, defined as the time necessary for one complete oscillation, is measured in time units and is given by the equation
T=21rf,f; where L is the length of the pendulum and a g is the acceleration due to gravity, which has units oflength divided by time squared. Check this equation for dimensional consistency.
CONCEPTUAL QUESTIONS 28. In a desperate attempt to come up with an equation to solve a problem during an examination, a student tries the following: (velocity in m/ s )2 = (acceleration in m/s2) x (time ins). Use dimensional analysis to determine whether this equation might be valid. 29. Estimate the number of breaths taken by a person during 70 years. 30. Estimate the number of times your heart beats in an average day. 31. Estimate the magnitude of your age, as measured in units of seconds. 32. An automobile tire is rated to last for 50 000 mi. Estimate the number of revolutions the tire will make in its lifetime. 33. Imagine that you are the equipment manager of a professional baseball team. One of your jobs is to keep a supply of baseballs for games in your home ballpark. Balls are sometimes lost when players hit them into the stands as either home runs or foul balls. Estimate how many baseballs you have to buy per season in order to make up for such losses. Assume your team plays an 81-game home schedule in a season.
34. A chain of hamburger restaurants advertises that it has sold more than 50 billion hamburgers over the years. Estimate how many pounds of hamburger meat must have been used by the restaurant chain to make 50 billion hamburgers and how many head of cattle were required to furnish the meat for these hamburgers.
35. Estimate the number of piano tuners living in New York City. (The population of New York City is approximately 8 million.) This problem was first proposed by the physicist Enrico Fermi, who was well known for his ability to quickly make order-of-magnitude calculations. 36. Estimate the number of table-tennis balls that would fit (without being crushed) into a room that is 4 m long, 4 m wide, and 3 m high. Assume that the diameter of a ball is 3.8 cm.
Chapter Review
29
Mixed Review 37. Calculate the circumference and area for the following circles. (Use the following formulas: circumference= 27rr and area= 7rr 2 .) a. a circle of radius 3.5 cm b. a circle of radius 4.65 cm
38. A billionaire offers to give you (1) $5 billion if you will count out the amount in $1 bills or (2) a lump sum of $5000. Which offer should you accept? Explain your answer. (Assume that you can count at an average rate of one bill per second, and be sure to allow for the fact that you need about 10 hours a day for sleeping and eating. Your answer does not need to be limited to one significant figure.)
39. Exactly 1 quart of ice cream is to be made in the form of a cube. What should be the length of one side in meters for the container to have the appropriate volume? (Use the following conversion: 4 qt = 3.786 X 10- 3 m 3 .) 40. You can obtain a rough estimate of the size of a molecule with the following simple experiment: Let a droplet of oil spread out on a fairly large but smooth water surface. The resulting “oil slick” that forms on the surface of the water will be approximately one molecule thick. Given an oil droplet with a mass of 9.00 x 10- 7 kgandadensityof918kg/ m 3 that spreads out to form a circle with a radius of 41.8 cm on the water surface, what is the approximate diameter of an oil molecule?
Mass Versus Length What is the relationship between the mass and length of three wires, each of which is made of a different substance? All three wires have the same diameter. Because the wires have the same diameter, their cross-sectional areas are the same. The cross-sectional area of any circle is equal to 1rr 2. Consider a wire with a diameter of 0.50 cm and a density of 8.96 g/cm3. The following equation describes the mass of the wire as a function of the length: Y1 = 8.96X x 7r(0.25)2 In this equation, Y1 represents the mass of the wire in grams, and X represents the length of the wire in centimeters. Each of the three wires is made of a different substance, so each wire has a different density and a different relationship between its mass and length.
30
Chapter 1
In this graphing calculator activity, you will • use dimensional analysis • observe the relationship between a mathematical function and a graph • determine values from a graph • gain a better conceptual understanding of density Go online to HMDScience.com to find this graphing calculator activity.
41. An ancient unit of length called the cubit was equal to approximately 50 centimeters, which is, of course, approximately 0.50 meters. It has been said that Noah’s ark was 300 cubits long, 50 cubits wide, and 30 cubits high. Estimate the volume of the ark in cubic meters. Also estimate the volume of a typical home, and compare it with the ark’s volume. 42. If one micrometeorite (a sphere with a diameter of 1.0 x 10- 6 m) struck each square meter of the moon each second, it would take many years to cover the moon with micrometeorites to a depth of 1.0 m. Consider a cubic box, 1.0 m on a side, on the moon. Estimate how long it would take to completely fill the box with micrometeorites. 43. One cubic centimeter (1.0 cm3 ) of water has a mass of 1.0 x 10-3 kg at 25°C. Determine the mass of 1.0 m 3 of water at 25°C.
ALTERNATIVE ASSESSMENT 1. Imagine that you are a member of your state’s highway board. In order to comply with a bill passed in the state legislature, all of your state’s highway signs must show distances in miles and kilometers. Two plans are before you . One plan suggests adding metric equivalents to all highway signs as follows: Dallas 300 mi (483 km). Proponents of the other plan say that the first plan makes the metric system seem more cumbersome, so they propose replacing the old signs with new signs every 50 km as follows: Dallas 300 km (186 mi). Participate in a class debate about which plan should be followed. 2. Can you measure the m ass of a five-cent coin with a bathroom scale? Record the mass in grams displayed by your scale as you place coins on the scale, one at a time. Then, divide each measurement by the number of coins to determine the approximate mass of a single five-cent coin, but remember to follow the rules for significant figures in calculations. Which estimate do you think is the most accurate? Which is the most precise?
44. Assuming biological substances are 90 percent water and the density of water is 1.0 x 103 kg/m3 , estimate the masses (density multiplied by volume) of the following: a. a spherical cell with a diameter of 1.0 µm (volume= m 3 ) b. a fly, which can be approximated by a cylinder 4.0 mm long and 2.0 mm in diameter 2 (volume= l7rr )
f
45. The radius of the planet Saturn is 6.03 x 107 m, and its mass is 5.68 x 1026 kg. a. Find the density of Saturn (its mass divided by its volume) in grams per cubic centimeter. (The volume of a sphere is given by 1rr 3 .) b. Find the surface area of Saturn in square meters. (The surface area ofa sphere is given by 4m 2 .)
f
3. Find out who were the Nobel laureates for physics last year, and research their work. Alternatively, explore the history of the Nobel Prizes. Who founded the awards? Why? Who delivers the award? Where? Document your sources and present your findings in a brochure, poster, or presentation. 4. You have a clock with a second hand, a ruler marked in millimeters, a graduated cylinder marked in milliliters, and scales sensitive to 1 mg. How would you measure the mass of a drop of water? How would you measure the period of a swing? How would you measure the volume of a paper clip? How can you improve the accuracy of your measurements? Write the procedures clearly so that a partner can follow them and obtain reasonable results. 5. Create a poster or other presentation depicting the
possible ranges of m easurement for a dimension, such as distance, time, temperature, speed, or mass. Depict examples ranging from the very large to the very small. Include several examples that are typical of your own experiences.
Chapter Review
31
MULTIPLE CHOICE 1. What area of physics deals with the subjects of heat and temperature? A. mechanics B. thermodynamics C. electrodynamics D. quantum mechanics 2. What area of physics deals with the behavior of
subatomic particles? F. mechanics G. thermodynamics H. electrodynamics J. quantum mechanics 3. What term describes a set of particles or interacting components considered to be a distinct physical entity for the purpose of study? A. system B. model C. hypothesis D. controlled experiment 4. What is the SI base unit for length? F. inch G. foot H. meter J. kilometer
5. A light-year (ly) is a unit of distance defined as the distance light travels in one year. Numerically, l ly = 9 500 000 000 000 km. How many meters are in a light-year? A. 9.5 x 1010 m B. 9.5 x 1012 m C. 9.5 x 1015 m D. 9.5 x 10 18 m
32
Chapter 1
6. If you do not keep your line of sight directly over a length measurement, how will your measurement most likely be affected? F. Your measurement will be less precise. G. Your measurement will be less accurate. H. Your measurement will have fewer significant figures. J. Your measurement will suffer from instrument error. 7. If you measured the length of a pencil by using the
meterstick shown in the figure below and you report your measurement in centimeters, how many significant figures should your reported measurement have?
I13 I
14 15 16 17 18
I 19 20 I
A. one B. two C. three D. four 8. A room is measured to be 3.6 m by 5.8 m. What is the area of the room? (Keep significant figures in mind.) F. 20.88 m 2 G. 2 x 101 m2 H. 2.0 x 101 m 2 J. 21 m 2 9. What technique can help you determine the power of 10 closest to the actual numerical value of a quantity? A. rounding B. order-of-magnitude estimation C. dimensional analysis D. graphical analysis
. TEST PREP
10. Which of the following statements is true of any valid physical equation? F. Both sides have the same dimensions. G. Both sides have the same variables. H. There are variables but no numbers. J. There are numbers but no variables. The graph below shows the relationship between time and distance for a ball dropped vertically from rest. Use the graph to answer questions 11-12.
Graph of experimental data
SHORT RESPONSE 13. Determine the number of significant figures in each
of the following measurements. A. 0.0057 kg B. 5.70 g C. 6070m D. 6.070 x 103 m 14. Calculate the following sum, and express the answer in meters. Follow the rules for significant figures. (25.873 km) + (1024 m) + (3.0 cm)
100.00
15. Demonstrate how dimensional analysis can be used to find the dimensions that result from dividing distance by speed.
90.00 80.00
e .e
70.00 60.00
Q)
50.00 40.00 i::i 30.00
EXTENDED RESPONSE
C
tl
20.00 10.00 0.00 0.100
0.200 Time (s)
0.300
11. About how far has the ball fallen after 0.200 s? A. 5.00cm B. 10.00cm C. 20.00 cm D. 30.00 cm
16. You have decided to test the effects of four different garden fertilizers by applying them to four separate rows of vegetables. What factors should you control? How could you measure the results? 17. In a paragraph, describe how you could estimate the
number of blades of grass on a football field.
12. Which of the following statements best describes the relationship between the variables? F. For equal time intervals, the change in position is increasing. G. For equal time intervals, the change in position is decreasing. H. For equal time intervals, the change in position is constant. J. There is no clear relationship between time and change in position.
Test Tip If more than one answer to a multiplechoice question seems to be correct, pick the answer that is most correct or that most directly answers the question.
Standards-Based Assessment
33
(9(D
SECTION 1 Objectives

Describe motion in terms of frame of reference, displacement, time, and velocity.

Calculate the displacement of an object traveling at a known velocity for a specific time interval.

Construct and interpret graphs of position versus time.
Displacement and Velocity Key Terms frame of reference displacement
average velocity instantaneous velocity
Motion Motion happens all around us. Every day, we see objects such as cars, people, and soccer balls move in different directions with different speeds. We are so familiar with the idea of motion that it requires a special effort to analyze motion as a physicist does.
One-dimensional motion is the simplest form of motion. One way to simplify the concept of motion is to consider only the kinds of motion that take place in one direction. An example of this onedimensional motion is the motion of a commuter train on a straight track, as in Figure 1.1.
frame of reference a system for specifying the precise location of objects in space and time
In this one-dimensional motion, the train can move either forward or backward along the tracks. It cannot move left and right or up and down. This chapter deals only with one-dimensional motion. In later chapters, you will learn how to describe more complicated motions such as the motion of thrown baseballs and other projectiles.
Frames of Reference The motion of a commuter train traveling along a straight route is an example of one-dimensional motion. Each train can move only forward and backward along the tracks.
Motion takes place over time and depends upon the frame of reference. It seems simple to describe the motion of the train. As the train in Figure 1.1 begins its route, it is at the first station. Later, it will be at another station farther down the tracks. But Earth is spinning on its axis, so the train, stations, and the tracks are also moving around the axis. At the same time, Earth is moving around the sun. The sun and the rest of the solar system are moving through our galaxy. This galaxy is traveling through space as well. When faced with a complex situation like this, physicists break it down into simpler parts. One key approach is to choose a frame of reference against which you can measure changes in position. In the case of the train, any of the stations along its route could serve as a convenient frame of reference. When you select a reference frame, note that it remains fixed for the problem in question and has an origin, or starting point, from which the motion is measured.
36
Chapter 2
C:
.2,
“” 0
e
a,
E ~ :::,
:a:
If an object is at rest (not moving), its position does not change
with respect to a fixed frame of reference. For example, the benches on the platform of one subway station never move down the tracks to another station.
Space Shuttle A space shuttle
In physics, any frame of reference can be chosen as long as it is used consistently. If you are consistent, you will get the same results, no matter which frame of reference you choose. But some frames of reference can make explaining things easier than other frames of reference. For example, when considering the motion of the gecko in Figure 1.2, it is useful to imagine a stick marked in centimeters placed under the gecko’s feet to define the frame of reference. The measuring stick serves as an x-axis. You can use it to identify the gecko’s initial position and its final position.
takes off from Florida and circles Earth several times, finally landing in California. While the shuttle is in flight, a photographer flies from Florida to California to take pictures of the astronauts when they step off the shuttle. Who undergoes the greater displacement, the photographer or the astronauts? Roundtrip What is the
Displacement As any object moves from one position to another, the length of the straight line drawn from its initial position to the object’s final position is called the displacement of the object.
Displacement is a change in position. The gecko in Figure 1.2 moves from left to right along the x-axis from an initial position, xi, to a final position, xf” The gecko’s displacement is the difference between its final and initial coordinates, or x – x i. In this case, 1 the displacement is about 61 cm (85 cm – 24 cm). The Greek letter delta (.6.) before the x denotes a change in the position of an object.
l
! Displacement – ~X=Xf-Xi
1

~
splacement = change in position = final position – initial positio:J
Measuring Displacement A gecko moving along thex-axis from X ; to displacement of ~ = X;.
x, –
difference between the displacement of the photographer flying from Florida to California and the displacement of the astronauts flying from California back to Florida?
displacement the change in position of an object
Tips and Tricks
A change in any quantity, indicated by the Greek symbol delta (~), is equal to the final value minus the initial value. When calculating displacement, always be sure to subtract the initial position from the final position so that your answer has the correct sign.
x, undergoes a ~x
~
C,
“‘
§
>-
~
~
~:;; ~
.£: “’16
C.
@ “‘
:e:
0
1
10
130
I 20
t
xi
140
I so
I 60
I 10
I so
190
t
Xf
Motion in One Dimension
37
Now suppose the gecko runs up a tree, as shown in Figure 1.3. In this case, we place the measuring stick parallel to the tree. The measuring stick can serve as the y-axis of our coordinate system. The gecko’s initial and final positions are indicated by Yi and y , respectively, and the gecko’s 1 displacement is denoted as 6.y.
Comparing Displacement and Distance When the gecko is climbing a tree, the displacement is measured on the y-axis. Again, the gecko’s position is determined by the position of the same point on its body.
Displacement is not always equal to the distance traveled. Displacement does not always tell you the distance an object has moved. For example, what if the gecko in Figure 1.3 runs up the tree from the 20 cm marker (its initial position) to the 80 cm marker. After that, it retreats down the tree to the 50 cm marker (its final position). It has traveled a total distance of 90 cm. However, its displacement is only 30 cm (y1 – Yi= 50 cm – 20 cm= 30 cm). If the gecko were to return to its starting point, its displacement would be zero because its initial position and final position would be the same.
Cl
co
Yr- –
Cl
in
Displacement can be positive or negative.
.,,
Displacement also includes a description of the direction of motion. In one-dimensional motion, there are only two directions in which an object can move, and these directions can be described as positive or negative.
Cl
Cl
M
Y;-
In this book, unless otherwise stated, the right (or east) will be considered the positive direction and the left (or west) will be considered the negative direction. Similarly, upward (or north) will be considered positive, and downward (or south) will be considered negative. Figure 1.4 gives examples of determining displacements for a variety of situations.
~


Cl
;,, I
. , ,t ff.f’L,f-rL.teL.,lllL,~. . .1~,,,,~ “”~ ‘~~-xi
n/ ,,I 4/Ji
soj J ,~ ~ ~ ‘”‘
d’f ((ttfff¾ llttftl, l lllll lll~ IUl/111 111111111 IIIIUll\\ 1111111~~\ll\\\ll\r \\\~ \nmm
Xf
6.x=x1 – xi= 80 cm – 10 cm= +70 cm
(lril.1hhlrL!L1bb.1\11h~i~1~i~\\~.. X;
Xf
Xf
6.x = x1 – xi = 12 cm – 3 cm = + 9 cm
.;:tiMk(dtMi?I~1A~~’~’t>s Xi
N “
~
Xf
6.x = x1 – xi= 6 cm – ( – 10 cm) = + 16 cm 38
Chapter 2
X;
6.x = x1 – xi = 0 cm – 15 cm = -15 cm

LFd’dkfd’bbL•IH ~hl, “”‘l””‘”‘ Xf
X;
6.x = x1 – xi= – 20 cm – ( – 10 cm) = – 10 cm
Velocity Where an object started and where it stopped does not completely describe the motion of the object. For example, the ground that you’re standing on may move 8.0 cm to the left. This motion could take several years and be a sign of the normal slow movement of Earth’s tectonic plates. If this motion takes place in just a second, however, you may be experiencing an earthquake or a landslide. Knowing the speed is important when evaluating motion.
. Did YOU Know?. – – – – – – – – – – – , : : : : ,I
The branch of physics concerned with ‘ motion and forces is called mechanics. , The subset of mechanics that describes motion without regard to its , causes is called kinematics.
Average velocity is displacement divided by the time interval. Consider the car in Figure 1.5. The car is moving along a highway in a straight line (the x-axis). Suppose that the positions of the car are xi at time ti and x1 at time 1. In the time interval /}.t = t_r- t i , the displacement of the car is /}.x = x1 – x i . The average velocity, vavg ‘ is defined as the displacement divided by the time interval during which the displacement occurred. In SI, the unit of velocity is meters per second, abbreviated asm/ s. Average Velocity
~x
xf-xi V ——avg – ~t – tf- ti
. change in position average velocity = h . . c angemtime
displacement time interval
J
The average velocity of an object can be positive or negative, depending on the sign of the displacement. (The time interval is always positive.) As an example, consider a car trip to a friend’s house 370 km to the west (the negative direction) along a straight highway. If you left your house at 10 A.M. and arrived at your friend’s house at 3 P.M., your average velocity would be as follows : v avg
i}.x
= LJ.t = A
-370km h 5.0
average velocity the total displacement divided by the time interval during which the displacement occurred
= – 74 km/ h = 74 km/h west
Tips and Tricks Average velocity is not always equal to the average of the initial and final velocities. For instance, if you drive first at 40 km/h west and later at 60 km/h west, your average velocity is not necessarily 50 km/h west.
Average Velocity The average velocity of this car tells you how fast and in which direction it is moving.
This value is an average. You probably did not travel exactly 74 km/ h at every moment. You may have stopped to buy gas or have lunch. At other times, you may have traveled more slowly as a result of heavy traffic. To make up for such delays, when you were traveling slower than 74 km/ h, there must also have been other times when you traveled faster than 74 km/h. The average velocity is equal to the constant velocity needed to cover the given displacement in a given time interval. In the example above, if you left your house and maintained a velocity of 74 km/ h to the west at every moment, it would take you 5.0 h to travel 370 km.
Motion in One Dimension
39
Average Velocity and Displacement Sample Problem A During a race on level ground, Andrea runs with an average velocity of 6.02 m/ s to the east. What is Andrea’s displacement after 137 s?
0
ANALYZE
Given:
vavg = 6.02 m/s /::i.t = 137 s
Unknown:
E)
SOLVE
/::i.x = ?
Rearrange the average velocity equation to solve for displacement.
vavg =
Tips and Tricks The calculator answer is 824.74 m, but both the values for velocity and time have three significant figures, so the displacement must be reported as 825 m.
/::i.x /::i.t
/::i.x = Vavg/::i. t /::i.x = vavg/::i.t = (6.02 m/s)(l37 s) =
Practice 1. Heather and Matthew walk with an average velocity of0.98 m / s eastward. Ifit
takes them 34 min to walk to the store, what is their displacement? 2. If Joe rides his bicycle in a straight line for 15 min with an average velocity of 12.5 km/ h south, how far has he ridden? 3. It takes you 9.5 min to walk with an average velocity of 1.2 m / s to the north from the bus stop to the museum entrance. What is your displacement?
4. Simpson drives his car with an average velocity of 48.0 km/h to the east. How long will it take him to drive 144 km on a straight highway? 5. Look back at item 4. How much time would Simpson save by increasing his average velocity to 56.0 km/ h to the east?
6. A bus travels 280 km south along a straight path with an average velocity of 88 km/ h to the south. The bus stops for 24 min. Then, it travels 210 km south with an average velocity of 75 km/ h to the south.
a. How long does the total trip last? b. What is the average velocity for the total trip?
40
Chapter 2
Velocity is not the same as speed. In everyday language, the terms speed and velocity are used interchangeably. In physics, however, there is an important distinction between these two terms. As we have seen, velocity describes motion with both a direction and a numerical value (a magnitude) indicating how fast something moves. However, speed has no direction, only magnitude. An object’s average speed is equal to the distance traveled divided by the time interval for the motion. average speed = dis~ance/ravel~d time o trave
Velocity can be interpreted graphically.
Position-Time Graph The motion of
The velocity of an object can be determined if the object’s position is known at specific times along its path. One way to determine this is to make a graph of the motion. Figure 1.6 represents such a graph. Notice that time is plotted on the horizontal axis and position is plotted on the vertical axis. The object moves 4.0 min the time interval between t = 0 s and t = 4.0 s. Likewise, the object moves an additional 4.0 min the time interval between t = 4.0 s and t = 8.0 s. From these data, we see that the average velocity for each of these time intervals is+ 1.0 mi s (because vavg= b..xl b..t = 4.0 m l 4.0 s). Because the average velocity does not change, the object is moving with a constant velocity of+ 1.0 mi s, and its motion is represented by a straight line on the position-time graph. For any position-time graph, we can also determine the average velocity by drawing a straight line between any two points on the graph. The slope of this line indicates the average velocity between the positions and times represented by these points. To better understand this concept, compare the equation for the slope of the line with the equation for the average velocity. Slope of a Line
slo e p
= rise = run
an object moving with constant velocity will provide a straight-line graph of position versus time. The slope of this graph indicates the velocity.
Position versus Time of an Object at Constant Velocity 16.0
g 12.0 C 0
~ ~
8.0 4.0
2.0
4.0 6.0 Time (s)
8.0
Average Velocity
change in vertical coordinates change in horizontal coordinates
Book on a Table A book is moved once around the edge of a tabletop
Travel Car A travels from New York to Miami at a speed of 25 m/ s.
with dimensions 1.75 m x 2.25 m. If the book ends up at its initial position, what is its displacement? If it completes its motion in 23 s, w hat is its average velocity? What is its average speed?
Car B travels from New York to Chicago, also at a speed of 25 m/ s. Are the velocities of the cars equal? Explain.
Motion in One Dimension
41
Figure 1.7 represents straight-line graphs of position versus
Position-Time Graphs These position-versus-time graphs show that Object 1 moves with a constant positive velocity. Object 2 is at rest. Object 3 moves with a constant negative velocity.
Position versus Time of Three Objects
time for three different objects. Object 1 has a constant positive velocity because its position increases uniformly with time. Thus, the slope of this line is positive. Object 2 has zero velocity because its position does not change (the object is at rest). Hence, the slope of this line is zero. Object 3 has a constant negative velocity because its position decreases with time. As a result, the slope of this line is negative.
Instantaneous velocity may not be the same as average velocity. Now consider an object whose position-versus-time graph is not a straight line, but a curve, as in Figure 1.8. The object moves through larger and larger displacements as each second passes. Thus, its velocity increases with time. For example, between t = 0 s and t = 2.0 s, the object moves 8.0 m, and its average velocity in this time interval is 4.0 m / s (because v avg = 8.0 m / 2.0 s ). However, between t = 0 s and t = 4.0 s, it moves 32 m, so its average velocity in this time interval is 8.0 mis (because vavg = 32 m / 4.0 s). We obtain different average velocities, depending on the time interval we choose. But how can we find the velocity at an instant of time?
Time
instantaneous velocity the velocity of an object at some instant or at a specific point in the object’s path
To determine the velocity at some instant, such as t = 3.0 s, we study a small time interval near that instant. As the intervals become smaller and smaller, the average velocity over that interval approaches the exact velocity at t = 3.0 s. This is called the instantaneous velocity. One way to determine the instantaneous velocity is to construct a straight line that is tangent to the position-versus-time graph at that instant. The slope of this tangent line is equal to the value of the instantaneous velocity at that point. For example, the instantaneous velocity of the object in Figure 1.8 at t = 3.0 sis 12 m / s. The table lists the instantaneous velocities of the object described by the graph in Figure 1.8. You can verify some of these values by measuring the slope of the curve. Position versus Time of an Object Showing Instantaneous Velocity
Finding Instantaneous Velocity The instantaneous velocity at a given time can be determined by measuring the slope of the line that is tangent to that point on the position-versus-time graph.
VELOCITY-TIME DATA
30
e C 0
20
E
42
t (s)
v (m/s)
0.0
0.0
1.0
4.0
2.0
8.0
3.0
12.0
4.0
16.0
Chapter 2
“‘ 0 Cl. 10
0 .__..:::::;;___’ – – – – – ‘ – – – – – – – J ‘ – – – – – – – J ‘ – – – 0 1.0 2.0 3.0 4.0 Time (s)

SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What is the shortest possible time in which a bacterium could travel a distance of8.4 cm across a Petri dish at a constant speed of3.5 mm/ s? 2. A child is pushing a shopping cart at a speed of 1.5 m / s. How long will it take this child to push the cart down an aisle with a length of9.3 m? 3. An athlete swims from the north end to the south end of a 50.0 m pool in 20.0 s and makes the return trip to the starting position in 22.0 s. a. What is the average velocity for the first half of the swim? b. What is the average velocity for the second half of the swim? c. What is the average velocity for the roundtrip? 4. Two students walk in the same direction along a straight path, at a constant speed-one at 0.90 mi s and the other at 1.90 m / s. a. Assuming that they start at the same point and the same time, how much sooner does the faster student arrive at a destination 780 m away? b. How far would the students have to walk so that the faster student arrives 5.50 min before the slower student?
Critical Thinking 5. Does knowing the distance between two objects give you enough information to locate the objects? Explain.
Interpreting Graphics 6. Figure 1.9 shows position-time graphs of the straight-line movement of two brown bears in a wildlife preserve. Which b ear has the greater average velocity over the entire period? Which bear has the greater velocity at t = 8.0 min? Is the velocity of bear A always positive? Is the velocity of bear B ever negative?
•iidihllli• Bear A
Bear B
3000
3000
2500
2500
g 2000
E 2000
~ 1500
C
.!2
:c: Cl)
·.;;
1500
0
:_ 1000
a. 1000
500
500 10
20
30
40
Time (min)
50
60
10
20
30
40
50
60
Time (min)
Motion in One Dimension
43
SECTION 2 Objectives ►
I
Describe motion in terms changing velocity.

Compare graphical representations of accelerated and nonaccelerated motions.

Apply kinematic equations to calculate distance, time, or velocity under conditions of constant acceleration.
Acceleration Key Term acceleration
Changes in Velocity Many bullet trains have a top speed of about 300 km/h. Because a train stops to load and unload passengers, it does not always travel at that top speed. For some of the time the train is in motion, its velocity is either increasing or decreasing. It loses speed as it slows down to stop and gains speed as it pulls away and heads for the next station.
Acceleration is the rate of change of velocity with respect to time. Similarly, when a shuttle bus approaches a stop, the driver begins to apply the brakes to slow down 5.0 s before actually reaching the stop. The speed changes from 9.0 mis to 0 mis over a time interval of 5.0 s. Sometimes, however, the shuttle stops much more quickly. For example, if the driver slams on the brakes to avoid hitting a dog, the bus slows from 9.0 mis to 0 mis in just 1.5 s. Clearly, these two stops are very different, even though the shuttle’s velocity changes by the same amount in both cases. What is different in these two examples is the time interval during which the change in velocity occurs. As you can imagine, this difference has a great effect on the motion of the bus, as well as on the comfort and safety of the passengers. A sudden change in velocity feels very different from a slow, gradual change. acceleration the rate at which velocity changes over time; an object accelerates if its speed, direction, or both change
The quantity that describes the rate of change of velocity in a given time interval is called acceleration. The magnitude of the average acceleration is calculated by dividing the total change in an object’s velocity by the time interval in which the change occurs. Average Acceleration
aavg=
b..v b..t
. average acceleration =
change in velocity . . d ti h time reqmre or c ange
Acceleration has dimensions of length divided by time squared. The units of acceleration in SI are meters per second per second, which is written as meters per second squared, as shown below. When m easured in these units, acceleration describes how much the velocity changes in each second. (mi s) m 1 m –= S
44
Chapter 2
S
X – = S
s2
PREMIUM CONTENT
~ interactive Demo \::,/ HMDScience.com
Sample Problem B A shuttle bus slows down with an average acceleration of – 1.8 m/ s2 • How long does it take the bus to slow from 9.0 m/s to a complete stop?
0
ANALYZE
vi= 9.0 mis
Given:
v1 = 0mls aavg= -1.8 mls Unknown:
f:}
SOLVE
2
fit = ?
Rearrange the average acceleration equation to solve for the time interval.
aavg=
!:iv fit
0 mis – 9.0 mis -l.8mls2
Practice 1. As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m / s2 as it slows from 9.0 mi s to 0.0 m/ s. Find the time interval of
acceleration for the bus.
2. A car traveling at 7.0 m / s accelerates uniformly at 2.5 m / s2 to reach a speed of 12.0 m / s. How long does it take for this acceleration to occur? 2
3. With an average acceleration of – 1.2 m / s , how long will it take a cyclist to bring a bicycle with an initial speed of 6.5 m / s to a complete stop? 4. Turner’s treadmill runs with a velocity of – 1.2 m /s and speeds up at regular intervals during a half-hour workout. After 25 min, the treadmill has a velocity of -6.5 m / s. What is the average acceleration of the treadmill during this period? 5. Suppose a treadmill has an average acceleration of 4.7 x 10- 3 m/ s2 .
a. How much does its speed change after 5.0 min? b. If the treadmill’s initial speed is 1.7 m / s, what will its final speed be?
Motion in One Dimension
45
Acceleration has direction and magnitude. High-Speed Train Highspeed trains such as this one can travel at speeds of about 300 km/h (186 mi/h).
shows a high-speed train leaving a station. Imagine that the train is moving to the right so that the displacement and the velocity are positive. The velocity increases in magnitude as the train picks up speed. Therefore, the final velocity will be greater than the initial velocity, and b..v will be positive. When b..v is positive, the acceleration is positive. Figure 2.1
On long trips with no stops, the train may travel for a while at a constant velocity. In this situation, because the velocity is not changing, b..v = 0 m / s. When the velocity is constant, the acceleration is equal to zero. Imagine that the train, still traveling in the positive direction, slows down as it approaches the next station. In this case, the velocity is still positive, but the initial velocity is larger than the final velocity, so b..v will be negative. When b..v is negative, the acceleration is negative.
The slope and shape of the graph describe the object’s motion. As with all motion graphs, the slope and shape of the velocity-time graph in Figure 2.2 allow a detailed analysis of the train’s motion over time. When the train leaves the station, its speed is increasing over time. The line on the graph plotting this motion slopes up and to the right, as at point A on the graph. When the train moves with a constant velocity, the line on the graph continues to the right, but it is horizontal, with a slope equal to zero. This indicates that the train’s velocity is constant, as at point B on the graph.
Fly Ball If a baseball has zero
velocity at some instant, is the acceleration of the baseball necessarily zero at that instant? Explain, and give examples. Runaway Train If a passenger
train is t raveling on a st raight track with a negative velocity and a positive acceleration , is it speed ing up or slowing down? Hike-and-Bike Trail When
Jennifer is out for a ride, she slows down on her bike as she approaches a group of hikers on a trail. Explain how her acceleration can be positive even t hough her speed is decreasing.
46
Chapter 2
Finally, as the train approaches the station, its velocity decreases over time. The graph segment representing this motion slopes down to the right, as at point C on the graph. This downward slope indicates that the velocity is decreasing over time. A negative value for the acceleration does not always indicate a decrease in speed. For example, if the train were moving in the negative direction, the acceleration would be negative when the train gained speed to leave a station and positive when the train lost speed to enter a station.
Velocity-Time Graphs When the velocity in the positive direction is increasing, the acceleration is positive, as at point A. When the velocity is constant, there is no acceleration, as at point B. When the velocity in the positive direction is decreasing, the acceleration is negative, as at point C.
Velocity-Time Graph of a Train
B
Time
“cc
Figure 2.3 shows how the signs
of the velocity and acceleration can be combined to give a description of an object’s motion. From this table, you can see that a negative acceleration can describe an object that is speeding up (when the velocity is negative) or an object that is slowing down (when the velocity is positive). Use this table to check your answers to problems involving acceleration. For example, in Figure 2.2 the initial velocity vi of the train is positive. At point A on the graph, the train’s velocity is still increasing, so its acceleration is positive as well. The first entry in Figure 2.3 shows that in this situation, the train is speeding up. At point C, the velocity is still positive, but it is decreasing, so the train’s acceleration is negative. Figure 2.3 tells you that in this case, the train is slowing down.
Motion of a Falling Ball
+
a
Motion
+
speeding up
The motion in this picture took place in about 1.00 s. In this short time interval, your eyes could only detect a blur. This photo shows what really happens within that time.
speeding up slowing down
+
+
slowing down
-or+
0
constant velocity
0
– or+
0
0
speeding up from rest remaining at rest
Motion with constant acceleration. Figure 2.4 is
a strobe photograph of a ball moving in a straight line with constant acceleration. While the ball was moving, its image was captured ten times in one second, so the time interval between successive images is 0.10 s. As the ball’s velocity increases, the ball travels a greater distance during each time interval. In this example, the velocity increases by exactly the same amount during each time interval. Thus, the acceleration is constant. Because the velocity increases for each time interval, the successive change in displacement for each time interval increases. You can see this in the photograph by noting that the distance between images increases while the time interval between images rem ains constant. The relationships between displacement, velocity, and constant acceleration are expressed by equations that apply to any object moving with constant acceleration.
Motion in One Dimension
47
Velocity versus Time of a Falling Ball
Constant Acceleration and Average Velocity If a ball moved for the same time with a constant velocity equal to vavg• it would have the same displacement as the ball in Figure 2.4 moving with constant acceleration.
120 110 100 90 ci, 80
E 7o ‘-‘
;:: 60 ~ 50 0 ;g! 40 30 20 10 0 – —-+—-+–l—–+—-+–l—-+—-+–1——-1 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 Time (s)
Displacement depends on acceleration, initial velocity, and time. Figure 2.5 is
a graph of the ball’s velocity plotted against time. The initial, final, and average velocities are marked on the graph. We know that the average velocity is equal to displacement divided by the time interval.
b.x Vavg = b.t For an object moving with constant acceleration, the average velocity is equal to the average of the initial velocity and the final velocity.
..Did YOU Know?__ ______ __. Decreases in speed are sometimes called decelerations. Despite the sound of the name, decelerations are really a special case of acceleration in which the magnitude of the velocity- and thus the speed- decreases with time.
Vi+ VJ Vavg = – – 2 –
. initial velocity + final velocity average velocity = 2
To find an expression for the displacement in terms of the initial and final velocity, we can set the expressions for average velocity equal to each other. b.x vi + vJ b.t = Vavg = – 2displacement time interval
initial velocity + final velocity 2
Multiplying both sides of the equation by b.t gives us an expression for the displacement as a function of time. This equation can be used to find the displacement of any object moving with constant acceleration. Displacement with Constant Acceleration
displacement = ½(initial velocity+ final velocity)(time interval)
48
Chapter 2
PREMIUM CONTENT
~ interactive Demo \::,/ HMDScience.com
Sample Problem C A racing car reaches a speed of 42 m/s. It then begins a uniform negative acceleration, using its parachute and braking system, and comes to rest 5.5 slater. Find the distance that the car travels during braking.
0
ANALYZE
Given:
= 42m/s v = Om/s 1
v.l
bi.t = 5.5 s Unknown:
E)
SOLVE
bi.x =? Tips and Tricks Remember that this equation applies only when acceleration is constant. In this problem, you know that acceleration is constant by the phrase “uniform negative acceleration.” All of the kinematic equations introduced in this chapter are valid only for constant acceleration.
Use the equation that relates displacement, initial and final velocities, and the time interval.
bi.x =
1
(vi + v1 )6i..t
2 bi.x = ~ (42 mis+ Om/s)(5.5 s)
Calculator Solution The calculator answer is 115.5. However, the velocity and time values have only two significant figures each, so the answer must be reported as 120 m.
Practice 1. A car accelerates uniformly from rest to a speed of 6.6 m /s in 6.5 s. Find the distance the car travels during this time. 2. When Maggie applies the brakes of her car, the car slows uniformly from 15.0 m / s to 0.0 m i s in 2.50 s. How many meters before a stop sign must she apply her brakes in order to stop at the sign? 3. A driver in a car traveling at a speed of 21.8 m / s sees a cat 101 m away on the road. How long will it take for the car to accelerate uniformly to a stop in exactly 99 m?
4. A car enters the freeway with a speed of 6.4 m / s and accelerates uniformly for 3.2 km in 3.5 min. How fast (in m / s) is the car moving after this time?
Motion in One Dimension
49
Final velocity depends on initial velocity, acceleration, and time. What if the final velocity of the ball is not known but we still want to calculate the displacement? If we know the initial velocity, the acceleration, and the elapsed time, we can find the final velocity. We can then use this value for the final velocity to find the total displacement of the ball. By rearranging the equation for acceleration, we can find a value for the final velocity.
ab,.t= vJ- vi By adding the initial velocity to both sides of the equation, we get an equation for the final velocity of the ball.
ab,.t + vi = vJ Velocity with Constant Acceleration
final velocity= initial velocity+ (acceleration x time interval)
You can use this equation to find the final velocity of an object after it has accelerated at a constant rate for any time interval. If you want to know the displacement of an object moving with
constant acceleration over some certain time interval, you can obtain another useful expression for displacement by substituting the expression for vJinto the expression for fu.
b,.x =½(vi+ v}.6.t b,.x =½(vi+ vi+ a.6.t).6.t b,.x = ½[2vi .6.t + a(.6.t)2] Displacement with Constant Acceleration
displacement = (initial velocity x time interval) + acceleration x (time interval) 2
½
This equation is useful not only for finding the displacement of an object moving with constant acceleration but also for finding the displacement required for an object to reach a certain speed or to come to a stop. For the latter situation, you need to use both this equation and the equation given above.
50
Chapter 2
PREMIUM CONTENT
g \ Interactive Demo
Velocity and Displacement with Constant Acceleration
~ HMDScience.com
Sample Problem D A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s2 for 15 s before takeoff. What is its speed at takeoff? How long must the runway be for the plane to be able to take off?
0
ANALYZE
Given:
vi= 0 mis a= 4.8mls
2
6..t = 15 s Unknown:
E)
SOLVE
First, use the equation for the velocity of a uniformly accelerated object.
v1 = vi+ a6..t
Tips and Tricks Because you now know v,, you could also use the equation
v1 = 0 mis+ (4.8 mls2)(15 s)
& = ½(v; + v,)(.6.t), or & = ½(72 m/s)(15 s) = 540 m.
Then, use the displacement equation that contains the given variables.
6..x
= vi6..t + ; a(6..t)
6..x
= (0 mls)(l5 s) + ; (4.8
2
2 mls )(15
2 s)
16..x= 540 m I Practice 1. A car with an initial speed of 6.5 m / s accelerates at a uniform rate of
0.92 m / s2 for 3.6 s. Find the final speed and the displacement of the car during this time. 2. An automobile with an initial speed of 4.30 m/ s accelerates uniformly at the rate
of3.00 m / s2 • Find the final speed and the displacement after 5.00 s. 3. A car starts from rest and travels for 5.0 s with a constant acceleration of – 1.5 m / s2 • What is the final velocity of the car? How far does the car travel in this time interval? 4. A driver of a car traveling at 15.0 m /s applies the brakes, causing a uniform acceleration of -2.0 m / s 2 • How long does it take the car to accelerate to a final speed of 10.0 m / s? How far h as the car moved during the braking period?
Motion in One Dimension
51
Final velocity depends on initial velocity, acceleration, and displacement.
, Did YOU Know?_- – – – – – – – – – – , The word physics comes from the ancient Greek word for “nature.” According to Aristotle, who assigned the name, physics is the study of natural events. Aristotle believed that the study of motion was the basis of physics. Galileo developed the foundations for the modern study of motion using mathematics. In 1632, Galileo published the first mathematical treatment of motion.
So far, all of the equations for motion under uniform acceleration have required knowing the time interval. We can also obtain an expression that relates displacement, velocity, and acceleration without using the time interval. This method involves rearranging one equation to solve for b..t and substituting that expression in another equation, making it possible to find the final velocity of a uniformly accelerated object without knowing how long it has been accelerating. Start with the following equation for displacement: b..x =½(vi+ vJ)b..t
Now multiply both sides by 2. Next, divide both sides by (vi + vJ) to solve for b..t.
Now that we have an expression for b..t, we can substitute this expression into the equation for the final velocity. vJ= vi+ a(b..t)
In its present form, this equation is not very helpful because vJ appears on both sides. To solve for vJ, first subtract vi from both sides of the equation.
Next, multiply both sides by (vi+ vJ) to get all the velocities on the same side of the equation. (vJ- v) (vJ+ v)
Add
= 2ab..x = vf- v/
v/ to both sides to solve for v/
Final Velocity After Any Displacement vf
= v/ + 2ab..x
(final velocity) 2 = (initial velocity) 2 + 2(acceleration)(displacement~
When using this equation, you must take the square root of the right side of the equation to find the final velocity. Remember that the square root may be either positive or negative. If you have been consistent in your use of the sign convention, you will be able to determine which value is the right answer by reasoning based on the direction of the motion.
52
Chapter 2
l
Sample Problem E A person pushing a stroller starts from rest, uniformly accelerating at a rate of 0.500 m/ s2 • What is the velocity of the stroller after it has traveled 4. 75 m 1
0
ANALYZE
Given:
vi= 0 mis
a= 0.500 m/s ~X=
Unknown:
2
4.75m

Diagram:
-x
+x
Choose a coordinate system. The most convenient one has an origin at the initial location of the stroller. The positive direction is to the right.
E}
PLAN
Choose an equation or situation: Because the initial velocity, acceleration, and displacement are known, the final velocity can be found by using the following equation:
v/ = v/ + 2a~x Rearrange the equation to isolate the unknown: Take the square root of both sides to isolate vi’
vJ=
E)
SOLVE
Substitute the values into the equation and solve:
vJ=
0
CHECKYOUR WORK
±y(vJ + 2a~x 2
Tips and Tricks Think about the physical situation to determine whether to keep the positive or negative answer from the square root. In this case, the stroller is speeding up because it starts from rest and ends with a speed of 2.18 m/s. An object that is speeding up and has a positive acceleration must have a positive velocity, as shown in Figure 2.3. So, the final velocity must be positive.
±y(o m/s)2 + 2(0.500 m/s )(4.75 m) 2
The stroller’s velocity after accelerating for 4.75 mis 2.18 m / s to the right.
Motion in One Dimension
53
Final Velocity After Any Displacement (continued) I
Practice 1. Find the velocity after the stroller in Sample Problem E has traveled 6.32 m.
2. A car traveling initially at+ 7.0 m /s accelerates uniformly at the rate of +0.80 m /s 2
for a distance of 245 m. a. What is its velocity at the end of the acceleration? b. What is its velocity after it accelerates for 125 m?
c. What is its velocity after it accelerates for 67 m?
3. A car accelerates uniformly in a straight line from rest at the rate of 2.3 m / s2 •
a. What is the speed of the car after it has traveled 55 m? b. How long does it take the car to travel 55 m?
4. A motorboat accelerates uniformly from a velocity of 6.5 m/ s to the west to a velocity of 1.5 m/ s to the west. If its acceleration was 2.7 m / s2 to the east, how far did it travel during the acceleration? 5. An aircraft has a liftoff speed of 33 m/s. What minimum constant acceleration
does this require if the aircraft is to be airborne after a takeoff run of 240 m? 6. A certain car is capable of accelerating at a uniform rate of0.85 m /s2 • What is the
magnitude of the car’s displacement as it accelerates uniformly from a speed of 83 km/ h to one of94 km/ h?
With the four equations presented in this section, it is possible to solve any problem involving one-dimensional motion with uniform acceleration. For your convenience, the equations that are used most often are listed in Figure 2.6. The first column of the table gives the equations in their standard form. For an object initially at rest, vi= 0. Using this value for vi in the equations in the first column will result in the equations in the second column. It is not necessary to memorize the equations in the second column. If vi= 0 in any problem, you will naturally derive this form of the equation. Referring back to the sample problems in this chapter will guide you through using these equations to solve many problems.
Form to use when accelerating object has an initial velocity
Form to use when object accelerating starts from rest fl.x = ½v1fl.t
v1 = vi+ afl.t fl.x = vifl.t + ½a(fl.t) v/
54
Chapter 2
= v/ + 2afl.x
VJ= afl.t 2
fl.x = la(fl.t) 2 2
v/
= 2afl.x

SECTION 2 FORMATIVE ASSESSMENT
0 Reviewing Main Ideas 1. Marissa’s car accelerates uniformly at a rate of +2.60 m/s2 . How long does it take for Marissa’s car to accelerate from a speed of 24.6 m/s to a speed of26.8 m/s?
2. A bowling ball with a negative initial velocity slows down as it rolls down the lane toward the pins. Is the bowling ball’s acceleration positive or negative as it rolls toward the pins? 3. Nathan accelerates his skateboard uniformly along a straight path from rest to 12.5 m/s in 2.5 s. a. What is Nathan’s acceleration? b. What is Nathan’s displacement during this time interval? c. What is Nathan’s average velocity during this time interval?
Critical Thinking 4. Two cars are moving in the same direction in parallel lanes along a highway. At some instant, the instantaneous velocity of car A exceeds the instantaneous velocity of car B. Does this mean that car Ns acceleration is greater than car B’s? Explain, and use examples.
Interpreting Graphics 5. The velocity-versus-time graph for a shuttle bus moving along a straight path is shown in Figure 2.7. a. Identify the time intervals during which the velocity of the shuttle bus is constant. Velocity Versus Time of a Shuttle Bus b. Identify the time intervals during which the acceleration of the shuttle bus is constant. c. Find the value for the average velocity of the shuttle bus during each time interval identified in b. d. Find the acceleration of the shuttle bus during each time interval identified in b. 0 400 500 600 e. Identify the times at which the velocity of the shuttle bus is zero. f. Identify the times at which the acceleration of the shuttle bus is zero. Time (s) g. Explain what the slope of the graph reveals about the acceleration in each time interval.
6. Is the shuttle bus in item 5 always moving in the same direction? Explain, and refer to the time intervals shown on the graph.
Motion in One Dimension
55
SECTION 3 Objectives ►

Relate the motion of a freely falling body to motion with constant acceleration. Calculate displacement, velocity, and time at various points in the motion of a free ly falling object.
Free Fall in a Vacuum When there is no air resistance, all objects fall with the same acceleration regardless of their masses.
Falling Objects Key Term free fall
Free Fall On August 2, 1971, a demonstration was conducted on the moon by astronaut David Scott. He simultaneously released a hammer and a feather from the same height above the moon’s surface. The hammer and the feather both fell straight down and landed on the lunar surface at exactly the same moment. Although the hammer is more massive than the feather, both objects fell at the same rate. That is, they traveled the same displacement in the same amount of time.
Freely falling bodies undergo constant acceleration. In Figure 3.1, a feather and an apple are released from rest in a vacuum chamber. The two objects fell at exactly the same rate, as indicated by the horizontal alignment of the multiple images. The amount of time that passed between the first and second images is equal to the amount of time that passed between the fifth and sixth images. The picture, however, shows that the displacement in each time interval did not remain constant. Therefore, the velocity was not constant. The apple and the feather were accelerating. Compare the displacement between the first and second images to the displacement between the second and third images. As you can see, within each time interval the displacement of the feather increased by the same amount as the displacement of the apple. Because the time intervals are the same, we know that the velocity of each object is increasing by the same amount in each time interval. In other words, the apple and the feather are falling with the same constant acceleration.
free fall the motion of a body when only the force due to gravity is acting on the body
56
Chapter 2
If air resistance is disregarded, all objects dropped near the surface of a planet fall with the same constant acceleration. This acceleration is due to gravitational force, and the motion is referred to as free fall. The acceleration due to gravity is denoted with the symbols a g (generally) or g (on Earth’s surface). The magnitude of g is about 9.81 m/s2 , or 32 ft/ s2 . Unless stated otherwise, this book will use the value 9.81 m/s2 for calculations. This acceleration is directed downward, toward the center of Earth. In our usual choice of coordinates, the downward direction is negative. Thus, the acceleration of objects in free fall near the surface of Earth is a g = – g = -9.81 m/s2 . Because an object in free fall is acted on only by gravity, a g is also known as free-fall acceleration.
Acceleration is constant during upward and downward motion. Figure 3.2 is
a strobe photograph of a ball thrown up into the air with an initial upward velocity of+ 10.5 m / s. The photo on the left shows the ball moving up from its release toward the top of its path, and the photo on the right shows the ball falling back down. Everyday experience shows that when we throw an object up in the air, it will continue to move upward for some time, stop momentarily at the peak, and then change direction and begin to fall. Because the object changes direction, it may seem that the velocity and acceleration are both changing. Actually, objects thrown into the air have a downward acceleration as soon as they are released.
Motion of a Tossed Ball At the very top of its path, the ball’s velocity is zero, but the ball’s acceleration is – 9.81 m/s 2 at every point- both when it is moving up (a) and when it is moving down (b). (a)
(b)
In the photograph on the left, the upward displacement of the ball between each successive image is smaller and smaller until the ball stops and finally begins to move with an increasing downward velocity, as shown on the right. As soon as the ball is released with an initial upward velocity of+ 10.5 m / s, it has an acceleration of -9.81 m / s2 . After 1.0 s (.6.t = 1.0 s), the ball’s velocity will change by-9.81 m / s to 0.69 m / s upward. After 2.0 s (.6.t = 2.0 s), the ball’s velocity will again change by -9.81 m / s, to -9.12 m / s. The graph in Figure 3.3 shows the velocity of the ball plotted against time. As you can see, there is an instant when the velocity of the ball is equal to 0 m / s. This happens at the instant when the ball reaches the peak of its upward motion and is about to begin moving downward. Although the velocity is zero at the instant the ball reaches the peak, the acceleration is equal to -9.81 m / s2 at every instant regardless of the magnitude or direction of the velocity. It is important to note that the acceleration is -9.81 m / s2 even at the peak where the velocity is zero. The straight-line slope of the graph indicates that the acceleration is constant at every moment. Velocity versus Time of a Dropped Ball
Slope of a Velocity-Time Graph On this velocity-time graph, the slope of the line, which is equal to the ball’s acceleration, is constant from the moment the ball is released (t = 0.00 s) and throughout its motion.
‘l:’
“‘
.t=
·@”‘ a:
12 10 8 6 4 2 u, 0 g -2 -4 ~ -6 .:; > -8 -10 -12 -14 -16 -18 -20

‘iii
Time (s)
Motion in One Dimension
57
Freely falling objects always have the same downward acceleration. It may seem a little confusing to think of something that is moving upward, like the ball in the example, as having a downward acceleration. Thinking of this motion as motion with a positive velocity and a negative acceleration may help. The downward acceleration is the same when an object is moving up, when it is at rest at the top of its path, and when it is moving down. The only things changing are the position and the magnitude and direction of the velocity. When an object is thrown up in the air, it has a positive velocity and a negative acceleration. From Figure 2.3, we see that this means the object is slowing down as it rises in the air. From the example of the ball and from everyday experience, we know that this makes sense. The object continues to move upward but with a smaller and smaller speed. In the photograph of the ball, this decrease in speed is shown by the smaller and smaller displacements as the ball moves up to the top of its path. At the top of its path, the object’s velocity has decreased until it is zero. Although it is impossible to see this because it happens so quickly, the object is actually at rest at the instant it reaches its peak position. Even though the velocity is zero at this instant, the acceleration is 2 still-9.81 m / s . When the object begins moving down, it has a negative velocity and its acceleration is still negative. From Figure 2.3, we see that a negative acceleration and a negative velocity indicate an object that is speeding up. In fact, this is what happens when objects undergo free-fall acceleration. Objects that are falling toward Earth move faster and faster as they fall. In the photograph of the ball in Figure 3.2 (on the previous page), this increase in speed is shown by the greater and greater displacements between the images as the ball falls. Knowing the free-fall acceleration makes it easy to calculate the velocity, time, and displacement of many different motions using the equations for constantly accelerated motion. Because the acceleration is the same throughout the entire motion, you can analyze the motion of a freely falling object during any time interval.
Quick LAB Your reaction time affects your performance in all kinds of activities-from sports to driving to catching something that you drop. Your reaction time is the time interval between an event and your response to it. Determine your reaction time by having a friend hold a meterstick vertically between the thumb and index finger of your open hand. The meterstick should be held
58
Chapter 2
so that the zero mark is between your fingers with the 1 cm mark above it. You should not be touching the meterstick, and your catching hand must be resting on a table. Without warning you, your friend should release the meterstick so that it falls between your thumb and your finger. Catch the meterstick as quickly as you can. You can calculate your reaction time
from the free-fall acceleration and the distance the meterstick has fallen through your grasp.
MATERIALS • meterstick or ruler SAFETY ~
‘iii”‘
Avoid eye injury; do not swing metersticks.
PREMIUM CONTENT
~ interactive Demo \::,/ HMDScience.com
Sample Problem F Jason hits a volleyball so that it moves with an initial velocity of 6.0 m/s straight upward. If the volleyball starts from 2.0 m above the floor, how long will it be in the air before it strikes the floor?
0
ANALYZE
vi=+ 6.0 m/s
Given:
a= -g = -9.81 m/s
2
~y= -2.0m Unknown:
~t=?
Diagram:
Place the origin at the starting point of the ball (yi = 0 at ti = O).
+ y
f 6.0 mis X
2.0 m
E)
PLAN
Choose an equation or situation: Both tlt and vJ are unknown. Therefore, first solve for vJ using the equation that does not require time. Then the equation for vJ that does involve time can be used to solve for tlt.
v/= v/+2a~y Rearrange the equations to isolate the unknown: Take the square root of the first equation to isolate vf’ The second equation must be rearranged to solve for tlt.
vJ= ±yv/ + 2a~y
E)
SOLVE
Tips and Tricks When you take the square root to find vf’ select the negative answer because the ball will be moving toward the floor in the negative direction.
Q
CHECKYOUR WORK
Cd·i ,rn ,\114-►
Substitute the values into the equations and solve: First find the velocity of the ball at the moment that it hits the floor.
vJ= ±yv/ + 2a~y = ±y(6.0 m/s) 2 + 2(-9.81 m/s 2)(-2.0 m) 2
2
2
2
2
2
vJ= ±V36 m / s = 39 m / s = ±V75 m /s = -8.7 m/s Next, use this value of
vJ- vi ~t = – – a
vin the second equation to solve for tlt.
-8.7 m /s – 6.0 m /s 2 -9.81 m /s
-14.7 m /s 2 -9.81 m/s
The solution, 1.50 s, is a reasonable amount of time for the ball to be in the air.
Motion in One Dimension
59
Falling Object (continued) I
Practice 1. A robot probe drops a camera off the rim of a 239 m high cliff on Mars, where the free-fall acceleration is -3.7 m/s2 •
a. Find the velocity with which the cam era hits the ground. b. Find the time required for it to hit the ground. 2. A flowerpot falls from a windowsill 25.0 m above the sidewalk. a. How fast is the flowerpot moving when it strikes the ground? b. How much time does a passerby on the sidewalk below have to move out of the
way before the flowerpot hits the ground? 3. A tennis ball is thrown vertically upward with an initial velocity of +8.0 m/s.
a. What will the ball’s speed be when it returns to its starting point? b. How long will the ball take to reach its starting point?
4. Calculate the displacement of the volleyball in Sample Problem F when the volleyball’s final velocity is 1.1 m / s upward.
Sky Diving hen these sky divers jump from an airplane, they plummet toward the ground. If Earth had no atmosphere, the sky divers would accelerate with the free-fall acceleration, g, equal to 9.81 m/s2. They would not slow down even after opening their parachutes. Fortunately, Earth does have an atmosphere, and the acceleration of the sky divers does not remain constant. Instead, because of air resistance, the acceleration decreases as they fall. After a few seconds, the acceleration drops to zero and the speed becomes constant. The constant speed an object reaches when falling through a resisting medium is called terminal velocity. The terminal velocity of an object depends on the object’s mass, shape, and size. When a sky diver is spread out horizontally to the ground, the sky diver’s
60
Chapter 2
terminal velocity is typically about 55 m/s (123 mi/h). If the sky diver curls into a ball, the terminal velocity may increase to close to 90 mis (200 mi/h). When the sky diver opens the parachute, air resistance increases, and the sky diver decelerates to a new, slower terminal velocity. For a sky diver with an open parachute, the terminal velocity is typically about 5 m/s (11 mi/h).

SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. A coin is tossed vertically upward. a. What happens to its velocity while it is in the air? b. Does its acceleration increase, decrease, or remain constant while it is in the air?
2. A pebble is dropped down a well and hits the water 1.5 slater. Using the equations for motion with constant acceleration, determine the distance from the edge of the well to the water’s surface. 3. A ball is thrown vertically upward. What are its velocity and acceleration when it reaches its maximum altitude? What is its acceleration just before it hits the ground? 4. Two children are bouncing small rubber balls. One child simply drops a ball. At the same time, the second child throws a ball downward so that it has an initial speed of 10 m/ s. What is the acceleration of each ball while in motion?
Critical Thinking 5. A gymnast practices two dismounts from the high bar on the uneven parallel bars. During one dismount, she swings up off the bar with an initial upward velocity of +4.0 m/s. In the second, she releases from the same height but with an initial downward velocity of – 3.0 m/ s. What is her acceleration in each case? How does the first final velocity as the gymnast reaches the ground differ from the second final velocity?
Interpreting Graphics 6. Figure 3.4 is a position-time graph of the motion of a basketball thrown straight up. Use the graph to sketch the path of the basketball and to sketch a velocity-time graph of the basketball’s motion.
Velocity-Time Graph of a Basketball
0.5 0.0 – 0.5
-i’–+—+~
Time (s) 1.4
‘lr—+—+—-
Time
Time
1. Which graph represents an object moving with a
constant positive velocity? A. I B. II C. III D. IV 2. Which graph represents an object at rest? F. I G. II H. III J. IV
3. Which graph represents an object moving with constant positive acceleration? A. I B. II C. III D. IV
4. A bus travels from El Paso, Texas, to Chihuahua, Mexico, in 5.2 h with an average velocity of73 km/h to the south. What is the bus’s displacement? F. 73 km to the south G. 370 km to the south H. 380 km to the south J. 14 km/ h to the south
76
Chapter 2
4.0

5.0
5. What is the squirrel’s displacement at time t = 3.0 s? A. -6.0m B. -2.0m C. +0.8m D. +2.0m 6. What is the squirrel’s average velocity during the time interval between 0.0 sand 3.0 s? F. -2.0 m /s G. -0.67 mis H. 0.0m/s J. +0.53m/s 7. Which of the following statements is true of
acceleration? A. Acceleration always has the same sign as displacement. B. Acceleration always has the same sign as velocity. C. The sign of acceleration depends on both the direction of motion and how the velocity is changing. D. Acceleration always has a positive sign. 8. A ball initially at rest rolls down a hill and has an
acceleration of 3.3 m /s2 • If it accelerates for 7.5 s, how far will it move during this time? F. 12m G. 93m H. 120m J. 190m
. TEST PREP
9. Which of the following statements is true for a ball thrown vertically upward? A. The ball has a negative acceleration on the way up and a positive acceleration on the way down. B. The ball has a positive acceleration on the way up and a negative acceleration on the way down. C. The ball has zero acceleration on the way up and a positive acceleration on the way down. D. The ball has a constant acceleration throughout its flight.
SHORT RESPONSE 10. In one or two sentences, explain the difference between displacement and distance traveled. 11. The graph below shows the position of a runner at different times during a run. Use the graph to determine the runner’s displacement and average velocity: a. for the time interval from t = 0.0 min to t= 10.0min b. for the time interval from t = 10.0 min to t= 20.0min c. for the time interval from t = 20.0 min to t= 30.0min d. for the entire run
4.0
… 0
For both graphs, assume the object starts with a positive velocity and a positive displacement from the origin. 13. A snowmobile travels in a straight line. The snow-
mobile’s initial velocity is +3.0 m / s. a. If the snowmobile accelerates at a rate of +0.50 m/ s2 for 7.0 s, what is its final velocity? b. If the snowmobile accelerates at the rate of – 0.60 m / s2 from its initial velocity of +3.0 m / s, how long will it take to reach a complete stop?
EXTENDED RESPONSE 14. A car moving eastward along a straight road
increases its speed uniformly from 16 m / s to 32 m/ s in 10.0 s. a. What is the car’s average acceleration? b. What is the car’s average velocity? c. How far did the car move while accelerating? Show all of your work for these calculations. 15. A ball is thrown vertically upward with a speed of 25.0 m / s from a height of2.0 m. a. How long does it take the ball to reach its highest point? b. How long is the ball in the air? Show all of your work for these calculations.
5.0
~
12. For an object moving with constant negative acceleration, draw the following: a. a graph of position versus time b. a graph of velocity versus time
3.0
C
0
:;:::
·;;;
2.0
0
Cl.
1.0 0.0 0.0
10.0
20.0 Time (min)
30.0
40.0
Test Tip When filling in your answers on an answer sheet, always check to make sure you are filling in the answer for the right question. If you have to change an answer, be sure to completely erase your previous answer.
Standards-Based Assessment
77
Vx
r
– Vy
rryVx
Vx
Vx
Vy
\ Vy

\
\
V
SECTION 1 Objectives ►
I ► I ►
Distinguish between a scalar and a vector. Add and subtract vectors by using the graphical method. Multiply and divide vectors by scalars.
Introduction to Vectors Key Terms scalar
vector
resultant
Scalars and Vectors In the chapter “Motion in One Dimension;’ our discussion of motion was limited to two directions, forward and backward. Mathematically, we described these directions of motion with a positive or negative sign. That method works only for motion in a straight line. This chapter explains a method of describing the motion of objects that do not travel along a straight line.
Vectors indicate direction; scalars do not. scalar a physical quantity that has magnitude but no direction vector a physical quantity that has both magnitude and direction
Each of the physical quantities encountered in this book can be categorized as either a scalar quantity or a vector quantity. A scalar is a quantity that has magnitude but no direction. Examples of scalar quantities are speed, volume, and the number of pages in this textbook. A vector is a physical quantity that has both direction and magnitude. Displacement is an example of a vector quantity. An airline pilot planning a trip must know exactly how far and which way to fly. Velocity is also a vector quantity. If we wish to describe the velocity of a bird, we must specify both its speed (say, 3.5 mis) and the direction in which the bird is flying (say, northeast). Another example of a vector quantity is acceleration.
Vectors are represented by boldface symbols.
Length of Vector Arrows The lengths of the vector arrows represent the magnitudes of these two soccer players’ velocities.
In physics, quantities are often represented by symbols, such as t for time. To help you keep track of which symbols represent vector quantities and which are used to indicate scalar quantities, this book will use boldface type to indicate vector quantities. Scalar quantities will be in italics. For example, the speed of a bird is written as v = 3.5 mis. But a velocity, which includes a direction, is written as v = 3.5 mis to the northeast. When writing a vector on your paper, you can distinguish it from a scalar by drawing an arrow above the abbreviation for a quantity, such as = 3.5 mis to the northeast.
u
One way to keep track of vectors and their directions is to use diagrams. In diagrams, vectors are shown as arrows that point in the direction of the vector. The length of a vector arrow in a diagram is proportional to the vector’s magnitude. For example, in Figure 1.1 the arrows represent th e velocities of the two soccer players running toward the soccer ball.
80
Chapter 3
A resultant vector represents the sum of two or more vectors.
‘ .Did YOU Know? : The word vector is also used by airline I
When adding vectors, you must make certain that they have the same units and describe similar quantities. For example, it would be meaningless to add a velocity vector to a displacement vector because they describe different physical quantities. Similarly, it would be meaningless, as well as incorrect, to add two displacement vectors that are not expressed in the same units. For example, you cannot add meters and feet together.
pilots and navigators. In this context, , a vector is the particular path followed : or to be followed, given as a compass ‘ heading.
The chapter “Motion in One Dimension” covered vector addition and subtraction in one dimension. Think back to the example of the gecko that ran up a tree from a 20 cm marker to an 80 cm marker. Then the gecko reversed direction and ran back to the 50 cm marker. Because the two parts of this displacement are each vectors, they can be added together to give a total displacement of 30 cm. The answer found by adding two vectors in this way is called the resultant.
resultant a vector that represents the sum of two or more vectors
Vectors can be added graphically. Consider a student walking 1600 m to a friend’s house and then 1600 m to school, as shown in Figure 1.2. The student’s total displacement during his walk to school is in a direction from his house to the school, as shown by the dotted line. This direct path is the vector sum of the student’s displacement from his house to his friend’s house and his displacement from the friend’s house to school. How can this resultant displacement be found? One way to find the magnitude and direction of the student’s total displacement is to draw the situation to scale on paper. Use a reasonable scale, such as 50 m on land equals 1 cm on paper. First draw the vector representing the student’s displacement from his house to his friend’s house, giving the proper direction and scaled magnitude. Then draw the vector representing his walk to the school, starting with the tail at the head of the first vector. Again give its scaled magnitude and the right direction. The magnitude of the resultant vector can then be determined by using a ruler. Measure the length of the vector pointing from the tail of the first vector to the head of the second vector. The length of that vector can then be multiplied by 50 (or whatever scale you have chosen) to get the actual magnitude of the student’s total displacement in meters. The direction of the resultant vector may be determined by using a protractor to measure the angle between the resultant and the first vector or between the resultant and any chosen reference line.
Graphical Method of Vector Addition A student walks from his house to his friend’s house (a) , then from his friend’s house to the school (b). The student’s resultant displacement (c) can be found by using a ruler and a protractor.
I
I I
(C)
Two-Dimensional Motion and Vectors
81
Properties of Vectors Now consider a case in which two or more vectors act at the same point. When this occurs, it is possible to find a resultant vector that has the same net effect as the combination of the individual vectors. Imagine looking down from the second level of an airport at a toy car moving at 0.80 m / s across a walkway that moves at 1.5 m/ s. How can you determine what the car’s resultant velocity will look like from your view point?
Vectors can be moved parallel to themselves in a diagram. Note that the car’s resultant velocity while moving from one side of the walkway to the other will be the combination of two independent motions. Thus, the moving car can be thought of as traveling first at 0.80 m/s across the walkway and then at 1.5 m / s down the walkway. In this way, we can draw a given vector anywhere in the diagram as long as the vector is parallel to its previous alignment (so that it still points in the same direction).
Triangle Method of Addition The resultant velocity (a) of a toy car moving at a velocity of 0.80 m/s (b) across a moving walkway with a velocity of 1.5 m/s (c) can be found using a ruler and a protractor.
Thus, you can draw one vector with its tail starting at the tip of the other as long as the size and direction of each vector do not change. This process is illustrated in Figure 1.3. Although both vectors act on the car at the same point, the horizontal vector has been moved up so that its tail begins at the tip of the vertical vector. The resultant vector can then be drawn from the tail of the first vector to the tip of the last vector. This method is known as the triangle (or polygon) method of addition.
vwalkway= 1.5 m/s
“‘
—8 0 00
0
II

0.5c 0.5
1.0
1.5
2.0
Kinetic energy (KE/mc 2)
176
Einstein’s relativistic expression for kinetic energy has been confirmed by experiments in which electrons are accelerated to extremely high speeds in particle accelerators. In all cases, the experimental data correspond to Einstein’s equation rather than to the classical equation. Nonetheless, the difference between the two theories at low speeds (relative to c) is so minimal that the classical equation can be used in all such cases when the speed is much less than c.
Chapter 5
This equation shows that an object has a certain amount of energy (ER), known as rest energy, simply by virtue of its mass. The rest energy of a body is equal to its mass, m, multiplied by the speed of light squared, c2-. Thus, the mass of a body is a measure of its rest energy. This equation is significant because rest energy is an aspect of special relativity that was not predicted by classical physics.
Stanford Linear Accelerator Electrons in the Stanford Linear Accelerator in California (SLAC) reach 99.999999967 percent of the speed of light. At such great speeds, the difference between classical and relativistic theories becomes significant.
Experimental Verification The magnitude of the conversion factor between mass and rest energy (2- = 9 x 1016 m 2 /s2 ) is so great that even a very small mass has a huge amount of rest energy. Nuclear reactions utilize this relationship by converting mass (rest energy) into other forms of energy. In nuclear fission, which is the energy source of nuclear power plants, the nucleus of an atom is split into two or more nuclei. Taken together, the mass of these nuclei is slightly less than the mass of the original nucleus, and a very large amount of energy is released. In typical nuclear reactions, about one-thousandth of the initial mass is converted from rest energy into other forms of energy. This change in mass, although very small, can be detected experimentally. Another type of nuclear reaction that converts mass into energy is fusion, which is the source of energy for our sun and other stars. About 4.5 million tons of the sun’s mass is converted into other forms of energy every second, by fusing hydrogen into helium. Fortunately, the sun has enough mass to continue to fuse hydrogen into helium for approximately 5 billion more years.
Nuclear Fusion in the Sun Our sun uses a nuclear reaction called fusion to convert mass to energy. About 90 percent of the stars, including our sun, fuse hydrogen, and some older stars fuse helium.
Most of the energy changes encountered in your typical experiences are much smaller than the energy changes that occur in nuclear reactions and are far too small to be detected experimentally. Thus, for typical cases, the classical equation still holds, and mass and energy can be thought of as separate. Before Einstein’s theory of relativity, conservation of energy and conservation of mass were regarded as two separate laws. The equivalence between mass and energy reveals that in fact these two laws are one. In the words of Einstein, “Prerelativity physics contains two conservation laws of fundamental importance . .. . Through relativity theory, they melt together into one principle:’
Work and Energy
177

__ = __ – – – – –
Roller Coaster Designer s the name states, the cars of a roller coaster really do coast along the tracks. A motor pulls the cars up a high hill at the beginning of the ride. After the hill, however, the motion of the car is a result of gravity and inertia. As the cars roll down the hill, they must pick up the speed that they need to whiz through the rest of the curves, loops, twists, and bumps in the track. To learn more about designing roller coasters, read the interview with Steve Okamoto.
The roller coaster pictured here is named Wild Thing and is located in Minnesota. The highest point on the track is 63 m off the ground and the cars’ maximum speed is 118 km/h.
How did you become a roller coaster designer?
I have been fascinated with roller coasters ever since my first ride on one. I remember going to Disneyland as a kid. My mother was always upset with me because I kept looking over the sides of the rides, trying to figure out how they worked. My interest in finding out how things worked led me to study mechanical engineering. What sort of training do you have?
I earned a degree in product design. For this degree, I studied mechanical engineering and studio art. Product designers consider an object’s form as well as its function. They also take into account the interests and abilities of the product’s consumer. Most rides and parks have some kind of theme, so I must consider marketing goals and concerns in my designs. What is the nature of your work?
To design a roller coaster, I study site maps of the location. Then, I go to the amusement park to look at the actual site. Because most rides I design are for older parks (few parks are built from scratch), fitting a coaster around, above, and in between existing rides and buildings is one of my biggest challenges. I also have to design how the parts of the ride will work together. The towers and structures that support the ride have to be strong enough to hold up a track and speeding cars that are full of people. The cars themselves need special wheels to keep them locked onto the track and
seat belts or bars to keep the passengers safely inside. It’s like putting together a puzzle, except the pieces haven’t been cut out yet. What advice do you have for a student who is interested in designing roller coasters?
Studying math and science is very important. To design a successful coaster, I have to understand how energy is converted from one form to another as the cars move along the track. I have to calculate speeds and accelerations of the cars on each part of the track. They have to go fast enough to make it up the next hill! I rely on my knowledge of geometry and physics to create the roller coaster’s curves, loops, and dips.
Steve Okamoto
SECTION 1
Work
,: ,
1
i
Ir:
work
• Work is done on an object only when a net force acts on the object to displace it in the direction of a component of the net force. • The amount of work done on an object by a force is equal to the component of the force along the direction of motion times the distance the object moves.
SECTION 2
Energy
f ::_, T[I , ·
• Objects in motion have kinetic energy because of their mass and speed.
kinetic energy
• The net work done on or by an object is equal to the change in the kinetic energy of the object.
work-kinetic energy theorem
• Potential energy is energy associated with an object’s position. Two forms of potential energy discussed in this chapter are gravitational potential energy and elastic potential energy.
gravitational potential energy
SECTION 3
Conservation of Energy
potential energy elastic potential energy spring constant
,c
• Energy can change form but can never be created or destroyed.
1
1 1- 1 ·.-
mechanical energy
• Mechanical energy is the total kinetic and potential energy present in a given situation. • In the absence of friction, mechanical energy is conserved, so the amount of mechanical energy remains constant.
SECTION 4
Power
, c:: • TU ;:
• Power is the rate at which work is done or the rate of energy transfer.
power
• Machines with different power ratings do the same amount of work in different time intervals.
VARIABLE SYMBOLS
Quantities
Units
w
work
J
joule
= Nern
KE
kinetic energy
J
joule
= kgem 2/s2
PEg
gravitational potential energy
J
joule
PEelastic elastic potential energy
J
joule
p
w
watt
power
Conversions
Problem Solving
= J/s
See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
Chapter Summary
179
Work REVIEWING MAIN IDEAS
1. Can the speed of an object change if the net work done on it is zero? 2. Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative. a. a chicken scratching the ground b. a person reading a sign c. a crane lifting a bucket of concrete d. the force of gravity on the bucket in (c)
3. Furniture movers wish to load a truck using a ramp from the ground to the rear of the truck. One of the movers claims that less work would be required if the ramp’s length were increased, reducing its angle with the horizontal. Is this claim valid? Explain.
CONCEPTUAL QUESTIONS 4. A pendulum swings back and forth, as shown at right. Does the tension force in the string do work on the pendulum bob? Does the force of gravity do work on the bob? Explain your answers. 5. The drivers of two identical cars heading toward each other apply the brakes at the same instant. The skid marks of one of the cars are twice as long as the skid marks of the other vehicle. Assuming that the brakes of both cars apply the same force, what conclusions can you draw about the motion of the cars? 6. When a punter kicks a football, is he doing work on the ball while his toe is in contact with it? Is he doing work on the ball after the ball loses contact with his toe? Are any forces doing work on the ball while the ball is in flight?
180
Chapter 5
PRACTICE PROBLEMS For problems 7-10, see Sample Problem A.
7. A person lifts a 4.5 kg cement block a vertical distance of 1.2 m and then carries the block horizontally a distance of 7.3 m. Determine the work done by the person and by the force of gravity in this process. 8. A plane designed for vertical takeoff has a mass of
8.0 x 103 kg. Find the net work done by all forces on the plane as it accelerates upward at 1.0 m/ s2 through a distance of 30.0 m after starting from rest. 9. When catching a baseball, a catcher’s glove moves by 10 cm along the line of motion of the ball. If the baseball exerts a force of 475 N on the glove, how much work is done by the ball? 10. A flight attendant pulls her 70.0 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52.0° above the horizontal. Find the following: a. the work she does on the flight bag b. the work done by the force of friction on the flight bag c. the coefficient of kinetic friction between the flight bag and the floor
Energy REVIEWING MAIN IDEAS 11. A person drops a ball from the top of a building while another person on the ground observes the ball’s motion. Each observer chooses his or her own location as the level for zero potential energy. Will they calculate the same values for: a. the potential energy associated with the ball? b. the change in potential energy associated with the ball? c. the ball’s kinetic energy?
12. Can the kinetic energy of an object be negative? Explain your answer. 13. Can the gravitational potential energy associated with an object be negative? Explain your answer. 14. Two identical objects move with speeds of 5.0 m /s and 25.0 m/s. What is the ratio of their kinetic energies?
CONCEPTUAL QUESTIONS 15. A satellite is in a circular orbit above Earth’s surface. Why is the work done on the satellite by the gravitational force zero? What does the work-kinetic energy theorem predict about the satellite’s speed? 16. A car traveling at 50.0 km/h skids a distance of 35 m after its brakes lock. Estimate how far it will skid if its brakes lock when its initial speed is 100.0 km/h. What happens to the car’s kinetic energy as it comes to rest? 17. Explain why more energy is needed to walk up stairs than to walk horizontally at the same speed. 18. How can the work- kinetic energy theorem explain why the force of sliding friction reduces the kinetic energy of a particle?
22. In a circus performance, a monkey on a sled is given an initial speed of 4.0 m/s up a 25° incline. The combined mass of the monkey and the sled is 20.0 kg, and the coefficient of kinetic friction between the sled and the incline is 0.20. How far up the incline does the sled move? For problems 23-25, see Sample Problem D. 23. A 55 kg skier is at the top of a slope, as shown in the illustration below. At the initial point A, the skier is 10.0 m vertically above the final point B. a. Set the zero level for gravitational potential energy at B, and find the gravitational potential energy associated with the skier at A and at B. Then find the difference in potential energy between these two points. b. Repeat this problem with the zero level at point A. c. Repeat this problem with the zero level midway down the slope, at a height of 5.0 m. A
T
IO.Om
l
B
PRACTICE PROBLEMS For problems 19-20, see Sample Problem B. 19. What is the kinetic energy of an automobile with a mass of 1250 kg traveling at a speed of 11 m / s?
20. What speed would a fly with a mass of 0.55 g need in order to have the same kinetic energy as the automobile in item 19? For problems 21-22, see Sample Problem C. 21. A 50.0 kg diver steps off a diving board and drops straight down into the water. The water provides an upward average net force of 1500 N. If the diver comes to rest 5.0 m below the water’s surface, what is the total distance between the diving board and the diver’s stopping point underwater?
24. A 2.00 kg ball is attached to a ceiling by a string. The distance from the ceiling to the center of the ball is 1.00 m, and the height of the room is 3.00 m. What is the gravitational potential energy associated with the ball relative to each of the following? a. the ceiling b. the floor c. a point at the same elevation as the ball 25. A spring has a force constant of 500.0 N/ m. Show that the potential energy stored in the spring is as follows: a. 0.400 J when the spring is stretched 4.00 cm from equilibrium b. 0.225 J when the spring is compressed 3.00 cm from equilibrium c. zero when the spring is unstretched
Chapter Review
181
Conservation of Mechanical Energy REVIEWING MAIN IDEAS 26. Each of the following objects possesses energy. Which forms of energy are mechanical, which are nonmechanical, and which are a combination? a. glowing embers in a campfire b. a strong wind c. a swinging pendulum d. a person sitting on a mattress e. a rocket being launched into space 27. Discuss the energy transformations that occur during the pole-vault event shown in the photograph below. Disregard rotational motion and air resistance.
..·r:
;f

·’ i
. a,J ,. . a..
.,
.
28. A strong cord suspends a bowling ball from the center of a lecture hall’s ceiling, forming a pendulum. The ball is pulled to the tip of a lecturer’s nose at the front of the room and is then released. If the lecturer remains stationary, explain why the lecturer is not struck by the ball on its return swing. Would this person be safe if the ball were given a slight push from its starting position at the person’s nose?
CONCEPTUAL QUESTIONS 29. Discuss the work done and change in mechanical energy as an athlete does the following: a. lifts a weight b. holds the weight up in a fixed position c. lowers the weight slowly
182
Chapter 5
31. Advertisements for a toy ball once stated that it would rebound to a height greater than the height from which it was dropped. Is this possible?
32. A weight is connected to a spring that is suspended vertically from the ceiling. If the weight is displaced downward from its equilibrium position and released, it will oscillate up and down. How many forms of potential energy are involved? If air resistance and friction are disregarded, will the total mechanical energy be conserved? Explain.
PRACTICE PROBLEMS For problems 33-34, see Sample Problem E.
. .. . .. . .
·r~
30. A ball is thrown straight up. At what position is its kinetic energy at its maximum? At what position is gravitational potential energy at its maximum?
33. A child and sled with a combined mass of 50.0 kg slide down a frictionless hill that is 7.34 m high. If the sled starts from rest, what is its speed at the bottom of the hill?
34. Tarzan swings on a 30.0 m long vine initially inclined at an angle of 37.0° with the vertical. What is his speed at the bottom of the swing if he does the following? a. starts from rest b. starts with an initial speed of 4.00 m / s
Power REVIEWING MAIN IDEAS PRACTICE PROBLEMS For problems 35-36, see Sample Problem F.
35. If an automobile engine delivers 50.0 hp of power, how much time will it take for the engine to do 6.40 x 105 J of work? (Hint: Note that one horsepower, 1 hp, is equal to 746 watts.) 36. Water flows over a section of Niagara Falls at the rate of 1.2 x 106 kg/ sand falls 50.0 m. How much power is generated by the falling water?
Mixed Review
43. Three identical balls, all with the same initial speed, are thrown by a juggling clown on a tightrope. The first ball is thrown horizontally, the second is thrown at some angle above the horizontal, and the third is thrown at some angle below the horizontal. Disregarding air resistance, describe the motions of the three balls, and compare the speeds of the balls as they reach the ground.
REVIEWING MAIN IDEAS 37. A 215 g particle is
released from rest at point A inside a smooth A hemispherical bowl of radius 30.0 cm, as shown at right. B Calculate the following: a. the gravitational potential energy at A relative to B b. the particle’s kinetic energy at B c. the particle’s speed at B d. the potential energy and kinetic energy at C

T IR 3
I
38. A person doing a chin-up weighs 700.0 N, disregard-
ing the weight of the arms. During the first 25.0 cm of the lift, each arm exerts an upward force of 355 N on the torso. If the upward movement starts from rest, what is the person’s speed at this point?
39. A 50.0 kg pole vaulter running at 10.0 m / s vaults over the bar. If the vaulter’s horizontal component of velocity over the bar is 1.0 m/ s and air resistance is disregarded, how high was the jump? 40. An 80.0 N box of clothes is pulled 20.0 m up a 30.0°
ramp by a force of 115 N that points along the ramp. If the coefficient of kinetic friction between the box and ramp is 0.22, calculate the change in the box’s kinetic energy.
41 . Tarzan and Jane, whose total mass is 130.0 kg, start their swing on a 5.0 m long vine when the vine is at an angle of 30.0° with the horizontal. At the bottom of the arc, Jane, whose mass is 50.0 kg, releases the vine. What is the maximum height at which Tarzan can land on a branch after his swing continues? (Hint: Treat Tarzan’s and Jane’s energies as separate quantities.) 42. A 0.250 kg block on a vertical spring with a spring constant of 5.00 x 103 N/m is pushed downward, compressing the spring 0.100 m. When released, the block leaves the spring and travels upward vertically. How high does it rise above the point of release?
44. A 0.60 kg rubber ball has a speed of2.0 m / s at point A and kinetic energy of7.5 J at point B. Determine the following: a. the ball’s kinetic energy at A b. the ball’s speed at B c. the total work done on the ball from A to B 45. Starting from rest, a 5.0 kg block slides 2.5 m down a rough 30.0° incline in 2.0 s. Determine the following: a. the work done by the force of gravity b. the mechanical energy lost due to friction c. the work done by the normal force between the block and the incline 46. A skier of mass 70.0 kg is pulled up a slope by a motor-driven cable. How much work is required to pull the skier 60.0 m up a 35° slope (assumed to be frictionless) at a constant speed of2.0 m / s? 47. An acrobat on skis starts from rest 50.0 m above the
ground on a frictionless track and flies off the track at a 45.0° angle above the horizontal and at a height of 10.0 m. Disregard air resistance. a. What is the skier’s speed wh en leaving the track? b. What is the maximum height attained? 48. Starting from rest, a 10.0 kg suitcase slides 3.00 m
down a frictionless ramp inclined at 30.0° from the floor. The suitcase then slides an additional 5.00 m along the floor before coming to a stop. Determine the following: a. the suitcase’s speed at the bottom of the ramp b. the coefficient of kinetic friction between the suitcase and the floor c. the change in mechanical energy due to friction
49. A light horizontal spring has a spring constant of 105 N/ m. A 2.00 kg block is pressed against one end of the spring, compressing the spring 0.100 m. After the block is released, the block moves 0.250 m to the right before coming to rest. What is the coefficient of kinetic friction between the horizontal surface and the block?
Chapter Review
183
50. A 5.0 kg block is pushed 3.0 mat a F constant velocity up a vertical wall by a constant force applied at an angle of 30.0° with the horizontal, as shown at right. If the coefficient of kinetic friction between the block and the wall is 0.30, determine the following: a. the work done by the force on the block b. the work done by gravity on the block c. the magnitude of the normal force between the block and the wall
51 . A 25 kg child on a 2.0 m long swing is released from rest when the swing supports make an angle of 30.0° with the vertical. a. What is the maximum potential energy associated with the child? b. Disregarding friction, find the child’s speed at the lowest position. c. What is the child’s total mechanical energy? d. If the speed of the child at the lowest position is 2.00 m/ s, what is the change in mechanical energy due to friction?
52. A ball of mass 522 g starts at rest and slides down a frictionless track, as shown in the diagram. It leaves the track horizontally, striking the ground. a. At what height above the ground does the ball start to move? b. What is the speed of the ball when it leaves the track? c. What is the speed of the ball when it hits the ground?
‘ 1.25m

\ \
I I
f- 1.00 m -i
Work of Displacement Work done, as you learned earlier in this chapter, is a result of the net applied force, the distance of the displacement, and the angle of the applied force relative to the direction of displacement. Work done is described by in the following equation:
Wnet = Fn81 dcos 0 The equation for work done can be represented on a graphing calculator as follows: Y1 = FXCOS(0)
184
Chapter 5
In this activity, you will use this equation and your graphing calculator to produce a table of results for various values of 0. Column one of the table will be the displacement (X) in meters, and column two will be the work done (Y1) in joules. Go online to HMDScience.com to find this graphing calculator activity.
ALTERNATIVE ASSESSMENT 1. Design experiments for measuring your power output
when doing pushups, running up a flight of stairs, pushing a car, loading boxes onto a truck, throwing a baseball, or performing other energy-transferrin g activities. What data do you need to measure or calculate? Form groups to present and discuss your plans. If your teacher approves your plans, perform the experiments. 2. Investigate the amount of kinetic energy involved when your car’s speed is 60 km/ h, 50 km/ h, 40 km/ h, 30 km/ h, 20 km/ h, and 10 km/ h. (Hint: Find your car’s mass in the owner’s manual.) How much work does the brake system have to do to stop the car at each speed? If the owner’s manual includes a table of braking
distances at different speeds, determine the force the braking system must exert. Organize your findings in charts and graphs to study the questions and to present your conclusions. 3. Investigate the energy transformations of your body as you swing on a swing set. Working with a partner, measure the height of the swing at the high and low points of your motion. What points involve a maximum gravitational potential energy? What points involve a maximum kinetic energy? For three other points in the path of the swing, calculate the gravitational potential energy, the kinetic energy, and the velocity. Organize your findings in bar graphs.
4. Design an experiment to test the conservation of mechanical energy for a toy car rolling down a ramp. Use a board propped up on a stack of books as the ramp. To find the final speed of the car, use the equation: final speed= 2(average speed)= 2(length/ time) Before beginning the experiment, make predictions about what to expect. Will the kinetic energy at the bottom equal the potential energy at the top? If not, which might be greater? Test your predictions with various ramp heights, and write a report describing your experiment and your results.
5. In order to save fuel, an airline executive recom-
mended the following changes in the airline’s largest jet flights: a. restrict the weight of personal luggage b. remove pillows, blankets, and magazines from the cabin c. lower flight altitudes by 5 percent d. reduce flying speeds by 5 percent Research the information necessary to calculate the approximate kinetic and potential energy of a large passenger aircraft. Which of the measures described above would result in significant savings? What might be their other consequences? Summarize your conclusions in a presentation or report. 6. Make a chart of the kinetic energies your body can
have. First, measure your mass. Then, measure your speed when walking, running, sprinting, riding a bicycle, and driving a car. Make a poster graphically comparing these findings. 7. You are trying to find a way to bring electricity to a
remote village in order to run a water-purifying device. A donor is willing to provide battery chargers that connect to bicycles. Assuming the water-purification device requires 18.6 kW•h daily, how many bicycles would a village need if a person can average 100 W while riding a bicycle? Is this a useful way to help the village? Evaluate your findings for strengths and weaknesses. Summarize your comments and suggestions in a letter to the donor. 8. Many scientific units are named after famous scientists or inventors. The SI unit of power, the watt, was named for the Scottish scientist James Watt. The SI unit of energy, the joule, was named for the English scientist James Prescott Joule. Use the Internet or library resources to learn about the contributions of one of these two scientists. Write a short report with your findings, and then present your report to the class.
Chapter Review
185
MULTIPLE CHOICE 1. In which of the following situations is work not being done? A. A chair is lifted vertically with respect to the floor. B. A bookcase is slid across carpeting. C. A table is dropped onto the ground. D. A stack of books is carried at waist level across a room. 2. Which of the following equations correctly describes the relation between power, work, and time? p F. W=t t
G. W=p
w H.P=t
4. What is the speed of the yo-yo after 4.5 s? F. 3.1 m/s G. 2.3m/ s H. 3.6m/ s J. l.6m/s 5. What is the maximum height of the yo-yo? A. 0.27m B. 0.54m C. 0.75m D. 0.82m
6. A car with mass m requires 5.0 kJ of work to move from rest to a final speed v. If this same amount of work is performed during the same amount of time on a car with a mass of 2m, what is the final speed of the second car? F. 2v
t
J. P= W Use the graph below to answer questions 3-5. The graph shows the energy of a 75 g yo-yo at different times as the yo-yo moves up and down on its string.
G. ‘✓’2.v
Energy of Yo-Yo versus Time Potential energy Kinetic energy
600
0:, UJ
.,; c:-
“‘”‘ Ee ::,; ~ @ C:
0
.s0
~
if
Q)
~
0
~
~ E .0
~
8
e
“‘ .c –, ‘-‘ C: C:
., C:
co ·.:::: .c “‘
,Sl E
cf ~ N …. “‘ :: -§
_g ·E -:z: “‘”‘ ~ c “‘ ‘-‘
“‘ =
-~ ~ 1i5 :.2 “‘ ‘-‘ . , C:
.0
.,
cn ·a
C: “‘ c: E
-§ @ –, .s ~~
294
1747
1752
1770
Contrary to the favored idea that heat is a fluid, Russian chemist Mikhail V. Lomonosov publishes his hypothesis that heat is the result of motion. Several years later, Lomonosov formulates conservation laws for mass and energy.
Benjamin Franklin performs the dangerous “kite experiment,” in which he demonstrates that lightning consists of electric charge. He would build on the first studies of electricity performed earlier in the century by describing electricity as having positive and negative charge.
Antoine Laurent Lavoisier begins his research on chemical reactions, notably oxidation and combustion.
1775 The American Revolution ~ – – – – – – – – – ‘ begins.
-;::
e. i,f

8
>,
©l ~
g§ -ri- ~ !::I ~ -o
B~ ~~ 0
§:c:-
kf~ .~ ~ ‘ u, C,
“‘
~@
g ‘o .s :: 0 ,::
.,
“-
C,
“‘
. “‘ -1: E
~~
!~ – ., “‘ > ~
:;; ~
QJ
:.c: ~
§
~
E~
8 §
~~ “‘= Cl 8
.E
cc
~ ~
1756 The Seven Year’s War begins. British general James Wolfe leads the capture of Fort Louisburg, in Canada, in 1758.
1757
1772
1785
German musician William Herschel emigrates to England to avoid fighting in the Seven Year’s War. Over the next 60 years, he pursues astronomy, constructing the largest reflecting telescopes of the era and discovering new objects, such as binary stars and the planet Uranus.
Caroline Herschel, sister of astronomer William Herschel, joins her brother in England. She compiles the most comprehensive star catalog of the era and discovers several nebulae-regions of glowing gas-within our galaxy.
Charles Augustin de Coulomb publishes the results of experi ments that will systematically and conclusively prove the inverse-square law for electric force. The law has been suggested for over 30 years by other scientists, such as Daniel Bernoulli, Joseph Priestly, and Henry Cavendish.
‘C “”
~ @
8~ .–:.. ‘&. “‘r–co
.
C,
cc
:c., ·:;
~ ~
E .lJ ·.:::: . ~~
a5 “E
-“‘cc ., “‘ ~
;:;
cc “‘ ~ ~ u.. >,
gI
295
SECTION 1 Objectives ►
I
Relate temperature to the kinetic energy of atoms and molecules.

Describe the changes in the temperatures of two objects reaching thermal equilibrium.

Identify the various temperature scales, and convert from one scale to another.
Temperature and Thermal Eauilibrium Key Terms temperature
internal energy
thermal equilibrium
Defining Temperature When you hold a glass of lemonade with ice, such as that shown in Figure 1.1, you feel a sharp sensation in your hand that we describe as “cold:’ Likewise, you experience a “hot” feeling when you touch a cup of hot chocolate. We often associate temperature with how hot or cold an object feels when we touch it. Our sense of touch serves as a qualitative indicator of temperature. However, this sensation of hot or cold also depends on the temperature of the skin and therefore can be misleading. The same object may feel warm or cool, depending on the properties of the object and on the conditions of your body.
Hot and Cold Objects at low temperatures feel cold to the touch, while objects at high temperatures feel hot. However, the sensation of hot and cold can be misleading.
Determining an object’s temperature with precision requires a standard definition of temperature and a procedure for making measurements that establish how “hot” or “cold” objects are.
Adding or removing energy usually changes temperature. Consider what happens when you use an electric range to cook food. By turning the dial that controls the electric current delivered to the heating element, you can adjust the element’s temperature. As the current is increased, the temperature of the element increases. Similarly, as the current is reduced, the temperature of the element decreases. In general, energy must be either added to or removed from a substance to change its temperature.
Quick LAB Fill one basin with hot tap water. Fill another with cold tap water and add ice until about one-third of the mixture is ice. Fill the third basin with an equal mixture of hot and cold tap water.
298
Chapter 9
Place your left hand in the hot water and your right hand in the cold water for 15 s. Then place both hands in the basin of lukewarm water for 15 s. Describe whether the water feels hot or cold to either of your hands.
MATERIALS
• 3 identical basins • hot and cold tap water • ice SAFETY
Use only hot tap water. The temperature of the hot water must not exceed 50°C (122°F).
Temperature is proportional to the kinetic energy of atoms and molecules. The temperature of a substance is proportional to the average kinetic energy of particles in the substance. A substance’s temperature increases as a direct result of added energy being distributed among the particles of the substance, as shown in Figure 1.2.
Temperature and Kinetic Energy When energy is added to the gas in (a) , its low average kinetic energy, and thus its temperature, increases as shown in (b).
A monatomic gas contains only one type of
atom. For a monatomic gas, temperature can be understood in terms of the translational kinetic energy of the atoms in the gas. For other kinds of substances, molecules can rotate or vibrate, so other types of energy are also present, as shown in Figure 1.3. The energies associated with atomic motion are referred to as internal energy, which is proportional to the substance’s temperature (assuming no phase change). For an ideal gas, the internal energy depends only on the temperature of the gas. For nonideal gases, as well as for liquids and solids, other properties contribute to the internal energy. The symbol U stands for internal energy, and ~ U stands for a change in internal energy.
temperature a measure of the average kinetic energy of the particles in a substance internal energy the energy of a substance due to both the random motions of its particles and to the potential energy that results from the distances and alignments between the particles
Form of energy
Macroscopic examples
Microscopic examples
Translational
airplane in flight, roller coaster at bottom of rise
CO2 molecule in linear motion
kinetic energy
Rotational
spinning top
CO 2 molecule spinning about its center of mass
kinetic energy
Vibrational
plucked guitar string
bending and stretching of bonds between atoms in a CO2 molecule
kinetic and potential energy
Energy type
A —A
-~-~ ~ ~ Heat
299
Temperature is meaningful only when it is stable.
thermal equilibrium the state in which two bodies in physical contact w ith each other have identical temperatures
. _Did YOU Know? – – – – – – – – – , As a thermometer comes into thermal equilibrium with an object, the object’s temperature changes slightly. In most cases the object is so massive compared with the thermometer that the object’s temperature change is insignificant.
Imagine a can of warm fruit juice immersed in a large beaker of cold water. After about 15 minutes, the can of fruit juice will be cooler and the water surrounding it will be slightly warmer. Eventually, both the can of fruit juice and the water will be at the same temperature. That temperature will not change as long as conditions remain unchanged in the beaker. Another way of expressing this is to say that the water and can of juice are in thermal equilibrium with each other. Thermal equilibrium is the basis for measuring temperature with thermometers. By placing a thermometer in contact with an object and waiting until the column ofliquid in the thermometer stops rising or falling, you can find the temperature of the object. The reason is that the thermometer is in thermal equilibrium with the object. Just as in the case of the can of fruit juice in the cold water, the temperature of any two objects in thermal equilibrium always lies between their initial temperatures.
Matter expands as its temperature increases.
1. Hot Chocolate If two cups of
Increasing the temperature of a gas at constant pressure causes the volume of the gas to increase. This increase occurs not only for gases, but also for liquids and solids. In general, if the temperature of a substance increases, so does its volume. This phenomenon is known as thermal expansion.
hot chocolate, one at 50°C and the other at 60°C, are poured together in a large container, w ill the final temperature of the double batch be
You may have noticed that the concrete roadway segments of a sidewalk are separated by gaps. This is necessary because concrete expands with increasing temperature. Without these gaps, thermal expansion would cause the segments to push against each other, and they would eventually buckle and break apart.
a. less than 50°C?
Different substances undergo different amounts of expansion for a given temperature change. The thermal expansion characteristics of a material are indicated by a quantity called the coefficient of volume expansion. Gases have the largest values for this coefficient. Liquids have much smaller values.
b. between 50°C and 60°C?
c. greater than 60°C? Explain your answer. 2. Hot and Cold Liquids A cup
of hot tea is poured from a teapot, and a swimming pool is filled with cold water. Which one has a higher total internal energy? Which has a higher average kinetic energy? Explain.
300
Chapter 9
In general, the volume of a liquid tends to decrease with decreasing temperature. However, the volume of water increases with decreasing temperature in the range between 0°c and 4°C. Also, as the water freezes, it forms a crystal that has more empty space between the molecules than does liquid water. This explains why ice floats in liquid water. It also explains why a pond freezes from the top down instead of from the bottom up. If this did not happen, fish would likely not survive in freezing temperatures. Solids typically have the smallest coefficient of volume expansion values. For this reason, liquids in solid containers expand more than the container. This property allows some liquids to be used to m easure changes in temperature.
Measuring Temperature In order for a device to be used as a thermometer, it must make use of a change in some physical property that corresponds to changing temperature, such as the volume of a gas or liquid, or the pressure of a gas at constant volume. The most common thermometers use a glass tube containing a thin column of mercury, colored alcohol, or colored mineral spirits. When the thermometer is heated, the volume of the liquid expands. (The cross-sectional area of the tube remains nearly constant during temperature changes.) The change in length of the liquid column is proportional to the temperature change, as shown in Figure 1.4.
, Calibrating thermometers requires fixed temperatures. A thermometer must be more than an unmarked, thin glass tube of liquid; the length of the liquid column at different temperatures must be known. One reference point is etched on the tube and refers to when the thermometer is in thermal equilibrium with a mixture of water and ice at one atmosphere of pressure. This temperature is called the ice point or melting point of water and is defined as zero degrees Celsius, or 0°C. A second reference mark is made at the point when the thermometer is in thermal equilibrium with a mixture of steam and water at one atmosphere of pressure. This temperature is called the steam point or boiling point of water and is defined as 100°c.
Mercury Thermometer The volume of mercury in this thermometer increases slightly when the mercury’s temperature increases from 0°c (a) to 50°C (b).
(a)
Volume of m ercury at 0°C = 0.100 mL = “‘;
o·c
A temperature scale can be made by dividing the distance between the reference marks into equally spaced units, called degrees. This process is based on the assumption that the expansion of the mercury is linear (proportional to the temperature difference), which is a very good approximation.
so·c
(b)
Temperature units depend on the scale used. The temperature scales most widely used today are the Fahrenheit, Celsius, and Kelvin scales. The Fahrenheit scale is commonly used in the United States. The Celsius scale is used in countries that have adopted the metric system and by the scientific community worldwide. Celsius and Fahrenheit temperature measurements can be converted to each other using this equation.
Volume of mercury at 50°C = 0.101 mL = V; + 0.001 mL
o·c
Celsius-Fahrenheit Temperature Conversion
Tp=fTc+32.0 Fahrenheit temperature = (¾x Celsius temperature) + 32.0
The number 32.0 in the equation indicates the difference between the ice point value in each scale. The point at which water freezes is 0.0 degrees on the Celsius scale and 32.0 degrees on the Fahrenheit scale.
Heat
301
Determining Absolute Zero for an Ideal Gas This graph suggests that if the gas’s temperature could be lowered to -273.15°C, or 0 K, the gas’s pressure would be zero. Pressure-Temperature Graph for an Ideal Gas
I!:?
iil
“‘ E Cl..
-100
0 100 Temperature (C)
200
300
Temperature values in the Celsius and Fahrenheit scales can have positive, negative, or zero values. But because the kinetic energy of the atoms in a substance must be positive, the absolute temperature that is proportional to that energy should be positive also. A temperature scale with only positive values is suggested in the graph of pressure versus temperature for an ideal gas at constant volume, shown in Figure 1.5. As the gas’s temperature decreases, so does its pressure. The graph suggests that if the temperature could be lowered to -273.15°C, the pressure of the sample would be zero. This temperature is designated in the Kelvin scale as 0.00 K, where K is the symbol for the temperature unit called the kelvin. Temperatures in this scale are indicated by the symbol T.
A temperature difference of one degree is the same on the Celsius and Kelvin scales. The two scales differ only in the choice of zero point. Thus, the ice point (0.00°C) equals 273.15 K, and the steam point (l00.00°C) equals 373.15 K (see Figure 1.6). The Celsius temperature can therefore be converted to the Kelvin temperature by adding 273.15. Celsius-Kelvin Temperature Conversion
T= Tc+ 273.15 Kelvin temperature = Celsius temperature + 273.15 Kelvin temperatures for various physical processes can range from around 1000000 000 K (10 9 K), which is the temperature of the interiors of the most massive stars, to less than 1 K, which is slightly cooler than the boiling point of liquid helium. The temperature 0 K is often referred to as absolute zero. Absolute zero has never been reached, although laboratory experiments have reached temperatures of just a half-billionth of a degree above absolute zero.
Scale
Ice point
Steam point
Applications
Fahrenheit
32°F
212°F
meteorology, medicine, and nonscientific uses (United States)
Celsius
0°c
100°c
meteorology, medicine, and nonscientific uses (outside United States); other sciences (international)
Kelvin (absolute)
273.15 K
373.15 K
physical chemistry, gas laws, astrophysics, thermodynamics, low-temperature physics
302
Chapter 9
PREMIUM CONTENT
®
Temperature Conversion
Interactive Demo HMDScience.com
Sample Problem A What are the equivalent Celsius and Kelvin temperatures of 50.0°F1
0
ANALYZE
Given:
Tp= 50.0°F
Unknown:
Tc = ? T=?
f:)
SOLVE
Use the Celsius-Fahrenheit equation to convert Fahrenheit into Celsius.
Tp=t Tc +32.0 Tc =¾(Tp-32.0) Tc =¾ (50.0 – 32.0)°C = 10.0°c
Use the Celsius-Kelvin equation to convert Celsius into Kelvin.
T = Tc + 273.15 T = (10.0 + 273.15)K = 283.2 K IT= 283.2K I Practice 1. The lowest outdoor temperature ever recorded on Earth is -128.6°F, recorded at Vostok Station, Antarctica, in 1983. What is this temperature on the Celsius and Kelvin scales?
2. The temperatures of one northeastern state range from 105°F in the summer to -25°F in winter. Express this temperature range in degrees Celsius and in kelvins.
3. The normal human body temperature is 98.6°F. A person with a fever may record 102°F. Express these temperatures in degrees Celsius. 4. A pan of water is heated from 23°C to 78°C. What is the change in its temperature on the Kelvin and Fahrenheit scales? 5. Liquid nitrogen is used to cool substances to very low temperatures. Express the boiling point ofliquid nitrogen (77.34 Kat 1 atm of pressure) in degrees Celsius and in degrees Fahrenheit.
Heat
303

SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. A hot copper pan is dropped into a tub of water. If the water’s temperature rises, what happens to the temperature of the pan? How will you know when the water and copper pan reach thermal equilibrium?
2. Oxygen condenses into a liquid at approximately 90.2 K. To what temperature does this correspond on both the Celsius and Fahrenheit temperature scales? 3. The boiling point of sulfur is 444.6°C. Sulfur’s melting point is 586.1 °F lower than its boiling point. a. Determine the melting point of sulfur in degrees Celsius. b. Find the melting and boiling points in degrees Fahrenheit. c. Find the melting and boiling points in kelvins. 4. Which of the following is true for popcorn kernels and the water molecules inside them during popping? a. The temperature of the kernels increases. b. The water molecules are destroyed. c. The kinetic energy of the water molecules increases. d. The mass of the water molecules changes.
Interpreting Graphics 5. Two gases that are in physical contact with each other consist of particles of identical mass. In what order should the images shown in Figure 1.7 be placed to correctly describe the changing distribution of kinetic energy among the gas particles? Which group of particles has the highest temperature at any time? Explain.
Kinetic Energy Distribution
Critical Thinking 6. Have you ever tried to make popcorn and found that most of the kernels did not pop? What might be the reason that they did not pop? What could you do to try to make more of the kernels pop?
304
Chapter 9
SECTION 2
Defining Heat
Objectives

Explain heat as the energy transferred between substances that are at different temperatures.

Relate heat and temperature change on the macroscopic level to particle motion on the microscopic level.

Apply the principle of energy conservation to calculate changes in potential, kinetic, and internal energy.
Key Term heat
Heat and Energy Thermal physics often appears mysterious at the macroscopic level. Hot objects become cool without any obvious cause. To understand thermal processes, it is helpful to shift attention to the behavior of atoms and molecules. Mechanics can be used to explain much of what is happening at the molecular, or microscopic, level. This in turn accounts for what you observe at the macroscopic level. Throughout this chapter, the focus will shift between these two viewpoints. What happens when you immerse a warm fruit juice bottle in a container of ice water, as shown in Figure 2.1? As the temperatures of the bottle and of the juice decrease, the water’s temperature increases slightly until both final temperatures are the same. Energy is transferred from the bottle of juice to the water because the two objects are at different temperatures. This energy that is transferred is defined as heat. The word heat is sometimes used to refer to the process by which energy is transferred between objects because of a difference in their temperatures. This textbook will use heat to refer only to the energy itself.
heat the energy transferred between objects because of a difference in their temperatures
Energy is transferred between substances as heat. From a macroscopic viewpoint, energy transferred as heat tends to move from an object at higher temperature to an object at lower temperature. This is similar to the mechanical behavior of objects moving from a higher gravitational potential energy to a lower gravitational potential energy. Just as a pencil will drop from your desk to the floor but will not jump from the floor to your desk, so energy will travel spontaneously from an object at higher temperature to one at lower temperature and not the other way around.
Energy Transfer as Heat Energy is transferred as heat from objects with higher temperatures (the fruit juice and bottle) to those with lower temperatures (the cold water).
Heat
305
Transfer of Particles’ Kinetic Energy as Heat Energy is transferred as heat from the higher-energy particles to lower-energy particles (a). The net energy transferred is zero when thermal equilibrium is reached (b). Twater= Direction of energy transfer
(a) Tjuice
s·c
= 45°C
Direction of energy transfer
Twater = 11·c (b) Tjuice
= 11·c
The direction in which energy travels as heat can be explained at the atomic level. Consider a warm can of fruit juice in ice water. At first, the molecules in the fruit juice have a higher average kinetic energy than do the water molecules that surround the can, as shown in Figure 2.2(a). This energy is transferred from the juice to the can by the juice molecules , colliding with the metal atoms of the can. The atoms vibrate more because of their increased energy. This energy is then transferred to the surrounding water molecules, as shown in Figure 2.2(b). As the energy of the water molecules gradually increases, the energy of the fruit juice’s molecules and of the can’s atoms decreases until all of the particles have, on the average, equal kinetic energies. In individual collisions, energy may be transferred from the lower-energy water molecules to the higher-energy metal atoms and fruit juice particles. That is, energy can be transferred in either direction. However, because the average kinetic energy of particles is higher in the object at higher temperature, more energy moves out of the object as heat than moves into it. Thus, the net transfer of energy as heat is in only one direction.
The transfer of energy as heat alters an object’s temperature. Equilibrium At thermal equilibrium, the net energy exchanged between two objects equals zero. Energy transferred into can from water
Twater
= ll°C
T=w • n ·c \ Energy transferred out of can into water
306
Chapter 9
Thermal equilibrium may be understood in terms of energy exchange between two objects at equal temperature. When the can of fruit juice and the surrounding water are at the same temperature, as depicted in Figure 2.3, the quantity of energy transferred from the can of fruit juice to the water is the same as the energy transferred from the water to the can of juice. The net energy transferred between the two objects is zero. This reveals the difference between temperature and heat. The atoms of all objects are in continuous motion, so all objects have some internal energy. Because temperature is a measure of that energy, all objects have some temperature. Heat, on the other hand, is the energy transferred from one object to another because of the temperature difference between them. When there is no temperature difference between a substance and its surroundings, no net energy is transferred as heat. Energy transfer as heat depends on the difference of the temperatures of the two objects. The greater the temperature difference is between two objects, the greater the rate of energy transfer between them as heat (other factors being the same).
For example, in winter, energy is transferred as heat from a car’s surface at 30°C to a cold raindrop at 5°C. In the summer, energy is transferred as heat from a car’s surface at 45°C to a warm raindrop at 20°c. In each case, the amount of energy transferred each second is the same, because the substances and the temperature difference (25°C) are the same. See Figure 2.4. The concepts of heat and temperature help to explain why hands held in separate bowls containing hot and cold water subsequently sense the temperature oflukewarm water differently. The nerves in the outer skin of your hand detect energy passing through the skin from objects with temperatures different from your body temperature. If one hand is at thermal equilibrium with cold water, more energy is transferred from the outer layers of your hand than can be replaced by the blood, which has a temperature of about 37.0°C (98.6°F). When the hand is immediately placed in water that is at a higher temperature, energy is transferred from the water to the cooler hand. The energy transferred into the skin causes the water to feel warm. Likewise, the hand that has been in hot water temporarily gains energy from the water. The loss of this energy to the lukewarm water makes that water feel cool.
Rate of Energy Transfer The energy transferred each second as heat from the car’s surface to the raindrop is the same for low temperatures (a) as for high temperatures (b), provided the temperature differences are the same. T raindrop =
s•c
(a)
T raindrop = 20 ° C
(b)
Heat has the units of energy. Before scientists arrived at the modern model for heat, several different units for measuring heat had already been developed. These units are still widely used in many applications and therefore are listed in Figure 2.5. Because heat, like work, is energy in transit, all heat units can be converted to joules, the SI unit for energy. Just as other forms of energy have a symbol that identifies them (PE for potential energy, KE for kinetic energy, U for internal energy, W for work), heat is indicated by the symbol Q.
Heat unit
Equivalent value
Uses
joule (J)
equal to 1 kg•(::)
SI unit of energy
calorie (cal)
4.186 J
non-SI unit of heat; found especially in older works of physics and chemistry
kilocalorie (kcal)
4.186 x 1o3 J
non-SI unit of heat
Calorie, or dietary Calorie
4.186 x 103 J
British thermal unit (Btu)
1.055 x
103 J
therm
1.055 x 108 J
= 1 kcal
food and nutritional science English unit of heat; used in engineering, airconditioning, and refrigeration equal to 100 000 Btu; used to measure natural-gas usage
Heat
307
Thermal Conduction When you first place an iron skillet on a stove, the metal handle feels comfortable to the touch. After a few minutes, the handle becomes too hot to touch without a cooking mitt, as shown in Figure 2.6. The handle is hot because energy was transferred from the high-temperature burner to the skillet. The added energy increased the temperature of the skillet and its contents. This type of energy transfer is called thermal conduction.
Conduction After this burner has been turned on, the skillet’s handle heats up because of conduction. An oven mitt must be used to remove the skillet safely.
The rate of thermal conduction depends on the substance. Thermal conduction can be understood by the behavior of atoms in a metal. As the skillet is heated, the atoms nearest to the burner vibrate with greater energy. These vibrating atoms jostle their less energetic neighbors and transfer some of their energy in the process. Gradually, iron atoms farther away from the element gain more energy. The rate of thermal conduction depends on the properties of the substance being heated. A metal ice tray and a cardboard package of frozen food removed from the freezer are at the same temperature. However, the metal tray feels colder than the package because metal conducts energy more easily and more rapidly than cardboard does. Substances that rapidly transfer energy as heat are called thermal conductors. Substances that slowly transfer energy as heat are called thermal insulators. In general, metals are good thermal conductors. Materials such as asbestos, cork, ceramic, cardboard, and fiberglass are poor thermal conductors (and therefore good thermal insulators).
Convection and radiation also transfer energy. . Did YOU Know?. – – – – – – – – – – – , Although cooking oil is no better a thermal conductor than most nonmetals are, it is useful for transferring energy uniformly around the surface of the food being cooked. When popping popcorn, for instance, coating the kernels with oil improves the energy transfer to each kernel, so a higher percentage of them pop.
308
Chapter 9
There are two other mechanisms for transferring energy between places or objects at different temperatures. Convection involves the movement of cold and hot matter, such as hot air rising upward over a flame. This mechanism does not involve heat alone. Instead, it uses the combined effects of pressure differences, conduction, and buoyancy. In the case of air over a flame, the air is heated through particle collisions (conduction), causing it to expand and its density to decrease. The warm air is then displaced by denser, colder air. Thus, the flame heats the air faster than by conduction alone. The other principal energy transfer mechanism is electromagnetic radiation. Unlike convection, en ergy in this form does not involve the transfer of matter. Instead, objects reduce their internal energy by giving off electromagnetic radiation of particular wavelengths or are heated by electromagnetic radiation like a car is h eated by the absorption of sunlight.
Heat and Work Hammer a nail into a block of wood. After several minutes, pry the nail loose from the block and touch the side of the nail. It feels warm to the touch, indicating that energy is being transferred from the nail to your hand. Work is done in pulling the nail out of the wood. The nail encounters friction with the wood, and most of the energy required to overcome this friction is transformed into internal energy. The increase in the internal energy of the nail raises the nail’s temperature, and the temperature difference between the nail and your hand results in the transfer of energy to your hand as heat. Friction is just one way of increasing a substance’s internal energy. In the case of solids, internal energy can be increased by deforming their structure. Common examples of this deformation are stretching a rubber band or bending a piece of metal.
Total energy is conserved. When the concept of mechanical energy was introduced, you discovered that whenever friction between two objects exists, not all of the work done appears as mechanical energy. Similarly, when objects collide inelastically, not all of their initial kinetic energy remains as kinetic energy after the collision. Some of the energy is absorbed as internal energy by the objects. For this reason, in the case of the nail pulled from the wood, the nail ( and if you could touch it, the wood inside the hole) feels warm. If changes in internal energy are taken into account along with changes in mechanical energy, the total energy is a universally conserved property. In other words, the sum of the changes in potential, kinetic, and internal energy is equal to zero.
Conservation of Energy
ilPE + ilKE + ilU = 0 the change in potential energy + the change in kinetic energy + the change in internal energy = 0
QuickLAB Hold the rubber band between your thumbs. Touch the middle section of the rubber band to your lip and note how it feels. Rapidly stretch the rubber band and keep it stretched. Touch the middle section of the rubber band to your
lip again. Notice whether the rubber band’s temperature has changed. (You may have to repeat this procedure several times before you can clearly distinguish the temperature difference.)
MATERIALS
• 1 large rubber band about 7-10 mm wide SAFETY
To avoid breaking the rubber band, do not stretch it more than a few inches. Do not point a stretched rubber band at another person.
Heat
309

Conservation of Energy Sample Problem B An arrangement similar to the one used to demonstrate energy conservation is shown at right. A vessel contains water. Paddles that are propelled by falling masses turn in the water. This agitation warms the water and increases its internal energy. The temperature of the water is then measured, giving an indication of the water’s internal-energy increase. If a total mass of 11.5 kg falls 1.3 m and all of the mechanical energy is converted to internal energy, by how much will the internal energy of the water increase? (Assume no energy is transferred as heat out of the vessel to the surroundings or from the surroundings to the vessel’s interior.)
0
ANALYZE
Given:
Unknown:
E)
PLAN
m
Joule’s Apparatus
= 11.5 kg
h
= 1.3 m
g
= 9.81 m /s
2
!:iU = ?
Choose an equation or situation: Use the conservation of en ergy equation, and solve for .6.U
!:iPE + !:iKE + !:iU = O
Tips and Tricks Don’t forget that a change in any quantity, indicated by the symbol .6., equals the final value minus the initial value.
(PE – PE)+ (KE – KE)+ !:iU = 0 1 1 !:iU = -PE1 + PEi-KE1 + KEi Because the m asses begin at rest, KEi equals zero. If we assume that KE1 is sm all compared to the loss of PE, we can set KE equal to zero also. 1
KEL. =0 Because all of the potential energy is assumed to be converted to internal energy, PEi can be set equal to m gh if PE is set equal to zero. 1
PEi = mgh Substitute each quantity into the equation for .6. U:
!:i U = O + mgh
E)
SOLVE
Substitute the values into the equation and solve:
Calculator Solution Because the minimum number of significant figures in the data is two, the calculator answer, 146.6595 J, should be rounded to two digits.
310
Chapter 9
+ O + O = mgh
!:iU = (11.5 kg)(9.81 m /s2 )(1.3 m)
I!:iU = 1.5 X
2
10
JI
CR·fo!i,\114-►

Conservation of Energy (continued)
0
CHECK YOUR WORK
The answer can be estimated using rounded values for m and g. If m:::::: 10 kg and g::::: 10 m/s2, then b..U ::::: 130 J, which is close to the actual value calculated.
Practice 1. In the arrangement described in Sample Problem B, how much would the water’s internal energy increase if the mass fell 6.69 m? 2. A worker drives a 0.500 kg spike into a rail tie with a 2.50 kg sledgehammer. The hammer hits the spike with a speed of 65.0 m / s. If one-third of the hammer’s kinetic energy is converted to the internal energy of the hammer and spike, how much does the total internal energy increase? 3. A 3.0 x 10- 3 kg copper penny drops a distance of 50.0 m to the ground. If 65 percent of the initial potential energy goes into increasing the internal energy of the penny, determine the magnitude of that increase. 4. The amount of internal energy needed to raise the temperature of 0.25 kg of water by 0.2°C is 209.3 J. How fast must a 0.25 kg baseball travel in order for its kinetic energy to equal this internal energy?
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Use the microscopic interpretations of temperature and heat to explain how you can blow on your hands to warm them and also blow on a bowl of hot soup to cool it.
2. If a bottle of water is shaken vigorously, will the internal energy of the water change? Why or why not?
3. At Niagara Falls, if 505 kg of water fall a distance of 50.0 m, what is the increase in the internal energy of the water at the bottom of the falls? Assume that all of the initial potential energy goes into increasing the water’s internal energy and that the final kinetic energy is zero.
Critical Thinking 4. A bottle of water at room temperature is placed in a freezer for a short time. An identical bottle of water that has been lying in the sunlight is placed in a refrigerator for the same amount of time. What must you know to determine which situation involves more energy transfer?
5. On a camping trip, your friend tells you that fluffing up a down sleeping bag before you go to bed will keep you warmer than sleeping in the same bag when it is still crushed from being in its storage sack. Explain why this happens. Heat
311
Climate and Clothing o remain healthy, the human body must maintain a temperature of about 37.0°C (98.6°F), which becomes increasingly difficult as the surrounding air becomes hotter or colder than body temperature.
Li
Unless the body is properly insulated, its temperature will drop in its attempt to reach thermal equilibrium with very cold surroundings. If this situation is not corrected in time, the body will enter a state of hypothermia, which lowers pulse, blood pressure, and respiration. Once body temperature reaches 32.2°C (90.0°F), a person can lose consciousness. When body temperature reaches 25.6°C (78.0°F), hypothermia is almost always fatal. To prevent hypothermia, the transfer of energy from the human body to the surrounding air must be hindered, which is done by surrounding the body with heat-insulating material. An extremely effective and common thermal insulator is air. Like most gases, air is a very poor thermal conductor, so even a thin layer of air near the skin provides a barrier to energy transfer. The lnupiat people of northern Alaska have designed clothing to protect them from the severe Arctic climate ‘ where average air temperatures range from 10°c (50°F) to -37°C (-35°F). The lnupiat clothing is made from animal skins that make use of air’s insulating properties. Until recently, the traditional parka (atigi) was made from caribou skins. Two separate parkas are worn in layers, with the fur
The lnupiat parka, called an atigi, consists today of a canvas shell over sheepskin. The wool provides layers of insulating air between the wearer and the cold.
The Bedouin headcloth, called a kefiyah, employs evaporation to remove energy from the air close to the head, which cools the wearer. lining the inside of the inner parka and the outside of the outer parka. Insulation is provided by air that is trapped between the short inner hairs and within the long, hollow hairs of the fur. Today, inner parkas are made from sheepskin, as shown on the left. At the other extreme, the Bedouins of the Arabian Desert have developed clothing that permits them to survive another of the harshest environments on Earth. Bedouin garments cover most of the body, which protects the wearer from direct sunlight and prevents excessive loss of body water from evaporation. These clothes are also designed to cool the wearer. The Bedouins must keep their body temperatures from becoming too high in desert temperatures, which often are in excess of 38°C(100°F). Heat exhaustion or heatstroke will result if the body’s temperature becomes too high. Although members of different tribes, as well as men and women within the same tribes, wear different types of clothing, a few basic garments are common to all Bedouins. One such garment is the kefiyah, a headcloth worn by Bedouin men, as shown in the photograph above. A similar garment made of two separate cloths, which are called a mandil and a hatta, is worn by Bedouin women. Firmly wrapped around the head of the wearer, the cloth absorbs perspiration and cools the wearer during evaporation. This same garment is also useful during cold periods in the desert. The garment, wound snugly around the head, has folds that trap air and provide an insulating layer to keep the head warm.
Changes in Temperature and Phase Key Terms specific heat capacity calorimetry phase change latent heat
Temperature Differences The air around the pool and the water in the pool receive energy from sunlight. However, the increase in temperature is greater for the air than for the water.
Specific Heat Capacity On a hot day the water in a swimming pool, such as the one shown in Figure 3.1, may be cool, even if the air around it is hot. This may seem odd, because both the air and water receive energy from sunlight. One reason that the water may be cooler than the air is evaporation, which is a cooling process. However, evaporation is not the only reason for the difference. Experiments have shown that the change in temperature due to adding or removing a given amount of energy depends on the particular substance. In other words, the same change in energy will cause a different temperature change in equal masses of different substances. The specific heat capacity of a substance is defined as the energy required to change the temperature of 1 kg of that substance by 1°C. (This quantity is also sometimes known as just specific heat.) Every substance has a unique specific heat capacity. This value tells you how much the temperature of a given mass of that substance will increase or decrease, based on how much energy is added or removed as heat. This relationship is expressed mathematically as follows:
Specific Heat Capacity C
p
specific heat capacity the quantity of heat required to raise a unit mass of homogeneous material 1 K or 1°C in a specified way given constant pressure and volume
Q = —
specific heat capacity =
m.b..T
energy transferred as heat . mass x change m temperature
The subscript p indicates that the specific heat capacity is measured at constant pressure. Maintaining constant pressure is an important detail when determining certain thermal properties of gases, which are much more affected by changes in pressure than are solids or liquids. Note that a temperature change of 1 °C is equal in magnitude to a temperature change of 1 K, so b..Tgives the temperature change in either scale. Heat
313
Substance
cP (J/kg•°C)
Substance
cP (J/kg•°C)
Substance
cP (J/kg•°C)
aluminum
8.99 X 102
ice
2.09 X 103
silver
2.34 X 102
copper
3.87 X 102
iron
4.48 X 102
steam
2.01 X 103
glass
8.37 X 102
lead
1.28 X 102
water
4.186 X 103
gold
1.29 X 102
mercury
1.38 X 102
The equation for specific heat capacity applies to both substances that absorb energy from their surroundings and those that transfer energy to their surroundings. When the temperature increases, 6-Tand Qare taken to be positive-which corresponds to energy transferred into the substance. Likewise, when the temperature decreases, 6- T and Q are negative and energy is transferred from the substance. Figure 3.2 lists specific heat capacities that have been determined for several substances.
Calorimetry is used to determine specific heat capacity. A Simple Calorimeter Acalorimeter allows the specific heat capacity of a substance to be determined. Stirrer
~
Insulated outer container
Thermometer
/
lid
/ Water
To measure the specific heat capacity of a substance, it is necessary to measure mass, temperature change, and energy transferred as heat. Mass and temperature change are directly measurable, but the direct measurement of heat is difficult. However, the specific heat capacity of water is known, so the energy transferred as heat between an object of unknown specific heat capacity and a known quantity of water can be measured.
If a hot substance is placed in an insulated container of cool water, energy conservation requires that the energy the substance gives up must equal the energy absorbed by the water. Although some energy is transferred to the surrounding container, this effect is small and will be ignored. Energy conservation can be used to calculate the specific heat capacity, cp,x’ of the substance (indicated by the subscriptx) as follows: energy absorbed by water = energy released by the substance
cp,wmw6.Tw = – cp,xmx6. Tx
Inner Test substance
calorimetry an experimental procedure used to measure the energy transferred from one substance to another as heat
314
Chapter 9
For simplicity, a subscript w will always stand for “water” in specific heat capacity problems. The energy gained by a substance is expressed as a positive quantity, and the energy released is expressed as a negative quantity. The first equation above can be rewritten as Qw + Qx = 0, which , shows that the net change in energy transferred as heat equals zero. Note that 6- T equals the final temperature minus the initial temperature. This approach to determining a substance’s specific heat capacity is called calorimetry, and devices that are used for making this measurement are called calorimeters. A calorimeter, shown in Figure 3.2, contains both a thermometer to measure the final temperature of substances at thermal equilibrium and a stirrer to ensure the uniform mixture of energy.
PREMIUM CONTENT
Interactive Demo
~
~ HMDScience.com
Sample Problem C A 0.050 kg metal bolt is heated to an unknown initial temperature. It is then dropped into a calorimeter containing 0.15 kg of water with an initial temperature of 21.0°C. The bolt and the water then reach a final temperature of 25.0°C. If the metal has a specific heat capacity of 899 J/ kg•°C, find the initial temperature of the metal.
0
ANALYZE
Given:
m metal = m m = 0.050 kg
cp,m = 899 J/ kge°C
m water = m w = 0.15 kg
cp,w = 4186 J/ kg•°C
Twater = Tw = 21.ooc
-Tm ? Tmetal- ·
Unknown: Diagram:
Before placing hot sample in calorimeter
After thermal equilibrium has been reached
_Jo] tf m m=0.050kg
mw= 0.15 kg

~
T f = 25.0° C
Tw = 21.0° C
f:)
PLAN
Choose an equation or situation:
Th e energy absorbed by the water equals the energy removed from the bolt.
Qw = -Qm cp,wm w.6.Tw = – cp,mm m.6.Tm cp,wm w(Tf- Tw)
= -cp,mm m( Tf- Tm)
Rearrange the equation to isolate the unknown:
cp,wm w(Tf – Tw) Tm = c m p,m m
E)
SOLVE
+ Tf
Tips and Tricks Because Twisless than r,,you know that Tm must be greater than r,.
Substitute the values into the equation and solve:
(4186 J/ kg• °C)(0.15 kg)(25.0°C – 21.0°C) T = – – – – – – – – – – – – – + 25.0°C m (899 J/ kg• °C)(0.050 kg)
I Tm= 310c I
0
CHECKYOUR WORK
Tm is greater than T1 , as expected.
G·iti!i ,M§. ► Heat
315
Practice 1. What is the final temperature when a 3.0 kg gold bar at 99°C is dropped into 0.22 kg of water at 25°C? 2. A 0.225 kg sample of tin initially at 97.5°C is dropped into 0.115 kg of water. The initial temperature of the water is l0.0°C. If the specific heat capacity of tin is 230 J/ kg•°C, what is the final equilibrium temperature of the tin-water mixture?
3. Brass is an alloy made from copper and zinc. A 0.59 kg brass sample at 98.0°C is dropped into 2.80 kg of water at 5.0°C. If the equilibrium temperature is 6.8°C, what is the specific heat capacity of brass? 4. A hot, just-minted copper coin is placed in 101 g of water to cool. The water temperature changes by 8.39°C, and the temperature of the coin changes by 68.0°C. What is the mass of the coin?
ST.EM
Earth-Coupled Heat Pumps s the earliest cave dwellers knew, a good way to stay warm in the winter and cool in the summer is to go underground. Now, scientists and engineers are using the same premise-and using existing technology in a new, more efficient way-to heat and cool above-ground homes for a fraction of the cost of conventional systems.
&
The average specific heat capacity of earth is smaller than the average specific heat capacity of air. However, earth has a greater density than air does, which means that near a house, there are more kilograms of earth than of air. So, a 1°C change in temperature involves transferring more energy to or from the ground than to or from the air. Thus, the temperature of the ground in the winter will probably be higher than the temperature of the air above it. In the summer, the temperature of the ground will likely be lower than the temperature of the air.
316
Chapter 9
An earth-coupled heat pump enables homeowners to tap the temperature just below the ground to heat their homes in the winter or cool them in the summer. The system includes a network of plastic pipes placed in trenches or inserted in holes drilled 2 to 3 m (6 to 1Oft) beneath the ground’s surface. To heat a home, a fluid circulates through the pipe, absorbs energy from the surrounding earth, and transfers this energy to a heat pump inside the house. Although the system can function anywhere on Earth’s surface, it is most appropriate in severe climates, where dramatic temperature swings may not be ideal for air-based systems.
Heating Curve of Water
Heating Curve of Water
125 100
p ~
::,
4§ 50 Cl>
Water + steam
C>.
E
Cl>
I-
0
A
B
Steam
This idealized graph shows the temperature change of 10.0 g of ice as it is heated from -25°C in the ice phase to steam above 125°C at atmospheric pressure. (Note that the horizontal scale of the graph is not uniform.)
Water
Ice+ water
-25 Ice
3.85
Heat(10 3J)
30.6
8.04
31.1
—–+
Latent Heat Suppose you place an ice cube with a temperature of -25°C in a glass, and then you place the glass in a room. The ice cube slowly warms, and the temperature of the ice will increase until the ice begins to melt at 0°C. The graph in Figure 3.4 and data in Figure 3.5 show how the temperature of 10.0 g of ice changes as energy is added. You can see that temperature steadily increases from – 25°C to 0°C (segment A of the graph). You could use the mass and the specific heat capacity of ice to calculate how much energy is added to the ice during this segment. At 0°C, the temperature stops increasing. Instead, the ice begins to melt and to change into water (segment B). The ice-and-water mixture remains at this temperature until all of the ice melts. Suppose that you now heat the water in a pan on a stovetop. From 0°c to 100°c , the water’s temperature steadily increases (segment C). At 100°C1 however, the temperature stops rising, and the water turns into steam (segment D). Once the water has completely vaporized, the temperature of the steam increases (segment E).
Segment of Graph
Type of Change
Amount of Energy Transferred as Heat
Temperature Range of Segment
A
temperature of ice increases
522 J
-25°C to 0°c
B
ice melts; becomes water
3.33 X 103 J
0°c
C
temperature of water increases
4.19 X 103 J
0°c to 100°c
D
water boils; becomes steam
2.26 X 104 J
100°c
E
temperature of steam increases
500 J
100°c to 125°c
Heat
317
phase change the physical change of a substance from one state (solid, liquid, or gas) to another at constant temperature and pressure
When substances melt, freeze, boil, condense, or sublime (change from a solid to vapor or from vapor to a solid), the energy added or removed changes the internal energy of the substance without changing the substance’s temperature. These changes in matter are called phase changes.
Latent heat is energy transferred during phase changes. To understand the behavior of a substance undergoing a phase change, you need to consider the changes in potential energy. Potential energy is present among a collection of particles in a solid or in a liquid in the form of attractive bonds. These bonds result from the charges within atoms and molecules. Potential energy is associated with the electric forces between these charges.
latent heat the energy per unit mass that is transferred during a phase change of a substance
Phase changes result from a change in the potential energy between particles of a substance. When energy is added to or removed from a substance that is undergoing a phase change, the particles of the substance rearrange themselves to make up for their change of energy. This rearrangement occurs without a change in the average kinetic energy of the particles. The energy that is added or removed per unit mass is called latent heat, abbreviated as L . Note that according to this definition, the energy transferred as heat during a phase change simply equals the mass multiplied by the latent heat, as follows: Q=mL
During m elting, the energy that is added to a substance equals the difference between the total potential energies for particles in the solid and the liquid phases. This type oflatent heat is called the heat offusion. During vaporization, the energy that is added to a substance equals the difference in the potential energy of attraction between the liquid particles and between the gas particles. In this case, the latent heat is called the heat of vaporization. The heat of fusion and the heat of vaporization are abbreviated as L1 and Lv, respectively. Figure 3.6 lists latent heats for a few substances.
Substance
Melting Point (°C)
L, (J/kg)
Boiling Point (°C)
LV (J/kg)
nitrogen
-209.97
2.55 X 104
-195.81
2.01 X 105
oxygen
– 218.79
1.38 X 104
– 182.97
2.13 X 105
ethyl alcohol
– 114
1.04 X 105
78
8.54 X 105
water
318
Chapter 9
0.00
3.33 X 105
100.00
2.26 X 106
lead
327.3
2.45 X 104
1745
8.70 X 105
aluminum
660.4
3.97 X 105
2467
1.14 X 107

SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. A jeweler working with a heated 47 g gold ring must lower the ring’s temperature to make it safe to handle. If the ring is initially at 99°C, what mass of water at 25°C is needed to lower the ring’s temperature to 38°C?
2. How much energy must be added to a bowl of 125 popcorn kernels in order for them to reach a popping temperature of l 75°C? Assume that their initial temperature is 21 °C, that the specific heat capacity of popcorn is 1650 J/kg•°C, and that each kernel has a mass of0.105 g. 3. Because of the pressure inside a popcorn kernel, water does not vaporize at 100°c . Instead, it stays liquid until its temperature is about l 75°C, at which point the kernel ruptures and the superheated water turns into steam. How much energy is needed to pop 95.0 g of corn if 14 percent of a kernel’s mass consists of water? Assume that the latent heat of vaporization for water at l 75°C is 0.90 times its value at 100°C and that the kernels have an initial temperature of l 75°C.
Critical Thinking 4. Using the concepts of latent heat and internal energy, explain why it is difficult to build a fire with damp wood. 5. Why does steam at 100°c cause more severe burns than does liquid water at 100°C?
Interpreting Graphics 6. From the heating curve for a 15 g sample, as shown in Figure 3.7, estimate the following properties of the substance. a. the specific heat capacity of the liquid b. the latent heat of fusion c. the specific heat capacity of the solid d. the specific heat capacity of the vapor e. the latent heat of vaporization
Heating Curve for 15 g of an Unknown Substance 400 Gas –
300
iE
200
E 8.
~ 100
Solid
I-
Solid + liquid
1.27
Liquid+ gas
8.37
15.8
795 796
Heat(kJ) –
Heat
319
HVAC Technician VAC stands for heating, ventilation, and air conditioning. An HVAC technician knows what it [}{] takes to keep buildings warm in winter and cool in summer. To learn more about working with HVAC as a career, read the interview with contractor and business owner Doug Garner.
Doug Garner is checking a potential relay. This relay is connected to a capacitor that starts the compressor.
What does an HVAC technician do?
Basically, we sell, replace, and repair air-conditioning and heating equipment. We replace obsolete A/C and heating units in older homes and buildings, we install new units in new homes and buildings, and we repair units when they break down. How did you become an HVAC technician?
There are numerous ways to get into the business. When I was about 17 years old, I was given an opportunity to work for a man with whom I went to church. I worked as an apprentice for three years after high school, and I learned from him and a couple of very good technicians. I also took some business courses at a local community college to help with the business end. What about HVAC made it more interesting than other fields?
There were other things that I was interested in doing, but realistically HVAC was more practical. In other words, that’s where the money and opportunities were for me. What is the nature of your work?
I have a company with two service technicians and an apprentice. Most of my duties involve getting jobs secured, bidding on and designing the different systems to suit the needs of the customer. I have to have a basic understanding of advertising, marketing, and sales as well as of the technical areas as they apply to this field. Our technicians must be able to communicate well and have a good mechanical aptitude.
What do you like most about your job?
You get to work in a lot of different places and situations. It is never boring, and you meet a lot of people. You can make as much money as you are willing to work for. What advice would you give to students who are interested in your field?
Take a course in HVAC at a technical institute or trade school, and then work as an apprentice for a few years. Mechanical engineering, sales, communication, and people skills are all important in this field; the more education you have, the more attractive you can be to a company.
I
SECTION 1
Temperature and Thermal Equilibrium
,: ,
• Temperature can be changed by transferring energy to or from a substance. • Thermal equilibrium is the condition in which the temperature of two objects in physical contact with each other is the same.
1 ,
1
r: ,
temperature internal energy thermal equilibrium
• The most common temperature scales are the Fahrenheit, Celsius, and Kelvin (or absolute) scales.
SECTION 2
Defining Heat
, ,-
• Heat is energy that is transferred from objects at higher temperatures to objects at lower temperatures.
1
l , ·:
heat
• Energy is transferred by thermal conduction through particle collisions. • Energy is conserved when mechanical energy and internal energy are taken into account. Thus, for a closed system, the sum of the changes in kinetic energy, potential energy, and internal energy must equal zero.
SECTION 3
Changes in Temperature and Phase
11
• Specific heat capacity is a measure of the energy needed to change a substance’s temperature.
1
1 1 1 ·.-
specific heat capacity calorimetry
• By convention, the energy that is gained by a substance is positive, and the energy that is released by a substance is negative.
phase change latent heat
• Latent heat is the energy required to change the phase of a substance.
VARIABLE SYMBOLS
Quantities
Units
T
temperature (Kelvin)
K
Tc
temperature (Celsius)
oc
degrees Celsius
TF
temperature (Fahrenheit)
OF
degrees Fahrenheit
6.U Q
kelvins
change in internal energy
J
joules
heat
J
joules
cP
specific heat capacity at constant pressure
L
latent heat
J kge°C J kg
—- – — –
Problem Solving See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
Chapter Summary
321
Temperature and Thermal Equilibrium REVIEWING MAIN IDEAS 1. What is the relationship between temperature and
PRACTICE PROBLEMS For problems 9-1 0, see Sample Problem A. 9. The highest recorded temperature on Earth was 136°F, at Azizia, Libya, in 1922. Express this temperature in degrees Celsius and in kelvins.
internal energy?
2. What must be true of two objects if the objects are in a state of thermal equilibrium? 3. What are some physical properties that could be used in developing a temperature scale?
10. The melting point of gold is 1947°F. Express this temperature in degrees Celsius and in kelvins.
Defining Heat REVIEWING MAIN IDEAS
CONCEPTUAL QUESTIONS 4. What property must a substance have in order to be used for calibrating a thermometer?
5. Which object in each of the following pairs has greater total internal energy, assuming that the two objects in each pair are in thermal equilibrium? Explain your reasoning in each case. a. a metal knife in thermal equilibrium with a hot griddle b. a 1 kg block of ice at -25°C or seven 12 g ice cubes at -25°C 6. Assume that each pair of objects in item 5 has the same internal energy instead of the same temperature. Which item in each pair will have the higher temperature? 7. Why are the steam and ice points of water better fixed
points for a thermometer than the temperature of a human body? 8. How does the temperature of a tub of hot water as measured by a thermometer differ from the water’s temperature before the measurement is made? What property of a thermometer is necessary for the difference between these two temperatures to be minimized?
322
Chapter 9
11. Which drawing below shows the direction in which net energy is transferred as heat between an ice cube and the freezer walls when the temperature of both is – l0°C? Explain your answer.
(a)

(b)

(c)

12. A glass of water has an initial temperature of8°C. In which situation will the rate of energy transfer be greater, when the air’s temperature is 25°C or 35°C? 13. How much energy is transferred between a piece of toast and an oven when both are at a temperature of 55°C? Explain. 14. How does a metal rod conduct energy from one end, which has been placed in a fire, to the other end, which is at room temperature?
15. How does air within winter clothing keep you warm on cold winter days?
CONCEPTUAL QUESTIONS 16. If water in a sealed, insulated container is stirred, is its temperature likely to increase slightly, decrease slightly, or stay the same? Explain your answer.
17. Given your answer to item 16, why does stirring a hot cup of coffee cool it down? 18. Given any two bodies, the one with the higher temperature contains more heat. What is wrong with this statement? 19. Explain how conduction causes water on the surface of a bridge to freeze sooner than water on the road surface on either side of the bridge.
20. A tile floor may feel uncomfortably cold to your bare feet, but a carpeted floor in an adjoining room at the same temperature feels warm. Why? 21 . Why is it recommended that several items of clothing be worn in layers on cold days? 22. Why does a fan make you feel cooler on a hot day? 23. A paper cup is filled with water and then placed over an open flame, as shown at right. Explain why the cup does not catch fire and burn.
PRACTICE PROBLEMS
25. A 0.75 kg spike is hammered into a railroad tie. The initial speed of the spike is equal to 3.0 m / s. a. If the tie and spike together absorb 85 percent of the spike’s initial kinetic energy as internal energy, calculate the increase in internal energy of the tie and spike. b. What happens to the remaining energy?
Changes in Temperature and Phase REVIEWING MAIN IDEAS 26. What principle permits calorimetry to be used to determine the specific heat capacity of a substance? Explain. 27. Why does the temperature of melting ice not change even though energy is being transferred as heat to the ice?
CONCEPTUAL QUESTIONS 28. Why does the evaporation of water cool the air near the water’s surface?
29. Until refrigerators were invented, many people stored fruits and vegetables in underground cellars. Why was this more effective than keeping them in the open air? 30. During the winter, the people mentioned in item 29 would often place an open barrel of water in the cellar alongside their produce. Explain why this was done and why it would be effective.
For problems 24-25, see Sample Problem B.
PRACTICE PROBLEMS
24. A force of 315 N is applied horizontally to a crate in order to displace the crate 35.0 m across a level floor at a constant velocity. As a result of this work, the crate’s internal energy is increased by an amount equal to 14 percent of the crate’s initial internal energy. Calculate the initial internal energy of the crate. (Disregard the work done on the floor, and assume that all work goes into the crate.)
For problems 31-32, see Sample Problem C. 31. A 25.5 g silver ring (cp = 234 J/kg•°C) is heated to a temperature of 84.0°C and then placed in a calorimeter containing 5.00 x 10- 2 kg of water at 24.0°C. The calorimeter is not perfectly insulated, however, and 0.140 kJ of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?
Chapter Review
323
32. When a driver brakes an automobile, friction between the brake disks and the brake pads converts part of the car’s translational kinetic energy to internal energy. If a 1500 kg automobile traveling at 32 m /s comes to a halt after its brakes are applied, how much can the temperature rise in each of the four 3.5 kg steel brake disks? Assume the disks are made of iron (cp = 448 J/ kg•°C) and that all of the kinetic energy is distributed in equal parts to the internal energy of the brakes.
Mixed Review REVIEWING MAIN IDEAS 33. Absolute zero on a temperature scale called the Rankine scale is TR= 0°R, and the scale’s unit is the same size as the Fahrenheit degree. a. Write a formula that relates the Rankine scale to the Fahrenheit scale. b. Write a formula that relates the Rankine scale to the Kelvin scale.
34. A 3.0 kg rock is initially at rest at the top of a cliff. Assuming the rock falls into the sea at the foot of the cliff and that its kinetic energy is transferred entirely to the water, how high is the cliff if the temperature of 1.0 kg of water is raised 0.l0°C? (Neglect the heat capacity of the rock.) 35. The freezing and boiling points of water on the imaginary “Too Hot” temperature scale are selected to be exactly 50 and 200 degrees TH. a. Derive an equation relating the Too Hot scale to the Celsius scale. (Hint: Make a graph of one temperature scale versus the other, and solve for the equation of the line.) b. Calculate absolute zero in degrees TH. 36. A hot-water heater is operated by solar power. If the solar collector has an area of 6.0 m 2 and the power delivered by sunlight is 550 W/ m 2 , how long will it take to increase the temperature of 1.0 m 3 of water from 21 °C to 61 °C?
Specific Heat Capacity Specific heat capacity (cp), as you learned earlier in this chapter, is equal to the amount of energy required to change the temperature of 1 kg of a substance by 1°C. This relationship is expressed by the following equation: Q
f}..T =
me
p
In this equation, f}.. Tis the change in temperature, Q is the amount of energy absorbed by the substance as heat, cP is the specific heat capacity of the substance, and m is the mass of the substance. This equation can be represented on a graphing calculator as follows: Y1 = T + (X/(MC))
324
Chapter 9
A graph of this equation will illustrate the relationship between energy absorbed as heat and temperature. In this graphing calculator activity, you will enter various values for the energy absorbed and will determine the resulting temperature. Then, you can explore how changing the specific heat capacity, mass, and initial temperature changes your results. Go online to HMDScience.com to find this graphing calculator activity.
37. A student drops two metallic objects into a 120 g steel container holding 150 g of water at 25°C. One object is a 253 g cube of copper that is initially at 85°C, and the other is a chunk of aluminum that is initially at 5°C. To the surprise of the student, the water reaches a final temperature of 25°C, its initial temperature. What is the mass of the aluminum chunk? 38. At what Fahrenheit temperature are the Kelvin and Fahrenheit temperatures numerically equal?
ALTERNATIVE ASSESSMENT 1. According to legend, Archimedes determined whether the king’s crown was pure gold by comparing its water displacement with the displacement of a piece of pure gold of equal mass. But this procedure is difficult to apply to very small objects. Use the concept of specific heat capacity to design a method for determining whether a ring is pure gold. Present your plan to the class, and ask others to suggest improvements to your design. Discuss each suggestion’s advantages and disadvantages. 2. The host of a cooking show on television claims that you can greatly reduce the baking time for potatoes by inserting a nail through each potato. Explain whether this advice has a scientific basis. Would this approach be more efficient than wrapping the potatoes in aluminum foil? List all arguments and discuss their strengths and weaknesses.
3. The graph of decreasing temperature versus time of a hot object is called its cooling curve. Design and perform an experiment to determine the cooling curve of water in containers of various materials and shapes. Draw cooling curves for each one. Which trends represent good insulation? Use your findings and graphs to design a lunch box that keeps food warm or cold.
39. A 250 g aluminum cup holds and is in thermal equilibrium with 850 g of water at 83°C. The combination of cup and water is cooled uniformly so that the temperature decreases by l.5°C per minute. At what rate is energy being removed? 40. A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 32°C. In an attempt to cool the liquid, which has a mass of 180 g, 112 g of ice at 0°C is added. At the time at which the temperature of the tea (and melted ice) is 1s°C, determine the mass of the remaining ice in the jar. Assume the specific heat capacity of the tea to be that of pure liquid water.
4. Research the life and work of James Prescott Joule, who is best known for his apparatus demonstrating the equivalence of work and heat and the conservation of energy. Many scientists initially did not accept Joule’s conclusions. Research the reasoning behind their objections. Prepare a presentation for a class discussion either supporting the objections ofJoule’s critics or defending Joule’s conclusion before England’s Royal Academy of Sciences. 5. Research how scientists measure the temperature of the following: the sun, a flame, a volcano, outer space, liquid hydrogen, mice, and insects. Find out what instruments are used in each case and how they are calibrated to known temperatures. Using what you learn, prepare a chart or other presentation on the tools used to measure temperature and the limitations on their ranges.
6. Get information on solar water heaters that are available where you live. How does each type work? Compare prices and operating expenses for solar water heaters versus gas water heaters. What are some of the other advantages and limitations of solar water heaters? Prepare an informative brochure for homeowners who are interested in this technology.
Chapter Review
325
MULTIPLE CHOICE
5. A cup of hot chocolate with a temperature of 40°C is placed inside a refrigerator at S C. An identical cup of hot chocolate at 90°C is placed on a table in a room at 25°C. A third identical cup of hot chocolate at 80°C is placed on an outdoor table, where the surrounding air has a temperature of 0°C. For which of the three cups has the most energy been transferred as heat when equilibrium has been reached? A. The first cup has the largest energy transfer. B. The second cup has the largest energy transfer. C. The third cup has the largest energy transfer. D. The same amount of energy is transferred as heat for all three cups. 0
1. What must be true about two given objects for energy to be transferred as heat between them? A. The objects must be large. B. The objects must be hot. C. The objects must contain a large amount of energy. D. The objects must have different temperatures. 2. A metal spoon is placed in one of two identical cups of hot coffee. Why does the cup with the spoon have a lower temperature after a few minutes? F. Energy is removed from the coffee mostly by conduction through the spoon. G. Energy is removed from the coffee mostly by convection through the spoon. H. Energy is removed from the coffee mostly by radiation through the spoon . J. The metal in the spoon has an extremely large specific heat capacity.
6. What data are required in order to determine the specific heat capacity of an unknown substance by means of calorimetry? F. cp,water’ Twater’ ~ubstance’ Tfinal’ V water’ ½ubstance
G.
cp,substance’ T water’ ~ubstance’ Tfinal’ mwater’ msubstance
H. cp,water’ ~ubstance’ m water’ msubstance
Use the passage below to answer questions 3-4.
J. cp,water’ Twater’ ~ubstance’ Tfinal’ m water’ m substance
The boiling point ofliquid hydrogen is -252.87°C. 7. During a cold spell, Florida orange growers often
3. What is the value of this temperature on the Fahrenheit scale? A. 20.28°F B. -220.87°F C. -423.2°F D. 0°F 4. What is the value of this temperature in kelvins? F. 273K G. 20.28K H. -423.2K J. OK
326
Chapter 9
spray a mist of water over their trees during the night. Why is this done? A. The large latent heat of vaporization for water keeps the trees from freezing. B. The large latent heat of fusion for water prevents it and thus the trees from freezing. C. The small latent heat of fusion for water prevents the water and thus the trees from freezing. D. The small heat capacity of water makes the water a good insulator.
. TEST PREP
Use the heating curve below to answer questions 8-10. The graph shows the change in temperature of a 23 g sample of a substance as energy is added to the substance as heat. 600
13. Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and
E 450 e = e a,

300
Solid
a.
E
a,
I-
12. If Lake Superior were still liquid at 0°C, how much energy would need to be removed from the lake for it to become completely frozen?
water in separate beakers at the same temperature are supplied with the same amount of energy, which will have the higher final temperature?
Solid+ liquid
150
14. A 0.200 kg glass holds 0.300 kg of hot water, as O 1.85
12.0 16.6 855 857 Heat (kJ) —–+
8. What is the specific heat capacity of the liquid? F. 4.4 X 105 J/kg•°C G. 4.0 x 102 J/kge°C H. 5.0 x 102 J/kg•°C J. 1.1 X 103 J/kg•°C
shown below. The glass and water are set on a table to cool. After the temperature has decreased by 2.0°C, how much energy has been removed from the water and glass? (The specific heat capacity of glass is 837 J/ kg•°C, and that of water is 4186 J/ kg• C.) 0
9. What is the latent heat of fusion? A. 4.4 x 105 J/kg B. 4.0 x 102 J/kg•°C C. 10.15 x 103 J D. 3.6 x 107 J/kg 10. What is the specific heat capacity of the solid? F. 1.85 x 103 J/ kg•°C G. 4.0 x 102 J/kg•°C H. 5.0 x 102 J/kge°C J. 1.1 X 103 J/kg•°C
Energy transferred as heat
EXTENDED RESPONSE 15. How is thermal energy transferred by the process of convection?
SHORT RESPONSE Base your answers to questions 11-12 on the information below.
The largest of the Great Lakes, Lake Superior, contains 1.20 x 10 16 kg of fresh water, which has a specific heat capacity of 4186 J/ kg•°C and a latent heat of fusion of 3.33 x 105 J/ kg.
16. Show that the temperature – 40.0° is unique in that it has the same numerical value on the Celsius and Fahrenheit scales. Show all of your work.
11. How much energy would be needed to increase the temperature of Lake Superior by l.0°C?
Test Tip
Use dimensional analysis to check your work when solving mathematical problems. Include units in each step of your calculation. If you do not end up with the correct unit in your answer, check each step of your calculation for errors.
Standards-Based Assessment
327
Global Warming Data recorded from various locations around the world over the past century indicate that the average atmospheric temperature is currently 0.6°C higher than it was 100 years ago. However, historical studies indicate that some short-term fluctuations in climate are natural, such as the Little Ice Age of the 17th century. Does this recent increase represent a trend toward global warming or is it simply part of a natural cyclic variation in climate? Although the answer cannot be determined with certainty, most scientists now believe that global warming is a significant issue that requires worldwide attention. Identify a Problem: The Greenhouse Effect
Global warming may be due, in part, to the greenhouse effect. The glass of a greenhouse traps sunlight inside the greenhouse, and thus a warm environment is created-even in the winter. Earth’s atmosphere functions in a similar way, as the diagram below shows. Molecules of “greenhouse gases,” primarily carbon dioxide and methane, absorb energy that radiates from Earth’s surface. These molecules then release energy as heat, causing the atmosphere to be warmer than it would be without these gases. The greenhouse effect is beneficial-without it, Earth would be far too cold to support life. However, increased levels of greenhouse gases in the
f)
0
328
Solar radiation passes through the atmosphere and warms Earth’s surface.
atmosphere are recognized by scientists as contributing to global warming. Carbon dioxide and methane are natural components of our atmosphere. However, the levels of atmospheric carbon dioxide and methane have increased rapidly during the last 100 years. This increase has been determined by analyzing air trapped in the ice layers of Greenland. Deeper sections of the ice contain air from earlier times. During the last ice age, our atmosphere contained about 185 ppm (parts per million) of carbon dioxide, CO 2. The levels 130 years ago were about 300 ppm. Today, the levels are about 390 ppm. This increase can be attributed to the increase in combustion reactions, due primarily to coal and petroleum burning, and to deforestation, which has decreased the number of trees that consume CO2. Brainstorm Solutions
What are some solutions for reducing greenhouse gases? One way is to decrease the output of greenhouse gases. Many technologies promising to do this are currently available. More energy-efficient cars and light bulbs reduce the amount of energy used. This decreases the amount of gasoline or coal consumed and lowers carbon dioxide emissions. Another option is to replace coal-based power plants with solar, wind, and nuclear power plants that produce almost no carbon
Energy from the sun is absorbed by Earth’s surface and then is radiated into the atmosphere as heat, some of which escapes into
Greenhouse gases also absorb some of the energy from Earth and radiate it back toward the lower atmosphere and Earth’s surface.
dioxide. Reducing output alone is not enough, however. We can also capture and store gases already in the atmosphere. By planting trees and reducing deforestation, carbon dioxide can be captured and stored. Select a Solution
Scientists, policy makers, businesses, and citizens need to work together to decide which solutions make most sense. There are many questions to consider. Which solutions are most effective? What are the costs associated with each technology? How easy is it to apply each? Different groups have different views. Some groups might be more worried about the economic costs. Other groups might be more concerned in reducing carbon dioxide output quickly. Selecting a solution requires balancing the costs and benefits not only in the United States but also all over the world.
This false-color image shows the energy radiating from Earth’s upper atmosphere. The blue areas are the coldest. The American southwest is in the upper right-hand corner.
Communicate
New technologies that reduce carbon emissions are not effective if people don’t use and implement them. Even if car companies produce efficient cars, nothing changes unless people buy them. Homeowners need to be willing to spend time and money to renovate their homes to be more energyefficient. Because many of these new technologies are more expensive than current technologies, they can be difficult to implement on a large scale. Companies and people are sometimes encouraged to use emission-reducing technologies through financial incentives, education, and outreach programs. People will use these technologies only if they understand the benefits and if the costs are not too high.
Design Your Own Conduct Research
Carbon dioxide levels in the atmosphere have varied throughout Earth’s history. Research the roles of volcanoes, plants, and limestone formation, and determine whether these processes have any bearing on the current increase in CO2 concentrations. Brainstorm Solutions Can you think of any practical means of using these formation processes to reduce CO 2 concentrations? Evaluate
Pick one of the solutions you brainstormed. What would be the advantages and disadvantages? Can it be easily implemented? Would people support it?
329
SECTION 1 Objectives

Recognize that a system can absorb or release energy as heat in order for work to be done on or by the system and that work done on or by a system can result in the transfer of energy as heat.

Compute the amount of work done during a thermodynamic process.

Distinguish between isovolumetric, isothermal, and adiabatic thermodynamic processes.
Steam Doing Work Energy transferred as heat turns water into steam. Energy from the steam does work on the air outside the balloon.
Relationships Between Heat and Work Key Terms system environment
isovolumetric process isothermal process
adiabatic process
Heat, Work, and Internal Energy Pulling a nail from a piece of wood causes the temperature of the nail and the wood to increase. Work is done by the frictional forces between the nail and the wood fibers. This work increases the internal energy of the iron atoms in the nail and the molecules in the wood. The increase in the nail’s internal energy corresponds to an increase in the nail’s temperature, which is higher than the temperature of the surrounding air. Thus, energy is transferred as heat from the nail to the air. When they are at the same temperature, this energy transfer stops.
Internal energy can be used to do work. The example of the hammer and nail illustrates that work can increase the internal energy of a substance. This internal energy can then decrease through the transfer of energy as heat. The reverse is also possible. Energy can be transferred to a substance as heat, and this internal energy can then be used to do work. Consider a flask of water. A balloon is placed over the mouth of the flask, and the flask is heated until the water boils. Energy transferred as heat from the flame of the gas burner to the water increases the internal energy of the water. When the water’s temperature reaches the boiling point, the water changes phase and becomes steam. At this constant temperature, the volume of the steam increases. This expansion provides a force that pushes the balloon outward and does work on the atmosphere, as shown in Figure 1.1. Thus, the steam does work, and the steam’s internal energy decreases as predicted by the principle of energy conservation.
Heat and work are energy transferred to or from a system. On a microscopic scale, heat and work are similar. In this textbook, both are defined as energy that is transferred to or from a substance. This changes the substance’s internal energy (and thus its temperature or phase). In other words, the terms heat and work always refer to energy in transit. An object never has “heat” or “work” in it; it has only internal energy. 332
Chapter 10
In the previous examples, the internal energy of a substance or combination of substances has been treated as a single quantity to which energy is added or from which energy is taken away. Such a substance or combination of substances is called a system. An example of a system would be the flask, balloon, water, and steam
system a set of particles or interacting components considered to be a distinct physical entity for the purpose of study
that were heated over the burner. As the burner transferred energy as heat to the system, the system’s internal energy increased. When the expanding steam did work on the air outside the balloon by pushing it back (as the balloon expanded), the system’s internal energy decreased. Some of the energy transferred to the system as heat was transferred out of the system as work done on the air. A system is rarely completely isolated from its surroundings. In the example above, a heat interaction occurs between the burner and the system, and work is done by the system on the surroundings (the balloon moves the outside air outward). Energy is also transferred as heat to the air surrounding the flask because of the temperature difference between the flask and the surrounding air. In such cases, we must account for all of the interactions between the system and its environment that could affect the system’s internal energy.
environment t he combination of conditions and influences outside a system that affect the behavior of the system
Work done on or by a gas is pressure multiplied by volume change. In thermodynamic systems, work is defined in terms of pressure and volume change. Pressure is a measure of how much force is applied over a given area (P = Fl A). Change in volume is equal to area multiplied by displacement(~ V = Ad). These expressions can be substituted into the definition of work introduced in the chapter “Work and Energy” to derive a new definition for the work done on or by a gas, as follows : W=Fd W = Fd (~)
= (~) (Ad) = P~ V Gas Expanding Work done on
Work Done by a Gas
work = pressure x volume change
or by the gas is the product of the volume change (area A multiplied by the displacement d) and the pressure of the gas.
This chapter will use only this new definition of work. Note that this definition assumes that Pis constant. If the gas expands, as shown in Figure 1.2, ~ Vis positive, and the work done by the gas on the piston is positive. If the gas is compressed, ~Vis negative, and the work done by the gas on the piston is negative. (In other words, the piston does work on the gas.) When the gas volume remains constant, there is no displacement and no work is done on or by the system.
Although the pressure can change during a process, work is done only if the volume changes. A situation in which pressure increases and volume remains constant is comparable to one in which a force does not displace a mass even as the force is increased. Work is not done in either situation. Thermodynamics
333
PREMIUM CONTENT
A:\
Work Done on or by a Gas
Interactive Demo
~ HMDScience.com
Sample Problem A An engine cylinder has a cross-sectional area of 0.010 m 2 • How much work can be done by a gas in the cylinder if the gas exerts a constant pressure of 7 .5 x 105 Pa on the piston and moves the piston a distance of 0.040 m?
0
ANALYZE
Given:
A= 0.010 rn2
d
= 0.040 rn
P = 7.5 x 105 Pa= 7.5 x 105 N/rn2 Unknown:
E)
SOLVE
W
=?
Use the equation for the work done on or by a gas.
W = P.6.V= PAd W
= (7.5 x
105 N/rn2 ) (0.010 rn 2) (0.040 rn)
I W = 3.0 x 10 J I 2
I
Tips and Tricks Because Wis positive, we can conclude that the work is done by the gas rather than on the gas.
Practice 1. Gas in a container is at a pressure of 1.6 x 105 Pa and a volume of 4.0 m 3 • What is the work done by the gas if a. it expands at constant pressure to twice its initial volume? b. it is compressed at constant pressure to one-quarter of its initial volume?
2. A gas is enclosed in a container fitted with a piston. The applied pressure is maintained at 599.5 kPa as the piston moves inward, which changes the volume of the gas from 5.317 x 10- 4 m 3 to 2.523 x 10- 4 m 3 • How much work is done? Is the work done on or by the gas? Explain your answer. 3. A balloon is inflated with helium at a constant pressure that is 4.3 x 105 Pa in excess of atmospheric pressure. If the balloon inflates from a volume of 1.8 x 10- 4 m 3 to 9.5 x 10- 4 m 3 , how much work is done on the surrounding air by the helium-filled balloon during this expansion?
4. Steam moves into the cylinder of a steam engine at a constant pressure and does 0.84 J of work on a piston. The diameter of the piston is 1.6 cm, and the piston travels 2.1 cm. What is the pressure of the steam?
334
Chapter 10
Thermodynamic Processes In this section, three distinct quantities have been related to each other: internal energy (U), heat ( Q), and work (W). Processes that involve only work or only heat are rare. In most cases, energy is transferred as both heat and work. However, in many processes, one type of energy transfer is dominant and the other type negligible. In these cases, the real process can be approximated with an ideal process. For example, if the dominant form of energy transfer is work and the energy transferred as heat is extremely small, we can neglect the heat transfer and still obtain an accurate model. In this way, many real processes can be approximated by one of three ideal processes. Later, you will learn about ideal processes in gases. All objects have internal energy, which is the sum of the kinetic and potential energies of their molecules. However, monatomic gases present a simpler situation because their molecules are too far apart to interact with each other significantly. Thus, all of their internal energy is kinetic.
No work is done in a constant-volume process. In general, when a gas undergoes a change in temperature but no change in volume, no work is done on or by the system. Such a process is called a constant-volume process, or isovolumetric process. One example of an isovolumetric process takes place inside a bomb calorimeter, shown in Figure 1.3. In the container, a small quantity of a substance undergoes a combustion reaction. The energy released by the reaction increases the pressure and temperature of the gaseous products. Because the walls are thick, there is no change in the volume of the gas. Energy can be transferred to or from the container as only heat. The temperature increase of water surrounding the bomb calorimeter provides information for calculating the amount of energy produced by the reaction.
isovolumetric process a thermodynamic process that takes place at constant volume so that no work is done on or by the system
A Bomb Calorimeter The volume inside the bomb calorimeter is nearly constant, so most of the energy is transferred to or from the calorimeter as heat. Insulated calorimeter with water Bomb
\
Bomb lid with valve for introducing oxygen
\
T Thermometer
/
Electrodes
1—- Combustion crucible with reactants
Thermodynamics
335
Internal energy is constant in a constant-temperature process. isothermal process a thermodynamic process that takes place at constant temperature
An Isothermal Process An isothermal process can be approximated if energy is slowly removed from a system as work while an equivalent amount of energy is added as heat.
During an isothermal process, the temperature of the system does not change. In an ideal gas, internal energy depends only on temperature; therefore, if temperature does not change, then internal energy cannot change either. Thus, in an isothermal process, internal energy does not change when energy is transferred to or from the system as heat or work. One example of an isothermal process is illustrated in Figure 1.4. Although you may think of a balloon that has been inflated and sealed as a static system, it is subject to continuous thermodynamic effects. Consider what happens to such a balloon during an approaching storm. (To simplify this example, we will assume that the balloon is only partially inflated and thus does not store elastic energy.) During the few hours before the storm arrives, the barometric pressure of the atmosphere steadily decreases by about 2000 Pa. If you are indoors and the temperature of the building is controlled, any change in outside temperature will not occur indoors. But because no building is perfectly sealed, changes in the pressure of the air outside also occur inside. As the atmospheric pressure inside the building slowly decreases, the balloon expands and slowly does work on the air outside the balloon. At the same time, energy is slowly transferred into the balloon as heat. The net result is that the air inside the balloon stays at the same temperature as the air outside the balloon. Thus, the internal energy of the balloon’s air does not change. The energy transferred out of the balloon as work is matched by the energy transferred into the balloon as heat. You may wonder how energy can be transferred as heat from the air outside the balloon to the air inside when both gases are at the same constant temperature. The reason is that energy can be transferred as heat in an isothermal process if you consider the process as consisting of a large number of very gradual, very small, sequential changes, as shown in Figure 1.5.
Small Energy Transfers In an isothermal process in a partially inflated balloon, (a) small amounts of energy are removed as work. (b) Energy is added to the gas within the balloon’s interior as heat so that (c) thermal equilibrium is quickly restored.
336
Chapter 10

Energy is not transferred as heat in an adiabatic process. When a tank of compressed gas is opened to fill a toy balloon, the process of inflation occurs rapidly. The internal energy of the gas does not remain constant. Instead, as the pressure of the gas in the tank decreases, so do the gas’s internal energy and temperature. If the balloon and the tank are thermally insulated, no energy can be
transferred from the expanding gas as heat. A process in which changes occur but no energy is transferred to or from a system as heat is called an adiabatic process. The decrease in internal energy must therefore be equal to the energy transferred from the gas as work. This work is done by the confined gas as it pushes the wall of the balloon outward, overcoming the pressure exerted by the air outside the balloon. As a result, the balloon inflates, as shown in Figure 1.6. Note that unlike an isothermal process, which must happen slowly, an adiabatic process must happen rapidly.
adiabatic process a thermodynamic process during which no energy is transferred to or from the system as heat
•iMihllU Adiabatic Process As the gas inside the tank and balloon rapidly expands, its internal energy decreases. This energy leaves the system by means of work done against the outside air.
As mentioned earlier, the three processes described here rarely occur ideally, but many situations can be approximated by one of the three processes. This allows you to make predictions. For example, both refrigerators and internal-combustion engines require that gases be compressed or expanded rapidly. By making the approximation that these processes are adiabatic, one can make quite good predictions about how these machines will operate. Tank
SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. In which of the situations listed below is energy being transferred as heat to the system in order for the system to do work? a. Two sticks are rubbed together to start a fire. b. A firecracker explodes. c. A red-hot iron bar is set aside to cool. 2. A gasoline vapor and air mixture is placed in an engine cylinder. The piston has an area of7.4 x 10-3 m 2 and is displaced inward by 7.2 x 10-2 m. If 9.5 x 105 Pa of pressure is placed on the piston, how much work is done during this process? Is work being done on or by the gas mixture?
3. A weather balloon slowly expands as energy is transferred as heat from the outside air. If the average net pressure is 1.5 x 103 Pa and the balloon’s volume increases by 5.4 x 10- 5 m 3, how much work is done by the expanding gas?
Critical Thinking 4. Identify the following processes as isothermal, isovolumetric, or adiabatic: a. a tire being rapidly inflated b. a tire expanding gradually at a constant temperature c. a steel tank of gas being heated Thermodynamics
337
SECTION 2
The First Law ol Thermodynamics
Objectives ►
Illustrate how the first law of thermodynamics is a statement of energy conservation.

Calculate heat, work, and the change in internal energy by applying the first law of thermodynamics.

Key Term cyclic process
Energy Conservation
Apply the first law of thermodynamics to describe cyclic processes.
Imagine a roller coaster that operates without friction. The car is raised against gravitational force by work. Once the car is freely moving, it will have a certain kinetic energy (KE) and a certain potential energy (PE) . Because there is no friction, the mechanical energy (KE + PE) remains constant throughout the ride’s duration. Thus, when the car is at the top of the rise, it moves relatively slowly (larger PE+ smaller KE ). At lower points in the track, the car has less potential energy and so moves more quickly (smaller PE + larger KE). If friction is taken into account, mechanical energy is no longer
conserved, as shown in Figure 2.1. A steady decrease in the car’s total mechanical energy occurs because of work being done against the friction between the car’s axles and its bearings and between the car’s wheels and the coaster track. Mechanical energy is tran sferred to the atoms and molecules throughout the entire roller coaster (both the car and the track). Thus, the roller coaster’s internal energy increases by an amount equal to the decrease in the mechanical energy. Most of this energy is then gradually dissipated to the air surrounding the roller coaster as heat. If the internal energy for the roller coaster (the system) and the energy dissipated to the surrounding air (the environment) are taken into account, then the total energy will be constant.
tl@i j;j flt Conservation of Total Energy W=mgh+work
required to overcome friction
b
In t he presence of friction, the internal energy (U) of the roller coaster increases as KE + PE decreases.

a

KE PE U
338
Chapter 10
KE PE U
KE PE U
KE PE U
KE PE
U
The principle of energy conservation that takes into account a system’s internal energy as well as work and heat is called the first law of thermodynamics. Imagine that the isothermally expanding toy balloon in the previous section is squeezed rapidly. The process is no longer isothermal. Instead, it is a combination of two processes. On the one hand, work (W) is done on the system. The balloon and the air inside it (the system) are compressed, so the air’s internal energy and temperature increase. Work is being done on the system, so Wis a negative quantity. The rapid squeezing of the balloon can be treated as an adiabatic process, so Q = 0 and, therefore, flU = -W
‘ Did YOU Know? – – – – – – – – – – , ‘ Not all ways of transferring energy can , be classified simply by work or by heat. Other processes that can change the , internal energy of a substance include ‘ changes in the chemical and magnetic : properties of the substance.
After the compression step, energy is transferred from the system as heat ( Q). Some of the internal energy of the air inside the balloon is transferred to the air outside the balloon. During this step, the internal energy of the gas decreases, so fl Uhas a negative value. Similarly, because energy is removed from the system, Q has a negative value. The change in internal energy for this step can be expressed as -fl U = -Q, or fl U = Q. The signs for heat and work for a system are summarized in Figure 2.2. To remember whether a system’s internal energy increases or decreases, you may find it helpful to visualize the system as a circle, as shown in Figure 2.3. When work is done on the system or energy is transferred as heat into the system, an arrow points into the circle. This shows that internal energy increases. When work is done by the system or energy is transferred as heat out of the system, the arrow points out of the circle. This shows that internal energy decreases.
0>0
energy added to system as heat
00
work done by system (expansion of gas)
W< 0
work done on system (compression of gas)
W= 0
no work done
Representing a System If visualizing a system as a circle, arrows represent work and heat.
Q>O
System
W 0) increases the system’s and W= 0; internal energy. therefore, D-U = Q Energy removed from the system as heat (Q < 0) decreases the system’s internal energy.
Isothermal
Adiabatic
no change in temperature or internal energy
no energy transferred as heat
D. T = 0, so D.U = 0; therefore, D-U= Q- W= 0, or 0 = W
Q = 0, so D.U = – W
Energy added to the system as heat is removed from the system as work done by the system. Energy added to the system by work done on it is removed from the system as heat. Work done on the system (W < 0) increases the system’s internal energy. Work done by the system (W > 0) decreases the system’s internal energy.
Isolated system
340
Chapter 10
no energy transferred as heat and no work done on or by the system
Q = 0 and W = 0, so
D-U = 0 and U; = u,
There is no change in the system’s internal energy.
PREMIUM CONTENT
~ Interactive Demo \.:::,/ HMDScience.com
The First Law of Thermodynamics Sample Problem B A total of 135 J of work is done on a gaseous refrigerant as it undergoes compression. If the internal energy of the gas increases by 114 J during the process, what is the total amount of energy transferred as heat? Has energy been added to or removed from the refrigerant as heat?
0
ANALYZE
Given:
W= -135 J
llU=ll4J Unknown:
Tips and Tricks Work is done on the gas, so work (W) has a negative value. The internal energy increases during the process, so the change in internal energy (.6.U) has a positive value.
Q=?
Diagram:
11U = 114 J
E)
PLAN
Choose an equation or situation: Apply the first law of thermodynamics using the values for ~ U and W in order to fin d the value for Q.
llU= Q -W Rearrange the equation to isolate the unknown:
E)
SOLVE
Substitute the values into the equation and solve:
Q = 114 J + (-135 J) = -21 J
IQ = – 21J I
0
CHECKYOUR WORK
Tips and Tricks The sign for the value of Q is negative. From Figure 2.2, Q < O indicates that energy is transferred as heat from the refrigerant.
Although the internal en ergy of the refrigerant increases under compression, more energy is added as work than can be accounted for by the increase in the internal energy. This energy is removed from the gas as heat, as indicated by the minus sign preceding the value for Q.
G·Mii,\it4- ► Thermodynamics
341
The First Law of Thermodynamics I
(continued)
Practice 1. Heat is added to a system, and the system does 26 J of work. If the internal energy
increases by 7 J, how much heat was added to the system? 2. The internal energy of the gas in a gasoline engine’s cylinder decreases by 195 J. If 52.0 J of work is done by the gas, how much energy is transferred as heat? Is this energy added to or removed from the gas? 3. A 2.0 kg quantity of water is held at constant volume in a pressure cooker and heated by a range element. The system’s internal energy increases by 8.0 x 103 J. However, the pressure cooker is not well insulated, and as a result, 2.0 x 103 J of energy is transferred to the surrounding air. How much energy is transferred from the range element to the pressure cooker as heat? 4. The internal energy of a gas decreases by 344 J. If the process is adiabatic, how much energy is transferred as heat? How much work is done on or by the gas? 5. A steam engine’s boiler completely converts 155 kg of water to steam. This process involves the transfer of 3.50 x 108 J as heat. If steam escaping through a safety
valve does 1.76 x 108 J of work expanding against the outside atmosphere, what is the net change in the internal energy of the water-steam system?
Cyclic Processes A refrigerator performs mechanical work to create temperature
cyclic process a thermodynamic process in which a system returns to the same conditions under which it started
differences between its closed interior and its environment (the air in the room). This process leads to the transfer of energy as heat. A heat engine does the opposite: it uses heat to do mechanical work. Both of these processes have something in common: they are examples of cyclic processes. In a cyclic process, the system’s properties at the end of the process are identical to the system’s properties before the process took place. The final and initial values of internal energy are the same, and the change in internal energy is zero . .6.Unet
= O and Qnet = W net
A cyclic process resembles an isothermal process in that all energy is transferred as work and heat. But now the process is repeated with no net change in the system’s internal energy.
Heat engines use heat to do work. A heat engine is a device that uses heat to do mechanical work. A heat engine is similar to a water wheel, which uses a difference in potential energy to do work. A water wheel uses the energy of water falling from one level above Earth’s surface to another. The change in potential energy increases the water’s kinetic energy so that the water can do work on one side of the wheel and thus turn it.
342
Chapter 10
Instead of using the difference in potential energy to do work, heat engines do work by transferring energy from a high-temperature substance to a lower-temperature substance, as indicated for the steam engine shown in Figure 2.5. For each complete cycle of the heat engine, the net work done will equal the difference between the energy transferred as heat from a high-temperature substance to the engine ( Qh) and the energy transferred as heat from the engine to a lower-temperature substance ( QJ Wnet= Qh- Qc
The larger the difference between the energy transferred as heat into the engine and out of the engine, the more work it can do in each cycle. The internal-combustion engine found in most vehicles is an example of a heat engine. Internal-combustion engines burn fuel within a closed chamber (the cylinder). The potential energy of the chemical bonds in the reactant gases is converted to kinetic energy of the particle products of the reaction. These gaseous products push against a piston and thus do work on the environment. In this case, a crankshaft transforms the linear motion of the piston to the rotational motion of the axle and wheels. Although the basic operation of any internal-combustion engine resembles that of an ideal cyclic heat engine, certain steps do not fit the idealized model. When gas is taken in or removed from the cylinder, matter enters or leaves the system so that the matter in the system is not isolated. No heat engine operates perfectly. Only part of the available internal energy leaves the engine as work done on the environment; most of the energy is removed as heat.
Heat Engine A heat engine is able to do work (b) by transferring energy from a high-temperature substance (the boiler) at Th (a) to a substance at a lower temperature (the air surrounding the engine) at Tc (c).
Heat engine
(b)
Thermodynamics
343
ST.EM
Gasoline Engines gasoline engine is one type of internal-combustion engine. The diagram below illustrates the steps in one cycle of operation for a gasoline engine. During compression, shown in (a), work is done by the piston as it adiabatically compresses the fuel-and-air mixture in the cylinder. Once maximum compression of the gas is reached, combustion takes place. The chemical potential energy released during combustion increases the internal energy of the gas, as shown in (b). The hot, high-pressure gases from the combustion reaction expand in volume, pushing the
piston and turning the crankshaft, as shown in (c). Once all of the work is done by the piston, some energy is transferred as heat through the walls of the cylinder. Even more energy is transferred by the physical removal of the hot exhaust gases from the cylinder, as shown in (d). A new fuel-air mixture is then drawn through the intake valve into the cylinder by the downward-moving piston, as shown in (e).
Spark plug Exhaust valve
Fuel-
Expanding combustionproduct gases
Cylinder Piston
{b) Ignition
Connecting rod
{a) Compression
Intake valve open
{c) Expansion
Intake valve
Exhaust valve open
Fuelproduct gases
{e) Fuel intake
344
Chapter 10
{d) Exhaust

SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Use the first law of thermodynamics to show that the internal energy of an isolated system is always conserved.
2. In the systems listed below, identify where energy is transferred as heat and work and where changes in internal energy occur. Is energy conserved in each case? a. the steam in a steam engine consisting of a boiler, a firebox, a cylinder, a piston, and a flywheel b. the drill bit of a power drill and a metal block into which a hole is being drilled 3. Express the first law of thermodynamics for the following processes: a. isothermal b. adiabatic c. isovolumetric 4. A compressor for a jackhammer expands the air in the hammer’s cylinder at a constant pressure of 8.6 x 105 Pa. The increase in the cylinder’s volume is 4.05 x 10-4 m 3 . During the process, 9.5 J of energy is transferred out of the cylinder as heat. a. What is the work done by the air? b. What is the change in the air’s internal energy? c. What type of ideal thermodynamic process does this approximate? 5. A mixture of fuel and air is enclosed in an engine cylinder fitted with a piston. The gas pressure is maintained at 7.07 x 105 Pa as the piston moves slowly inward. If the gas volume decreases by 1.1 x 10-4 m 3 and the internal energy of the gas increases by 62 J, how much energy is added to or removed from the system as heat? 6. Over several cycles, a refrigerator does 1.51 x 104 J of work on the refrigerant. The refrigerant in turn removes 7.55 x 104 J as heat from the air inside the refrigerator. a. How much energy is transferred as heat to the outside air? b. What is the net change in the internal energy of the refrigerant? c. What is the amount of work done on the air inside the refrigerator? d. What is the net change in the internal energy of the air inside the refrigerator? 7. If a weather balloon in flight gives up 15 J of energy as heat and the gas within it does 13 J of work on the outside air, by how much does its internal energy change?
Critical Thinking 8. After reading the feature on the n ext page, explain why opening the refrigerator door on a hot day does not cause your kitchen to become cooler.
Thermodynamics
345
S.T.E.M. Refrigerators s shown in the photograph below, a refrigerator can be represented schematically as a system that transfers energy from a body at a low temperature (c) to one at a high temperature (a). The refrigerator uses work performed by an electric motor to compress the refrigerant, which is a substance that evaporates at a very low temperature. In the past, ammonia was used as a refrigerant in home refrigerators. However, ammonia leaks pose a risk because pure ammonia is highly toxic to people. In the 1930s, home refrigerators began using a newly developed, nontoxic class of refrigerants called CFCs (chlorofluorocarbons). Today, it is known that CFCs damage the ozone layer. Since the 1990s, home refrigerators have used refrigerants that are less harmful to the ozone layer. The process by which a refrigerator operates consists of four basic steps, as illustrated in the diagram on the next page. The system to and from which energy is transferred is defined here as the refrigerant contained within the inner surface of the tubing. Initially, the liquid refrigerant is at a low temperature and pressure so that it is colder than the air inside the refrigerator. The refrigerant absorbs energy from
(c)
inside the refrigerator and lowers the refrigerator’s interior temperature. This transfer of energy as heat increases the temperature of the liquid refrigerant until it begins to boil, as shown in (a). The refrigerant continues to absorb energy until it has completely vaporized. Once it is in the vapor phase, the refrigerant is passed through a compressor. The compressor does work on the gas by decreasing its volume without transferring energy as heat, as shown in (b). This adiabatic process increases the pressure and internal energy (and thus the temperature) of the gaseous refrigerant.
(c)
Refrigerator (b)
(a) (a)
A refrigerator does work (b) in order to transfer energy as heat from the inside of the refrigerator (c) to the air outside the refrigerator (a).
346
Next, the refrigerant is moved to the outer parts of the refrigerator, where thermal contact is made with the air. The refrigerant loses energy to the air, which is at a lower temperature, as in (c). The gaseous refrigerant at high pressure condenses at a constant temperature to a liquid. The liquefied refrigerant is then brought back into the refrigerator. Just outside the low-temperature interior of the refrigerator, the refrigerant goes through an expansion valve and expands without absorbing energy as heat. The liquid then does work as it moves from a high-pressure region to a low-pressure region, and its volume increases, as shown in (d). In doing so, the gas expands and cools.
THERMODYNAMICS OF A REFRIGERANT The refrigerant now has the same internal energy and phase as it did at the start of the process. If the temperature of the refrigerant is still lower than the temperature of the air inside the refrigerator, the cycle will repeat. Because the final internal energy is equal to the initial internal energy, this process is cyclic. The first law of thermodynamics can be used to describe the signs of each thermodynamic quantity in the four steps listed, as shown in the table.
Step
Q
w
au
A
+
0
+
B
0 0
C
D
+
0
+
\ Outside of refrigerator
W(-)
Outside refrigerator
Inside refrigerator
Key higher-pressure gas ~
~
higher-pressure liquid lower- pressure liquid
~
lower-pressure ~ gas
Expansion valve
Evaporator
In each of the four steps of a refrigeration cycle, energy is transferred to or from the refrigerant either by heat or by work.
347
SECTION 3 Objectives ►
Recognize why the second law of thermodynamics requires two bodies at different temperatures for work to be done.

I
Calculate the efficiency of a heat engine.

Relate the disorder of a system to its ability to do work or transfer energy as heat.
The Second Law of Thermodynamics Key Term entropy
Efficiency of Heat Engines In the previous section, you learned how a heat engine absorbs a quantity of energy from a high-temperature body as heat, does work on the environment, and then gives up energy to a low-temperature body as heat. The work derived from each cycle of a heat engine equals the difference between the heat input and heat output during the cycle, as follows : Wnet = Qnet = Qh – Qc
This equation, obtained from the first law of thermodynamics, indicates that all energy entering and leaving the system is accounted for and is thus conserved. The equation also suggests that more work is gained by taking more energy at a higher temperature and giving up less energy at a lower temperature. If no energy is given up at the lower temperature ( Qc = 0), then it seems that work could be obtained from energy transferred as heat from any body, such as the air around the engine. Such an engine would be able to do more work on hot days than on cold days, but it would always do work as long as the engine’s temperature was less than the temperature of the surrounding air.
A heat engine cannot transfer all energy as heat to do work. Unfortunately, it is impossible to make such an engine. As we have seen, a heat engine carries some substance through a cyclic process during which (1) the substance absorbs energy as heat from a high-temperature reservoir, (2) work is done by the engine, and (3) energy is expelled as heat to a lower-temperature reservoir. In practice, all heat engines operating in a cycle must expel some energy to a lower-temperature reservoir. In other words, it is impossible to construct a heat engine that, operating in a cycle, absorbs energy from a hot reservoir and does an equivalent amount of work. The requirement that a heat engine give up some energy at a lower temperature in order to do work does not follow from the first law of thermodynamics. This requirement is the basis of what is called the second law of thermodynamics. The second law of thermodynamics can be stated as follows: No cyclic process that converts heat entirely into work is possible. According to the second law of thermodynamics, W can never be equal to Qh in a cyclic process. In other words, some en ergy must always be transferred as heat to the system’s surroundings (Qc > 0). 348
Chapter 10
Efficiency measures how well an engine operates. A cyclic process cannot completely convert energy transferred as heat into work, nor can it transfer energy as heat from a low-temperature body to a high-temperature body without work being done in the process. However, we can measure how closely a cyclic process approaches these ideal situations. A measure of how well an engine operates is given by the engine’s efficiency (ejf). In general, efficiency is a measure of the useful energy taken out of a process relative to the total energy that is put into the process. Efficiencies for different types of engines are listed in Figure 3.1 . Recall from the first law of thermodynamics that the work done on the environment by the engine is equal to the difference between the energy transferred to and from the system as heat. For a heat engine, the efficiency is the ratio of work done by the engine to the energy added to the system as heat during one cycle. Equation for the Efficiency of a Heat Engine
eff =
Wnet =
Qh
Qh – QC = l _ QC Qh Qh
FIGURE 3.1
TYPICAL EFFICIENCIES FOR ENGINES Engine type
eff (calculated maximum values)
steam engine
0.29
steam turbine
0.40
gasoline engine
0.60
diesel engine
0.56
Engine type
eff (measured values)
steam engine
0.17
steam turbine
0.30
gasoline engine
0.25
diesel engine
0.35
. net work done by engine efficiency=————energy added to engine as heat energy added as heat – energy removed as heat energy added as heat
=
energy removed as heat I – ———energy added as heat
Notice that efficiency is a unitless quantity that can be calculated using only the magnitudes for the energies added to and taken away from the engine. This equation confirms that a heat engine has 100 percent efficiency ( eff = 1) only if there is no energy transferred away from the engine as heat ( Qc = 0). Unfortunately, there can be no such heat engine, so the efficiencies of all engines are less than 1.0. The smaller the fraction of usable energy that an engine can provide, the lower its efficiency is.
1. Cooling Engines Use the second
law of thermodynamics to explain why an automobile engine requires a cooling system to operate. 2. Power Plants Why are many
coal-burning and nuclear power plants located near rivers?
Thermodynamics
349
The equation also provides some important information for increasing engine efficiency. If the amount of energy added to the system as heat is increased or the amount of energy given up by the system is reduced, the ratio of Q/ Qh becomes much smaller and the engine’s efficiency comes closer to 1.0. The efficiency equation gives only a maximum value for an engine’s efficiency. Friction, thermal conduction, and the inertia of moving parts in the engine hinder the engine’s performance, and experimentally measured efficiencies are significantly lower than the calculated efficiencies. Several examples of these differences can be found in Figure 3.1.
Heat-Engine Efficiency Sample Problem C Find the efficiency of a gasoline engine that, during one cycle, receives 204 Jof energy from combustion and loses 153 Jas heat to the exhaust.
0
ANALYZE
Given: Unknown:
Diagram:
@
PLAN
Qc = 153 J
Choose an equation or situation:
The efficiency of a heat engine is the ratio of the work done by the engine to the energy transferred to it as heat.
eff =
wnet
=1-
Qh
E)
SOLVE
Substitute the values into the equation and solve:
ff= e 1
0
CHECK YOUR WORK
QC Qh
153 1J = 0.250 204 J
eff = 0.250
1
Only 25 percent of the energy added as heat is used by the engine to do work. As expected, the efficiency is less than 1.0.
,a.,,iii ,\114- ► 350
Chapter 10
Heat-Engine Efficiency (continued) Practice 1. If a steam engine takes in 2.254 x 104 kJ from the boiler and gives up 1.915 x 104 kJ in exhaust during one cycle, what is the engine’s efficiency?
2. A test model for an experimental gasoline engine does 45 J of work in one cycle and gives up 31 J as heat. What is the engine’s efficiency? 3. A steam engine absorbs 1.98 x 105 J and expels 1.49 x 105 Jin each cycle.
Assume that all of the remaining energy is used to do work.
a. What is the engine’s efficiency? b. How much work is done in each cycle?
4. If a gasoline engine has an efficiency of 21 percent and loses 780 J to the cooling system and exhaust during each cycle, how much work is done by the engine? 5. A certain diesel engine performs 372 J of work in each cycle with an efficiency of 33.0 percent. How much energy is transferred from the engine to the exhaust and cooling system as heat? 2
6. If the energy removed from an engine as heat during one cycle is 6.0 x 10 J, how much energy must be added to the engine during one cycle in order for it to operate at 31 percent efficiency?
Entropy When you shuffle a deck of cards, it is highly improbable that the cards will end up separated by suit and in numerical sequence. Such a highly ordered arrangement can be formed in only a few ways, but there are more than 8 x 1067 ways to arrange 52 cards. In thermodynamics, a system left to itself tends to go from a state with a very ordered set of energies to one in which there is less order. In other words, the system tends to go from one that has only a small probability of being randomly formed to one that has a high probability of being randomly formed. The measure of a system’s disorder is called the entropy of the system. The greater the entropy of a system is, the greater the system’s disorder.
entropy a measure of the randomness or disorder of a system
The greater probability of a disordered arrangement indicates that an ordered system is likely to become disordered. Put another way, the entropy of a system tends to increase. This greater probability also reduces the chance that a disordered system will become ordered at random. Thus, once a system has reached a state of the greatest disorder, it will tend to remain in that state and have maximum entropy.
Did YOU Know?. – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – – . Entropy decreases in many systems on Earth. For example, atoms and molecules become incorporated into complex and orderly biological structures such as cells and tissues. These appear to be spontaneous because we think of the Earth as a closed system. So much energy comes from the sun that the disorder in chemical and biological systems is reduced, while the total entropy of the Earth, sun, and intervening space increases.
Thermodynamics
351
Greater disorder means there is less energy to do work. Low and High Entropy Systems If all gas particles moved toward the piston, all of the internal energy could be used to do work. This extremely well ordered situation is highly improbable.
a)
Well ordered; high efficiency and highly improbable distribution of velocities
b)
Highly disordered; average efficiency and highly probable distribution of velocities
Heat engines are limited because only some of the energy added as heat can be used to do work. Not all of the gas particles move in an orderly fashion toward the piston and give up all of their energy in collision with the piston, as shown in Figure 3.2(a). Instead, they move in all available directions, as shown in Figure 3.2(b). They transfer energy through collisions with the walls of the engine cylinder as well as with each other. Although energy is conserved, not all of it is available to do useful work. The motion of the particles of a system is not well ordered and therefore is less useful for doing work. Because of the connection between a system’s entropy, its ability to do work, and the direction of energy transfer, the second law of thermodynamics can also be expressed in terms of entropy change. This law applies to the entire universe, not only to a system that interacts with its environment. So, the second law can be stated as follows: The entropy of the universe increases in all natural processes. Note that entropy can decrease for parts of systems, such as the water in the freezer shown in Figure 3.3, provided this decrease is offset by a greater increase in entropy elsewhere in the universe. The water’s entropy decreases as it becomes ice, but the entropy of the air in the room is increased by a greater amount as energy is transferred by heat from the refrigerator. The result is that the total entropy of the refrigerator and the room together has increased.
Entropy in a Refrigerator Because of the refrigerator’s imperfect efficiency, the entropy of the outside air molecules increases more than the entropy of the freezing water decreases.
Ice tray
Water before freezing
Air before water freezes
Ice after freezing
Air after ice is frozen
Small decrease in entropy
352
Chapter 10
Large increase in entropy

QuickLAB Take two dice from a board game. Record all the possible ways to obtain the numbers 2 through 12 on the sheet of paper. How many possible dice combinations can be rolled? How many combinations of both dice will produce the number 5? the number 8? the number 11? Which number(s) from 2 through 12 is most probable? How many ways out of the total number of
ways can this number(s) be rolled? Which number(s) from 2 through 12 is least probable? How many ways out of the total number of ways can this number(s) be rolled?
MATERIALS • 3 dice • a sheet of paper • a pencil
Repeat the experiment with three dice. Write down all of the possible combinations that will produce the numbers 3 through 18. What number is most probable?
SECTION 3 FORMATIVE ASSESSMENT 1. Is it possible to construct a heat engine that doesn’t transfer energy to its surroundings? Explain. 2. An engineer claims to have built an engine that takes in 7.5 x 104 J and
expels 3.5 x 104 J. a. How much energy can the engine provide by doing work? b. What is the efficiency of the engine? c. Is this efficiency possible? Explain your answer. 3. Of the items listed below, which ones have high entropy? a. papers scattered randomly across a desk b. papers organized in a report c. a freshly opened pack of cards d. a mixed deck of cards e. a room after a party f. a room before a party 4. Some compounds have been observed to form spontaneously, even though they are more ordered than their components. Explain how this is consistent with the second law of thermodynamics. 5. Discuss three common examples of natural processes that involve decreases in entropy. Identify the corresponding entropy increases in the environments of these processes.
Critical Thinking 6. A steam-driven turbine is one major component of an electric power
plant. Why is it advantageous to increase the steam’s temperature as much as possible? 7. Show that three purple marbles and three light blue marbles in two groups of three marbles each can be arranged in four combinations: two with only one possible arrangement each and two with nine possible arrangements each. Thermodynamics
353
S.T.E.M. Deep-Sea Air Conditioning eep beneath the ocean, about half a mile down, sunlight barely penetrates the still waters. Scientists at Makai Ocean Engineering in Hawaii are now tapping into that pitch-dark region as a resource for air conditioning. In tropical locations where buildings are cooled yearround, air-conditioning systems operate with cold water. Refrigeration systems cool the water, and pumps circulate it throughout the walls of a building, where the water absorbs heat from the rooms. Unfortunately, powering these compressors is neither cheap nor efficient. Instead of cooling the water in their operating systems, the systems designed by Makai use frigid water from the ocean’s depths. First, engineers install a pipeline that reaches deep into the ocean, where the water is nearly freezing. Then, powerful pumps on the shoreline move the water directly into a building’s air-conditioning system. There, a system of heat exchangers uses the sea water to cool the fresh water in the air-conditioning system.
-+–
Circulation pump for ‘-… fresh water ‘-…
Building
Warm fresh
l
water
-+-Discharge of sea water / into ground
Warm sea water
Intake of sea water from ocean-
Cool fresh water
Heat exchanger
l
Cool sea water
\
Sea-water pump
One complicating factor is that the water must also be returned to the ocean in a manner that will not disrupt the local ecosystem. It must be either piped to a depth of a few hundred feet, where its temperature is close to that of the ocean at that level, or poured into onshore pits, where it eventually seeps through the land and comes to an acceptable temperature by the time it reaches the ocean. “This deep-sea air conditioning benefits the environment by operating with a renewable resource instead of freon,” said Dr. Van Ryzin, the president of Makai. “Because the system eliminates the need for compressors, it uses only about 1o percent of the electricity of current methods, saving fossil fuels and a lot of money.” However, deep-sea airconditioning technology works only for buildings within a few kilometers of the shore and carries a hefty installation cost of several million dollars. For this reason, Dr. Van Ryzin thinks this type of system is most appropriate for large central air-conditioning systems, such as those necessary to cool resorts or large manufacturing plants, where the electricity savings can eventually make up for the installation costs. Under the right circumstances, air conditioning with sea water can be provided at one-third to one-half the cost of conventional air conditioning.
(/)
1i5 a::
0
~ :c
i
:. a
1820 Hans Christian Oersted demonstrates that an electric current produces a magnetic field. (Gian Dominico Romagnosi, an amateur scientist, discovered the effect 18 years earlier, but at the time attracted no attention.) Andre-Marie Ampere repeats Oersted’s experiment and formulates the law of electromagnetism that today bears his name. Fmagnetic = BU
1826 Katsushika Hokusai begins his series of prints Thirty-Six Views of Mount Fuji.
401
‘ I
SECTION 1 Objectives

I
Explain how sound waves are produced.
Key Terms
►I
Relate frequency to pitch.

Compare the speed of sound in various media.
I ► I ►
Sound Waves
Relate plane waves to spherical waves. Recognize the Doppler effect, and determine the direction of a frequency shift when there is relative motion between a source and an observer.
compression the region of a longitudinal wave in which the density and pressure are at a maximum rarefaction the region of a longitudinal wave in which the density and pressure are at a minimum
Compressions and Rarefactions (a) The sound from a tuning fork is produced by (b) the vibrations of each of its prongs. (c) When a prong swings to the
right, there is a region of high density and pressure. (d) When the prong swings back to the left, a region of lower density and pressure exists.
404
Chapter 12
compression rarefaction
pitch Doppler effect
The Production of Sound Waves Whether a sound wave conveys the shrill whine of a jet engine or the melodic whistling of a bird, it begins with a vibrating object. We will explore how sound waves are produced by considering a vibrating tuning fork, as shown in Figure 1.1(a). The vibrating prong of a tuning fork, shown in Figure 1.1 (b), sets the air molecules near it in motion. As the prong swings to the right, as in Figure 1.1 (c), the air molecules in front of the movement are forced closer together. (This situation is exaggerated in the figure for clarity.) Such a region of high molecular density and high air pressure is called a compression. As the prong moves to the left, as in Figure 1.1 (d), the molecules to the right spread apart, and the density and air pressure in this region become lower than normal. This region of lower density and pressure is called a rarefaction. As the tuning fork continues to vibrate, a series of compressions and rarefactions forms and spreads away from each prong. These compressions and rarefactions spread out in all directions, like ripple waves on a pond. When the tuning fork vibrates with simple harmonic motion, the air molecules also vibrate back and forth with simple harmonic motion.
Representing Sound Waves
(a) As this tuning fork vibrates, (b) a series of compressions and rarefactions moves away from each prong. (c) The crests of this sine wave correspond to compressions, and the troughs correspond to rarefactions.
(a}
(b}
(c}
Sound waves are longitudinal. In sound waves, the vibrations of air molecules are parallel to the direction of wave motion. Thus, sound waves are longitudinal. The simplest longitudinal wave produced by a vibrating object can be represented by a sine curve. In Figure 1.2, the crests correspond to compressions (regions of higher pressure), and the troughs correspond to rarefactions (regions of lower pressure). Thus, the sine curve represents the changes in air pressure due to the propagation of the sound waves. Note that Figure 1.2 shows an idealized case. This example disregards energy losses that would decrease the wave amplitude.
Characteristics of Sound Waves As discussed earlier, frequency is defined as the number of cycles per unit of time. Sound waves that the average human ear can hear, called audible sound waves, have frequencies between 20 and 20 000 Hz. (An individual’s hearing depends on a variety of factors, including age and experiences with loud noises.) Sound waves with frequencies less than 20 Hz are called infrasonic waves, and those above 20 000 Hz are called ultrasonic waves. It may seem confusing to use the term sound waves for infrasonic or ultrasonic waves because humans cannot hear these sounds, but these waves consist of the same types of vibrations as the sounds that we can hear. The range of audible sound waves depends on the ability of the average human ear to detect their vibrations. Dogs can hear ultrasonic waves that humans cannot.
‘.Did YOU Know?_- – – – – – – – – – – , ‘ Elephants use infrasonic sound waves : to communicate with one another. Their large ears enable them to detect , these low-frequency sound waves, : which have relatively long wavelengths. ‘ Elephants can effectively communicate , in this way, even when they are , separated by many kilometers.
Sound
, : : ‘
405
Frequency determines pitch. pitch a measure of how high or low a sound is perceived to be, depending on the frequency of the sound wave
The frequency of an audible sound wave determines how high or low we perceive the sound to be, which is known as pitch. As the frequency of a sound wave increases, the pitch rises. The frequency of a wave is an objective quantity that can be measured, while pitch refers to how different frequencies are perceived by the human ear. Pitch depends not only on frequency but also on other factors, such as background noise and loudness.
Speed of sound depends on the medium. Sound waves can travel through solids, liquids, and gases. Because waves consist of particle vibrations, the speed of a wave depends on how quickly one particle can transfer its motion to another particle. For example, solid particles respond more rapidly to a disturbance than gas particles do because the molecules of a solid are closer together than those of a gas are. As a result, sound waves generally travel faster through solids than through gases. Figure 1.3 shows the speed of sound waves in various media.
ST.E.M.
Ultrasound Images ltrasonic waves can be used to produce images of objects inside the body. Such imaging is possible because sound waves are partially reflected when they reach a boundary between two materials of different densities. The images produced by ultrasonic waves are clearer and more detailed than those that can be produced by lower-frequency sound waves because the short wavelengths of ultrasonic waves are easily reflected off small objects. Audible and infrasonic sound waves are not as effective because their longer wavelengths pass around small objects. In order for ultrasonic waves to “see” an object inside the body, the wavelength of the waves used must be about the same size as or smaller than the object. A typical frequency used in an ultrasonic device is about 10 MHz. The speed of an ultrasonic wave in human tissue is about 1500 mis, so the wavelength of 10 MHz waves is A = v/f = 0.15 mm. A 10 MHz ultrasonic device will not detect objects smaller than this size.
406
Chapter 12
Physicians commonly use ultrasonic waves to observe fetuses. In this process, a crystal emits ultrasonic pulses. The same crystal acts as a receiver and detects the reflected sound waves. These reflected sound waves are converted to an electrical signal, which forms an image on a fluorescent screen. By repeating this process for different portions of the mother’s abdomen, a physician can obtain a complete picture of the fetus, as shown above. These images allow doctors to detect some types of fetal abnormalities.
The speed of sound also depends on the temperature of the medium. As temperature rises, the particles of a gas collide more frequently. Thus, in a gas, the disturbance can spread faster at higher temperatures than at lower temperatures. In liquids and solids, the particles are close enough together that the difference due to temperature changes is less noticeable.
FIGURE 1.3
SPEED OF SOUND IN VARIOUS MEDIA Medium
v (m/s)
Gases air (0°C)
331
Sound waves propagate in three dimensions.
air (25°C)
346
Sound waves actually travel away from a vibrating source in all three dimensions. When a musician plays a saxophone in the middle of a room, the resulting sound can be heard throughout the room because the sound waves spread out in all directions. The wave fronts of sound waves spreading in three dimensions are approximately spherical. To simplify, we shall assume that the wave fronts are exactly spherical unless stated otherwise.
air (100°C)
366
helium (0°C)
972
Spherical waves can be represented graphically in two dimensions with a series of circles surrounding the source, as shown in Figure 1.4. The circles represent the centers of compressions, called wave fronts. Because we are considering a three-dimensional phenomenon in two dimensions, each circle represents a spherical area. Because each wave front locates the center of a compression, the distance between adjacent wave fronts is equal to one wavelength, A. The radial lines perpendicular to the wave fronts are called rays.
hydrogen (0°C)
1290
oxygen (0°C)
317
Liquids at 25°C methyl alcohol
1140
sea water
1530
water
1490
Solids aluminum
5100
copper
3560
iron
5130
lead
1320
vulcanized rubber
54
Spherical Waves In this representation of a spherical wave, the wave fronts represent compressions, and the rays show the direction of wave motion. Each wave front corresponds to a crest of the sine curve. In turn, the sine curve corresponds to a single ray.
Music from a Trumpet Suppose you hear
Lightning and Thunder Light waves travel
music being played from a trumpet that is across the room from you. Compressions and rarefactions from the sound wave reach your ear, and you interpret these vibrations as sound. Were the air particles that are vibrat-
nearly 1 million times faster than sound waves in air. With this in mind, explain how the distance to a lightning bolt can be determined by counting the seconds between the flash and the sound of the thunder.
ing near your ear carried across the room by the sound wave? How do you know? -c-
a
Sound
407
Spherical Waves Spherical wave fronts that are a great distance from the source can be approximated with parallel planes known as plane waves.
Rays

Wave fronts/
Rays indicate the direction of the wave motion. The sine curve used in our previous representation of sound waves, also shown in Figure 1.4, corresponds to a single ray. Because crests of the sine curve represent compressions, each wave front crossed by this ray corresponds to a crest of the sine curve. Consider a small portion of a spherical wave front that is many wavelengths away from the source, as shown in Figure 1.5. In this case, the rays are nearly parallel lines, and the wave fronts are nearly parallel planes. Thus, at distances from the source that are great relative to the wavelength, we can approximate spherical wave fronts with parallel planes. Such waves are called plane waves. Any small portion of a spherical wave that is far from the source can be considered a plane wave. Plane waves can be treated as one-dimensional waves all traveling in the same direction, as in the chapter “Vibrations and Waves:’
The Doppler Effect If you stand on the street while an ambulance speeds by with its siren on,
you will notice the pitch of the siren change. The pitch will be higher as the ambulance approaches and will be lower as it moves away. As you read earlier in this section, the pitch of a sound depends on its frequency. But in this case, the siren is not changing its frequency. How can we account for this change in pitch?
Relative motion creates a change in frequency. If a siren sounds in a parked ambulance, an observer standing on the
street hears the same frequency that the driver hears, as you would expect. When an ambulance is moving, as shown in Figure 1.6, there is relative motion between the moving ambulance and a stationary observer. This relative motion affects the way the wave fronts of the sound waves produced by the siren are perceived by an observer. (For simplicAs this ambulance moves to the left, Observer A hears the siren ity’s sake, assume that the sound waves at a higher frequency than the driver does, while Observer B hears a lower frequency. produced by the siren are spherical.) Although the frequency of the siren remains constant, the wave fronts reach an observer in front of the ambulance (Observer A) more often than they would if the ambulance were stationary. The reason is that the source of the sound waves is moving toward the observer. The speed of sound in the air doesnotchange,becausethespeed depends only on the temperature of the air. Thus, the product of wavelength and frequency remains constant. Because the wavelength is less, the frequency heard by Observer A is greater than the source frequen cy. 408
Chapter 12

For the same reason, the wave fronts reach an observer behind the ambulance (Observer B) less often than they would if the ambulance were stationary. As a result, the frequency heard by Observer B is less than the source frequency. This frequency shift is known as the Doppler effect. The Doppler effect is named for the Austrian physicist Christian Doppler (1803-1853), who first described it.
Doppler effect an observed change in frequency when there is relative motion between the source of waves and an observer
We have considered a moving source with respect to a stationary observer, but the Doppler effect also occurs when the observer is moving with respect to a stationary source or when both are moving at different velocities. In other words, the Doppler effect occurs whenever there is relative motion between the source of waves and an observer. (If the observer is moving instead of the source, the wavelength in air does not change, but the frequency at which waves arrive at the ear is altered by the motion of the ear relative to the medium.) Although the Doppler effect is most commonly experienced with sound waves, it is a phenomenon common to all waves, including electromagnetic waves, such as visible light.
SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What is the relationship between frequency and pitch?
2. Dolphin echolocation is similar to ultrasound. Reflected sound waves allow a dolphin to form an image of the object that reflected the waves. Dolphins can produce sound waves with frequencies ranging from 0.25 kHz to 220 kHz, but only those at the upper end of this spectrum are used in echolocation. Explain why high-frequency waves work better than low-frequency waves. 3. Sound pulses emitted by a dolphin travel through 20°C ocean water at a rate of 1450 m/s. In 20°c air, these pulses would travel 342.9 m/s. How can you account for this difference in speed?
Interpreting Graphics 4. Could a portion of the innermost wave front shown in Figure 1.7 be approximated by a plane wave? Why or why not? Figure 1.7
5.
Figure 1.8 is
a diagram of the Doppler effect in a ripple tank. In which direction is the source of these ripple waves moving?
6. If the source of the waves in Figure 1.8 is stationary, which way must the ripple tank be moving?
Critical Thinking 7. As a dolphin swims toward a fish, the dolphin sends out sound waves to determine the direction the fish is moving. If the frequency of the reflected waves is higher than that of the emitted waves, is the dolphin catching up to the fish or falling behind?
Figure 1.8
Sound
409
SECTION 2
Relate intensity, decibel level, and perceived loudness.
Sound Intensity and Resonance
Explain why resonance occurs.
Key Terms
Objectives ►
I ► I ►
Calculate the intensity of sound waves.
intensity
decibel
resonance
Sound Intensity Inside a Piano As a piano wire vibrates, it transfers energy to the piano’s soundboard, which in turn transfers energy into the air in the form of sound.
When a piano player strikes a piano key, a hammer inside the piano strikes a wire and causes it to vibrate, as shown in Figure 2.1. The wire’s vibrations are then transferred to the piano’s soundboard. As the soundboard vibrates, it exerts a force on air molecules around it, causing the air molecules to move. Because this force is exerted through displacement of the soundboard, the soundboard does work on the air. Thus, as the soundboard vibrates back and forth, its kinetic energy is converted into sound waves. This is one reason that the vibration of the soundboard gradually dies out.
Intensity is the rate of energy flow through a given area. As described in Section 1, sound waves traveling in air are longitudinal waves. As the sound waves travel outward from the source, energy is transferred from one air molecule to the next. The rate at which this energy is transferred through a unit area of the plane wave is called the intensity of the wave. Because power, P, is defined as the rate of energy transfer, intensity can also be described in terms of power. . . b..E/ b..t P mtensity = area = area intensity the rate at which energy flows through a unit area perpendicular to the direction of wave motion
The SI unit for power is the watt. Thus, intensity has units of watts per square meter (W/ m 2) . In a spherical wave, energy propagates equally in all directions; no one direction is preferred over any other. In this case, the power emitted by the source (P) is distributed over a spherical surface ( area = 4-rrr2), assuming that there is no absorption in the medium.
Intensity of a Spherical Wave
. . p mtens1ty = —-:::;:41rr
. . {power) mtens1ty = – – – – – – – – – – – – ( 41r) (distance from the source) 2
This equation shows that the intensity of a sound wave decreases as the distance from the source (r) increases. This occurs because the same amount of en ergy is spread over a larger area.
410
Chapter 12
PREMIUM CONTENT
t:’\
Intensity of Sound Waves
Interactive Demo
\.::,/ HMDScience.com
Sample Problem A What is the intensity of the sound waves produced by a trumpet at a distance of 3.2 m when the power output of the trumpet is 0.20 W? Assume that the sound waves are spherical.
0
SOLVE
Given:
P= 0.20W
Unknown:
Intensity=?
r= 3.2m
:r
Use the equation for the intensity of a spherical wave.
Intensity = Intensity =
Calculator Solution
IIntensity =
4
O.ZO W 47r(3.2m) 2 3
1.6 x 10- W / m
2
I
The calculator answer for intensity is 0.0015542. This is rounded to 1.6 x 10-3 because each of the given quantities has two significant figures.
Practice 1. Calculate the intensity of the sound waves from an electric guitar’s amplifier at a
distance of 5.0 m when its power output is equal to each of the following values: a. 0.25W b. 0.50W
c. 2.0W 2. At a maximum level of loudness, the power output of a 75-piece orchestra radiated
as sound is 70.0 W. What is the intensity of these sound waves to a listener who is sitting 25.0 m from the orchestra? 3. If the intensity of a person’s voice is 4.6 x 10- 7 W/ m 2 at a distance of 2.0 m, how much sound power does that person generate? 4. How much power is radiated as sound from a band whose intensity is 1.6 x 10- 3 W/m2 at a distance of 15 m ? 5. The power output of a tuba is 0.35 W. At what distance is the sound intensity of the tuba 1.2 x 10- 3 w/m2 ?
Sound
411
•i@i;liit Range of Human Hearing Human hearing depends on both the frequency and the intensity of sound waves. Sounds in the middle of the spectrum of frequencies can be heard more easily (at lower intensities) than those at lower and higher frequencies.
Range of Audibility of an Average Human Ear
1.0
10 0
1.0
10-2
“‘E
1.0
10-4
.e:,
1.0
10-6
1.0
10-8
1.0
1 o-1 o
3
‘iii C
.l!!
.E
1.0
10-12
– lL
\(
Music region
,,,,,,,.,,,…–
-…….._

‘- …….
“‘


—- –…..—
/~
\
Threshold ~f pain
Speech region
…… r—…….

“‘
“‘) — ………… _,I I

Area of sound
Threshold of hearing~

I
1/
/
Frequency (Hz)
Intensity and frequency determine which sounds are audible. The frequency of sound waves heard by the average human ranges from 20 to 20 000 Hz. Intensity is also a factor in determining which sound waves are audible. Figure 2.2 shows how the range of audibility of the average human ear depends on both frequency and intensity. Sounds at low frequencies (those below 50 Hz) or high frequencies (those above 12 000 Hz) must be relatively intense to be h eard, whereas sounds in the middle of the spectrum are audible at lower intensities. The softest sounds that can be heard by the average human ear occur at a frequency of about 1000 Hz and an intensity of 1.0 x 10- 12 W/ m 2 . Such a sound is said to be at the threshold ofhearing. The threshold of hearing at each frequency is represented by the lowest curve in Figure 2.2. For frequencies near 1000 Hz and at the threshold of hearing, the changes in pressure due to compressions and rarefactions are about three ten-billionths of atmospheric pressure. The maximum displacement of an air molecule at the threshold of hearing is approximately 1 x 10- 11 m. Comparing this number to the diameter of a typical air molecule (about 1 x 10- 10 m) reveals that the ear is an extremely sensitive detector of sound waves.
Did YOU Know? A 75-piece orchestra produces about 75 Wat its loudest. This is comparable to the power required to keep one medium-sized electric light bulb burning. Speech has even less power. It would take the conversation of about 2 million people to provide the amount of power required to keep a 50 W light bulb burning.
412
Chapter 12
, :
‘ ,
The loudest sounds that the human ear can tolerate have an intensity of about 1.0 W/ m 2 . This is known as the threshold of pain because sounds with greater intensities can produce pain in addition to hearing. The highest curve in Figure 2.2 represents the threshold of pain at each frequency. Exposure to sounds above the threshold of pain can cause immediate damage to the ear, even if no p ain is felt. Prolonged exposure to sounds oflower intensities can also damage the ear. Note that the threshold of hearing and the threshold of pain merge at both high and low ends of the spectrum.
Relative intensity is measured in decibels. Just as the frequency of a sound wave determines its pitch, the intensity of a wave approximately determines its perceived loudness. However, loudness is not directly proportional to intensity. The reason is that the sensation of loudness is approximately logarithmic in the human ear. Relative intensity is the ratio of the intensity of a given sound wave to the intensity at the threshold of hearing. Because of the logarithmic dependence of perceived loudness on intensity, using a number equal to 10 times the logarithm of the relative intensity provides a good indicator for human perceptions of loudness. This measure of loudness is referred to as the decibel level. The decibel level is dimensionless because it is proportional to the logarithm of a ratio. A dimensionless unit called the decibel (dB) is used for values on this scale.
The conversion of intensity to decibel level is shown in Figure 2.3. Notice in Figure 2.3 that when the intensity is multiplied by ten, 10 dB are added to the decibel level. A given difference in decibels corresponds to a fixed difference in perceived loudness. Although much more intensity (0.9 W/ m 2) is added between 110 and 120 dB than between 10 and 20 dB (9 x 10-11 W/m 2) , in each case the perceived loudness increases by the same amount.
decibel a dimensionless unit that describes the ratio of two intensities of sound ; the threshold of hearing is commonly used as the reference intensity
. .Did YOU Know?. – – – – – – – – – – – , Intensity (W/m 2)
Decibel level (dB)
Examples
1.0
X
10-12
0
1.0
X
10-ll
10
rustling leaves
1.0
X
10- 10
20
quiet whisper
1.0
X
10-9
30
whisper
1.0
X
10-8
40
mosquito buzzing
1.0
X
10-?
50
normal conversation
1.0
X
10- 6
60
air conditioner at 6 m
1.0
X
10- 5
70
vacuum cleaner
1.0
X
10-4
80
busy traffic, alarm clock
1.0
X
10- 3
90
lawn mower
1.0
X
10- 2
100
subway, power motor
1.0
X
10-l
110
auto horn at 1 m
1.0
X
10°
120
threshold of pain
1.0
X
101
130
thunderclap, machine gun
1.0
X
103
150
nearby jet airplane
threshold of hearing
: , : : ‘
The original unit of decibel level is the be/, named in honor of Alexander Graham Bell, the inventor of the telephone. The decibel is equivalent to0.1 bel.
Sound
413
Forced Vibrations and Resonance Forced Vibrations
If one blue pendulum is set in motion, only the other blue pendulum, whose length is the same, wi ll eventually oscillate with a large amplitude, or resonate.
When an isolated guitar string is held taut and plucked, hardly any sound is heard. When the same string is placed on a guitar and plucked, the intensity of the sound increases dramatically. What is responsible for this difference? To find the answer to this question, consider a set of pendulums suspended from a beam and bound by a loose rubber band, as shown in Figure 2.4. If one of the pendulums is set in motion, its vibrations are transferred by the rubber band to the other pendulums, which will also begin vibrating. This is called a forced vibration. The vibrating strings of a guitar force the bridge of the guitar to vibrate, and the bridge in turn transfers its vibrations to the guitar body. These forced vibrations are called sympathetic vibrations. Because the guitar body has a larger area than the strings do, it enables the strings’ vibrations to be transferred to the air more efficiently. As a result, the intensity of the sound is increased, and the strings’ vibrations die out faster than they would if they were not attached to the body of the guitar. In other words, the guitar body allows the energy exchange between the strings and the air to happen more efficiently, thereby increasing the intensity of the sound produced. In an electric guitar, string vibrations are translated into electrical impulses, which can be amplified as much as desired. An electric guitar can produce sounds that are much more intense than those of an unamplified acoustic guitar, which uses only the forced vibrations of the guitar’s body to increase the intensity of the sound from the vibrating strings.
Vibration at the natural frequency produces resonance. As you saw in the chapter on waves, the frequency of a pendulum depends on its string length. Thus, every pendulum will vibrate at a certain frequency, known as its natural frequency. In Figure 2.4, the two blue pendulums have the same natural frequency, while the red and green pendulums have different natural frequencies. When the first blue pendulum is set in motion, the red and green pendulums will vibrate only slightly, but the second blue pendulum will oscillate with a much larger amplitude because its natural frequency matches the frequency of the pendulum that was initially set in motion. This system is said to be in
QuickLAB Go to a playground, and swing on one of the swings. Try pumping (or being pushed) at different ratesfaster than, slower than, and equal to the natural frequency of the swing. Observe whether the rate at which you pump (or are pushed) affects how easily the amplitude of
414
Chapter 12
the vibration increases. Are some rates more effective at building your amplitude than others? You should find that the pushes are most effective when they match the swing’s natural frequency. Explain how your results support the statement that resonance
works best when the frequency of the applied force matches the system’s natural frequency.
MATERIALS • swing set
Because energy is transferred from one pendulum to the other, the amplitude of vibration of the first blue pendulum will decrease as the second blue pendulum’s amplitude increases. resonance.
A striking example of structural resonance occurred in 1940, when the Tacoma Narrows bridge in Washington, shown in Figure 2.5, was set in motion by the wind. High winds set up standing waves in the bridge, causing the bridge to oscillate at one of its natural frequencies. The amplitude of the vibrations increased until the bridge collapsed. A more recent example of structural resonance occurred during the Loma Prieta earthquake near Oakland, California, in 1989, when part of the upper deck of a freeway collapsed. The collapse of this particular section of roadway has been traced to the fact that the earthquake waves had a frequency of 1.5 Hz, very close to the natural frequency of that section of the roadway.
resonance a phenomenon that occurs when the frequency of a force applied to a system matches the natural frequency of vibration of the system , resulting in a large amplitude of vibration
Effects of Resonance On November 7, 1940, the Tacoma Narrows suspension bridge collapsed, just four months after it opened. Standing waves caused by strong winds set the bridge in motion and led to its collapse.
Concert If a 15-person musical ensemble gains 15
Broken Crystal Opera singers have been known
new members, so that its size doubles, will a listener perceive the music created by the ensemble to be twice as loud? Why or why not?
to set crystal goblets in vibration with their powerful voices. In fact, an amplified human voice can shatter the glass, but only at certain fundamental frequen cies. Speculate about why only certain
gJ C,
“‘ .E 0.. «
@
u
.!:
~.r:::;
i::
ill
gJ a:
fg ff.
A Noisy Factory Federal regulations require that no
office or factory worker be exposed to noise levels that average above 90 dB over an 8 h day. Thus, a factory that currently averages 100 dB must reduce its noise level by 10 dB. Assuming that each piece of machinery produces the same amount of noise, what percentage of eq uipment must be removed? Explain your answer.
fundamental frequencies will break the glass. Electric Guitars Electric guitars, which use electric amplifiers to magnify their
sound, can have a variety of shapes, but acoustic guitars all have the same basic shape. Explain why.
@
Sound
415
The human ear transmits vibrations that cause nerve impulses. The Human Ear Sound waves travel through the three regions of the ear and are then transmitted to the brain as impulses through nerve endings on the basilar membrane. Inner ear
Middle ear
Outer ear
Cochlea
Basilar membrane
Eardrum
The human ear is divided into three sections-outer, middle, and inner-as shown in Figure 2.6. Sound waves travel down the ear canal of the outer ear. The ear canal terminates at a thin, flat piece of tissue called the eardrum. The eardrum vibrates with the sound waves and transfers these vibrations to the three small bones of the middle ear, known as the hammer, the anvil, and the stirrup. These bones in turn transmit the vibrations to the inner ear, which contains a snail-shaped tube about 2 cm long called the cochlea. The basilar membrane runs through the coiled cochlea, dividing it roughly in half. The basilar membrane has different natural frequencies at different positions along its length, according to the width and thickness of the membrane at that point. Sound waves of varying frequencies resonate at different spots along the basilar membrane, creating impulses in hair cells-specialized nerve cells-embedded in the membrane. These impulses are then sent to the brain, which interprets them as sounds of varying frequencies.
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. When the decibel level of traffic in the street goes from 40 to 60 dB, how much greater is the intensity of the noise?
2. If two flutists play their instruments together at the same intensity, is the sound twice as loud as that of either flutist playing alone at that intensity? Why or why not? 3. A tuning fork consists of two metal prongs that vibrate at a single frequency when struck lightly. What will happen if a vibrating tuning fork is placed near another tuning fork of the same frequency? Explain. 4. A certain microphone placed in the ocean is sensitive to sounds emitted by dolphins. To produce a usable signal, sound waves striking the microphone must have a decibel level of 10 dB. If dolphins emit sound waves with a power of 0.050 W, how far can a dolphin be from the microphone and still be heard? (Assume the sound waves propagate spherically, and disregard absorption of the sound waves.)
Critical Thinking 5. Which of the following factors change when a sound gets louder? Which change when a pitch gets higher? a. intensity b. speed of the sound waves c. frequency d. decibel level e. wavelength f. amplitude 416
Chapter 12
Hearing Loss bout 10 percent of all Americans have some degree of hearing loss. There are three basic types of hearing loss. Conductive hearing loss is an impairment of the transmission of sound waves in the outer ear or transmission of vibrations in the middle ear. Conductive hearing loss is most often caused by improper development of the parts of the outer or middle ear or by damage to these parts of the ear by physical trauma or disease. Conductive hearing loss can often be corrected with medicine or surgery. Neural hearing loss is caused by problems with the auditory nerve, which carries signals from the inner ear to the brain. One common cause of neural hearing loss is a tumor pressing against the auditory nerve. Sensory hearing loss is caused by damage to the inner ear, particularly the microscopic hair cells in the cochlea. Sensory hearing loss can be present at birth and may be genetic or due to disease or developmental disorders. However, the most common source of damage to hair cells is exposure to loud noise. Short-term exposure to loud noise can cause ringing in the ears and temporary hearing impairment. Frequent or long-term exposure to noise above 80 dBincluding noise from familiar sources such as hair dryers or lawn mowers-can damage the hair cells permanently. The hair cells in the cochlea are not like the hair on your head or skin. They are highly specialized nerve cells that cannot be repaired or replaced by the body when they are severely damaged or destroyed. Cochlear hair cells can recover from minor damage, but if the source of the damage recurs frequently, even if it is only moderately loud noise, the hair cells can become permanently damaged. It is therefore important to protect yourself from sensory hearing loss by reducing your exposure to loud noise or by using a noise-
To prevent damage to their ears, people should wear ear protection when working with power tools.
dampening headset or earplugs that fully block the ear canal when you must be exposed to loud noise. Permanent sensory hearing loss usually occurs gradually, sometimes over 20 years or more. Because the hair cells that respond to higher-pitched sounds are smaller and more delicate, sensitivity to sounds with frequencies around 20 kHz is usually the first to be lost. Loss of sensitivity to sounds with frequencies around 4 kHz is often the first to be noticed because these frequencies are in the upper range of human speech. People who are starting to lose their hearing often have trouble hearing higher-pitched voices or hearing consonant sounds such as s, t, p, d, and f. As the hearing loss advances, loss of sensitivity to a wider range of sounds follows. Birds can regrow damaged hair cells. Scientists are studying this process to see if a similar process can be triggered in humans. For now, however, there is no “cure” for hearing loss, but some remedies are available. Hearing aids make any sounds that reach the ear louder. Assistive listening devices serve to amplify a specific small range of frequencies for people who have only partial hearing loss in that range. Cochlear implants use an electrode that is surgically implanted into the cochlea through a hole behind the outer ear. Electrical signals to the electrode stimulate the auditory nerve directly, in effect bypassing the hair cells altogether.
417
SECTION 3 Objectives ►
Differentiate between the harmonic series of open and closed pipes.

Calculate the harmonics of a vibrating string and of open and closed pipes.

Relate harmonics and timbre.
I

Relate the frequency difference between two waves to the number of beats heard per second.
Harmonics Key Terms fundamental frequency harmonic series
timbre beat
Standing Waves on a Vibrating String As discussed in the chapter “Vibrations and Waves;’ a variety of standing waves can occur when a string is fixed at both ends and set into vibration. The vibrations on the string of a musical instrument, such as the violin in Figure 3.1, usually consist of many standing waves together at the same time, each of which has a different wavelength and frequency. So, the sounds you hear from a stringed instrument, even those that sound like a single pitch, actually consist of multiple frequencies. on the next page, shows several possible vibrations on an idealized string. The ends of the string, which cannot vibrate, must always be nodes (N). The simplest vibration that can occur is shown in the first row of Figure 3.2. In this case, the center of the string experiences the most displacement, and so it is an antinode (A). Because the distance from one node to the next is always half a wavelength, the string length (L) must equal J../2. Thus, the wavelength is twice the string length (J.. 1 = 2L). Figure 3.2,
Stringed Instruments
The
vibrating strings of a violin produce standing waves whose frequencies depend on the string lengths.
As described in the chapter on waves, the speed of a wave equals the frequency times the wavelength, which can be rearranged as shown. V
v =JJ.., sof= –
)..
By substituting the value for wavelength found above into this equation for frequency, we see that the frequency of this vibration is equal to the speed of the wave divided by twice the string length. fundamental frequency = Ji
={ =
:r,
This frequency of vibration is called the fundamental frequency of the vibrating string. Because frequency is inversely proportional to wavelength and because we are considering the greatest possible wavelength, the fundamental frequency is the lowest possible frequency of a standing wave on this string.
fundamental frequency the lowest freq uency of vibration of a standing wave
418
Chapter 12
Harmonics are integral multiples of the fundamental frequency. The next possible standing wave for a string is shown in the second row of Figure 3.2. In this case, there are three nodes instead of two, so the string length is equal to one wavelength. Because this wavelength is half the previous wavelength, the frequency of this wave is twice that of the fundamental frequency.
N @ N ~ N ~ N m
A
A
A
A
A
A
A
A
A
A
N N t N ! N
>-. 1 = 2L
,,
>-.2 = L
‘2 =
fundamental frequency, or first harmonic
2,,
second harmonic
>-.3=
L
f3 = 3t,
third harmonic
>-. 4 =
L
f4 = 4t,
fourth harmonic
This pattern continues, and the frequency of the standing wave shown in the third row of Figure 3.2 is three times the fundamental frequency. More generally, the frequencies of the standing wave patterns are all integral multiples of the fundamental frequency. These frequencies form what is called a harmonic series. The fundamental frequency (Ji) corresponds to the first harmonic, the next frequency (f2 ) corresponds to the second harmonic, and so on.
harmonic series a series of frequencies that includes the fundamental frequency and integral multiples of the fundamental frequency
Because each harmonic is an integral multiple of the fundamental frequency, the equation for the fundamental frequency can be generalized to include the entire harmonic series. Thus, fn = nf1, where Ji is the fundamental frequency (Ji= :i,) andfn is the frequency of the nth harmonic. The general form of the equation is written as follows :
‘ Did YOU Know? Harmonic Series of Standing Waves on a Vibrating String
fn
= n {i n = I, 2, 3, …
. (speed of waves on the string) . . . ) frequency = harmomc number x ( )( 2 length of vibratmg strmg
When a guitar player presses down on a guitar string at any point, that point becomes a node and only a portion of the string vibrates. As a result, a single string can be used to create a variety of fundamental frequencies. In the equation on this page, L refers to the portion of the string that is vibrating.
Note that v in this equation is the speed of waves on the vibrating string and not the speed of the resultant sound waves in air. If the string vibrates at one of these frequencies, the sound waves produced in the surrounding air will have the same frequency. However, the speed of these waves will be the speed of sound waves in air, and the wavelength of these waves will be that speed divided by the frequency. Sound
419
Standing Waves in an Air Column Waves in a Pipe The harmonic series present in each of these organ pipes depends on whether the end of the pipe is open or closed.
Standing waves can also be set up in a tube of air, such as the inside of a trumpet, the column of a saxophone, or the pipes of an organ like those shown in Figure 3.3. While some waves travel down the tube, others are reflected back upward. These waves traveling in opposite directions combine to produce standing waves. Many brass instruments and woodwinds produce sound by means of these vibrating air columns.
If both ends of a pipe are open, all harmonics are present. The harmonic series present in an organ pipe depends on whether the reflecting end of the pipe is open or closed. When the reflecting end of the pipe is open, as is illustrated in Figure 3.4, the air molecules have complete freedom of motion, so an antinode (of displacement) exists at this end. If a pipe is open at both ends, each end is an antinode. This situation is the exact opposite of a string fixed at both ends, where both ends are nodes. Because the distance from one node to the next(½,\) equals the distance from one antinode to the next, the pattern of standing waves that can occur in a pipe open at both ends is the same as that of a vibrating string. Thus, the entire harmonic series is present in this case, as shown in Figure 3.4, and our earlier equation for the harmonic series of a vibrating string can be used.
Did YOU Know?. – – – – – – – A flute is similar to a pipe open at both ends. When all keys of a fl ute are closed, the length of the vibrating , air column is approximately equal to the length of the flute. As the keys are : opened one by one, the length of the vibrating air column decreases, and the : fundamental frequency increases.
Harmonics in an Open-Ended Pipe In a pipe open at both ends, each end is an antinode of displacement, and all harmonics are present.
f
monic Series of~ Pi~e Op~n at Both—;;nds
fn – n ZL
. frequency= harmomc number x
IL
I l L
A
(2){length of vibrating air column)
A+
A+
+
N•
N• N•
A+ A ,=L
AI =2L
f 2 = !!.= 2!, L I
f, = :i,
(a) First harmonic
Chapter 12
( speed of sound in the pipe)
In this equation, L represents the length of the vibrating air column. Just as the fundamental frequency of a string instrument can be varied by changing the string length, the fundamental frequency of many woodwind and brass instruments can be varied by changing the length of the vibrating air column.
(a)
420
n – 1, 2, 3, …
(b) (b) Second harmonic
N•
A+ N•
A 3 =1-L 3
i,=~~=3f,
(c) (c) Third harmonic
QuickLAB
Harmonics in a Pipe Closed at One End In a pipe closed at one end, the closed end is a node of displacement and the open end is an antinode of displacement. In this case, only the odd harmonics are present.
MATERIALS
I
SAFETY
L
1
At
At
A1 =4L
A-3 =½L
N•
f, = :i
• straw • scissors
!3 = !~=31,
N•
At
As =:!L 5
♦ Always
use caution when working with scissors.
~ =~~=5f, A PIPE CLOSED AT ONE END
(b}
(a) (a) First harmonic
(c)
(b) Second harmonic
(c) Third harmonic
If one end of a pipe is closed, only odd harmonics are present. When one end of an organ pipe is closed, as is illustrated in Figure 3.5, the movement of air molecules is restricted at this end, making this end a node. In this case, one end of the pipe is a node and the other is an antinode. As a result, a different set of standing waves can occur. As shown in Figure 3.5(a), the simplest possible standing wave that can exist in this pipe is one for which the length of the pipe is equal to onefourth of a wavelength. Hence, the wavelength of this standing wave equals four times the length of the pipe. Thus, in this case, the fundamental frequency equals the velocity divided by four times the pipe length. V
f1
V
=-x-= 4L 1
For the case shown in Figure 3.5(b), the length of the pipe is equal to three-fourths of a wavelength, so the wavelength is four-thirds the length of the pipe (A 3 f L). Substituting this value into the equation for frequency gives the frequency of this harmonic.
=
V
V
/ 3 = A3 = i L
Snip off the corners of one end of the straw so that the end tapers to a point, as shown below. Chew on this end to flatten it, and you create a double-reed instrument! Put your lips around the tapered end of the straw, press them together tightly, and blow through the straw. When you hear a steady tone, slowly snip off pieces of the straw at the other end. Be careful to keep about the same amount of pressure with your lips. How does the pitch change as the straw becomes shorter? How can you account for this change in pitch? You may be able to produce more than one tone for any given length of the straw. How is this possible?
3V
= 4L =3f1
3
The frequency of this harmonic is three times the fundamental frequency. Repeating this calculation for the case shown in Figure 3.5(c) gives a frequency equal to five times the fundamental frequency. Thus, only the odd-numbered harmonics vibrate in a pipe closed at one end. We can generalize the equation for the harmonic series of a pipe closed at one end as follows: Harmonic Series of a Pipe Closed at One End
fn
= n :Z,
n = I, 3, 5, …
. ( speed of sound in the pipe) frequency = harmomc number x – – – – – – – – – – – – – ( 4){Iength of vibrating air column) Sound
421
PREMIUM CONTENT
A:\
Harmonics
Interactive Demo
~ HMDScience.com
Sample Problem B What are the first three harmonics in a 2.45 m long pipe that is open at both ends? What are the first three harmonics of this pipe when one end of the pipe is closed? Assume that the speed of sound in air is 345 m/ s.
0
ANALYZE
Given:
L = 2.45 m
Unknown:
Pipe open at both ends:I 1
v = 345m/s
Pipe closed at one end: I1
E)
PLAN
Choose an equation or situation: When the pipe is open at both ends, the fundamental frequency can be found by using the equation for the entire harmonic series:
In= n{L’ n = l, 2, 3, … When the pipe is closed at one end, use the following equation:
In= n
:L’ n =
l, 3, 5, …
In both cases, the second two harmonics can be found by multiplying the harmonic numbers by the fundamental frequency.
E)
SOLVE
Substitute the values into the equations and solve: For a pipe open at both ends:
J1 = Tips and Tricks Be sure to use the correct harmonic numbers for each situation. For a pipe open at both ends, n = 1, 2, 3, etc. For a pipe closed at one end, only odd harmonics are present, so n = 1, 3, 5, etc.
= (1) ( 345m/ s ) = 170.4 Hz l 2L (2)(2.45m) · ·
n__E_
The next two harmonics are the second and the third:
I2 = 2I1 =
(2)(70.4 Hz)= 1141 Hzl
I3 = 3I1 =
(3)(70.4 Hz)= 1211 Hzl
For a pipe closed at one end:
I1
I
v ( 345ml s ) = n 4L = (l) (2)( 2.4 m) = 35.2 Hz 5
The next possible harmonics are the third and the fifth:
422
Chapter 12
I3 = 3fi =
(3)(35.2 Hz)= 1106 Hzl
Is=5I1 =
(5)(35.2 Hz)= 1176 Hz l
I
Harmonics (continued)
0
CHECKYOUR WORK
In a pipe open at both ends, the first possible wavelength is 2L; in a pipe closed at one end, the first possible wavelength is 4L. Because frequency and wavelength are inversely proportional, the fundamental frequency of the open pipe should be twice that of the closed pipe, that is, 70.4 = (2)(35.2).
Practice 1. What is the fundamental frequency of a 0.20 m long organ pipe that is closed at one end, when the speed of sound in the pipe is 352 m i s? 2. A flute is essentially a pipe open at both ends. The length of a flute is
approximately 66.0 cm. What are the first three harmonics of a flute when all keys are closed, making the vibrating air column approximately equal to the length of the flute? The speed of sound in the flute is 340 m/ s. 3. What is the fundamental frequency of a guitar string when the speed of waves on the string is 115 m/ s and the effective string lengths are as follows? a. 70.0 cm
b. 50.0 cm
c. 40.0 cm
4. A violin string that is 50.0 cm long has a fundamental frequency of 440 Hz. What is the speed of the waves on this string?
Trumpets, saxophones, and clarinets are similar to a pipe closed at one end. For example, although the trumpet shown in Figure 3.6 has two open ends, the player’s mouth effectively closes one end of the instrument. In a saxophone or a clarinet, the reed closes one end.
~
~
~C!i
Despite the similarity between these instruments and a pipe closed at one end, our equation for the harmonic series of pipes does not directly apply to such instruments. One reason the equation does not apply is that any deviation from the cylindrical shape of a pipe affects the harmonic series of an instrument. Another reason is that the open holes in many instruments affect the harmonics. For example, a clarinet is primarily cylindrical, but there are some even harmonics in a clarinet’s tone at relatively small intensities. The shape of a saxophone is such that the harmonic series in a saxophone is similar to that in a cylindrical pipe open at both ends even though only one end of the saxophone is open. These deviations are in part responsible for the variety of sounds that can be produced by different instruments.
Shape and Harmonic Series Variations in shape give each instrument a different harmonic series.
;;::
., .c t: 0
a:;
@
Sound
423
Harmonics account for sound quality, or timbre. Figure 3. 7 shows the harmonics present in a
tuning fork, a clarinet, and a viola when each sounds the musical note A-natural. Each instrument has its own characteristic mixture of harmonics at varying intensities. The harmonics shown in the second column of Figure 3.7 add together according to the principle of superposition to give the resultant waveform shown in the third column. Since a tuning fork vibrates at only its fundamental frequency, its waveform is simply a sine wave. (Some tuning forks also vibrate at higher frequencies when they are struck hard enough.) The waveforms of the other instruments are more complex because they consist of many harmonics, each at different intensities. Each individual harmonic waveform is a sine wave, but the resultant wave is more complex than a sine wave because each individual waveform has a different frequency.
timbre the musical quality of a tone resulting from the combination of harmonics present at different intensities
In music, the mixture of harmonics that produces the characteristic sound of an instrument is referred to as the spectrum of the sound. From the perspective of the listener, this spectrum results in sound quality, or timbre. A clarinet sounds different from a viola because of differences in timbre, even when both instruments are sounding the same note at the same volume. The rich harmonics of most instruments provide a much fuller sound than that of a tuning fork. The intensity of each harmonic varies within a particular instrument, depending on frequency, amplitude of vibration, and a variety of other factors. With a violin, for example, the intensity of each harmonic
Tuning fork
.;:, “iii
-.. C a,
I\ I\ I\
vvvv
.E a, >
“‘ .; cc:
12345678910 Harmonics
Clarinet
Resultant waveform
.;:, “iii
Nth
C
J!l
.E a,
~ .; “‘ cc:
1 2 3 4 5 6 7 8 9 10
Harmonics
Resultant waveform
(/J
@
Q
1 0
Viola
~ ,::;
~
.,
“‘0
C
J!l
o/\ o/\ o/\
vvvvvv
.5
a,
~ .;
“‘
Cl
>,
.c
EC. .. 1 = 5.6 x 10 m I
).. = 3.00 x 10 2
1.7
>..2 = 1.8 x
X
8
mis
Calculator Solution Although the calculator solutions are 555.5555556 m and 176.470588 m, both answers must be rounded to two digits because the frequencies have only two significant figures.
106 Hz
102 m
G·I, iii, m&-► 444
Chapter 13
Electromagnetic Waves (continued) Practice 1. Gamma-ray bursters are objects in the universe that emit pulses of gamma rays
with high energies. The frequency of the most energetic bursts has been measured at around 3.0 x 1021 Hz. What is the wavelength of these gamma rays? 2. What is the wavelength range for the FM radio band (88 MHz-108 MHz)? 3. Shortwave radio is broadcast between 3.50 and 29.7 MHz. To what range of
wavelengths does this correspond? Why do you suppose this part of the spectrum is called shortwave radio? 4. What is the frequency of an electromagnetic wave if it has a wavelength of 1.0 km? 5. The portion of the visible spectrum that appears brightest to the human eye is
around 560 nm in wavelength, which corresponds to yellow-green. What is the frequency of 560 nm light? 6. What is the frequency of highly energetic ultraviolet radiation that has
a wavelength of 125 nm?
Waves can be approximated as rays. Consider an ocean wave coming toward the shore. The broad crest of the wave that is perpendicular to the wave’s motion consists of a line of water particles. Similarly, another line of water particles forms a low-lying trough in the wave, and still another line of particles forms the crest of a second wave. In any type of wave, these lines of particles are called wavefronts. All the points on the wave front of a plane wave can be treated as point sources, that is, coming from a source of negligible size. A few of these points are shown on the initial wave front in Figure 1.4. Each of these point sources produces a circular or spherical secondary wave, or wavelet. The radii of these wavelets are indicated by the blue arrows in Figure 1.4. The line that is tangent to each of these wavelets at some later time determines the new position of the initial wave front (the new wave front in Figure 1.4). This approach to analyzing waves is called Huygens’s principle, named for the physicist Christian Huygens, who developed it. Huygens’s principle can be used to derive the properties of any wave (including light) that interacts with matter, but the same results can be obtained by treating the propagating wave as a straight line perpendicular to the wave front. This line is called a ray, and this simplification is called the ray approximation.
Tangent line
Huygens’s Principle According to Huygens’s principle, a wave front can be divided into point sources. The line tangent to the wavelets from these sources marks the wave front’s new position. Initial wave front
Light and Reflection
445
llluminance decreases as the square of the distance from the source. You have probably noticed that it is easier to read a book beside a lamp using a 100 W bulb rather than a 25 W bulb. It is also easier to read nearer to a lamp than farther from a lamp. These experiences suggest that the intensity oflight depends on both the amount oflight energy emitted from a source and the distance from the light source. Light bulbs are rated by their power input (measured in watts) and their light output. The rate at which light is emitted from a source is called the luminous flux and is measured in lumens (Im). Luminous flux is a measure of power output but is weighted to take into account the response of the human eye to light. Luminous flux helps us understand why the illumination on a book page is reduced as you move away from a light. Imagine spherical surfaces of different sizes with a point light Less light falls on each unit source at the center of the sphere, shown in Figure 1.5. A point source provides light equally in all directions. The principle of conservation of energy requires that the luminous flux is the same on each sphere. However, the luminous flux divided by the area of the surface, which is called the illuminance (measured in lm/ m 2, or lux), decreases as the radius squared when you move away lm 2m 3m from a light source.

SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Identify which portions of the electromagnetic spectrum are used in each of the devices listed. a. a microwave oven b. a television set c. a single-lens reflex camera
2. If an electromagnetic wave has a frequency of 7 .57 x 1014 Hz, what is its wavelength? To what part of the spectrum does this wave belong? 3. Galileo performed an experiment to measure the speed of light by timing how long it took light to travel from a lamp he was holding to an assistant about 1.5 km away and back again. Why was Galileo unable to conclude that light had a finite speed?
Critical Thinking 4. How bright would the sun appear to an observer on Earth if the sun were four times farther from Earth than it actually is? Express your answer as a fraction of the sun’s brightness on Earth’s surface. 446
Chapter 13
Flat Mirrors Key Terms reflection angle of incidence
angle of reflection virtual image
Reflection of Light Suppose you have just had your hair cut and you want to know what the back of your head looks like. You can do this seemingly impossible task by using two mirrors to direct light from behind your head to your eyes. Redirecting light with mirrors reveals a basic property of light’s interaction with matter. Light traveling through a uniform substance, whether it is air, water, or a vacuum, always travels in a straight line. However, when the light encounters a different substance, its path will change. If a material is opaque to the light, such as the dark, highly polished surface of a wooden table, the light will not pass into the table more than a few wavelengths. Part of the light is absorbed, and the rest of it is deflected at the surface. This change in the direction of the light is called reflection. All substances absorb ‘ at least some incoming light and reflect the rest. A good mirror can reflect about 90 percent of the incident light, but no surface is a perfect reflector. Notice in Figure 2.1 that the images of the golf ball get successively darker.
reflection the change in direction of an electromagnetic wave at a surface that causes it to move away from the surface
Reflection Mirrors reflect nearly all incoming light, so multiple images of an object between two mirrors are easily formed.
The texture of a surface affects how it reflects light. The manner in which light is reflected from a surface depends on the surface’s smoothness. Light that is reflected from a rough, textured surface, such as paper, cloth, or unpolished wood, is reflected in many different directions, as shown in Figure 2.2(a). This type of reflection is called diffuse reflection and is covered later in the chapter. Light reflected from smooth, shiny surfaces, such as a mirror or water in a pond, is reflected in one direction only, as shown in Figure 2.2(b). This type of reflection is called specular reflection. A surface is considered smooth if its surface variations are small compared with the wavelength of the incoming light. For our discussion, reflection will be used to mean only specular reflection. §2 z uf .c
C.
I
0..
]9
., C:
E
“‘ ~
ir
ol C:
., ::,;
0,
Diffuse and Specular Reflection Diffusely reflected light is reflected in many directions (a), whereas specularly reflected light is reflected in the same forward direction only (b).
(a)
(b)
-c-
a
Light and Reflection
447
Symmetry of Reflected Light The symmetry of reflected light (a) is described by the law of reflection, which states that the angles of the incoming and reflected rays are equal (b).
Incoming light
Normal
Reflected light
I
Reflecting surface (b)
(a)
· Incoming and reflected angles are equal. You probably have noticed that when incoming rays oflight strike a smooth reflecting surface, such as a polished table or mirror, at an angle close to the surface, the reflected rays are also close to the surface. When the incoming rays are high above the reflecting surface, the reflected rays are also high above the surface. An example of this similarity between incoming and reflected rays is shown in Figure 2.3(a). If a straight line is drawn perpendicular to the reflecting surface at the
point where the incoming ray strikes the surface, the angle of incidence and the angle of reflection can be defined with respect to the line. Careful measurements of the incident and reflected angles 0 an 0′, respectively, reveal that the angles are equal, as illustrated in Figure 2.3(b).
angle of incidence the angle between a ray that strikes a surface and the line perpendicular to that surface at the point of contact angle of reflection the angle formed by the line perpendicular to a surface and the direction in which a reflected ray moves
0= 0′
angle of incoming light ray = angle of reflected light ray The line perpendicular to the reflecting surface is referred to as the normal to the surface. It therefore follows that the angle between the incoming ray and the surface equals 90° – 0, and the angle between the reflected ray and the surface equals 90° – 0′.
Flat Mirror Light reflecting off of a flat mirror creates an image that appears to be behind the mirror. Flat mirror
Object
Image
:~–P—- –q——,: Object distance
448
Chapter 13
=
Image distance
Flat Mirrors The simplest mirror is the flat mirror. If an object, such as a pencil, is placed at a distance in front of a flat mirror and light is bounced off the pencil, light rays will spread out from the pencil and reflect from the mirror’s surface. To an observer looking at the mirror, these rays appear to come from a location on the other side of the mirror. As a convention, an object’s image is said to be at this location behind the mirror because the light appears to come from that point. The relationship between the object distance from the mirror, represented as p, and the image distance, represented as q, is such that the object and image distances are equal, as shown in Figure 2.4. Similarly, the image of the object is the same size as the object.
The image formed by rays that appear to come from the image point behind the mirror-but never really do-is called a virtual image. As shown in Figure 2.5(a), a flat mirror always forms a virtual image, which always appears as if it is behind the surface of the mirror. For this reason, a virtual image can never be displayed on a physical surface.
virtual image an image from which light rays appear to diverge, even though they are not actually focused there; a virtual image cannot be projected on a screen
Image location can be predicted with ray diagrams. Ray diagrams, such as the one shown in Figure 2.5(b), are drawings that use simple geometry to locate an image formed by a mirror. Suppose you want to make a ray diagram for a pencil placed in front of a flat mirror. First, sketch the situation. Draw the location and arrangement of the mirror and the position of the pencil with respect to the mirror. Construct the drawing so that the object and the image distances (p and q, respectively) are proportional to their actual sizes. To simplify matters, we will consider only the tip of the pencil. To pinpoint the location of the pencil tip’s image, draw two rays on your diagram. Draw the first ray from the pencil tip perpendicular to the mirror’s surface. Because this ray makes an angle of 0° with a line perpendicular (or normal) to the mirror, the angle of reflection also equals 0°, causing the ray to reflect back on itself. In Figure 2.5(b), this ray is denoted by the number 1 and is shown with arrows pointing in both directions because the incident ray reflects back on itself. Draw the second ray from the tip of the pencil to the mirror, but this time place the ray at an angle that is not perpendicular to the surface of the mirror. The second ray is denoted in Figure 2.5(b) by the number 2. Then, draw the reflected ray, keeping in mind that it will reflect away from the surface of the mirror at an angle, 0′, equal to the angle of incidence, 0. Next, trace both reflected rays back to the point from which they appear to have originated, that is, behind the mirror. Use dotted lines when drawing these rays that appear to emerge from behind the mirror to distinguish them from the rays of light in front of the mirror. The point at which these dotted lines meet is the image point, which in this case is where the image of the pencil’s tip forms. By continuing this process for all of the other parts of the pencil, you can locate the complete virtual image of the pencil. Note that the pencil’s image appears as far behind the mirror as the pencil is in front of the mirror (p = q). Likewise, the object height, h, equals the image height, h’.
Ray Diagram The position and size of the virtual image that forms in a flat mirror (a) can be predicted by constructing a ray diagram (b).
Eye
I
~
/- / – / 7
//
T h’
Image
Mirror
(a)
(b)
Light and Reflection
449
Mirror Reversal The front of an object becomes the back of its image.
This ray-tracing procedure will work for any object placed in front of a flat mirror. By selecting a single point on the object (usually its uppermost tip or edge), you can use ray tracing to locate the same point on the image. The rest of the image can be added once the image point and image distance have been determined. The image formed by a flat mirror appears reversed to an observer in front of the mirror. You can easily observe this effect by placing a piece of writing in front of a mirror, as shown in Figure 2.6. In the mirror, each of the letters is reversed. You may also notice that the angle the word and its reflection make with respect to the mirror is the same.

SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Which of the following are examples of specular reflection, and which are examples of diffuse reflection? a. reflection of light from the surface of a lake on a calm day b. reflection of light from a plastic trash bag c. reflection of light from the lens of eyeglasses d. reflection oflight from a carpet
2. Suppose you are holding a flat mirror and standing at the center of a giant clock face built into the floor. Someone standing at 12 o’clock shines a beam of light toward you, and you want to use the mirror to reflect the beam toward an observer standing at 5 o’clock. What should the angle of incidence be to achieve this? What should the angle of reflection be? 3. Some department-store windows are slanted inward at the bottom. This is to decrease the glare from brightly illuminated buildings across the street, which would make it difficult for shoppers to see the display inside and near the bottom of the window. Sketch a light ray reflecting from such a window to show how this technique works.
Interpreting Graphics 4. The photograph in Figure 2.1 shows multiple images that were created by multiple reflections between two flat mirrors. What conclusion can you make about the relative orientation of the mirrors? Explain your answer.
Critical Thinking 5. If one wall of a room consists of a large flat mirror, how much larger will the room appear to be? Explain your answer. 6. Why does a flat mirror appear to reverse the person looking into a mirror left to right, but not up and down? 450
Chapter 13
Curved Mirrors Key Terms concave spherical mirror
real image
convex spherical mirror
Concave Spherical Mirrors Small, circular mirrors, such as those used on dressing tables, may appear at first glance to be the same as flat mirrors. However, the images they form differ from those formed by flat mirrors. The images for objects close to the mirror are larger than the object, as shown in Figure 3.1 (a), whereas the images of objects far from the mirror are smaller and upside down, as shown in Figure 3.1 (b). Images such as these are characteristic of curved mirrors. The image in Figure 3.1 (a) is a virtual image like those created by flat mirrors. In contrast, the image in Figure 3.1 (b) is a real image.
Concave mirrors can be used to form real images. One basic type of curved mirror is the spherical mirror. A spherical mirror, as its name implies, has the shape of part of a sphere’s surface. A spherical mirror with light reflecting from its silvered, concave surface (that is, the inner surface of a sphere) is called a concave spherical mirror. Concave mirrors are used whenever a magnified image of an object is needed, as in the case of the dressing-table mirror.
concave spherical mirror a mirror whose reflecting surface is a segment of the inside of a sphere
One factor that determines where the image will appear in a concave spherical mirror and how large that image will be is the radius of curvature, R, of the mirror. The radius of curvature is the same as the radius of the spherical shell of which the mirror is a small part; R is therefore the distance from the mirror’s surface to the center of curvature, C.
Concave Spherical Mirror Curved mirrors can be used to form images that are larger (a) or smaller (b) than the object.
“l:’
“”-‘
.t::
ii:
@
Light and Reflection
451
Images and Concave Mirrors (a) The rays from a light bulb converge
to form a real image in front of a concave mirror. Mirror Front of
T h Principal axis
(b) In this lab set up, the real image of a light-bulb filament
appears on a glass plate in front of a concave mirror. mirror _….___+R-j I
f—- f – i 1—–p——
1 6.
5. Mirror
Mirror 2
Back of mirror
Configuration: object at F Image: image at infinity (no image) 456
Chapter 13
mirror
Back of mirror
Configuration: object inside F Image: virtual, upright image at C with magnification > 1
PREMIUM CONTENT
~
Interactive Demo
\:;I
HMDScience.com
Sample Problem B A concave spherical mirror has a focal length of I 0.0 cm. Locate the image of a pencil that is placed upright 30.0 cm from the mirror. Find the magnification of the image. Draw a ray diagram to confirm your answer.
0
ANALYZE
Determine the sign and magnitude of the focal length and object size.
f=
p= +30.0cm
+10.0cm
The mirror is concave, so f is positive. The object is in front of the mirror, so p is positive. Unknown: Diagram:
Draw a ray diagram using the rules given in Figure 3.5.
1
2
f— f = 10.0 cm – – – l t – – – – – – – – – – – – p = 30.0 cm – – – – – – – – – – – – – – – i
1——q=?—–~
E)
PLAN
Use the mirror equation to relate the object and image distances to the focal length.
Use the magnification equation in terms of object and image distances.
q M = -p Rearrange the equation to isolate the image distance, and calculate. Subtract the reciprocal of the object distance from the reciprocal of the focal length to obtain an expression for the unknown image distance.
CS·i ,iii ,\114-►
1
1
1
q
f
p
Light and Reflection
457
Imaging with Concave Mirrors
E)
SOLVE
(continued)
Substitute the values forf and pinto the mirror equation and the magnification equation to find the image distance and magnification.
1 q
1 10.0 cm
1 30.0 cm
15 cm 30.0 cm
0
CHECKYOUR WORK
=-
0.100 1 cm
Evaluate your answer in terms of the image location and size. The image appears between the focal point (10.0 cm) and the center of curvature (20.0 cm), as confirmed by the ray diagram. The image is smaller than the object and inverted ( -1 < M < 0), as is also confirmed by the ray diagram. The image is therefore real.
1. Find the image distance and magnification of the mirror in the sample problem when the object distances are 10.0 cm and 5.00 cm. Are the images real or virtual? Are the images inverted or upright? Draw a ray diagram for each case to confirm your results.
2. A concave shaving mirror has a focal length of 33 cm. Calculate the image position of a cologne bottle placed in front of the mirror at a distance of 93 cm. Calculate the m agnification of the image. Is the image real or virtual? Is the image inverted or upright? Draw a ray diagram to show where the image forms and how large it is with respect to the object. 3. A concave makeup mirror is designed so that a person 25.0 cm in front of it sees an upright image at a distance of 50.0 cm behind the mirror. What is the radius of curvature of the mirror? What is the magnification of the image? Is the image real or virtual?
4. A pen placed 11.0 cm from a concave spherical mirror produces a real image 13.2 cm from the mirror. What is the focal length of the mirror? What is the magnification of the image? If the pen is placed 27.0 cm from the mirror, what is the new position of the image? What is the magnification of the new image? Is the new image real or virtual? Draw ray diagrams to confirm your results.
Chapter 13
0.067 1 cm
0.05 I
Practice
458
0.003 1cm
Convex Spherical Mirrors On recent models of automobiles, there is a side-view mirror on the passenger’s side of the car. Unlike the flat mirror on the driver’s side, which produces unmagnified images, the passenger’s mirror bulges outward at the center. Images in this mirror are distorted near the mirror’s edges, and the image is smaller than the object. This type of mirror is called a convex spherical mirror. A convex spherical mirror is a segment of a sphere that is silvered so that light is reflected from the sphere’s outer, convex surface. This type of mirror is also called a diverging mirror because the incoming rays diverge after reflection as though they were coming from some point behind the mirror. The resulting image is therefore always virtual, and the image distance is always negative. Because the mirrored surface is on the side opposite the radius of curvature, a convex spherical mirror also has a negative focal length. The sign conventions for all mirrors are summarized in Figure 3.8.
convex spherical mirror a mirror whose reflecting surface is an outward-curved segment of a sphere
The technique for drawing ray diagrams for a convex mirror differs slightly from that for concave mirrors. The focal point and center of curvature are situated behind the mirror’s surface. Dotted lines are extended along the reflected reference rays to points behind the mirror, as shown in Figure 3.7(a). A virtual, upright image forms where the three rays apparently intersect. Magnification for convex mirrors is always less than 1, as shown in Figure 3.7(b). Convex spherical mirrors take the objects in a large field of view and produce a small image, so they are well suited for providing a fixed observer with a complete view of a large area. Convex mirrors are often placed in stores to help employees monitor customers and at the intersections of busy hallways so that people in both hallways can tell when others are approaching. The side-view mirror on the passenger’s side of a car is another application of the convex mirror. This mirror usually carries the warning, “objects are closer than they appear:’ Without this warning, a driver might think that he or she is looking into a flat mirror, which does not alter the size of the image. The driver could therefore be fooled into believing that a vehicle is farther away than it is because the image is smaller than the actual object.
-1: ~
~ z
Reflection from a Convex Mirror Light rays diverge upon reflection from a convex mirror (a), forming a virtual image that is always smaller than the object (b).
uf .c
Eye
C.
I
.air• The frequency of the light does not change when the light passes from one medium to another.
. Did YOU Know?. – – – – – – – – – – – , : The speed of light in a vacuum, c, is an ‘ important constant used by physicists. ‘ It has been measured to be about 3.00 x 108 m/s. Inside of other ‘ mediums, such as air, glass, or water, ‘ the speed of light is different and is : less than c.
Refraction and the Wave Model of Light A plane wave traveling in air (a) has a wavelength of >.air and velocity
of vair• Each wave front turns as it strikes the glass. Because the speed of the wave fronts in the glass (b), vglass’ is slower, the wavelength of the light becomes shorter, and the wave fronts change direction.
Air
Glass
Refraction
483
The Law of Refraction index of refraction the ratio of the speed of light in a vacuum to the speed of light in a given transparent medium
An important property of transparent substances is the index of refraction.
The index of refraction for a substance is the ratio of the speed of light in a vacuum to the speed of light in that substance.
Index of Refraction
speed of light in vacuum index of refraction=———speed oflight in medium
. Did YOU Know?_- – – – – – – – – – – ‘ The index of refraction of any medium can also be expressed as the ratio of the wavelength of light in a vacuum, Ao, to the wavelength of light in that medium, An, as shown in the following relation. A
n= _Q_ An
From this definition, we see that the index of refraction is a dimensionless number that is always greater than 1 because light always travels slower in a substance than in a vacuum. Figure 1.4 lists the indices of refraction for different substances. Note that the larger the index of refraction is, the slower light travels in that substance and the more a light ray will bend when it passes from a vacuum into that material. Imagine, as an example, light passing between air and water. When light begins in the air (high speed of light and low index of refraction) and travels into the water (lower speed oflight and higher index ofrefraction), the light rays are bent toward the normal. Conversely, when light passes from the water to the air, the light rays are bent away from the normal. Note that the value for the index of refraction of air is nearly that of a vacuum. For simplicity, use the value n = 1.00 for air when solving problems.
Solids at 20°C
n
Liquids at 20°c
n
Cubic zirconia
2.20
Benzene
1.501
Diamond
2.419
Carbon disulfide
1.628
Fluorite
1.434
Carbon tetrachloride
1.461
Fused quartz
1.458
Ethyl alcohol
1.361
Glass, crown
1.52
Glycerine
1.473
Glass, flint
1.66
Water
1.333
Ice (at 0°C)
1.309
Gases at 0°C, 1 atm
n
Polystyrene
1.49
Air
1.000 293
Sodium chloride
1.544
Carbon dioxide
1.000 450
Zircon
1.923
*measured with light of vacuum wavelength 484
Chapter 14
= 589 nm
Image Position for Objects in Different Media (a) To the cat
Normal
on the pier, the fish looks closer to the surface than it really is. (b) To the fish , the cat seems to be farther from the surface than it actually is.
(a)
Vi< Vr
(b)
Vi> Vr
Objects appear to be in different positions due to refraction. When looking at a fish underwater, a cat sitting on a pier perceives the fish to be closer to the water’s surface than it actually is, as shown in Figure 1.5(a). Conversely, the fish perceives the cat on the pier to be farther from the water’s surface than it actually is, as shown in Figure 1.5(b). Because of the reversibility of refraction, both the fish and the cat see along the same path, as shown by the solid lines in both figures. However, the light ray that reaches the fish forms a smaller angle with respect to the normal than does the light ray from the cat to the water’s surface. The reason is that light is bent toward the normal when it travels from a medium with a lower index of refraction (the air) to one with a higher index of refraction (the water). Extending this ray along a straight line shows the cat’s image to be above the cat’s actual position. On the other hand, the light ray that reaches the cat from the water’s surface forms a larger angle with respect to the normal instead of a smaller one. This is because the light from the fish travels from a medium with a higher index of refraction to one with a lower index of refraction.
The Invisible Man H. G. Wells
Fishing When
wrote a famous novel about a man who made himself invisible by
trying to catch a fish, should a pelican
changing his index of refract ion. What would his index of refraction have to be to accomplish this?
dive into the water horizontally in front of or behind the image of the fish it sees?
Visibility for the Invisible Man Would the invisible man be
able to see anything? -c-
a
Refract ion
485
Note that the fish’s image is closer to the water’s surface than the fish actually is. An underwater object seen from the air above appears larger than its actual size because the image, which is the same size as the object, is closer to the observer.
Wavelength affects the index of refraction. Differences in Refraction According to Wavelength Different wavelengths of light refract different amounts. For example, blue refracts more than red. This difference leads to the separation of white light into different colors. White light
Note that the indices ofrefraction listed in Figure 1.4 are only valid for light that has a wavelength of 589 nm in a vacuum. The reason is that the amount that light bends when entering a different medium depends on the wavelength of the light as well as the speed, as shown in Figure 1.6. Each color of light has a different wavelength, so each color of the spectrum is refracted by a different amount. This explains why a spectrum is produced when white light passes through a prism.
Snell’s law determines the angle of refraction. The index of refraction of a material can be used to figure out how much a ray of light will be refracted as it passes from one medium to another. As mentioned, the greater the index of refraction, the more refraction occurs. But how can the angle of refraction be found? Blue
Red
In 1621, Willebrord Snell experimented with light passing through different media. He developed a relationship called Snell’s law, which can be used to find the angle of refraction for light traveling between any two media.
l
Snell’s Law
index of refraction of first medium x sine of the angle of incidence = index of refraction of second medium x sine of the angle of refractio:J
PREMIUM CONTENT
t!:’\
Snell’s Law
Interactive Demo
\::,/ HMDScience.com
I
Sample Problem A A light ray of wavelength 589 nm (produced by a sodium lamp) traveling through air strikes a smooth, flat slab of crown glass at an angle of 30.0° to the normal. Find the angle ofrefraction, 0r.
0
ANALYZE
Given:
0.l
= 30.0°
= 1.00 nr = 1.52 ni
Unknown:
0r
= ?.
— — — — – – ————————-· –

G·M!i,\114-►
486
Chapter 14

Snell’s Law (continued)
E)
Use the equation for Snell’s law.
SOLVE
ni sin ei = nr sin er er = sinI
1
er= 19.2°
[ ~:
(sin
0)] = sin- ~:~~ (sin 30.0°)] 1
[
1
Practice 1. Find the angle of refraction for a ray of light that enters a bucket of water from air
at an angle of25.0° to the normal. (Hint: Use Figure 1.4.) 2. For an incoming ray of light of vacuum wavelength 589 nm, fill in the unknown values in the following table. (Hint: Use Figure 1.4.)
to (medium)
0;
0r
crown glass
25.0°
?
b. air
?
14.5°
9.80°
c. air
diamond
31.6°
?
from (medium) a. flint glass
3. A ray of light of vacuum wavelength 550 nm traveling in air enters a slab of transparent material. The incoming ray makes an angle of 40.0° with the normal, and the refracted ray makes an angle of 26.0° with the normal. Find the index of refraction of the transparent material. (Assume that the index of refraction of air for light of wavelength 550 nm is 1.00.)
SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Sunlight passes into a raindrop at an angle of22.5° from the normal at one point on the droplet. What is the angle of refraction? 2. For each of the following cases, will light rays be bent toward or away from the normal?
a. n; > nr, where 0; = 20° b. n; < nr, where 0; = 20° c. from air to glass with an angle of incidence of 30° d. from glass to air with an angle of incidence of 30° 3. Find the angle of refraction of a ray of light that enters a diamond from air at an angle of 15.0° to the normal. (Hint: Use Figure 1.4.)
Critical Thinking 4. In which of the following situations will light from a laser be refracted? a. traveling from air into a diamond at an angle of 30° to the normal b. traveling from water into ice along the normal c. upon striking a metal surface d. traveling from air into a glass of iced tea at an angle of 25° to the normal Refraction
487
SECTION 2 Objectives


I ► I ►
Use ray diagrams to find the position of an image produced by a converging or diverging lens, and identify the image as real or virtual. Solve problems using the thin-lens equation. Calculate the magnification of lenses. Describe the positioning of lenses in compound microscopes and refracting telescopes.
Thin Lenses Key Term lens
Types of Lenses When light traveling in air enters a pane of glass, it is bent toward the normal. As the light exits the pane of glass, it is bent again. When the light exits, however, its speed increases as it enters the air, so the light bends away from the normal. Because the amount of refraction is the same regardless of whether light is entering or exiting a medium, the light rays are bent as much on exiting the pane of glass as they were on entering.
Curved surfaces change the direction of light.
lens a transparent object that refracts light rays such that they converge or diverge to create an image
With curved surfaces, the direction of the normal line differs for each spot on the medium. When light passes through a medium that has one or more curved surfaces, the change in the direction of the light rays varies from point to point. This principle is applied in media called lenses. Like mirrors, lenses form images, but lenses do so by refraction instead of reflection. The images formed can be real or virtual, depending on the type of lens and the placement of the object. Recall that real images form when rays oflight actually intersect to form the image. Virtual images form at a point from which light rays appear to come but do not actually come. Real images can be projected onto a screen; virtual images cannot be projected. Lenses are commonly used to form images in optical instruments, such as cameras, telescopes, and microscopes. In fact, transparent tissue in the front of the human eye acts as a lens, converging light toward the light-sensitive retina at the back of the eye. A typical lens consists of a piece of glass or plastic ground so that each of its surfaces is a segment of either a sphere or a plane. Figure 2.1 shows examples of lenses. Notice that the lenses are shaped differently. The lens
Converging and Diverging Lenses When rays of light pass through (a) a converging lens (thicker at the middle), they are bent inward. When they pass through (b) a diverging lens (thicker at the edge), they are bent outward.
:s
488
Chapter 14
that is thicker at the middle than it is at the rim, shown in Figure 2.1 (a), is a converging lens. The lens that is thinner at the middle than it is at the rim, shown in Figure 2.1 {b), is a diverging lens. The light rays show why the names converging and diverging are applied to these lenses.
Focal Points Both (a) converging lenses and (b) diverging lenses have two focal points but only one focal length.
Focal length is the image distance for an infinite object distance. As with mirrors, it is convenient to define a point called the focal point for a lens. Note that light rays from an object far away are nearly parallel. The focal point of a converging lens is the location where the image of an object at an infinite distance from the lens is focused. In Figure 2.2(a) a group of rays parallel to the principal axis passes through a focal point, F, after being bent inward by the lens. Unlike mirrors, lenses have two focal points, one on each side of the lens because light can pass through the lens from either side, as shown in Figure 2.2. The distance from the focal point to the center of the lens is called the focal length, f The focal length is the image distance that corresponds to an infinite object distance. Rays parallel to the principal axis diverge after passing through a diverging lens, as shown in Figure 2.2(b). In this case, the focal point is defined as the point from which the diverged rays appear to originate. Again, the focal length is defined as the distance from the center of the lens to the focal point.
F
F
(a)
(b)
1-f
Ray diagrams of thin-lens systems help identify image height and location. Earlier, we used a set of rays and ray diagrams to predict the images formed by mirrors. A similar approach can be used for lenses. As shown in Figure 2.1, refraction occurs at a boundary between two materials with different indexes of refraction. However, for thin lenses (lenses for which the thickness of the lens is small compared to the radius of curvature of the lens or the distance of the object from the lens), we can represent the front and back boundaries of the lens as a line segment passing through the center of the lens. To draw ray diagrams, we will use a line segment with arrow ends to indicate a converging lens, as in Figure 2.2(a). To show a diverging lens, we will draw a line segment with “upsidedown” arrow ends, as illustrated in Figure 2.2(b). We can then draw ray diagrams using the set of rules outlined in Figure 2.3.
Ray
From object to lens
From converging lens to image
From diverging lens to image
Parallel ray
parallel to principal axis
passes through focal point, F
directed away from focal point, F
Central ray
to the center of the lens
from the center of the lens
from the center of the lens
Focal ray
passes through focal point, F
parallel to principal axis
parallel to principal axis
Refraction
489
Quick LA.B MATERIALS
• magnifying glass • ruler SAFETY
~~
Care should be taken not to focus the sunlight onto a flammable surface or any body parts, such as hands or arms. Also, do not look at the sun through the magnifying glass because serious eye injury can result.
FOCAL LENGTH On a sunny day, hold the magnifying glass, which is a converging lens, above a nonflammable surface, such as a sidewalk, so that a round spot of light is formed on the surface. Move the magnifying glass up and down to find the height at which the spot formed by the lens is most distinct, or smallest. Use the ruler to measure the distance between the magnifying glass and the surface. This distance is the approximate focal length of the lens.
The reasons why these rules work relate to concepts already covered in this textbook. From the definition of a focal point, we know that light traveling parallel to the principal axis (parallel ray) will be focused at the focal point. For a converging lens, this means that light will come together at the focal point in back of the lens. (In this book, the front of the lens is defined as the side of the lens that the light rays first encounter. The back of the lens refers to the side of the lens opposite where the light rays first encounter the lens.) But a similar ray passing through a diverging lens will exit the lens as if it originated from the focal point in front of the lens. Because refraction is reversible, a ray entering a converging lens from either focal point will be refracted so that it is parallel to the principal axis. For both lenses, a ray passing through the center of the lens will continue in a straight line with no net refraction. This occurs because both sides of a lens are parallel to one another along any path through the center of the lens. As with a pane of glass, the exiting ray will be parallel to the ray that entered the lens. For ray diagrams, the usual assumption is that the lens is negligibly thin, so it is assumed that the ray is not displaced sideways but instead continues in a straight line.
Characteristics of Lenses on the next page summarizes the possible relationships between object and image positions for converging lenses. The rules for drawing reference rays were used to create each of these diagrams. Note that applications are listed along with each ray diagram to show the varied uses of the different configurations. Figure 2.4
Converging lenses can produce real or virtual images of real objects. An object infinitely far away from a converging len s will create a point
image at the focal point, as shown in the first diagram in Figure 2.4. This image is real, which means that it can be projected on a screen. As a distant object approaches the focal point, the image becomes larger and farther away, as shown in the second, third, and fourth diagrams in Figure 2.4. When the object is at the focal point, as shown in the fifth diagram, the light rays from the object are refracted so that they exit the lens parallel to each other. (Because the object is at the focal point, it is impossible to draw a third ray that passes through that focal point, the lens, and the tip of the object.) When the object is between a converging lens and its focal point, the light rays from the object diverge when they pass through the lens, as shown in the sixth diagram in Figure 2.4. This image appears to an observer in back of the lens as being on the same side of the lens as the object. In other words, the brain interprets these diverging rays as coming from an object directly along the path of the rays that reach the eye. The ray diagram for this final case is less straightforward than those drawn for the other cases in the table. The first two rays (parallel to the axis and through the center of the lens) are drawn in the usual fashion. The third ray, however, is drawn so that if it were extended, it would connect the focal point in front of the lens, the
490
Chapter 14
Ray diagrams
1.
2.
Front
F
Configuration: object at infinity; point image at F Applications: burning a hole with a magnifying glass
Configuration: object outside 2F; real, smaller image between F and 2F Applications: lens of a camera, human eyeball lens, and objective lens of a refracting telescope
3.
4.
Front
Front
Configuration: object at 2F; real image at 2F same size as object Applications: inverting lens of a field telescope
Back
Configuration: object between F and 2F; magnified real image outside 2F Applications: motion-picture or slide projector and objective lens in a compound microscope
6.
5.
Image
Front
Configuration: object at F; image at infinity Applications: lenses used in lighthouses and searchlights
Back
Front
,.
F
Configuration: object inside F; magnified virtual image on the same side of the lens as the object Applications: magnifying with a magnifying glass; eyepiece lens of microscope, binoculars, and telescope
Refraction
491
tip of the object, and the lens in a straight line. To determine where the image is, draw lines extending from the rays exiting the lens back to the point where they would appear to have originated to an observer on the back side of the lens (these lines are dashed in the sixth diagram in Figure 2.4 ) .
Diverging lenses produce virtual images from real objects. A diverging lens creates a virtual image of a real object placed anywhere with respect to the lens. The image is upright, and the magnification is always less than one; that is, the image size is reduced. Additionally, the image appears inside the focal point for any placement of the real object.
■ijtill;Jtf1 ‘ Did YOU Know?_- – – – – – – – – – – , The lens of a camera forms an inverted image on the film in the back of the camera. Two methods are used to view this image before taking a picture. In one, a system of mirrors and prisms reflects the image to the viewfinder, making the image upright in the process. In the other method, the viewfinder is a diverging lens that is separate from the main lens system. This lens forms an upright virtual image that resembles the image that will be projected onto the film.
Ray Diagram for a Diverging Lens The image created by a diverging lens is always a virtual, smaller image. Smaller, virtual image inside F
Object anywhere
Object
Front
— — —–
F
F Back
The ray diagram shown in Figure 2.5 for diverging lenses was created using the rules given in Figure 2.3. The first ray, parallel to the axis, appears to come from the focal point on the same side of the lens as the object. This ray is indicated by the oblique dashed line. The second ray passes through the center of the lens and is not refracted. The third ray is drawn as if it were going to the focal point in back of the lens. As this ray passes through the lens, it is refracted parallel to the principal axis and must be extended backward, as shown by the dashed line. The location of the tip of the image is the point at which the three rays appear to have originated.
The Thin-Lens Equation and Magnification Ray diagrams for lenses give a good estimate of image size and distance, but it is also possible to calculate these values. The equation that relates object and image distances for a lens is called the thin-lens equation because it is derived using the assumption that the lens is very thin. In other words, this equation applies when the lens thickness is much smaller than its focal length.
492
Chapter 14
Thin-Lens Equation
l+l=l p q f I
I
distance from object to lens
+ distance from image to lens I
focal length
When using the thin-lens equation, we often illustrate it using the ray diagram model in which we magnify the vertical axis and show the lens position as a thin line. Remember that actual light rays bend at the lens surfaces while our diagram shows bending at a single central line in an idealized model, which is quite good for thin lenses. But the model, and the equation, must be modified to deal properly with thick lenses, systems oflenses, and object and image points far from the principal axis. The thin-lens equation can be applied to both converging and diverging lenses if we adhere to a set of sign conventions. Figure 2.6 gives the sign conventions for lenses. Under this convention, an object in front of the lens has a positive object distance and an object in back of the lens, or a virtual object, has a negative object distance. Note that virtual objects only occur in multiple-lens systems. Similarly, an image in back of the lens (that is, a real image) has a positive image distance, and an image in front of the lens, or a virtual image, has a negative image distance. A converging lens has a positive focal length and a diverging lens has a negative focal len gth. Therefore, converging lenses are sometimes called positive lenses and diverging lenses are sometimes called negative lenses.
Magnification by a lens depends on object and image distances. Recall that magnification (M) is defined as the ratio of image height to object height. The following equation can be used to calculate the magnification of both converging and diverging lenses. Magnification of a Lens
h’ M=-= h
. . image height h .gh magmficatmn = b. o 1ect e1 t
+ p
real object in front of the lens
virtual object in back of the lens
q
real image in back of the lens
virtual image in front of the lens
f
converging lens
diverging lens
q
-p distance from image to lens distance from object to lens
If close attention is given to the sign conventions defined in Figure 2.6, then the magnification will describe the image’s size and orientation. When the magnitude of the magnification of an object is less than one, the image is smaller than the object. Conversely, when the magnitude of the magnification is greater than one, the image is larger than the object. Additionally, a negative sign for the magnification indicates that the image is real and inverted. A positive magnification signifies that the image is upright and virtual. Refraction
493
PREMIUM CONTENT
~ Interactive Demo \::,/ HMDScience.com
Lenses Sample Problem B An object is placed 30.0 cm in front of a converging lens and then 12.5 cm in front of a diverging lens. Both lenses have a focal length of 10.0 cm. For both cases, find the image distance and the magnification. Describe the images.
0
ANALYZE
Given:
!converging = 10.0 cm P converging
Unknown:
Diagrams:
= 3o.0cm
q converging — ?· ? q diverging – · p
! diverging =
-l0.0 cm
P diverging = lZ.S cm
M converging — ?· M diverging ? – · p = 12.5cm
= 30.0 cm
Object
F
f=-
E)
PLAN
10.0cm
Choose an equation or situation: The thin-lens equation can be used to find the image distance, and the equation for magnification will serve to describe the size and orientation of the image.
q M =-p
Rearrange the equation to isolate the unknown:
1 q
E)
SOLVE
1
1 p

!
For the converging lens:
1 q
1
1 p

!
1 10.0 cm
1 30.0 cm
2
30.0 cm
lq= 15.0 cm I M = _ !{ = _ 15.0cm P 30.0 cm
IM= – o.sool CS·M!i,\114-► 494
Chapter 14
Lenses (continued) For the diverging lens:
1 q
1 -1 –
J
1
P
-10.0cm
1 12.5 cm
22.5 125cm
lq= -5.56cm l M = _ :!_ = _ -5.56cm
P
12.5 cm
IM=0.445 0
I
These values and signs for the converging lens indicate a real, inverted, smaller image. This is expected because the object distance is longer than twice the focal length of the converging lens. The values and signs for the diverging lens indicate a virtual, upright, smaller image formed inside the focal point. This is the only kind of image diverging lenses form.
CHECKYOUR WORK
Practice 1. An object is placed 20.0 cm in front of a converging lens of focal length 10.0 cm. Find the image distance and the magnification. Describe the image. 2. Sherlock Holmes examines a clue by holding his magnifying glass at arm’s length and 10.0 cm away from an object. The magnifying glass has a focal length of 15.0 cm. Find the image distance and the magnification. Describe the image that he observes.
3. An object is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification. Describe the image.
4. Fill in the missing values in the following table. f
p
q
M
Converging lens
a. 6.0cm
?
-3.0cm
?
b. 2.9cm
?
7.0cm
?
Diverging lens
c. – 6.0cm
4.0cm
?
?
d. ?
5.0cm
?
0.50
Refraction
495
Eyeglasses and Contact Lenses The transparent front of the eye, called the cornea, acts like a lens, directing light rays toward the light-sensitive retina in the back of the eye. Although most of the refraction of light occurs at the cornea, the eye also contains a small lens, called the crystalline lens, that refracts light as well. When the eye attempts to produce a focused image of a nearby object but the image position is behind the retina, the abnormality is known as hyperopia, and the person is said to be farsighted. With this defect, distant objects are seen clearly, but near objects are blurred. Either the hyperopic eye is too short or the ciliary muscle that adjusts the shape of the lens cannot adjust enough to properly focus the image. Figure 2.7 shows how hyperopia can be corrected with a converging lens. Another condition, known as myopia, or nearsightedness, occurs either when the eye is longer than normal or when the maximum focal length of the lens is insufficient to produce a clear image on the retina. In this case, light from a distant object is focused in front of the retina. The distinguishing feature of this imperfection is that distant objects are not seen clearly. Nearsightedness can be corrected with a diverging lens, as shown in Figure 2.7. A contact lens is simply a lens worn directly over the cornea of the eye. The lens floats on a thin layer of tears.
Farsighted
Hyperopia
Corrected with a converging lens
Nearsighted
Myopia
496
Chapter 14
Corrected with a diverging lens
Combination of Thin Lenses If two lenses are used to form an image, the system can be treated in the following manner. First, the image of the first lens is calculated as though the second lens were not present. The light then approaches the second lens as if it had come from the image formed by the first lens. Hence, the image formed by the first lens is treated as the object for the second lens. The image formed by the second lens is the final image of the system. The overall magnification of a system of lenses is the product of the magnifications of the separate lenses. If the image formed by the first lens is in back of the second lens, then the image is treated as a virtual object for the second lens (that is, pis negative). The same procedure can be extended to a system of three or more lenses.
Compound microscopes use two converging lenses. A simple magnifier, such as a magnifying glass, provides only limited assistance when inspecting the minute details of an object. Greater magnification can be achieved by combining two lenses in a device called a compound microscope. It consists of two lenses: an objective lens (near the object) with a focal length ofless than 1 cm and an eyepiece with a focal length of a few centimeters. As shown in Figure 2.8, the object placed just outside the focal point of the objective lens forms a real, inverted, and enlarged image that is at or just inside the focal point of the eyepiece. The eyepiece, which serves as a simple magnifier, uses this enlarged image as its object and produces an even more enlarged virtual image. The image viewed through a microscope is upside down with respect to the actual orientation of the specimen, as shown in Figure 2.8.
QuickLAB MATERIALS
• several pairs of prescription eyeglasses
PRESCRIPTION GLASSES Hold a pair of prescription glasses at various distances from your eye, and look at different objects through the lenses. Try this with different types of glasses, such as those for farsightedness and nearsightedness, and describe what effect the differences have on the image you see. If you have bifocals, how do the images produced by the top and bottom portions of the bifocal lens compare?
The microscope has extended our vision into the previously unknown realm of incredibly small objects. A question that is often asked about microscopes is, “With extreme patience and care, would it be possible to construct a microscope that would enable us to see an atom?” As long as visible light is used to illuminate the object, the answer is no. In order to be seen, the object under a microscope must be at least as large as a wavelength of light. An atom is many times smaller than a wavelength of visible light, so its mysteries must be probed through other techniques.
Compound Microscope
In a compound microscope, the real, inverted image produced by the objective lens is used as the object for the eyepiece lens.
Objective
Eyepiece
Refraction
497
S.T.E.M. CAMERAS ameras come in many types and sizes, from the small and simple camera on your cell phone to the large and complex video camera used to film a Hollywood motion picture. Most cameras have at least one lens, and more complex cameras may have 30 or more lenses and may even contain mirrors and prisms. Up until relatively recently, cameras used film to record an image. The film would undergo a chemical change when exposed to light. Today, however, most cameras are digital and no longer require film. Instead, they use a charged-coupled device (CCD), an array of tiny electronic sensors that can sense light. The CCD lies on the wall opposite the lens and creates an electrical impulse when hit by incoming photons. A microchip in the camera then translates these data into an image that is then stored on a memory storage device like a hard drive. The simplest camera, called a pinhole camera, consists of a closed light-tight box with a small hole, about 0.5 mm, in it. A surprisingly good image can be made with a pinhole camera! The hole bends the light so that it forms an image on the back of the box that is captured by a light-sensing device, either film or a CCD. More sophisticated cameras use lenses. The simplest of these cameras, called a fixed-focus camera, includes a single, converging lens and a shutter, which opens and closes quickly to allow light to pass through the lens and expose the light sensor. Phones and webcams are of this kind. This type of camera usually gives good images only for objects far from the camera, but can’t focus on nearby objects. For this reason, fixed-focus cameras are of limited use. The simplest form of a camera consists of a box with a very small hole in the front. light is projected onto the inside back of the box.
498
Chapter 14
This cross-sectional view of a dSLR camera shows the many optical elements used to form an image on the CCD.
Even more sophisticated cameras, like point-and-shot cameras or digital single-lens-reflex cameras (dSLRfor short) include a series of lenses that can allow the user to zoom in and out and to focus on the object. They are able to focus the image of the object onto the CCD by making slight changes in the distance between the lenses. Zooming also works by moving the lenses in certain ways. The most complex lenses can be found on single-lens reflex (SLR) cameras. Although named single-lens, these cameras in fact have multiple lenses that are interchangeable, meaning that one can be removed and replaced with another. Some lenses are fixed, in that they don’t zoom. An example of such a lens is a normal lens that provides about the same field of view as a human eye. On the other hand, a wide-angle lens has a very short focal length and can capture a larger field of view than a normal lens. A telephoto lens has a long focal length and increases magnification. Telephoto lenses have a narrow angle of view. Sometimes, however, a photographer wants to photograph distant objects with more detail or capture a larger object without taking multiple shots. Zoom lenses allow the photographer to change the focal length without changing lenses. As you might imagine, zoom lenses require multiple lenses and are therefore bulkier and heavier than fixed lenses. High-quality cameras contain quite a few lenses, both converging and diverging, to minimize the distortions and aberrations, or imperfect focusing of light rays, that are created by a single converging lens. The most prevalent aberration occurs because lenses bend light of different colors by different amounts, causing, in effect, rainbows to appear in the image. Therefore the quality of the final image depends not only on the type of material used to manufacture the lens, but also in the design of lenses that reduce these aberrations.
Refracting telescopes also use two converging lenses. As mentioned in the chapter on light and reflection, there are two types of telescopes, reflecting and refracting. In a refracting telescope, an image is formed at the eye in much the same manner as is done with a microscope. A small, inverted image is formed at the focal point of the objective lens, F01 because the object is essentially at infinity. The eyepiece is positioned so that its focal point lies very close to the focal point of the objective lens, where the image is formed, as shown in Figure 2.9. Because the image is now just inside the focal point of the eyepiece, Fe, the eyepiece acts like a simple magnifier and allows the viewer to examine the object in detail.
Refracting Telescope The image
Objective
Eyepiece
produced by the objective lens of a refracting telescope is a real, inverted image that is at its focal point. This inverted image, in turn, is the object from which the eyepiece creates a magnified, virtual image.
t>
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What type of image is produced by the cornea and the lens on the retina? 2. What type of image, virtual or real, is produced in the following cases? a. an object inside the focal point of a camera lens b. an object outside the focal point of a refracting telescope’s objective lens c. an object outside the focal point of a camera’s viewfinder
3. Find the image position for an object placed 3.0 cm outside the focal point of a converging lens with a 4.0 cm focal length. 4. What is the magnification of the object from item 3?
Interpreting Graphics 5. Using a ray diagram, find the position and height of an image produced by a viewfinder in a camera with a focal length of 5.0 cm if the object is 1.0 cm tall and 10.0 cm in front of the lens. A camera viewfinder is a diverging lens.
Critical Thinking 6. Compare the length of a refracting telescope with the sum of the focal lengths of its two lenses. Refraction
499
SECTION 3 Objectives ►

I

Predict whether light will be refracted or undergo total internal reflection. Recognize atmospheric conditions that cause refraction. Explain dispersion and phenomena such as rainbows in terms of the relationship between the index of refraction and the wavelength.
Optical Phenomena Key Terms total internal reflection critical angle
dispersion chromatic aberration
Total Internal Reflection An interesting effect called total internal reflection can occur when light moves along a path from a medium with a higher index of refraction to one with a lower index of refraction. Consider light rays traveling from
water into air, as shown in Figure 3.1 (a). Four possible directions of the rays are shown in the figure. total internal reflection the complete reflection that takes place within a substance when the angle of incidence of light striking the surface boundary is greater than t he critical angle critical angle the angle of incidence at which the refracted light makes an angle of 90° with the normal
At some particular angle of incidence, called the critical angle, the refracted ray moves parallel to the boundary, making the angle of refraction equal to 90°, as shown in Figure 3.1(b). For angles of incidence greater than the critical angle, the ray is entirely reflected at the boundary, as shown in Figure 3.1. This ray is reflected at the boundary as though it had struck a perfectly reflecting surface. Its path and the path of all rays like it can be predicted by the law of reflection; that is, the angle of incidence equals the angle of reflection. In optical equipment, prisms are arranged so that light entering the prism is totally internally reflected off the back surface of the prism. Prisms are used in place of silvered or aluminized mirrors because they reflect light more efficiently and are more scratch resistant. Snell’s law can be used to find the critical angle. As mentioned above, when the angle of incidence, 0i, equals the critical angle, 0c, then the angle of refraction, 0,, equals 90°. Substituting these values into Snell’s law gives the following relation.
tm.4•1 ;J
••r
Internal Reflection (a) This photo demonstrates several different paths of light radiated from the bottom of an aquarium.
(b) At the critical angle, 0C’ a light ray will travel parallel to the boundary. Any rays with an angle of incidence greater than 0c will be totally internally reflected at the boundary. Normal I
Air
Water
500
Chapter 14
Because the sine of 90° equals 1, the following relationship results.
QuickLAB Critical Angle
. n, sm 0c= n.
MATERIALS
• two 90° prisms
l
) _ . ( •t· al index of refraction of second medium sme cri 1c ang1e .1nd ex o f refraction . o f fiirst med.1um
PERISCOPE Align the two prisms side by side as shown below.
but only if index of refraction of first medium > index of refraction of second medium
Note that this equation can be used only when ni is greater than n,. In other words, total internal reflection occurs only when light moves along a path from a medium of higher index of refraction to a medium oflower index of refraction. Ifni were less than n,, this equation would give sin 0c > 1, which is an impossible result because by definition the sine of an angle can never be greater than 1.
Note that this configuration can be used like a periscope to see an object above your line of sight if the configuration is oriented vertically and to see around a corner if it is oriented horizontally. How would you arrange the prisms to see behind you? Draw your design on paper and test it.
When the second substance is air, the critical angle is small for substances with large indices of refraction. Diamonds, which have an index ofrefraction of2.419, have a critical angle of24.4°. By comparison, the critical angle for crown glass, a very clear optical glass, where n = 1.52, is 41.0°. Because diamonds have such a small critical angle, most of the light that enters a cut diamond is totally internally reflected. The reflected light eventually exits the diamond from the most visible faces of the diamond. Jewelers cut diamonds so that the maximum light entering the upper surface is reflected back to these faces.
PREMIUM CONTENT
Critical Angle
~
Interactive Demo
\:;I
HMDScience.com
Sample Problem C Find the critical angle for a water-air boundary if the index ofrefraction of water is 1.333.
0
ANALYZE
Given: Unknown:
E)
SOLVE
ni = 1.333
n, = 1.000
0C =?•
Use the equation for critical angle on this page.
. n, sm 0c=n. l
. 0c -_ sm
1
(n’)- sm. ni –
– 1(
Tips and Tricks Remember that the critical angle equation is valid only if the light is moving from a higher to a lower index of refraction.
1.00) 1.333
,a., ,iii ,MA·► Refraction
501
Critical Angle (continued) Practice 1. Glycerine is used to make soap and other personal care products. Find the critical angle for light traveling from glycerine (n = 1.473) into air.
2. Calculate the critical angle for light traveling from glycerine (n = 1.473) into water (n = 1.333). 3. Ice has a lower index of refraction than water. Find the critical angle for light traveling from ice (n = 1.309) into air.
4. Which has a smaller critical angle in air, diamond (n = 2.419) or cubic zirconia (n = 2.20)? Show your work.
ST.E.M.
Fiber Optics nother interesting application of total internal reflection is the use of glass or transparent plastic rods, like the ones shown in the photograph, to transfer light from one place to another. As indicated in the illustration, light is confined to traveling within the rods, even around gentle curves, as a result of successive internal reflections. Such a light pipe can be flexible if thin fibers rather than thick rods are used. If a bundle of parallel fibers is used to construct an optical transmission line, images can be transferred from one point to another. This technique is used in a technology known as fiber optics. Very little light intensity is lost in these fibers as a result of reflections on the sides. Any loss of intensity is due essentially to reflections from the two ends and absorption by the fiber material. Fiber-optic devices are particularly useful for viewing images produced at inaccessible locations. For example, a fiber-optic cable can be threaded through the esophagus and into the stomach to look for ulcers. Fiber-optic cables are used in telecommunications because the fibers can carry much higher volumes of telephone calls and computer signals than can electrical wires.
502
Chapter 14
Light is guided along a fiber by multiple internal reflections.
Atmospheric Refraction We see an example of refraction every day: the sun can be seen even after it has passed below the horizon. Rays of light from the sun strike Earth’s atmosphere and are bent because the atmosphere has an index of refraction different from that of the near-vacuum atmosphere of space. The bending in this situation is gradual and continuous because the light moves through layers of air that have a continuously changing index of refraction. Our eyes follow them back along the direction from which they appear to have come. This effect is pictured in Figure 3.2 in the observation of a star.
Atmospheric Refraction The atmosphere of the Earth bends the light of a star and causes the viewer to see the star in a slightly different location.
Refracted light produces mirages. The mirage is another phenomenon of nature produced by refraction in the atmosphere. A mirage can be observed when the ground is so hot that the air directly above it is warmer than the air at higher elevations. These layers of air at different heights above Earth have different densities and different refractive indices. The effect this can have is pictured in Figure 3.3. In this situation, the observer sees a tree in two different ways. One group of light rays reaches the observer by the straight-line path A, and the eye traces these rays back to see the tree in the normal fashion. A second group of rays travels along the curved path B. These rays are directed toward the ground and are then bent as a result of refraction. Consequently, the observer also sees an inverted image of the tree by tracing these rays back to the point at which they appear to have originated. Because both an upright image and an inverted image are seen when the image of a tree is observed in a reflecting pool of water, the observer subconsciously calls upon this past experience and concludes that a pool of water must be in front of the tree.
Mirage A mirage is produced by the bending of light rays in the atmosphere when there are large temperature differences between the ground and the air.
f>
Dispersion An important property of the index of refraction is that its value in
anything but a vacuum depends on the wavelength oflight. Because the index of refraction is a function of wavelength, Snell’s law indicates that incoming light of different wavelengths is bent at different angles as it moves into a refracting material. This phenomenon is called dispersion. As mentioned in Section 1, the index of refraction decreases with increasing wavelength. For instance, blue light(,\ ::::: 470 nm) bends more than red light(,\ ::::: 650 nm) when passing into refracting material.
White light passed through a prism produces a visible spectrum. To understand how dispersion can affect light, consider what happens when light strikes a prism, as in Figure 3.4. Because of dispersion, the blue component of the incoming ray is bent more than the red component, and the rays that emerge from the second face of the prism fan out in a series of colors known as a visible spectrum. These colors, in order of decreasing wavelength, are red, orange, yellow, green, blue, and violet.
dispersion the process of separating polychromatic light into its component wavelengths
Dispersion When white light enters a prism, the blue light is bent more than the red, and the prism disperses the white light into its various spectral components.
White light
Red
~—-~
Refraction
Blue
503
Rainbows and Raindrops Rainbows (a) are produced because of dispersion of light in raindrops. Sunlight is spread into a spectrum upon entering a spherical raindrop (b), then internally
reflected on the back side of the raindrop. The perceived color of each water droplet then depends on the angle at which that drop is viewed.
Sunlight
(b)
(a)
Rainbows are created by dispersion of light in water droplets. The dispersion of light into a spectrum is demonstrated most vividly in nature by a rainbow, often seen by an observer positioned between the sun and a rain shower. When a ray of sunlight strikes a drop of water in the atmosphere, it is first refracted at the front surface of the drop, with the violet light refracting the most and the red light the least. Then, at the back surface of the drop, the light is reflected and returns to the front surface, where it again undergoes refraction as it moves from water into air. The rays leave the drop so that the angle between the incident white light and the returning violet ray is 40° and the angle between the white light and the returning red ray is 42°, as shown in Figure 3.5(b). Now, consider Figure 3.5(a). When an observer views a raindrop high in the sky, the red light reaches the observer, but the violet light, like the other spectral colors, passes over the observer because it deviates from the path of the white light more than the red light does. Hence, the observer sees this drop as being red. Similarly, a drop lower in the sky would direct violet light toward the observer and appear to be violet. (The red light from this drop would strike the ground and not be seen.) The remaining colors of the spectrum would reach the observer from raindrops lying between these two extreme positions. Note that rainbows are most commonly seen above the horizon, where the ends of the rainbow disappear into the ground. However, if an observer is at an elevated vantage point, such as on an airplane or at the rim of a canyon, a complete circular rainbow can be seen . 504
Chapter 14
Lens Aberrations One of the basic probl ems of lenses and lens systems is the imperfect quality of the images. The simple theory of mirrors and lenses assumes that rays make small angles with the principal axis and that all rays reaching the lens or mirror from a point source are focused at a single point, producing a sharp image. Clearly, this is not always true in the real world. Where the approximations used in this theory do not hold, imperfect images are formed.
Chromatic Aberration Because of dispersion, white light passing through a converging lens is focused at different focal points for each wavelength of light. (The angles in this figure are exaggerated for clarity.)
As with spherical mirrors, spherical aberration occurs for lenses also. It results from the fact that the focal points of light rays far from the principal axis of a spherical lens are different from the focal points of rays with the same wavelength passing near the axis. Rays near the middle of the lens are focused farther from the lens than rays at the edges. Another type of aberration, called chromatic aberration, arises from the wavelength dependence of refraction. Because the index of refraction of a ‘ material varies with wavelength, different wavelengths of light are focused at different focal points by a lens. For example, when white light passes through a lens, violet light is refracted more than red light, as shown in Figure 3.6; thus, the focal length for red light is greater than that for violet light. Other colors’ wavelengths have intermediate focal points. Because a diverging lens has the opposite shape, the chromatic aberration for a diverging lens is opposite that for a converging lens. Chromatic aberration can be greatly reduced by the use of a combination of converging and diverging lenses made from two different types of glass.
chromatic aberration the focusing of different colors of light at different distances behind a lens
SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Find the critical angle for light traveling from water (n = 1.333) into ice (n = 1.309).
2. Which of the following describe places where a mirage is likely to appear? a. above a warm lake on a warm day b. above an asphalt road on a hot day c. above a ski slope on a cold day d. above the sand on a beach on a hot day e. above a black car on a sunny day 3. When white light passes through a prism, which will be bent more, the red or green light?
Critical Thinking 4. After a storm, a man walks out onto his porch. Looking to the east, he sees a rainbow that has formed above his neighbor’s house. What time of day is it, morning or evening?
Refraction
505
Optometrist he job of an optometrist is to correct imperfect vision using optical devices such as eyeglasses or contact lenses. Optometrists also treat diseases of the eye such as glaucoma. To learn more about optometry as a career, read the interview with Dewey Handy, O.D.
Li
How did you decide to become an optometrist?
For a while, I didn’t know what career I was going to choose. In high school, I had a great love for geometry and an interest in science and anatomy. In college, I was looking for a challenge, so I ended up majoring in physics-almost by accident. In college, I decided to apply my abilities in science to directly help people. I wasn’t excited about dentistry or general medicine, but I was looking for something in a health career that would allow me to use physics. What education is required to become an optometrist?
I have a bachelor of science in physics, and I attended optometry school for four years. What sort of work does an optometrist do?
After taking a complete eye and medical history, the doctor may use prisms and/or lenses to determine the proper prescription for the patient. Then, a series of neurological, health, and binocular vision tests are done. After the history and data have been collected, a diagnosis and treatment plan are developed. This treatment may include glasses, contact lenses, low-vision aids, vision training, or medication for treatment of eye disease. What do you enjoy most about your job?
I like the problem-solving nature of the work, putting the data together to come up with solutions. We read the problem, compile data, develop a formula, and solve the problem-just as in physics, but with people instead of abstract problems. I also like helping people.
Dr. Dewey Handy uses optical devices to test the vision of a patient.
What advice do you have for students who are interested in optometry?
You definitely need to have a good background in basic science: chemistry, biology, and physics. Even if you don’t major in science, you need to have a good grasp of it by the time you get to optometry school. Being well rounded will help you get into optometry school-and get out, too. You have to be comfortable doing the science; you also have to be comfortable dealing with people.
SECTION 1
Refraction
,: ,
1 ,
1
r: .
refraction index of refraction
• According to Snell’s law, as a light ray travels from one medium into another medium where its speed is different, the light ray will change its direction unless it travels along the normal. • When light passes from a medium with a smaller index of refraction to one with a larger index of refraction, the ray bends toward the normal. For the opposite situation, the ray bends away from the normal.
SECTION 2
Thin Lenses
I
• The image produced by a converging lens is real and inverted when the object is outside the focal point and virtual and upright when the object is inside the focal point. Diverging lenses always produce upright, virtual images.
c, Tu ,,-
lens
• The location of an image created by a lens can be found using either a ray diagram or the thin-lens equation.
SECTION 3
Optical Phenomena
1,, •
• Total internal reflection can occur when light attempts to move from a material with a higher index of refraction to one with a lower index of refraction. If the angle of incidence of a ray is greater than the critical angle, the ray is totally reflected at the boundary. • Mirages and the visibility of the sun after it has physically set are natural phenomena that can be attributed to refraction of light in Earth’s atmosphere.
Quantities
degrees
0( angle of refraction
degrees
n
index of refraction
(unitless)
p
distance from object to lens
m meters
q
distance from image to lens
m meters —– —-
h
m meters
image height object height
m meters ————
0c critical angle
L-, ,,. –
total internal reflection critical angle dispersion chromatic aberration
Units
0; angle of incidence
h’
1
0
degrees
Problem Solving See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
Chapter Summary
507
Retraction and Snell’s Law REVIEWING MAIN IDEAS 1. Does a light ray traveling from one medium into another always bend toward the normal?
2. As light travels from a vacuum (n = 1) to a medium such as glass (n > 1), does its wavelength change? Does its speed change? Does its frequency change? 3. What is the relationship between the speed oflight and the index of refraction of a transparent substance? 4. Why does a clear stream always appear to be shallower than it actually is?
5. What are the three conditions that must be met for refraction to occur?
11. A ray of light enters the top of a glass of water at an
angle of 36° with the vertical. What is the angle between the refracted ray and the vertical? 12. A narrow ray of yellow light from glowing sodium
(>. 0 = 589 nm) traveling in air strikes a smooth surface of water at an angle of 0i = 35.0°. Determine the angle ofrefraction, 0,. 13. A ray of light traveling in air strikes a flat 2.00 cm thick
block of glass (n = 1.50) at an angle of 30.0° with the normal. Trace the light ray through the glass, and find the angles of incidence and refraction at each surface. 14. The light ray shown in the figure below makes an angle of20.0° with the normal line at the boundary of linseed oil and water. Determine the angles 01 and 02 . Note that n = 1.48 for linseed oil. Air
CONCEPTUAL QUESTIONS
Linseed oil
6. Two colors of light (X and Y) are sent through a glass prism, and Xis bent more than Y. Which color travels
more slowly in the prism?
Water
7. Why does an oar appear to be bent when part of it is
in the water? 8. A friend throws a coin into a pool. You close your eyes
and dive toward the spot where you saw it from the edge of the pool. When you reach the bottom, will the coin be in front of you or behind you? 9. The level of water (n = 1.33) in a clear glass container is easily observed with the naked eye. The level of liquid helium (n = 1.03) in a clear glass container is extremely difficult to see with the naked eye. Explain why.
PRACTICE PROBLEMS For problems 10-14, see Sample Problem A.
10. Light passes from air into water at an angle of incidence of 42.3°. Determine the angle of refraction in the water.
508
Chapter 14
Rav Diagrams and Thin Lenses REVIEWING MAIN IDEAS 15. Which type oflens can focus the sun’s rays? 16. Why is no image formed when an object is at the focal
point of a converging lens?
17. Consider the image formed by a thin converging lens.
Under what conditions will the image be a. inverted? b. upright? c. real? d. virtual? e. larger than the object? f. smaller than the object? 18. Repeat a-f of item 17 for a thin diverging lens. 19. Explain this statement: The focal point of a
converging lens is the location of an image of a point object at infinity. Based on this statement, can you think of a quick method for determining the focal length of a positive lens?
26. An object is placed in front of a converging lens with a focal length of20.0 cm. For each object distance, find the image distance and the magnification. Describe each image. a. 40.0 cm b. 10.0 cm
Total Internal Reflection, Atmospheric Refraction, and Aberrations REVIEWING MAIN IDEAS 27. Is it possible to have total internal reflection for light incident from air on water? Explain.
CONCEPTUAL QUESTIONS 28. What are the conditions necessary for the occurrence 20. If a glass converging lens is submerged in water, will
its focal length be longer or shorter than when the lens is in air?
ofa mirage? 29. On a hot day, what is it that we are seeing when we
observe a “water on the road” mirage? 21. In order to get an upright image, slides must be
placed upside down in a slide projector. What type of lens must the slide projector have? Is the slide inside or outside the focal point of the lens? 22. If there are two converging lenses in a compound microscope, why is the image still inverted?
23. In a Jules Verne novel, a piece of ice is shaped into the form of a magnifying lens to focus sunlight and thereby start a fire. Is this possible?
PRACTICE PROBLEMS For problems 24-26, see Sample Problem B.
24. An object is placed in front of a diverging lens with a focal length of20.0 cm. For each object distance, find the image distance and the magnification. Describe each image. a. 40.0cm b. 20.0cm c. 10.0cm
30. Why does the arc of a rainbow appear with red colors on top and violet colors on the bottom? 31. What type of aberration is involved in each of the
following situations? a. The edges of the image appear reddish. b. The central portion of the image cannot be clearly focused. c. The outer portion of the image cannot be clearly focused. d. The central portion of the image is enlarged relative to the outer portions.
CONCEPTUAL QUESTIONS 32. A laser beam passing through a nonhomogeneous sugar solution follows a curved path. Explain. 33. On a warm day, the image of a boat floating on cold water appears above the boat. Explain. 34. Explain why a mirror cannot give rise to chromatic
aberration.
25. A person looks at a gem using a converging lens with a focal length of 12.5 cm. The lens forms a virtual image 30.0 cm from the lens. Determine the magnification. Is the image upright or inverted?
35. Why does a diamond show flashes of color when
observed under ordinary white light?
Chapter Review
509
PRACTICE PROBLEMS For problems 36-38, see Sample Problem C. 36. Calculate the critical angle for light going from glycerine into air. 37. Assuming that>. = 589 nm, calculate the critical angles for the following materials when they are surrounded by air: a. zircon b. fluorite c. ice
38. Light traveling in air enters the flat side of a prism made of crown glass (n = 1.52), as shown at right. Will the light pass through the other side of the prism or will it be totally internally reflected? Be sure to show your work.
4-5•
– ~
Mixed Review REVIEWING MAIN IDEAS 39. The angle of incidence and the angle of refraction for light going from air into a material with a higher index of refraction are 63.5° and 42.9°, respectively. What is the index of refraction of this material? 40. A person shines a light at a friend who is swimming underwater. If the ray in the water makes an angle of 36.2° with the normal, what is the angle of incidence? 41. What is the index of refraction of a material in which the speed oflight is 1.85 x 108 m / s? Look at the indices of refraction in Figure 1.4 to identify this material.
42. Light moves from flint glass into water at an angle of incidence of 28. 7°. a. What is the angle of refraction? b. At what angle would the light have to be incident to give an angle ofrefraction of90.0°? 43. A magnifying glass has a converging lens of focal length 15.0 cm. At what distance from a nickel should you hold this lens to get an image with a magnification of +2.00?
510
Chapter 14
44. The image of the United States postage stamps in the figure above is 1.50 times the size of the actual stamps in front of the lens. Determine the focal length of the lens if the distance from the lens to the stamps is 2.84 cm. 45. Where must an object be placed to have a magnification of 2.00 in each of the following cases? Show your work. a. a converging lens of focal length 12.0 cm b. a diverging lens of focal length 12.0 cm 46. A diverging lens is used to form a virtual image of an object. The object is 80.0 cm in front of the lens, and the image is 40.0 cm in front of the lens. Determine the focal length of the lens. 47. A microscope slide is placed in front of a converging lens with a focal length of 2.44 cm. The lens forms an image of the slide 12.9 cm from the slide. a. How far is the lens from the slide if the image is real? b. How far is the lens from the slide if the image is virtual? 48. Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 40.0 cm? Determine the magnification of the image. 49. The index of refraction for red light in water is 1.331, and that for blue light is 1.340. If a ray of white light traveling in air enters the water at an angle of incidence of 83.0°, what are the angles of refraction for the red and blue components of the light?
50. A ray of light traveling in air strikes the surface of mineral oil at an angle of 23.1 ° with the normal to the surface. If the light travels at 2.1 7 x 108 m/ s through the oil, what is the angle of refraction? (Hint: Remember the definition of the index of refraction.) 51. A ray of light traveling in air strikes the surface of a liquid. If the angle of incidence is 30.0° and the angle of refraction is 22.0°1 find the critical angle for light traveling from the liquid back into the air.
52. The laws of refraction and reflection are the same for sound and for light. The speed of sound is 340 m/s in air and 1510 m/ s in water. If a sound wave that is traveling in air approaches a flat water surface with an angle of incidence of 12.0°, what is the angle of refraction?
56. A fiber-optic cable used for telecommunications
has an index of refraction of 1.53. For total internal reflection oflight inside the cable, what is the minimum angle of incidence to the inside wall of the cable if the cable is in the following: a. air b. water 57. A ray of light traveling in air strikes the midpoint of one face of an equiangular glass prism (n = 1.50) at an angle of exactly 30.0°1 as shown below. a. Trace the path of the light ray through the glass, and find the angle of incidence of the ray at the bottom of the prism. b. Will the ray pass through the bottom surface of the prism, or will it be totally internally reflected?
53. A jewel thief decides to hide a stolen diamond by placing it at the bottom of a crystal-clear fountain. He places a circular piece of wood on the surface of the water and anchors it directly above the diamond at the bottom of the fountain, as shown below. If the fountain is 2.00 m deep, find the minimum diameter of the piece of wood that would prevent the diamond from being seen from outside the water.
1-Diameter-j
I 1
58. Light strikes the surface of a prism, n = 1.8, as shown in the figure below. If the prism is surrounded by a fluid, what is the maximum index of refraction of the fluid that will still cause total internal reflection within the prism?
2.00m
54. A ray of light traveling in air strikes the surface of a block of clear ice at an angle of 40.0° with the normal. Part of the light is reflected, and part is refracted. Find the angle between the reflected and refracted light. 55. An object’s distance from a converging lens is
10 times the focal length. How far is the image from the lens? Express the answer as a fraction of the focal length.
59. A fiber-optic rod consists of a central strand of
material surrounded by an outer coating. The interior portion of the rod has an index of refraction of 1.60. If all rays striking the interior walls of the rod with incident angles greater than 59.5° are subject to total internal reflection, what is the index of refraction of the coating?
Chapter Review
511
60. A flashlight on the bottom of a 4.00 m deep swimming pool sends a ray upward and at an angle so that the ray strikes the surface of the water 2.00 m from the point directly above the flashlight. What angle (in air) does the emerging ray make with the water’s surface? (Hint: To determine the angle of incidence, consider the right triangle formed by the light ray, the pool bottom, and the imaginary line straight down from where the ray strikes the surface of the water.) 61. A submarine is 325 m horizontally out from the shore and 115 m beneath the surface of the water. A laser beam is sent from the submarine so that it strikes the surface of the water at a point 205 m from the shore. If the beam strikes the top of a building standing directly at the water’s edge, find the height of the building. (Hint: To determine the angle of incidence, consider the right triangle formed by the light beam, the horizontal line drawn at the depth of the submarine, and the imaginary line straight down from where the beam strikes the surface of the water.) 62. A laser beam traveling in air strikes the midpoint of one end of a slab of material, as shown in the figure in the next column. The index of refraction of the slab is 1.48. Determine the number of internal reflections of the laser beam before it finally emerges from the opposite end of the slab.
~ – ~Q_.Qo
42.0cm n = 1.48
T 3.1mm J_
63. A nature photographer is using a camera that has a lens with a focal length of 4.80 cm. The photographer is taking pictures of ancient trees in a forest and wants the lens to be focused on a very old tree that is 10.0maway. a. How far must the lens be from the film in order for the resulting picture to be clearly focused? b. How much would the lens have to be moved to take a picture of another tree that is only 1. 75 m away? 64. The distance from the front to the back of your eye is approximately 1.90 cm. If you can see a clear image of a book when it is 35.0 cm from your eye, what is the focal length of the lens/cornea system? 65. Suppose you look out the window and see your friend, who is standing 15.0 m away. To what focal length must your eye muscles adjust the lens of your eye so that you may see your friend clearly? Remember that the distance from the front to the back of your eye is about 1.90 cm.
Snell’s Law What happens to a light ray that passes from air into a medium whose index of refraction differs from that of air? Snell’s law, as you learned earlier in this chapter, describes the relationship between the angle of refraction and the index of refraction.
In this equation, n; is the index of refraction of the medium of the incident light ray, and 0; is the angle of incidence; nr is the index of refraction of the medium of the refracted light, and 0r is the angle of refraction. 512
Chapter 14
In this graphing calculator activity, you will enter the angle of incidence and will view a graph of the index of refraction versus the angle of refraction. You can use this graph to better understand the relationship between the index of refraction and the angle of refraction. Go online to HMDScience.com to find this graphing calculator activity.
ALTERNATIVE ASSESSMENT 1. Interview an optometrist, optician, or ophthalmologist. Find out what equipment and tools each uses. What kinds of eye problems is each able to correct? What training is n ecessary for each career? 2. Obtain permission to use a microscope and slides from your school’s biology teacher. Identify the optical components (lenses, mirror, object, and light source) and knobs. Find out how they function at different magnifications and what adjustments must be made to obtain a clear image. Sketch a ray diagram for the microscope’s image formation. Estimate the size of the images you see, and calculate the approximate size of the actual cells or microorganisms you observe. How closely do your estimates match the magnification indicated on the microscope? 3. Construct your own telescope with mailing tubes (one small enough to slide inside the other), two lenses, cardboard disks for mounting the lenses, glue, and masking tape. Test your instrument at night. Try to combine different lenses and explore ways to improve your telescope’s performance. Keep records of your results to make a brochure documenting the development of your telescope.
4. Study the history of the camera. Possible topics include the following: How did the camera obscura work? What discovery made the first permanent photograph possible? How do instant cameras work? How do modern digital cameras differ from film cameras? Give a short presentation to the class to share the information. 5. Create a pinhole camera with simple household materials. Find instructions for constructing a pinhole camera on the Internet, and follow them to make your own pinhole camera. Partner with a photography student to develop the pictures in your school’s darkroom. Create a visual presentation to share your photographs with the class.
6. Research how phone, television, and radio signals are transmitted over long distances through fiberoptic devices. Obtain information from companies that provide telephone or cable television service. What materials are fiber-optic cables made of? What are their most important properties? Are there limits on the kind of light that travels in these cables? What are the advantages of fiber-optic technology over broadcast transmission? Produce a brochure or informational video to explain this technology to consumers.
7. When the Indian physicist Venkata Raman first saw the Mediterranean Sea, he proposed that its blue color was due to the structure of water molecules rather than to the scattering of light from suspended particles. Later, he won the Nobel Prize for work relating to the implications of this hypothesis. Research Raman’s life and work. Find out about his background and the challenges and opportunities he met on his way to becoming a physicist. Create a presentation about him in the form of a report, poster, short video, or computer presentation. 8. Choose a radio telescope to research. Possibilities include the Very Large Array in New Mexico, the Arecibo telescope in Puerto Rico, or the Green Bank Telescope in West Virginia. Use the Internet to learn about observations that have been made with the telescope. How long has the telescope been operating? How large is the telescope? What discoveries have been made with it? Has the telescope been used for any SETI (Search for Extra-Terrestrial Intelligence) investigations? After your research is complete, write a list of questions that you still have about the telescope. If possible, call the observatory and interview a member of the staff. Write a magazine article with the results of your research.
Chapter Review
513
MULTIPLE CHOICE 1. How is light affected by an increase in the index of refraction? A. Its frequency increases. B. Its frequency decreases. C. Its speed increases. D. Its speed decreases. 2. Which of the following conditions is not necessary for refraction to occur? F. Both the incident and refracting substances must be transparent. G. Both substances must have different indices of refraction. H. The light must have only one wavelength. J. The light must enter at an angle greater than 0° with respect to the normal. Use the ray diagram below to answer questions 3-4.
p = 50.0cm q = -10.0 cm
3. What is the focal length of the lens? A. – 12.5 cm B. -8.33 cm C. 8.33 cm D. 12.5 cm
4. What is true of the image formed by the lens? F. real, inverted, and enlarged G. real, inverted, and diminished H. virtual, upright, and enlarged J. virtual, upright, and diminished
514
Chapter 14
5. A block of flint glass with an index of refraction of 1.66 is immersed in oil with an index of refraction of 1.33. How does the critical angle for a refracted light ray in the glass vary from when the glass is surrounded by air? A. It remains unchanged. B. It increases. C. It decreases. D. No total internal reflection takes place when the glass is placed in the oil. 6. Which color of light is most refracted during dispersion by a prism? F. red G. yellow H. green J. violet
7. If an object in air is viewed from beneath the surface of water below, where does the object appear to be? A. The object appears above its true position. B. The object appears exactly at its true position. C. The object appears below its true position. D. The object cannot be viewed from beneath the water’s surface. 8. The phenomenon called “looming” is similar to a mirage, except that the inverted image appears above the object instead of below it. What must be true if looming is to occur? F. The temperature of the air must increase with distance above the surface. G. The temperature of the air must decrease with distance above the surface. H. The mass of the air must increase with distance above the surface. J. The mass of the air must increase with distance above the surface.
. TEST PREP
9. Light with a vacuum wavelength of 500.0 nm passes into benzene, which has an index of refraction of 1.5. What is the wavelength of the light within the benzene? A. 0.0013nm B. 0.0030nm C. 330nm D. 750nm
EXTENDED RESPONSE 14. Explain how light passing through raindrops is reflected and dispersed so that a rainbow is produced. Include in your explanation why the lower band of the rainbow is violet and the outer band is red. Use the ray diagram below to answer questions 15-18.
10. Which of the following is not a necessary condition for seeing a magnified image with a lens? F. The object and image are on the same side of the lens. G. The lens must be converging. H. The observer must be placed within the focal length of the lens. J. The object must be placed within the focal length of the lens.
SHORT RESPONSE 11. In telescopes, at least two converging lenses are used: one for the objective and one for the eyepiece. These lenses must be positioned in such a way that the final image is virtual and very much enlarged. In terms of the focal points of the two lenses, how must the lenses be positioned? 12. A beam of light passes from the fused quartz of a bottle (n = 1.46) into the ethyl alcohol (n = 1.36) that is contained inside the bottle. If the beam of the light inside the quartz makes an angle of 25.0° with respect to the normal of both substances, at what angle to the normal will the light enter the alcohol? 13. A layer of glycerine (n = 1.47) covers a zircon slab (n = 1.92). At what angle to the normal must a beam of light pass through the zircon toward the glycerine so that the light undergoes total internal reflection?
p
A collector wishes to observe a coin in detail and so places it 5.00 cm in front of a converging lens. An image forms 7.50 cm in front of the lens, as shown in the figure below. 15. What is the focal length of the lens? 16. What is the magnification of the coin’s image?
17. If the coin has a diameter of2.8 cm, what is the diameter of the coin’s image? 18. Is the coin’s image virtual or real? Upright or inverted?
Test Tip When calculating the value of an angle by taking the arcsine of a quantity, recall that the quantity must be positive and no greater than 1.
Standards-Based Assessment
515
SECTION 1 Objectives ►

I

Describe how light waves interfere with each other to produce bright and dark fringes.
lnterlerence Key Terms
Identify the conditions required for interference to occur.
coherence
path difference
Predict the location of interference fringes using the equation for double-slit interference.
Combining Light Waves
order number
You have probably noticed the bands of color that form on the surface of a soap bubble, as shown in Figure 1.1. Unlike the colors that appear when light passes through a refracting substance, these colors are the result of light waves combining with each other.
Interference takes place only between waves with the same wavelength. Interference on a Soap Bubble Light waves interfere to form bands of color on a soap bubble’s surface.
To understand how light waves combine with each other, let us review how other kinds of waves combine. If two waves with identical wavelengths interact, they combine to form a resultant wave. This resultant wave has the same wavelength as the component waves, but according to the superposition principle, its displacement at any instant equals the sum of the displacements of the component waves. The resultant wave is the consequence of the interference between the two waves. Figure 1.2 can be used to
describe pairs of mechanical waves or electromagnetic waves with the same wavelength. A light source that has a single wavelength is called monochromatic, which means single colored. In the case of constructive interference, the component waves combine to form a resultant wave with the same wavelength but with an amplitude that is greater than the amplitude of either of the individual component waves. For light, the result of constructive interference is light that is brighter than the light from the contributing waves. In the case of destructive interference, the resultant amplitude is less than the amplitude of the larger component wave. For light, the result of destructive interference is dimmer light or dark spots.
Wave Interference Two waves can interfere (a) constructively or (b) destructively. In interference, energy is not lost but is instead redistributed.
First wave Second wave Resultant wave
~
(b)
518
Chapter 15
F;rat wa,e Resultant wave Second wave
Comparison of Waves In Phase and 180° Out of Phase (a) The features of two waves in phase completely match, whereas (b) they are opposite each other in waves that are 180° out of phase.
(a)
(b)
Waves must have a constant phase difference for interference to be observed. For two waves to produce a stable interference pattern, the phases of the individual waves must remain unchanged relative to one another. If the crest of one wave overlaps the crest of another wave, as in Figure 1.3(a), the two h ave a phase difference of 0° and are said to be in phase. If the crest of one wave overlaps the trough of the other wave, as in Figure 1.3(b), the two waves have a phase difference of 180° and are said to be out ofphase. Waves are said to have coherence when the phase difference between two waves is constant and the waves do not shift relative to each other as time passes. Sources of such waves are said to be coherent.
coherence t he correlation between the phases of two or more waves
When two light bulbs are placed side by side, no interference is observed. The reason is that the light waves from one bulb are emitted independently of the waves from the other bulb. Random changes occurring in the light from one bulb do not necessarily occur in the light from the other bulb. Thus, the phase difference between the light waves from the two bulbs is not constant. The light waves still interfere, but the conditions for the interference change with each phase change, and therefore, no single interference pattern is observed. Light sources of this type are said to be incoherent.
Demonstrating Interference Interference in light waves from two sources can be demonstrated in the following way. Light from a single source is passed through a narrow slit and then through two narrow parallel slits. The slits serve as a pair of coherent light sources because the waves emerging from them come from the same source. Any random change in the light emitted by the source will occur in the two separate beams at the same time.
-c-
a
Monochromatic Light Interference An interference pattern consists of alternating light and dark fringes.
If monochromatic light is used, the light from the two slits produces a series of bright and dark parallel bands, or fringes, on a distant viewing screen, as shown in Figure 1.4. When the light from the two slits arrives at a point on the viewing screen where constructive interference occurs, a bright fringe appears at that location. When the light from the two slits combines destructively at a point on the viewing screen, a dark fringe appears at that location. Interference and Diffraction
519
White Light Interference When waves of white light from two coherent sources interfere, the pattern is indistinct because different colors interfere constructively and destructively at different positions.
When a white-light source is used to observe interference, the situation becomes more complicated. The reason is that white light includes waves of many wavelengths. An example of a white-light interference pattern is shown in Figure 1.5. The interference pattern is stable or well defined at positions where there is constructive interference between light waves of the same wavelength. This explains the color bands on either side of the center band of white light. Figure 1.6 shows some of the ways that two coherent waves leaving the
slits can combine at the viewing screen. When the waves arrive at the central point of the screen, as in Figure 1.6(a), they have traveled equal distances. Thus, they arrive in phase at the center of the screen, constructive interference occurs, and a bright fringe forms at that location. When the two light waves combine at a specific point off the center of the screen, as in Figure 1.6{b), the wave from the more distant slit must travel one wavelength farther than the wave from the nearer slit. Because the second wave has traveled exactly one wavelength farther than the first wave, the two waves are in phase when they combine at the screen. Constructive interference therefore occurs, and a second bright fringe appears on the screen.
If the waves meet midway between the locations of the two bright fringes, as in Figure 1.6(c), the first wave travels half a wavelength farther than the second wave. In this case, the trough of the first wave overlaps the crest of the second wave, giving rise to destructive interference. Consequently, a dark fringe appears on the viewing screen between the bright fringes.
Conditions for Interference of Light Waves (a) When both waves of light travel the same distance (l 1), they arrive at the screen in phase and interfere constructively. (b) If the difference between the distances
traveled by the light from each source equals a whole wavelength (,\), the waves still interfere constructively. (c) If the distances traveled by the light differ by a half wavelength, the waves interfere destructively.
If–________ l / Slits
“‘
~ ·
(a)
520
Chapter 15
Bright area at center of screen
Predicting the location of interference fringes. Consider two narrow slits separated by a distance d, as shown in Figure 1.7, and through which two coherent, monochromatic light waves, 11 and 12 , pass and are projected onto a screen. If the distance from the slits to the screen is very large compared with the distance between the slits, then l 1 and 12 are nearly parallel. As a result of this approximation, l 1 and 12 make the same angle, 0, with the horizontal dotted lines that are perpendicular to the slits. Angle 0 also indicates the position where waves combine with respect to the central point of the screen. The difference in the distance traveled by the two waves is called their path difference. Study the right triangle shown in Figure 1.7, and note that the path difference between the two waves is equal to d sin 0. Note carefully that the value for the path difference varies with angle 0 and that each value of 0 defines a specific position on the screen. The value of the path difference determines whether the two waves are in or out of phase when they arrive at the viewing screen. If the path difference is either zero or some whole-number multiple of the wavelength, the two waves are in phase, and constructive interference results. The condition for bright fringes (constructive interference) is given by:
Path Difference for Light Waves from Two Slits The path difference for two light waves equals d sin 0. In order to emphasize the path difference, the figure is not drawn to scale.
d
l ——
r•n0
Equation for Constructive Interference
d sin 0 = ±m>.
m = 0, l, 2, 3, …
path difference the difference in the distance traveled by two beams when they are scattered in the same direction from different points
the path difference between two waves = an integer multiple of the wavelength
In this equation, mis the order number of the fringe. The central bright fringe at 0 = 0 (m = 0) is called the zeroth-order maximum, or the central maximum; the first maximum on either side of the central maximum, which occurs when m = 1, is called the.first-order maximum, and so forth.
order number the number assigned to interference fringes with respect to the central bright fringe
Similarly, when the path difference is an odd multiple of½..\, the two waves arriving at the screen are 180° out of phase, giving rise to destructive interference. The condition for dark fringes is given by the following equation: Equation for Destructive Interference
d sin 0 =
±(m + !)>-
m = 0, 1, 2, 3, …
the path difference between two waves = an odd number of half wavelengths If m = 0 in this equation, the path difference is ±½..\,which is the condition required for the first dark fringe on either side of the bright which central maximum. Likewise, if m = 1, the path difference is ± is the condition for the second dark fringe on each side of the central maximum, and so forth.
f ..\,
Interference and Diffraction
521
A representation of the interference pattern formed by double-slit interference is shown in Figure 1.8. The numbers indicate the two maxima (the plural of maximum) that form on either side of the central (zerothorder) maximum. The darkest areas indicate the positions of the dark fringes, or minima (the plural of minimum), that also appear in the pattern.
Position of Higher-Order Interference Fringes The higher-order (m = 1, 2) maxima appear on either side of the central maximum (m = 0). m 2
1 0
Because the separation between interference fringes varies for light of different wavelengths, double-slit interference provides a method of measuring the wavelength oflight. In fact, this technique was used to make the first measurement of the wavelength of light.
-1
-2 Viewing screen
PREMIUM CONTENT
/,C\ Inter active Demo
Interference
\.::.I
HMDScience.com
Sample Problem A The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is measured on a viewing screen at an angle of2.I5° from the central maximum. Determine the wavelength of the light.
0
ANALYZE
= 3.0 x
Given:
d
Unknown:
>-=?
10- 5 m
Diagram:
d = 0.030 m; …1…
m=2
0 = 2.15°
Second-order bright fringe
I
(m= 2)
• 8 = 2.15
Zeroth-order bright fringe (m= 0)
1
Diagram not to scale
E)
PLAN
Choose an equation or situation:
Use the equation for constructive interference.
dsin 0 = m>. Rearrange the equation to isolate the unknown:
A = dsin 0 m
E)
SOLVE
Substitute the values into the equation and solve:
>- =
Calculator Solution Because the minimum number of significant figures for the data is two, the calculator answer 5.627366 x 10-7 should be rounded to two significant figures.
0 522
CHECKYOUR WORK
Chapter 15
(3.0 x 10- 5 m)(sin 2.15°) 2
A = 5.6 x 10- 7 m = 5.6 x 10 2 nm 2
I A = 5.6 x 10 nm I
This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color.
G·M!i,\114-►

Interference I
(continued)
Practice 1. Lasers are devices that can emit light at a specific wavelength. A double-slit
interference experiment is performed with blue-green light from an argon-gas laser. The separation between the slits is 0.50 mm, and the first-order maximum of the interference pattern is at an angle of 0.059° from the center of the pattern. What is the wavelength of argon laser light? 2. Light falls on a double slit with slit separation of2.02 x 10- 6 m, and the first bright fringe is seen at an angle of 16.5° relative to the central maximum. Find the wavelength of the light. 3. A pair of narrow parallel slits separated by a distance of 0.250 mm is illuminated
by the green component from a mercury vapor lamp (..\ = 546.1 nm). Calculate the angle from the central maximum to the first bright fringe on either side of the central maximum. 4. Using the data from item 2, determine the angle between the central maximum and the second dark fringe in the interference pattern.
SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What is the necessary condition for a path length difference between two waves that interfere constructively? Destructively?
2. If white light is used instead of monochromatic light to demonstrate interference, how does the interference pattern change? 3. If the distance between two slits is 0.0550 mm, find the angle between the first-order and second-order bright fringes for yellow light with a wavelength of 605 nm.
Interpreting Graphics 4. Two radio antennas simultaneously transmit identical signals with a wavelength of3.35 m, as shown in Figure 1.9. A radio several miles away in a car traveling parallel to the straight line between the antennas receives the signals. If the second maximum is located at an angle of 1.28° north of the central maximum for the interfering signals, what is the distance, d, between the two antennas?
Two Radio Antennas
Ant~na
A. ~ m
Car ::.
J cY?:.—–r,–■ …1..
~
e = 1.2s·
Interference and Diffraction
523
SECTION 2 Objectives ►
Describe how light waves bend around obstacles and produce bright and dark fringes.

I
Calculate the positions of fringes for a diffraction grating.

Describe how diffraction determines an optical instrument’s ability to resolve images.
Dilfraction Key Terms diffraction
resolving power
The Bending of Light Waves If you stand near the corner of a building, you can hear someone who is talking around the corner, but you cannot see the person. The reason is that sound waves are able to bend around the corner. In a similar fashion, water waves bend around obstacles, such as the barriers shown in Figure 2.1. Light waves can also bend around obstacles, but because of their short wavelengths, the amount they bend is too small to be easily observed.
!flight traveled in straight lines, you would not be able to observe an interference pattern in the double-slit demonstration. Instead, you would see two thin strips of light where each slit and the source were lined up. The rest of the screen would be dark. The edges of the slits would appear on the screen as sharply defined shadows. But this does not happen. Some of the light bends to the right and left as it passes through each slit.
diffraction a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge
Water Waves and Diffraction A property of all waves is that they bend, or diffract, around objects.
524
Chapter 15
The bending of light as it passes through each of the two slits can be understood using Huygens’s principle, which states that any point on a wave front can be treated as a point source of waves. Because each slit serves as a point source of light, the waves spread out from the slits. The result is that light deviates from a straight-line path and enters the region that would otherwise be shadowed. This divergence of light from its initial direction of travel is called diffraction. In general, diffraction occurs when waves pass through small openings, around obstacles, or by sharp edges. When a wide slit (1 mm or more) is placed between a distant light source and a screen, the light produces a bright rectangle with clearly marked edges on the screen. But if the slit is
gradually narrowed, the light eventually begins to spread out and produce a diffraction pattern, such as that shown in Figure 2.2. Like the interference fringes in the double-slit demonstration, this pattern of light and dark bands arises from the combination of light waves.
Wavelets in a wave front interfere with each other. Diffraction patterns resemble interference patterns because they also result from constructive and destructive interference. In the case of interference, it is assumed that the slits behave as point sources of light. For diffraction, the actual width of a single slit is considered.
Diffraction of Light with Decreasing Slit Width Diffraction becomes more evident as the width of the slit is narrowed. Slit width W i d e – – – – – – – Narrow
According to Huygens’s principle, each portion of a slit acts as a source of waves. Thus, light from one portion of the slit can interfere with light from another portion of the slit. The resultant intensity of the diffracted light on the screen depends on the angle, 0, through which the light is diffracted. To understand the single-slit diffraction pattern, consider Figure 2.3(a), which shows an incoming plane wave passing through a slit of width a. Each point (or, more accurately, each infinitely thin slit) within the wide slit is a source of Huygens wavelets. The figure is simplified by showing only five among this infinite number of sources. As with double-slit interference, the viewing screen is assumed to be so far from the slit that the rays emerging from the slit are nearly parallel. At the viewing screen’s midpoint, all rays from the slit travel the same distance, so a bright fringe appears. The wavelets from the five sources can also interfere destructively when they arrive at the screen, as shown in Figure 2.3(b). When the extra distance traveled by the wave originating at point 3 is half a wavelength longer than the wave from point 1, these two waves interfere destructively at the screen. At the same time, the wave from point 5 travels half a wavelength farther than the wave from point 3, so these waves also interfere destructively. With all pairs of points interfering destructively, this point on the screen is dark. For angles other than those at which destructive interference completely occurs, some of the light waves remain uncanceled. At these angles light appears on the screen as part of a bright band. The brightest band appears in the pattern’s center, while the bands to either side are much dimmer.
Destructive Interference in Single-Slit Diffraction (a) By treating the light coming through
the slit as a line of infinitely thin sources along the slit’s width, one can determine (b) the conditions at which destructive interference occurs between the waves from the upper half of the slit and the waves from the lower half.
(a)
I T 1 \I –
a

(b)
1• –
2• –
3• –
-~ 4• 5• Incident wave
2•0
·—–
I
Central bright fringe /
Viewing screen
a3.q; /
15:e:r I 4
~
Interference and Diffraction
525
Diffraction Pattern from a Single Slit In a diffraction pattern, the central maximum is twice as wide as the secondary maxima.
Shadow of a Washer A diffraction pattern forms in the washer’s shadow when light is diffracted at the washer’s edge. Note the dark and light stripes both around the washer and inside the washer.
Light diffracted by an obstacle also produces a pattern. The diffraction pattern that results from monochromatic light passing through a single slit consists of a broad, intense central band-the central maximum-flanked by a series of narrower, less intense secondary bands (called secondary maxima), and a series of dark bands, or minima.
An example of such a pattern is shown in Figure 2.4. The
points at which maximum constructive interference occurs lie approximately halfway between the dark fringes. Note that the central bright fringe is quite a bit brighter and about twice as wide as the next brightest maximum. Diffraction occurs around the edges of all objects. Figure 2.5 shows the diffraction pattern that appears in the shadow of a washer. The pattern consists of the shadow and a series of bright and dark bands of light that continue around the edge of the shadow. The washer is large compared with the wavelength of the light, and a magnifying glass is required to observe the pattern.
Diffraction Gratings Constructive Interference on a CD Compact discs disperse light into its component colors in a manner similar to that of a diffraction grating.
You have probably noticed that if white light is incident on a compact disc, streaks of color are visible. These streaks appear because the digital information (alternating pits and smooth reflecting surfaces) on the disc forms closely spaced rows. These rows of data do not reflect nearly as much light as the thin portions of the disc that separate them. These areas consist entirely of reflecting material, so light reflected from them undergoes constructive interference in certain directions. This constructive interference depends on the direction of the incoming light, the orientation of the disc, and the light’s wavelength. Each wavelength of light can be seen at a particular angle with respect to the disc’s surface, causing you to see a “rainbow” of color, as shown in Figure 2.6.
~ 0
“‘”‘
:,:
.!: ~
::;; C:
.s
.c:
g, 0
:,:
@
This phenomenon has been put to practical use in a device called a diffraction grating. A diffraction grating, which can be constructed to either transmit or reflect light, uses diffraction and interference to disperse light into its component colors with an effect similar to that of a glass prism. A transmission grating consists of many equally spaced parallel slits. Gratings are made by ruling equally spaced lines on a piece
526
Chapter 15
Constructive Interference by a Diffraction Grating Light of a single wavelength passes through each of the slits of a diffraction grating to constructively interfere at a particular angle 0.
Spiked Stars Photographs
p
Diffraction grating

Screen
“‘-
of stars always show spikes extending from the stars. Given that the aperture of a camera’s rectangular shutter has straight edges, explain how diffraction accounts for the spikes. Radio Diffraction
Visible light waves are not observed diffracting around buildings or other obstacles. However, radio waves can
of glass using a diamond cutting point driven by an elaborate machine called a ruling engine. Replicas are then made by pouring liquid plastic on the grating and then peeling it off once it has set. This plastic grating is then fastened to a flat piece of glass or plastic for support.
be detected around buildings or mountains, even when the transmitter is not visible. Explain why diffraction is more evident for radio waves than for visible light.
Figure 2. 7 shows a schematic diagram of a section of a
diffraction grating. A monochromatic plane wave is incoming from the left, normal to the plane of the grating. The waves that emerge nearly parallel from the grating are brought together at a point Pon the screen by the lens. The intensity of the pattern on the screen is the result of the combined effects of interference and diffraction. Each slit produces diffraction, and the diffracted beams in turn interfere with one another to produce the pattern. For some arbitrary angle, 0, measured from the original direction of travel of the wave, the waves must travel different path lengths before reaching point Pon the screen. Note that the path difference between waves from any two adjacent slits is d sin 0. If this path difference equals one wavelength or some integral multiple of a wavelength, waves from all slits will be in phase at P, and a bright line will be observed. The condition for bright line formation at angle 0 is therefore given by the equation for constructive interference: d sin 0 = ±m>.
m = 0, l, 2, 3, . ..
This equation can be used to calculate the wavelength of light if you know the grating spacing and the angle of deviation. The integer m is the order number for the bright lines of a given wavelength. If the incident radiation contains several wavelengths, each wavelength deviates by a specific angle, which can be determined from the equation. Interference and Diffract ion
527
Maxima from a Diffraction Grating Light is dispersed by a diffraction grating. The angle of deviation for the first-order maximum is smaller for blue light than for yellow light. Diffraction grating
Second order (m = -2)
Spectrometer The spectrometer uses a grating to disperse the light from a source.
First order (m =-1)
Zeroth order (m = 0)
First order (m = 1)
Second order (m=2)
Note in Figure 2.8 that all wavelengths combine at 0 = 0, which corresponds to m = 0. This is called the zeroth-order maximum. The first-order maximum, corresponding to m = 1, is observed at an angle that satisfies the relationship sin 0 = )../ d. The second-order maximum, corresponding to m = 2, is observed at an angle where sin 0 = 2)../d. The sharpness of the principal maxima and the broad range of the dark areas depend on the number oflines in a grating. The number of lines per unit length in a grating is the inverse of the line separation d. For example, a grating ruled with 5000 lines/cm has a slit spacing, d, equal to the inverse of this number; hence, d = (1 /5000) cm= 2 x 10- 4 cm. The greater the number oflines per unit length in a grating, the less separation between the slits and the farther spread apart the individual wavelengths of light.
Lens
Grating
rnfll);l:fllil,
Spectrum of Mercury Vapor The light from mercury vapor is passed through a diffraction grating, producing the spectrum shown.
528
Chapter 15
Diffraction gratings are frequently used in devices called spectrometers, which separate the light from a source into its monochromatic components. A diagram of the basic components of a spectrometer is shown in Figure 2.9. The light to be analyzed passes through a slit and is formed into a parallel beam by a lens. The light then passes through the grating. The diffracted light leaves the grating at angles that satisfy the diffraction grating equation. A telescope with a calibrated scale is used to observe the first-order maxima and to measure the angles at which they appear. From these measurements, the wavelengths of the light can be determined and the chemical composition of the light source can be identified. An example of a spectrum produced by a spectrometer is shown in Figure 2.1 o. Spectrometers are used in astronomy to study the chemical compositions and temperatures of stars, interstellar gas clouds, and galaxies.
PREMIUM CONTENT
~
Interactive Demo
\:::,J
HMDScience.com
Sample Problem B Monochromatic light from a helium-neon laser(>. = 632.8 nm) shines at a right angle to the surface of a diffraction grating that contains 150 500 lines/m. Find the angles at which one would observe the first-order and second-order maxima.
0
ANALYZE
Given:
A = 632.8 nm= 6.328 x 10- 7 m 1
d=
m= 1 and2
1
150 500 m
150 500 lines
m
Unknown:
01–?·
Diagram:
Second-order maximum (m= 2)
First-order maximum (m= 1)
Screen
f:)
PLAN
Choose an equation or situation: Use the equation for a diffraction grating.
dsin 0= ±m.\ Rearrange the equation to isolate the unknown:
1 0 = sin – ( ~,\ )
E)
SOLVE
Substitute the values into the equation and solve: For the first-order maximum, m = 1:
01 = sin-1 ( ~) = sin-1 (6.328
r
7
10- m)
150500 m
101
= 5.465°
Calculator Solution
1
Because the minimum number of significant figures for the data is four, the calculator answers 5.464926226 and 10.98037754 should be rounded to four significant figures.
Form = 2:
0 = sin- 1 2dA 2
02
(
. -_ Slll
)
1(2(6.328
7
10- m)) 1 m X
150 500
102 = 10.98°
1
Interference and Diffraction
529
Diffraction Gratings (continued)
0
I
CHECKYOUR WORK
The second-order maximum is spread slightly more than twice as far from the center as the first-order maximum. This diffraction grating does not have high dispersion, and it can produce spectral lines up to the tenth-order maxima (where sin 0 = 0.9524).
Practice 1. A diffraction grating with 5.000 x 103 lines/ cm is used to examine the sodium
spectrum. Calculate the angular separation of the two closely spaced yellow lines of sodium (588.995 nm and 589.592 nm) in each of the first three orders.
2. A diffraction grating with 4525 lines/ cm is illuminated by direct sunlight. The first-order solar spectrum is spread out on a white screen hanging on a wall opposite the grating. a. At what angle does the first-order maximum for blue light with a wavelength of 422 nm appear? b. At what angle does the first-order maximum for red light with a wavelength of
655 nm appear? 3. A grating with 1555 lines/ cm is illuminated with light of wavelength 565 nm. What is the highest-order number that can be observed with this grating? (Hint: Remember that sin 0 can never be greater than 1 for a diffraction grating.)
4. Repeat item 3 for a diffraction grating with 15 550 lines/ cm that is illuminated with light of wavelength 565 nm. 5. A diffraction grating is calibrated by using the 546.1 nm line of mercury vapor. The first-order maximum is found at an angle of21.2°. Calculate the number oflines per centimeter on this grating.
Diffraction and Instrument Resolution Limits of an Optical System Each of two distant point sources produces a diffraction pattern.
Slit
530
Chapter 15
Screen
The ability of an optical system, such as a microscope or a telescope, to distinguish between closely spaced objects is limited by the wave nature of light. To understand this limitation, consider Figure 2.11, which shows two light sources far from a narrow slit. The sources can be taken as two point sources that are not coherent. For example, they could be two distant stars that appear close to each other in the night sky.
If no diffraction occurred, you would observe two distinct bright spots (or images) on the screen at the far right. However, because of diffraction, each source is shown to have a bright central region flanked by weaker bright and dark rings. What is observed on the screen is the resultant from the superposition of two diffraction patterns, one from each source.
Resolution depends on wavelength and aperture width. If the two sources are separated so that their central maxima do not overlap, as in Figure 2.12, their images can just be distinguished and are said to be barely resolved. To achieve high resolution or resolving power, the angle between the resolved objects, 0, should be as small as possible as shown in Figure 2.11. The shorter the wavelength of the incoming light or the wider the opening, or aperture, through which the light passes, the smaller the angle of resolution, 0, will be and the greater the resolving power will be. For visible-light telescopes, the aperture width, D, is approximately equal to the diameter of the mirror or lens. The equation to determine the limiting angle of resolution in radians for an optical instrument with a circular aperture is as follows:
resolving power the ability of an optical instrument to form separate images of two objects that are close together
0= 1.22; The constant 1.22 comes from the derivation of the equation for circular apertures and is absent for long slits. Note that one radian equals (180/7r) The equation indicates that for light with a short wavelength, such as an X ray, a small aperture is sufficient for high resolution. On the other hand, if the wavelength of the light is long, as in the case of a radio wave, the aperture must be large in order to resolve distant objects. This is one reason why radio telescopes have large dishlike antennas. 0

Yet, even with their large sizes, radio telescopes cannot resolve sources as easily as visible-light telescopes resolve visible-light sources. At the shortest radio wavelength (1 mm), the largest single antenna for a radio telescope-the 305 m dish at Arecibo, Puerto Rico-has a resolution angle of 4 x 10- 6 rad. The same resolution angle can be obtained for the longest visible light waves (700 nm) by an optical telescope with a 21 cm mirror.
Resolution Two point sources are barely resolved if the central maxima of their diffraction patterns do not overlap. E
~
C,
C:
§ a5
“-~
;,:,

> ‘E
:::,
“”8 E
,.,’
i
oil
,
I

j’
\
I
\. ,..__,\

,,,
. ‘
/.
These results apply to any capacitor. In practice, there is a limit to the maximum energy (or charge) that can be stored because electrical breakdown ultimately occurs between the plates of the capacitor for a sufficiently large potential difference. So, capacitors are usually labeled with a maximum operating potential difference. Electrical breakdown in a capacitor is like a lightning discharge in the atmosphere. Figure 2.4 shows a pattern created in a block of plastic resin that has undergone electrical breakdown. This book’s problems assume that all potential differences are below the maximum.
PREMIUM CONTENT
~
Capacitance
\::I
Interactive Demo HMDScie nce.com
Sample Problem B A capacitor, connected to a 12 V battery, holds 36 µC of charge on each plate. What is the capacitance of the capacitor? How much electrical potential energy is stored in the capacitor?
0
ANALYZE
Given:
Q = 36 µC
= 3.6 x 10- 5 C PEelectric = ?
Unknown:
E)
SOLVE
~ V= 12V
To determine the capacitance, use the definition of capacitance.
Q
C=
I C=
~v =
3.6 x 10- 5 C 12V
3.0 X 10- 6 F = 3.0 µF
I
To determine the potential energy, use the alternative form of the equation for the potential energy of a charged capacitor shown on this page: PEelectric PEelectric
= ½C(~ V)
2
= (0.5)(3.0 x
6
10- F)(l2 V)
2
I P Eelectric = 2.2 X 10- 4 J I
G·i,iii,M§.► 592
Chapter 17

Capacitance (continued) Practice 1. A 4.00 µF capacitor is connected to a 12.0 V battery.
a. What is the charge on each plate of the capacitor? b. If this same capacitor is connected to a 1.50 V battery, how much electrical potential energy is stored? 2. A parallel-plate capacitor has a charge of 6.0 µC when charged by a potential
difference of 1.25 V.
a. Find its capacitance. b. How much electrical potential energy is stored when this capacitor is connected to a 1.50 V battery? 3. A capacitor has a capacitance of2.00 pF. a. What potential difference would be required to store 18.0 pC?
b. How much charge is stored when the potential difference is 2.5 V?
4. You are asked to design a parallel-plate capacitor having a capacitance of 1.00 F and a plate separation of 1.00 mm. Calculate the required surface area of each plate. Is this a realistic size for a capacitor?
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Assume Earth and a cloud layer 800.0 m above the Earth can be treated as plates of a parallel-plate capacitor. a. If the cloud layer has an area of 1.00 x 106 m 2 , what is the capacitance? b. If an electric field strength of2.0 x 106 N/C causes the air to conduct charge (lightning), what charge can the cloud hold? c. Describe what must happen to its molecules for air to conduct electricity.
2. A parallel-plate capacitor has an area of2.0 cm2, and the plates are separated by 2.0 mm. a. What is the capacitance? b. How much charge does this capacitor store when connected to a 6.0 V battery?
3. A parallel-plate capacitor has a capacitance of 1.35 pF. If a 12.0 V battery is connected to this capacitor, how much electrical potential energy would it store?
Critical Thinking 4. Explain why two metal plates near each other will not become charged until connected to a source of potential difference.
Electrical Energy and Current
593
SECTION 3 Objectives

Describe the basic properties of electric current, and solve problems relating current, charge, and time.

Distinguish between the drift speed of a charge carrier and the average speed of the charge carrier between collisions.

Key Terms electric current
drift velocity
resistance
Current and Charge Movement
Calculate resistance, current, and potential difference by using the definition of
resistance. — – — —– — – — —– —

Current and Resistance
Distinguish between ohmic and non-ohmic materials, and learn what factors affect resistance.
electric current the rate at which electric c harges pass through a given area
Although many practical applications and devices are based on the principles of static electricity, electricity did not become an integral part of our daily lives until scientists learned to control the movement of electric charge, known as current. Electric currents power our lights, radios, television sets, air conditioners, and refrigerators. Currents also are used in automobile engines, travel through miniature components that make up the chips of computers, and perform countless other invaluable tasks. Electric currents are even part of the human body. This connection between physics and biology was discovered by Luigi Galvani (1737-1798). While conducting electrical experiments near a frog he had recently dissected, Galvani noticed that electrical sparks caused the frog’s legs to twitch and even convulse. After further research, Galvani concluded that electricity was present in the frog. Today, we know that electric currents are responsible for transmitting messages between body muscles and the brain. In fact, every function involving the nervous system is initiated by electrical activity.
Current is the rate of charge movement. Current The current in this wire is defined as the rate at which electric charges pass through a cross-sectional area of the wire.
—-0 +——-0-

:::::::========::>
0
o
A current exists whenever there is a net movement of electric charge through a medium. To define current more precisely, suppose electrons are moving through a wire, as shown in Figure 3.1. The electric current is the rate at which these charges move through the cross section of the wire. If ~Q is the amount of charge that passes through this area in a time interval, ~t, then the current, I, is the ratio of the amount of charge to the time interval. Note that the direction of current is opposite the movement of the negative charges. We will further discuss this detail later in this section. Electric Current
1
I . charge passing through a given area . . electric current = tunemterva1
The SI unit for current is the ampere, A. One ampere is equivalent to one coulomb of charge passing through a cross-sectional area in a time interval of one second (1 A = 1 C/s). 594
Chapter 17
PREMIUM CONTENT
JC\
Interactive Demo
\::,/ HMDScience.com
Sample Problem C The current in a light bulb is 0.835 A. How long does it take for a total charge of 1.67 C to pass through the filament of the bulb?
0
ANALYZE
Given:
flQ
= 1.67 C
l= 0.835A Unknown:
E)
SOLVE
flt=?
Use the definition of electric current. Rearrange to solve for the time interval.
flt = flQ
t
67
flt = l. C = 12.00 sl 0.835A Practice 1. If the current in a wire of a CD player is 5.00 mA, how long would it take for 2.00 C of charge to pass through a cross-sectional area of this wire? 2. In a particular television tube, the beam current is 60.0 µA. How long does it take for 3.75 x 10 14 electrons to strike the screen? (Hint: Recall that an electron has a charge of – 1.60 x 10- 19 C.)
3. If a metal wire carries a current of 80.0 mA, how long does it take for 3.00 x 1020 electrons to pass a given cross-sectional area of the wire?
4. The compressor on an air conditioner draws 40.0 A when it starts up. If the start-up time is 0.50 s, how much charge passes a cross-sectional area of the circuit in this time? 5. A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.5 s.
a. What is the current in the wire? b. How many electrons pass through the cross-sectional area in 10.0 s?
c. If the number of charges that pass through the cross-sectional area during the given time interval doubles, what is the resulting current?
Electrical Energy and Current
595
Conventional current is defined in terms of positive charge movement.
QuickLAB MATERIALS • lemon • copper wire • paper clip
A LEMON BATTERY Straighten the paper clip, and insert it and the copper wire into the lemon to construct a chemical cell. Touch the ends of both wires with your tongue. Because a potential difference exists across the two metals and because your saliva provides an electrolytic solution that conducts electric current, you should feel a slight tingling sensation on your tongue. CAUTION: Do not share battery set-ups with other students. Dispose of your materials according to your teacher’s instructions.
The moving charges that make up a current can be positive, negative, or a combination of the two. In a common conductor, such as copper, current is due to the motion of negatively charged electrons, because the atomic structure of solid conductors allows many electrons to be freed from their atoms and to move freely through the material. In contrast, the protons are relatively fixed inside the nucleus of the atom. In certain particle accelerators, a current exists when positively charged protons are set in motion. In some cases-in gases and dissolved salts, for examplecurrent is the result of positive charges moving in one direction and negative charges moving in the opposite direction. Positive and negative charges in motion are sometimes called charge carriers. Conventional current is defined in terms of the flow of positive charges. Thus, negative charge carriers, such as electrons, would have a conventional current in the direction opposite their physical motion. The three possible cases of charge flow are shown in Figure 3.2. We will use conventional current in this book unless stated otherwise.
First case Motion of charge carriers Equivalent conventional current
Second case
Third case
+–+—
+ —
As you have learned, an electric field in a material sets charges in motion. For a material to be a good conductor, charge carriers in the material must be able to move easily through the material. Many metals are good conductors because metals usually contain a large number of free electrons. Body fluids and saltwater are able to conduct electric charge because they contain charged atoms called ions. Because dissolved ions can move through a solution easily, they can be charge carriers. A solute that dissolves in water to give a solution that conducts electric current is called an electrolyte.
Drift Velocity When you turn on a light switch, the light comes on almost immediately. For this reason, many people think that electrons flow very rapidly from the switch to the light bulb. However, this is not the case. When you turn on the switch, electron motion near the switch changes the electric field there, and the change propagates throughout the wire very quickly. Such changes travel through the wire at nearly the speed of light. The charges themselves, however, travel much more slowly. 596
Chapter 17
Drift velocity is the net velocity of charge carriers. To see how the electrons move, consider a solid conductor in which the charge carriers are free electrons. When the conductor is in electrostatic equilibrium, the electrons move randomly, similar to the movement of molecules in a gas. When a potential difference is applied across the conductor, an electric field is set up inside the conductor. The force due to that field sets the electrons in motion, thereby creating a current. These electrons do not move in straight lines along the conductor in a direction opposite the electric field. Instead, they undergo repeated collisions with the vibrating metal atoms of the conductor. If these collisions were charted, the result would be a complicated zigzag pattern like the one shown in Figure 3.3. The energy transferred from the electrons to the metal atoms during the collisions increases the vibrational energy of the atoms, and the conductor’s temperature increases. The electrons gain kinetic energy as they are accelerated by the electric field in the conductor. They also lose kinetic energy because of the collisions described above. However, despite the internal collisions, the individual electrons move slowly along the conductor in a direction opposite the electric field, E, with a velocity known as the drift velocity, vdrift”
Drift Velocity When an electron moves through a conductor, collisions with the vibrating metal atoms of the conductor force the electron to change its direction constantly. – – – Vdrift
drift velocity the net velocity of a charge carrier moving in an electric field
Drift speeds are relatively small. The magnitudes of drift velocities, or drift speeds, are typically very small. In fact, the drift speed is much less than the average speed between collisions. For example, in a copper wire that has a current of 10.0 A, the drift speed of electrons is only 2.46 x 10-4 m /s. These electrons would take about 68 min to travel 1 m! The electric field, on the other hand, reaches electrons throughout the wire at a speed approximately equal to the speed of light.
Electric Field Inside a Conductor We concluded in our study of
Particle Accelerator The positively charged
electrostatics that the field inside a conductor is zero, yet we have seen that an electric field exists inside a conductor that carries a
dome of a Van de Graaff generator can be used to accelerate positively charged protons. A current exists due
current. How is this zero electric field possible?
to the motion of these protons. In this case, how does the direction of conventional
Turning on a Light If charges travel very slowly through a metal (ap”‘ .!: ~.r:::;
i::
ill
o-
proximately 1 4 m/s), why doesn’t it take several hours for a light to c ome on after you flip a switch?
current compare with the direction in which the charge carriers move?
gJ a:
fg ff.
©l
-c-
a
Electrical Energy and Current
597
Resistance to Current
resistance the opposition presented to electric current by a material or device
When a light bulb is connected to a battery, the current in the bulb depends on the potential difference across the battery. For example, a 9.0 V battery connected to a light bulb generates a greater current than a 6.0 V battery connected to the same bulb. But potential difference is not the only factor that determines the current in the light bulb. The materials that make up the connecting wires and the bulb’s filament also affect the current in the bulb. Even though most materials can be classified as conductors or insulators, some conductors allow charges to move through them more easily than others. The opposition to the motion of charge through a conductor is the conductor’s resistance. Quantitatively, resistance is defined as the ratio of potential difference to current, as follows :
Resistance
R=~V I potential difference current resistance =
The SI unit for resistance, the ohm, is equal to one volt per ampere and is represented by the Greek letter n (omega).
Resistance is constant over a range of potential differences. For many materials, including most metals, experiments show that the resistance is constant over a wide range ofapplied potential differences. This statement, known as Ohm’s law, is named for Georg Simon Ohm (1789-1854), who was the first to conduct a systematic study of electrical resistance. Mathematically, Ohm’s law is stated as follows:
~V = constant As can be seen by comparing the definition of resistance with Ohm’s law, the constant of proportionality in the Ohm’s law equation is resistance. It is common practice to express Ohm’s law as 6.. V = IR.
Ohm’s law does not hold for all materials. Ohm’s law is not a fundamental law of nature like the conservation of energy or the universal law of gravitation. Instead, it is a behavior that is valid only for certain materials. Materials that have a constant resistance over a wide range of potential differences are said to be ohmic. A graph of current versus potential difference for an ohmic material is linear, as shown in Figure 3.4(a). This is because the slope of such a graph (II 6.. V) is inversely proportional to resistance. When resistance is constant, the current is proportional to the potential difference and the resulting graph is a straight line.
598
Chapter 17
Materials that do not function according to Ohm’s law are said to be non-ohmic. Figure 3.4(b) shows a graph of current versus potential difference for a non-ohmic material. In this case, the slope is not constant because resistance varies. Hence, the resulting graph is nonlinear. One common semiconducting device that is non-ohmic is the diode. Its resistance is small for currents in one direction and large for currents in the reverse direction. Diodes are used in circuits to control the direction of current. This book assumes that all resistors function according to Ohm’s law unless stated otherwise.
Resistance depends on length, area, material, and temperature.
Comparing Ohmic and Non-Ohmic Materials (a) The current-potential difference
curve of an ohmic material is linear, and the slope is the inverse of the material’s resistance. (b) The current-potential difference curve of a non-ohmic material is nonlinear. Resistance of an Ohmic Material
Earlier in this section, you learned that electrons do not move in straightline paths through a conductor. Instead, they undergo repeated collisions with the metal atoms. These collisions affect the motion of charges somewhat as a force of internal friction would. This is the origin of a material’s resistance. Thus, any factors that affect the number of collisions will also affect a material’s resistance. Some of these factors are shown in Figure 3.5.
Potential difference
Two of these factors-length and cross-sectional area-are purely geometrical. It is intuitive that a longer length of wire provides more resistance than a shorter length of wire does. Similarly, a wider wire allows charges to flow more easily than a thinner wire does, much as a larger pipe allows water to flow more easily than a smaller pipe does. The material effects have to do with the structure of the atoms making up the material. Finally, for most materials, resistance increases as the temperature of the metal increases. When a material is hot, its atoms vibrate fast, and it is more difficult for an electron to flow through the material.
Resistance of a Non-Ohmic Material
Potential difference
Factor
Less resistance
Greater resistance
Length
Cross-sectional area
Material Copper
Iron
Temperature
Electrical Energy and Current
599
Resistors Resistors, such as those shown here, are used to control current. The colors of the bands represent a code for the values of the resistances.
Resistors can be used to control the amount of current in a conductor. One way to change the current in a conductor is to change the potential difference across the ends of the conductor. But in many cases, such as in household circuits, the potential difference does not change. How can the current in a certain wire be changed if the potential difference remains constant? According to the definition of resistance, if~ V remains constant, current decreases when resistance increases. Thus, the current in a wire can be decreased by replacing the wire with one of higher resistance. The same effect can be accomplished by making the wire longer or by connecting a resistor to the wire. A resistor is a simple electrical element that provides a specified resistance. Figure 3.6 shows a group of resistors in a circuit board. Resistors are sometimes used to control the current in an attached conductor because this is often more practical than changing the potential difference or the properties of the conductor. PREMIUM CONTENT
Al:\
Resistance
\::I
Interactive Demo HMDScience.com
Sample Problem D The resistance of a steam iron is 19.0 n. What is the current in the iron when it is connected across a potential difference of 120 V?
0 E)
ANALYZE
SOLVE
Given:
R = 19.0!1
Unknown:
I= ?
b. V= 120V
Use Ohm’s law to relate resistance to potential difference and current.
R= b.V I I = b.V = 120V = l6.32AI R 19.0 !1
600
Chapter 17
Resistance
(continued)
iPEMii+■ 1. A 1.5 V battery is connected to a small light bulb with a resistance of 3.5 n. What is
the current in the bulb? 2. A stereo with a resistance of 65 n is connected across a potential difference of
120 V. What is the current in this device? 3. Find the current in the following devices when they are connected across a
potential difference of 120 V. a. a hot plate with a resistance of 48
n
b. a microwave oven with a resistance of20 n
4. The current in a microwave oven is 6.25 A. If the resistance of the oven’s circuitry is 17.6 n, what is the potential difference across the oven? 5. A typical color television draws 2.5 A of current when connected across a potential
difference of 115 V. What is the effective resistance of the television set? 6. The current in a certain resistor is 0.50 A when it is connected to a potential difference of 110 V. What is the current in this same resistor if a. the operating potential difference is 90.0 V?
b. the operating potential difference is 130 V?
Saltwater and perspiration lower the body’s resistance. The human body’s resistance to current is on the order of 500 000 n when the skin is dry. However, the body’s resistance decreases when the skin is wet. If the body is soaked with saltwater, its resistance can be as low as 100 n. This is because ions in saltwater readily conduct electric charge. Such low resistances can be dangerous if a large potential difference is applied between parts of the body because current increases as resistance decreases. Currents in the body that are less than 0.01 A either are imperceptible or generate a slight tingling feeling. Greater currents are painful and can disturb breathing, and currents above 0.15 A disrupt the electrical activity of the heart and can be fatal. Perspiration also contains ions that conduct electric charge. In a galvanic skin response ( GSR) test, commonly used as a stress test and as part of some so-called lie detectors, a very small potential difference is set up across the body. Perspiration increases when a person is nervous or stressed, thereby decreasing the resistance of the body. In GSR tests, a state of low stress and high resistance, or “normal” state, is used as a control, and a state of higher stress is reflected as a decreased resistance compared with the normal state.
Electrical Energy and Current
601
Potentiometer Rotating the knob of a potentiometer changes the resistance.
~—
Resistive Element
Rotating Dial
……

Potentiometers have variable resistance. A potentiometer, shown in Figure 3.7, is a special type of resistor that has a fixed contact on one end and an adjustable, sliding contact that allows the user to tap off different potential differences. The sliding contact is frequently mounted on a rotating shaft, and the resistance is adjusted by rotating a knob. Potentiometers (frequently called pots for short) have many applications. In fact, most of the knobs on everyday items, such as the volume control on a stereo, are potentiometers. Potentiometers may also be mounted linearly. One example is a dimmer switch to control the light output of a light fixture. The joystick on a video game controller uses two potentiometers, one for motion in the x direction and one for motion in they direction, to tell the computer the movements that you make when playing a game.
SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Can the direction of conventional current ever be opposite the direction of charge movement? If so, when?
2. The charge that passes through the filament of a certain light bulb in 5.00 s is 3.0 C. a. What is the current in the light bulb? b. How many electrons pass through the filament of the light bulb in a time interval of 1.0 min? 3. How much current would a 10.2 n toaster oven draw when connected to a 120 V outlet? 4. An ammeter registers 2.5 A of current in a wire that is connected to a 9.0 V battery. What is the wire’s resistance? 5. In a particular diode, the current triples when the applied potential difference is doubled. What can you conclude about the diode? 6. What is the function of resistors in a circuit board? What is the function of diodes in a circuit board?
7. Calculate the current in a 75 n resistor when a potential difference of 115 Vis placed across it. What will the current be if the resistor is replaced with a 4 7 n resistor?
Critical Thinking 8. Which is less in a conductor that carries a current, the drift speed of an electron, or the average speed of the electron between collisions? Explain your answer.
9. You have only one type of wire. If you are connecting a battery to a light bulb with this wire, how could you decrease the current in the wire? 602
Chapter 17
S.T.E.M. Superconductors ake a moment to imagine the many things that could be created with materials that conduct electricity with zero resistance. There would be no heating or reduction in the current when conducting electricity with such a material. These materials exist and are called superconductors. Superconductors have zero resistance below a certain temperature, called the critical temperature. The graph of resistance as a function of temperature for a superconductor resembles that of a normal metal at temperatures well above the critical temperature. But when the temperature is near or below the critical temperature, the resistance suddenly drops to zero, as the graph below shows. This graph shows the resistance of mercury just above and below its critical temperature of 4.15 K. 0.150 0.125
S 0.100 ~ ~
1n
O.D75
“:lj 0.050 a:
Critical temperature
0.025 0.000 – l – – – – – — ” ” ===—+ – – – – – + – – – – – – – – I 4.3 4.1 4.4 4.2 4.0 Temperature (K)
Today, there are thousands of known superconductors, including common metals such as aluminum, tin, lead, and zinc. However, for common metals that exhibit superconductivity, the critical temperature is extremely low- near absolute zero. For example, aluminum reaches superconductivity at 1.19 K, just a little more than one degree above absolute zero. Temperatures near absolute zero are difficult to achieve and maintain. Interestingly, copper, silver, and gold, which are excellent conductors at room temperature, do not exhibit superconductivity. An important recent development in physics is the discovery of high-temperature superconductors. The excitement began with a 1986 publication by scientists at the IBM Zurich Research Laboratory in Switzerland. In this publication, scientists reported evidence for superconductivity at a temperature near 30 K. More recently, scientists have found superconductivity at temperatures as high as 150 K.
This express train in Shanghai, China, which utilizes the Meissner effect, levitates above the track and travels at up to 430 km/h in normal operations. However, 150 K is still -123°C, which is much colder than room temperature. The search continues for a material that has superconducting qualities at room temperature. This important search has both scientific and practical applications. One of the truly remarkable features of superconductors is that once a current is established in them, the current continues even if the applied potential difference is removed. In fact, steady currents have been observed to persist for many years in superconducting loops with no apparent decay. This feature makes superconducting materials attractive for a wide variety of applications. Because electric currents produce magnetic effects, current in a superconductor can be used to float a magnet in the air over a superconductor. This effect, known as the Meissner effect, is used with high-speed express trains, such as the one shown in the figure above. This type of train levitates a few inches above the track. One useful application of superconductivity is superconducting magnets. Such magnets are being considered for storing energy. The idea of using superconducting power lines to transmit power more efficiently is also being researched. Modern superconducting electronic devices that consist of two thin-film superconductors separated by a thin insulator have been constructed. They include magnetometers (magnetic-field measuring devices) and various microwave devices.
603
SECTION 4 Objectives ►
I
Differentiate between direct current and alternating current.

Relate electric power to the rate at which electrical energy is converted to other forms of energy.

Calculate electric power and the cost of running electrical appliances.
Electric Power Sources and Types of Current When you drop a ball, it falls to the ground, moving from a place of higher gravitational potential energy to one of lower gravitational potential energy. As discussed in Section 1, charges behave in similar ways. For example, free electrons in a conductor move randomly when all points in the conductor are at the same potential. But when a potential difference is applied across the conductor, they will move from a position of higher electric potential to a position of lower electric potential. Thus, a potential difference maintains current in a circuit.
Batteries and generators supply energy to charge carriers. Batteries maintain a potential difference across their terminals by converting chemical energy to electrical potential energy. Figure 4.1 shows students measuring the potential difference of a battery created using a lemon, copper, and tin.
Batteries Batteries maintain electric current by converting chemical energy into electrical energy.
As charge carriers move from higher to lower electrical potential energy, this energy is converted into kinetic energy. This motion allows collisions to occur between the moving charges and the remaining material in the circuit elements. These collisions transfer energy (in the form of heat) back to the circuit. A battery stores energy in the form of chemical energy, and its energy is released through a chemical reaction that occurs inside the battery. The battery continues to supply electrical energy to the charge carriers until its chemical energy is depleted. At this point, the battery must be replaced or recharged. Because batteries must often be replaced or recharged, generators are sometimes preferable. Generators convert m echanical energy into electrical energy. For example, a hydroelectric power plant converts the kinetic energy of falling water into electrical potential energy. Generators are the source of the current to a wall outlet in your home and supply the electrical energy to operate your appliances. When you plug an appliance into an outlet, an effective potential difference of 120 Vis applied to the device.
Current can be direct or alternating. There are two different types of current: direct current (de) and alternating current (ac). In direct current, charges move in only one direction with negative charges moving from a lower to higher electric potential. Hence, the conventional current is directed from the positive terminal to the negative terminal of a battery. Note, however, that the electrons actually move in the opposite direction.
604
Chapter 17
Alternating Current
Direct current
(a)
(a) The direction of direct current does not change, while (b) the direction of
~
alternating current continually changes.
Q,)
(b)
Alternating current
c

– – – – – – – – –
~ u ~——–Time (s)
Time (s)
Consider a light bulb connected to a battery. The potential difference between the terminals of a battery is fixed, so batteries always generate a direct current. In alternating current, the terminals of the source of potential difference are constantly changing sign. Hence, there is no net motion of the charge carriers in alternating current; they simply vibrate back and forth. If this vibration were slow enough, you would notice flickering in lights and similar effects in other appliances. To eliminate this problem, alternating current is made to change direction rapidly. In the United States, alternating current oscillates 60 times every second. Thus, its frequency is 60 Hz. The graphs in Figure 4.2 compare direct and alternating current. Alternating current has advantages that make it more practical for use in transferring electrical energy. For this reason, the current supplied to your home by power companies is alternating current rather than direct current.
Energy Transfer When a battery is used to maintain an electric current in a conductor, chemical energy stored in the battery is continuously converted to the electrical energy of the charge carriers. As the charge carriers move through the conductor, this electrical energy is converted to internal energy due to collisions between the charge carriers and other particles in the conductor. For example, consider a light bulb connected to a battery, as shown in Figure 4.3(a). Imagine a charge Q moving from the battery’s terminal to the light bulb and then back to the other terminal. The changes in electrical potential energy are shown in Figure 4.3(b). If we disregard the resistance of the connecting wire, no loss in energy occurs as the charge moves through the wire (A to B). But when the charge moves through the filament of the light bulb (B to C), which has a higher resistance than the wire has, it loses electrical potential energy due to collisions. This electrical energy is converted into internal energy, and the filament warms up and glows. When the charge first returns to the battery’s terminal (D), its potential energy is, by convention, zero, and the battery must do work on the charge. As the charge moves between the terminals of the battery (D to A), its electrical potential energy increases by QL1 V(where ~ Vis the potential difference across the two terminals). The battery’s chemical energy must decrease by the same amount.
Changes in Electrical Potential Energy A charge leaves the battery at A with a certain amount of electrical potential energy. The charge loses this energy while moving from B to C, and then regains the energy as it moves through the battery from D to A. (a)
(bl
~–el
A
A
B
B
a,
cii
~
~ C
D
Location of charge
Electrical Energy and Current
605
Electric power is the rate of conversion of electrical energy.
QuickLAB MATERIALS
• three small household appliances, such as a toaster, television, lamp, or stereo • household electric-company bill (optional)
Earlier in the text, power was described as the rate at which work is done. Electric power, then, is the rate at which charge carriers do work. Put another way, electric power is the rate at which charge carriers convert electrical potential energy to nonelectrical forms of energy.
Potential difference is the change in potential energy per unit of charge. .6. V= .6.PE
q
SAFETY ♦ Unplug appliances before
examination. Use extreme caution when handling electrical equipment.
ENERGY USE IN HOME APPLIANCES Look for a label on the back or bottom of each appliance. Record the power rating, which is given in units of watts (YV). Use the billing statement to find the cost of energy per kilowatt-hour. {If you don’t have a bill, choose a value between $0.05 and $0.20 per kilowatt-hour to use for your calculations.) Calculate the cost of running each appliance for 1 h. Estimate how many hours a day each appliance is used. Then calculate the monthly cost of using each appliance based on your daily estimate.
This equation can be rewritten in terms of potential energy. .6.PE = q.6. V
We can then substitute this expression for potential energy into the equation for power. .6.PE q.6. V P—— .6.t – .6.t
Because current, I, is defined as the rate of charge movement (qi .6.t), we can express electric power as current multiplied by potential difference.
Electric Power
P=lflV electric power = current x potential difference
This equation describes the rate at which charge carriers lose electrical potential energy. In other words, power is the rate of conversion of electrical energy. Recall that the SI unit of power is the watt, W. In terms of the dissipation of electrical energy, 1 Wis equivalent to 1 J of electrical energy being converted to other forms of energy per second. Most light bulbs are labeled with their power ratings. The amount of heat and light given off by a bulb is related to the power rating, also known as wattage, of the bulb. Because .6. V = IR for ohmic resistors, we can express the power dissipated by a resistor in the following alternative forms: P = l.6. V = I(IR)
= f2R
~r) .6.V = (.6.:)
P = l.6. V = (
2
The conversion of electrical energy to internal energy in a resistant material is called joule heating, also often referred to as an I 2 R loss.
606
Chapter 17
PREMIUM CONTENT
~ Interactive Demo \ : ; / HMDScience.com
Electric Power
Sample Problem E An electric space heater is connected across a 120 V outlet. The heater dissipates 1320 W of power in the form of electromagnetic radiation and heat. Calculate the resistance of the heater.
0
ANALYZE
Given:
6.V= 120V
P= 1320W Unknown:
E)
SOLVE
Because power and potential difference are given but resistance is unknown, use the form of the power equation that relates power to the other two variables.
(6. V)2
P=–R
Rearrange the equation to solve for resistance.
(6. V) 2 R= P
(120 V) 2 1320W
(120)2 J2;c2 1320 J/s
2 R = (l 0)2 J/C = 10.9V/ A 1320 C/s
Practice 1. A 1050 W electric toaster operates on a household circuit of 120 V. What is the resistance of the wire that makes up the h eating element of the toaster? 2. A small electronic device is rated at 0.25 W when connected to 120 V. What is the resistance of this device?
3. A calculator is rated at 0.10 Wand h as an internal resistance of22 n. What battery potential difference is required for this device? 4. An electric heater is operated by applying a potential difference of 50.0 V across a wire of total resistance 8.00 n. Find the current in the wire and the power rating of the heater. 5. What would the current in the heater in problem 4 be if the wire developed a short and the resistance was reduced to 0.100 0?
Electrical Energy and Current
607
Electric companies measure energy consumed in kilowatt-hours. Electric power, as discussed previously, is the rate of energy transfer. Power companies charge for energy, not power. However, the unit of energy used by electric companies to calculate consumption, the kilowatt-hour, is defined in terms of power. One kilowatt-hour (kW•h) is the energy delivered in 1 h at the constant rate of 1 kW. The following equation shows the relationship between the kilowatt-hour and the SI unit of energy, the joule: 3
60 minx~= 3.6 x 106 Wes = 3.6 x 106 J 1 kW•h x l0 W x lkW lh 1mm On an electric bill, the electrical energy used in a given period is usually stated in multiples of kilowatt-hours. An electric meter, such as the one outside your home, is used by the electric company to determine how much energy is consumed over some period of time. So, the electric company does not charge for the amount of power delivered but instead charges for the amount of energy used.
Household Appliance Power Usage he electrical energy supplied by power companies is used to generate electric currents. These currents are used to operate household appliances. When the charge carriers that make up an electric current encounter resistance, some of the electrical energy is converted to internal energy by collisions and the conductor warms up. This effect is used in many appliances, such as hair dryers, electric heaters, electric clothes dryers, steam irons, and toasters. Hair dryers contain a long, thin heating coil that becomes very hot when there is an electric current in the coil. This coil is commonly made of an alloy of the two metals nickel and chromium. This nickel chromium alloy conducts electricity poorly. In a hair dryer, a fan behind the heating coil blows air through the hot coils. The air is then heated and blown out of the hair dryer. The same principle is also used in clothes dryers and electric heaters.
608
Chapter 17
Hair dryers contain a resistive coil that becomes hot when there is an electric current in the coil. In a steam iron, a heating coil warms the bottom of the iron and also turns water into steam. An electric toaster has heating elements around the edges and in the center. When bread is loaded into the toaster, the heating coils turn on and a timer controls how long the elements remain on before the bread is popped out of the toaster. Appliances that use resistive heater coils consume a relatively large amount of electric energy. This energy consumption occurs because a large amount of current is required to heat the coils to a useful level. Because power is proportional to the current squared times the resistance, energy consumption is high.

Electrical energy is transferred at high potential differences to minimize energy loss. When transporting electrical energy by power lines, such as those shown in Figure 4.4, power companies want to minimize the I 2R loss and maximize the energy delivered to a consumer. This can be done by decreasing either current or resistance. Although wires have little resistance, recall that resistance is proportional to length. Hence, resistance becomes a factor when power is transported over long distances. Even though power lines are designed to minimize resistance, some energy will be lost due to the length of the power lines.
Electrical Power Lines Power companies transfer electrical energy at high potential differences in order to minimize the I 2R loss.
As expressed by the equation P = I 2R, energy loss is proportional to the square of the current in the wire. For this reason, decreasing current is even more important than decreasing resistance. Because P = lb. V, the same amount of power can be transported either at high currents and low potential differences or at low currents and high potential differences. Thus, transferring electrical energy at low currents, thereby minimizing the I 2R loss, requires that electrical energy be transported at very high potential differences. Power plants transport electrical energy at potential differences of up to 765 000 V. Locally, this potential difference is reduced by a transformer to about 4000 V. At your home, this potential difference is reduced again to about 120 V by another transformer.
SECTION 4 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What does the power rating on a light bulb describe?
2. If the resistance of a light bulb is increased, how will the electrical energy used by the light bulb over the same time period change? 3. The potential difference across a resting neuron in the human body is about 70 mV, and the current in it is approximately 200 µA. How much power does the neuron release? 4. How much does it cost to watch an entire World Series (21 h) on a 90.0 Wblack-and-white television set? Assume that electrical energy costs $0.070/kW•h. 5. Explain why it is more efficient to transport electrical energy at high potential differences and low currents rather than at low potential differences and high currents.
Electrical Energy and Current
609
Electron Tunneling Current is the motion of charge carriers, which can be treated as particles. But the electron has both particle and wave characteristics. The wave nature of the electron leads to some strange consequences that cannot be explained in terms of classical physics. One example is tunneling, a phenomenon whereby electrons can pass into regions that, according to classical physics, they do not have the energy to reach.
Probability Waves To see how tunneling is possible, we must explore matter waves in greater detail. De Broglie’s revolutionary idea that particles have a wave nature raised the question of how matter waves behave. In 1926, Erwin Schrodinger proposed a wave equation that described the manner in which de Broglie matter waves change in space and time. Two years later, in an attempt to relate the wave and particle natures of matter, Max Born suggested that the square of the amplitude of a matter wave is proportional to the probability of finding the corresponding particle at that location. This theory is called quantum mechanics.
Tunneling Born’s interpretation makes it possible for a particle to be found in a location that is not allowed by classical physics. Consider an electron with a potential energy of zero in the region between Oand L (region II) of Figure 1. We call this region the potential well. The electron has a potential energy of some finite value U outside this area (regions I and III). If the energy of the electron is less than U, then according to classical physics, the electron cannot escape the well without first acquiring additional energy.
Electron in a Potential Well An electron has
a potential energy of zero inside the well (region II) and a potential energy of U outside the well. According to classical physics, if the electron’s energy is less than U, it cannot escape the well without absorbing energy.
T
I
II
III
u
1 –
0
Potential
well
610
C hapter 17
L

The probability wave for this electron (in its lowest energy state) is shown in Figure 2 on the next page. Between any two points of this curve, the area under the corresponding part of the curve is proportional to the probability of finding the electron in that region. The highest point of the curve corresponds to the most probable location of the electron, while the lower points correspond to less probable locations. Note that the curve never actually meets the x-axis. This means that the electron has some finite probability of being anywhere in space. Hence, there is a probability that the electron will actually be found outside the potential well. In other words, according to quantum mechanics, the electron is no longer confined to strict boundaries because of its energy. When the electron is found outside the boundaries established by classical physics, it is said to have tunneled to its new location.
Scanning Tunneling Microscopes In 1981, Gerd Binnig and Heinrich Rohrer, at IBM Zurich, discovered a practical application of tunneling current: a powerful microscope called the scanning tunneling microscope, or STM. The STM can produce highly detailed images with resolution comparable to the size of a single atom. The image of the surface of graphite shown in Figure 3 demonstrates the power of the STM. Note that individual carbon atoms are recognizable. The smallest detail that can be discerned is about 0.2 nm, or approximately the size of an atom’s radius. A typical optical microscope has a resolution no better than 200 nm, or about half the wavelength of visible light, so it could never show the detail seen in Figure 3. In the STM, a conducting probe with a very sharp tip (about the width of an atom) is brought near the surface to be studied. According to classical physics, electrons cannot move between the surface and the tip because they lack the energy to escape either material. But according to quantum theory, electrons can tunnel across the barrier, provided the distance is small enough (about 1 nm). Scientists can apply a potential difference between the surface and the tip to make electrons tunnel preferentially from surface to tip. In this way, the tip samples the distribution of electrons just above the surface.
•iiftihlfl Probability Wave of an Electron in a Potential Well The probability curve for an electron in its lowest energy state shows that there is a certain probability of finding the electron outside the potential well.
I
I
I
I
——-r—–4——I
I I
II
I I
III
0 L Probability wave
The STM works because the probability of tunneling decreases exponentially with distance. By monitoring changes in the tunneling current as the tip is scanned over the surface, scientists obtain a sensitive measure of the topography of the electron distribution on the surface. The result is used to make images such as the one in Figure 3. The STM can measure the height of surface features to within 0.001 nm, approximately 1/ 100 of an atomic diameter. Although the STM was originally designed for imaging atoms, other practical applications are being developed. Engineers have greatly reduced the size of the STM and hope to someday develop a computer in which every piece of data is held by a single atom or by small groups of atoms and then read by an STM.
Surface of Graphite A scanning tunneling microscope (STM) was used to produce this image of the surface of graphite, a form of carbon. The contours represent the arrangement of individual carbon atoms on the surface. An STM enables scientists to see small details on surfaces with a lateral resolution of 0.2 nm and a vertical resolution of 0.001 nm.
Electrical Energy and Current
611
Superconductors and BCS Theory The resistance of many solids (other than semiconductors) increases with increasing temperature. The reason is that at a nonzero temperature, the atoms in a solid are always vibrating, and the higher the temperature, the larger the amplitude of the vibrations. It is more difficult for electrons to move through the solid when the atoms are moving with large amplitudes. This situation is somewhat similar to walking through a crowded room . It is much harder to do so when the people are in motion t han when they are standing still. If the resistance depended only on atomic vibrations, we would expect the resistance of the material that is cooled to absolute zero to go gradually to zero. Experiments have shown, however, that this does not happen. In fact, the resistances of very cold solids behave in two very different ways – either the substance suddenly begins superconducting at temperatures above absolute zero or it never superconducts, no matter how cold it gets.
Resistance from Lattice Imperfections
Resistance and Temperature The resistance of silver exhibits the behavior of a normal metal. The resistance of tin goes to zero at temperature Tc, the temperature at which tin becomes a superconductor. Temperature Dependence of Resistance
§:
“‘..,
~ 10
:ii “‘”‘ cc
Silver Object
612
Tc
10 Temperature (K)
C hapter 17
Part of the cause of this nonzero resistance, even at absolute zero, is lattice imperfection. The regular, geometric pattern of the crystal, or lattice, in a solid is often flawed. A lattice imperfection occurs when some of the atoms do not line up perfectly. Imagine you are walking through a crowded room in which the people are standing in perfect rows. It would be easy to walk through the room between two rows. Now imagine that occasionally one person stands in the middle of the aisle instead of in the row, making it harder for you to pass. This is similar to the effect of a lattice imperfection. Even in the absence of thermal vibrations, many materials exhibit a residual resistance due to the imperfect geometric arrangement of their atoms.
20
0
The graph in Figure 1 shows the temperature dependence of the resistance of two similar objects, one made of silver and the other made of tin. The temperature dependence of the resistance of the silver object is similar to that of a typical metal. At higher temperatures, the resistance decreases as the metal is cooled. This decrease in resistance suggests that the amplitude of the lattice vibrations is decreasing, as expected. But at a temperature of about 10 K, the curve levels off and the resistance becomes constant. Cooling the metal further does not appreciably lower the resistance, even though the vibrations of the metal’s atoms have been lessened.
20
Figure 1 shows that the resistance of tin jumps to zero below a certain
temperature that is well above absolute zero. A solid whose resistance is zero below a certain nonzero temperature is called a superconductor. The temperature at which the resistance goes to zero is the critical temperature of the superconductor.
BCS Theory Before the discovery of superconductivity, it was thought that all materials should have some nonzero resistance due to lattice vibrations and lattice imperfections, much like the behavior of the silver in Figure 1. The first complete microscopic theory of superconductivity was not developed until 1957. This theory is called BCS theory after the three scientists who first developed it: John Bardeen, Leon Cooper, and Robert Schrieffer. The crucial breakthrough of BCS theory is a new understanding of the special way that electrons traveling in pairs move through the lattice of a superconductor. According to BCS theory, electrons do suffer collisions in a superconductor, just as they do in any other material. However, the collisions do not alter the total momentum of a pair of electrons. The net effect is as if the electrons moved unimpeded through the lattice.
Cooper Pairs Imagine an electron moving through a lattice, such as electron 1 in Figure 2. There is an attractive force between the electron and the nearby positively charged atoms in the lattice. As the electron passes by, the attractive force causes the lattice atoms to be pulled toward the electron. The result is a concentration of positive charge near the electron. If a second electron is nearby, it can be attracted to this excess positive charge in the lattice before the lattice has had a chance to return to its equilibrium position. Through the process of deforming the lattice, the first electron gives up some of its momentum. The deformed region of the lattice attracts the second electron, transferring excess momentum to the second electron. The net effect of this two-step process is a weak, delayed attractive force between the two electrons, resulting from the motion of the lattice as it is deformed by the first electron. The two electrons travel through the lattice acting as if they were a single particle. This particle is called a Cooper pair. In BCS theory, Cooper pairs are responsible for superconductivity.
Cooper Pair The first electron deforms the lattice, and the deformation affects the second electron. The net result is as if the two electrons were loosely bound together. Such a two-electron bound state is called a Cooper pair. Lattice ion
\ Electron 2 /”

~
l
/”
____.
I
Electron 1
.,- ‘,\ ,, ‘ ,
0 -+
The reason superconductivity has been found at only low temperatures so far is that Cooper pairs are weakly bound. Random thermal motions in the lattice tend to destroy the bonds between Cooper pairs. Even at very low temperatures, Cooper pairs are constantly being formed, destroyed, and reformed in a superconducting material, usually with different pairings of electrons. Calculations of the properties of a Cooper pair have shown that this peculiar bound state of two electrons has zero total momentum in the absence of an applied electric field. When an external electric field is applied, the Cooper pairs move through the lattice under the influence of the field. However, the center of mass for every Cooper pair has exactly the same momentum. This crucial feature of Cooper pairs explains superconductivity. If one electron scatters, the other electron in a pair also scatters in a way that keeps the total momentum constant. The net result is that scattering due to lattice imperfections and lattice vibrations has no net effect on Cooper pairs. Electrical Energy and Current
613
Electrician lectricity enables us to see at night, to cook, to have heat and hot water, to communicate, to be entertained, and to do many other things. Without electricity, our lives would be unimaginably different. To learn more about being an electrician, read the interview with master electrician David Ellison. How did you become an electrician?
r—–=====:::’.=::::~==:::::::::::~!~ David Ellison teaches electrician skills to students at a local community college.
I went to junior college to learn electronics-everything from TVs and radios to radio towers and television stations. But I didn’t particularly like that sort of work. While working in a furniture factory, I got to know the master electrician for the factory, and I began working with him. Eventually he got me a job with a master electrician in town.
Since I own my business, I go from 6:00 in the morning until 9:00 or 10:00 at night. I am on call at the local hospital-I was there on Thanksgiving day. But that’s the nature of my relationship with my customers.
Most of my experience has been on the job-very little schooling. But back then, there wasn’t a lot of schooling. Now they have some good classes.
What advice do you have for a student who is interested in becoming an electrician?
What about electrical work made it more interesting than other fields?
I enjoy working with something you can’t see or smell-but if you do touch it, it’ll let you know. And if you flip a light switch, there it is. I also enjoy wiring up the switches and safeties, and solving problems when they don’t work. Where do you currently work?
I have been self-employed since 1989. About three years ago, I was invited to teach at the community college. I enjoy it. My students seem to relate better to the fact that I’m still working in the field. When I explain something to them, I can talk from recent experience. Teaching helps me stay on top of the field, too.
If you know a local electrical contractor, go talk or visit for the day. Or take a class at the local community college to see if it interests you. Some companies have their own classes, usually one night a week. Going to school gives you some technical knowledge, but getting out and doing it is still the best way to learn.
David Ellison Are there any drawbacks to your work?
Electricity is dangerous. I’ve been burned twice over 30 percent of my body. Also, the hours can be bad.
Electric Potential
SECTION 1
,: ,
1 ,
1
r: ·
• Electrical potential energy is energy that a charged object has because of its shape and its position in an electric field .
electrical potential energy electric potential
• Electric potential is electrical potential energy divided by charge.
potential difference
• Only differences in electric potential (potential differences) from one position to another are useful in calculations.
SECTION 2
Capacitance
1 ‘-, 1 L 1 • •
capacitance
• The capacitance, C, of an object is the magnitude of the charge, Q, on each of a capacitor’s plates divided by the potential difference, .6. 1/, between the plates. • A capacitor is a device that is used to store electrical potential energy. The potential energy stored in a charged capacitor depends on the charge and the potential difference between the capacitor’s two plates.
SECTION 3
Current and Resistance
“c Tc, r.-
• Current is the rate of charge movement.
electric current
• Resistance equals potential difference divided by current.
drift velocity
• Resistance depends on length, cross-sectional area, temperature, and material.
resistance
SECTION 4
Electric Power
• In direct current, charges move in a single direction; in alternating current, the direction of charge movement continually alternates. • Electric power is the rate of conversion of electrical energy. • The power dissipated by a resistor equals current squared times resistance. • Electric companies measure energy consumed in kilowatt-hours.
VARIABLE SYMBOLS
Quantities
Units
DIAGRAM SYMBOLS
Conversions
::::=====~>
Electric field
——- -+——
PEe,ectric electrical potential energy
~v
potential difference
j
joule
V volt
= Nern = kgem 2/s2 = J/C
E
: c:::::::::==::: > I
Current
+
Positive charge Negative charge
C
capacitance
F farad
= CN
current
A ampere
= C/s
R
resistance
n
ohm
= VIA
p
electric power
w watt
= J/s
Problem Solving See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
Chapter Summary
615
Electrical Potential Energy and Potential Difference REVIEWING MAIN IDEAS 1. Describe the motion and explain the energy conversions that are involved when a positive charge is placed in a uniform electric field. Be sure your discussion includes the following terms: electrical potential energy, work, and kinetic energy. 2. If a point charge is displaced perpendicular to a uniform electric field, which of the following expressions is likely to be equal to the change in electrical potential energy? a. -qEd
(q
b. O c. -k C
2 )
r2
3. Differentiate between electrical potential energy and electric potential. 4. Differentiate between electric potential and potential difference.
5. At what location in relationship to a point charge is the electric potential considered by convention to be zero?
CONCEPTUAL QUESTIONS 6. If the electric field in some region is zero, must the electric potential in that same region also be zero? Explain your answer.
7. If a proton is released from rest in a uniform electric field, does the corresponding electric potential at the proton’s changing locations increase or decrease? What about the electrical potential energy?
PRACTICE PROBLEMS For problems 8-9, see Sample Problem A.
8. The magnitude of a uniform electric field between two plates is about 1.7 x 106 N/C. If the distance between these plates is 1.5 cm, find the potential difference between the plates.
9. In the figure below, find the electric potential at point P due to the grouping of charges at the other corners of the rectangle.
PD T 8.0 µC

0.20m
_l
-8.0µC I- 0 _35 m —, -1 2µC
Capacitance REVIEWING MAIN IDEAS 10. What happens to the charge on a parallel-plate capacitor if the potential difference doubles? 11. You want to increase the maximum potential difference of a parallel-plate capacitor. Describe how you can do this for a fixed plate separation. 12. Why is Earth considered a “ground” in electric terms? Can any other object act as a ground?
CONCEPTUAL QUESTIONS 13. If the potential difference across a capacitor is doubled, by what factor is the electrical potential energy stored in the capacitor multiplied? 14. Two parallel plates are uncharged. Does the set of plates have a capacitance? Explain. 15. If you were asked to design a small capacitor with high capacitance, what factors would be important in your design? 16. A parallel-plate capacitor is charged and then disconnected from a battery. How much does the stored energy change when the plate separation is doubled?
17. Why is it dangerous to touch the terminals of a high-voltage capacitor even after the potential difference has been removed? What can be done to make the capacitor safe to handle?
PRACTICE PROBLEMS For problems 18-19, see Sample Problem B. 18. A 12.0 Vbattery is connected to a 6.0 pF parallel-plate capacitor. What is the charge on each plate?
616
Chapter 17
PRACTICE PROBLEMS
19. Two devices with capacitances of25 µF and 5.0 µFare each charged with separate 120 V power supplies. Calculate the total energy stored in the two capacitors.
For problems 32-33, see Sample Problem C.
Electric Current
32. How long does it take a total charge of 10.0 C to pass through a cross-sectional area of a copper wire that carries a current of 5.0 A?
REVIEWING MAIN IDEAS 20. What is electric current? What is the SI unit for electric current? 21. 1n a metal conductor, current is the result of moving electrons. Can charge carriers ever be positive?
22. What is meant by the term conventional current? 23. What is the difference between the drift speed of an electron in a metal wire and the average speed of the electron between collisions with the atoms of the metal wire?
24. There is a current in a metal wire due to the motion of electrons. Sketch a possible path for the motion of a single electron in this wire, the direction of the electric field vector, and the direction of conventional current.
33. A hair dryer draws a current of 9. 1 A. a. How long does it take for 1.9 x 103 C of charge to pass through the hair dryer? b. How many electrons does this amount of charge represent?
Resistance REVIEWING MAIN IDEAS 34. What factors affect the resistance of a conductor? 35. Each of the wires shown below is made of copper. Assuming each piece of wire is at the same temperature, which has the greatest resistance? Which has the least resistance? (al
25. What is an electrolyte?
(bl
26. What is the direction of conventional current in each case shown below?
(cl
(dl
36. Why are resistors used in circuit boards? (al
(bl
CONCEPTUAL QUESTIONS 27. In an analogy between traffic flow and electric current, what would correspond to the charge, Q? What would correspond to the current, I?
28. Is current ever “used up”? Explain your answer. 29. Why do wires usually warm up when an electric current is in them?
30. When a light bulb is connected to a battery, charges begin moving almost immediately, although each electron travels very slowly across the wire. Explain why the bulb lights up so quickly. 31. What is the net drift velocity of an electron in a wire that has alternating current in it?
CONCEPTUAL QUESTIONS 37. For a constant resistance, how are potential difference and current related? 38. If the potential difference across a conductor is constant, how is current dependent on resistance? 39. Using the atomic theory of matter, explain why the resistance of a material should increase as its temperature increases.
PRACTICE PROBLEMS For problems 40-42, see Sample Problem D.
40. A nichrome wire with a resistance of 15 n is connected across the terminals of a 3.0 V flashlight battery. How much current is in the wire?
Chapter Review
617
41. How much current is drawn by a television with a resistance of 35 that is connected across a potential difference of 120 V?
n
42. Calculate the current that each resistor shown below would draw when connected to a 9.0 V battery. (a)
5.0!1
(b)
2.0!1
(c)
20.0!1
53. It is estimated that in the United States (population 250 million) there is one electric clock per person, with each clock using energy at a rate of2.5 W. Using this estimate, how much energy is consumed by all of the electric clocks in the United States in a year?
54. When a small lamp is connected to a battery, the filament becomes hot enough to emit electromagnetic radiation in the form of visible light, while the wires do not. What does this tell you about the relative resistances of the filament and the wires?
PRACTICE PROBLEMS
Electric Power REVIEWING MAIN IDEAS 43. Why must energy be continuously pumped into a circuit by a battery or a generator to maintain an electric current? 44. Name at least two differences between batteries and generators.
45. What is the difference between direct current and alternating current? Which type of current is supplied to the appliances in your home? 46. Compare and contrast mechanical power with electric power. 47. What quantity is measured in kilowatt-hours? What quantity is measured in kilowatts? 48. If electrical energy is transmitted over long distances, the resistance of the wires becomes significant. Why?
49. How many joules are in a kilowatt-hour?
CONCEPTUAL QUESTIONS 50. A student in your class claims that batteries work by supplying the charges that move in a conductor, generating a current. What is wrong with this reasoning? 51 . A 60 W light bulb and a 75 W light bulb operate from 120 V Which bulb has a greater current in it? 52. Two conductors of the same length and radius are connected across the same potential difference. One conductor has twice as much resistance as the other. Which conductor dissipates more power?
618
Chapter 17
For problems 55-56, see Sample Problem E.
55. A computer is connected across a 110 V power supply. The computer dissipates 130 W of power in the form of electromagnetic radiation and heat. Calculate the resistance of the computer. 56. The operating potential difference of a light bulb is 120 V The power rating of the bulb is 75 W. Find the current in the bulb and the bulb’s resistance.
Mixed Review REVIEWING MAIN IDEAS 57. At some distance from a point charge, the electric potential is 600.0 V and the magnitude of the electric field is 200.0 N/C. Determine the distance from the charge and the charge. 58. A circular parallel-plate capacitor with a spacing of 3.0 mm is charged to produce a uniform electric field with a strength of3.0 x 106 N/C. What plate radius is required if the stored charge is -1 .0 µC? 59. A 12 V battery is connected across two parallel metal plates separated by 0.30 cm. Find the magnitude of the electric field. 60. A parallel-plate capacitor has an area of 5.00 cm2, and the plates are separated by 1.00 mm. The capacitor stores a charge of 400.0 pC. a. What is the potential difference across the plates of the capacitor? b. What is the magnitude of the uniform electric field in the region that is located between the plates?
61. A proton is accelerated from rest through a potential difference of 25 700 V. a. What is the kinetic energy of this proton in joules after this acceleration? b. What is the speed of the proton after this acceleration?
62. A proton is accelerated from rest through a potential difference of 120 V. Calculate the final speed of this proton. 63. A pair of oppositely charged parallel plates are separated by 5.33 mm. A potential difference of 600.0 V exists between the plates. a. What is the magnitude of the electric field strength in the region that is located between the plates? b. What is the m agnitude of the force on an electron that is in the region between the plates at a point that is exactly 2.90 mm from the positive plate? c. The electron is moved to the negative plate from an initial position 2.90 mm from the positive plate. What is the change in electrical potential energy due to the movement of this electron? 64. The three charges shown below are located at the vertices of an isosceles triangle. Calculate the electric potential at the midpoint of the base if each one of the charges at the corners has a magnitude of 5.0 x 10- 9 C.
67. A proton is accelerated through a potential difference of 4.5 X 106 V. a. How much kinetic energy has the proton acquired? b. If the proton started at rest, how fast is it moving? 68. Each plate on a 3750 pF capacitor carries a charge with a magnitude of 1.75 x 10- 8 C. a. What is the potential difference across the plates when the capacitor has been fully charged? b. If the plates are 6.50 x 10- 4 m apart, what is the magnitude of the electric field between the two plates? 69. A net charge of 45 mC passes through the crosssectional area of a wire in 15 s. a. What is the current in the wire? b. How many electrons pass the cross-sectional area in 1.0 min? 70. The current in a lightning bolt is 2.0 x 105 A. How many coulombs of charge pass through a crosssectional area of the lightning bolt in 0.50 s?
71 . A person notices a mild shock if the current along a path through the thumb and index finger exceeds 80.0 µA. Determine the maximum allowable potential difference without shock across the thumb and index finger for the following: a. a dry-skin resistance of 4.0 x 105 n b. a wet-skin resistance of2.0 x 103 n 72. A color television has a power rating of 325 W. How much current does this set draw from a potential difference of 120 V?
-q -q l-2.0 cm ‘”1 65. A charge of -3.00 x 10- 9 C is at the origin of a coordinate system, and a charge of 8.00 x 10- 9 C is on the x-axis at 2.00 m. At what two locations on the x-axis is the electric potential zero? (Hint: One location is between the charges, and the other is to the left of the y-axis.)
66. An ion is displaced through a potential difference of 60.0 V and experiences an increase of electrical potential energy of 1.92 x 10- 17 J. Calculate the charge on the ion.
73. An X-ray tube used for cancer therapy operates at 4.0 MV with a beam current of 25 mA striking a metal target. Calculate the power of this beam. 74. The mass of a gold atom is 3.27 x 10- 25 kg. If 1.25 kg of gold is deposited on the negative electrode of an electrolytic cell in a period of 2.78 h, what is the current in the cell in this period? Assume that each gold ion carries one elementary unit of positive charge. 75. The power supplied to a typical black-and-white television is 90.0 W when the set is connected across a potential difference of 120 V. How much electrical energy does this set consume in 1.0 h?
Chapter Review
619
76. A color television set draws about 2.5 A of current when connected to a potential difference of 120 V. How much time is required for it to consume the same energy that the black-and-white model described in item 75 consumes in 1.0 h? 77. The headlights on a car are rated at 80.0 W. If they are connected to a fully charged 90.0 A•h, 12.0 V battery, how long does it take the battery to completely discharge?
78. The current in a conductor varies over time as shown in the graph below. a. How many coulombs of charge pass through a cross section of the conductor in the time interval t = Oto t = 5.0 s? b. What constant current would transport the same total charge during the 5.0 s interval as does the actual current?
79. Birds resting on high-voltage power lines are a common sight. A certain copper power line carries a current of 50.0 A, and its resistance per unit length is 1.12 x 10- 5 O/m. If a bird is standing on this line with its feet 4.0 cm apart, what is the potential difference across the bird’s feet?
80. An electric car is designed to run on a bank of batteries with a total potential difference of 12 V and a total energy storage of2.0 x 107 J. a. If the electric motor draws 8.0 kW, what is the current delivered to the motor? b. If the electric motor draws 8.0 kW as the car moves at a steady speed of20.0 m / s, how far will the car travel before it is “out of juice”?
Current in a Conductor 6
I \ I \ 0
0
1
2
3
4
5
Time (s)
Resistance and Current When you install a 100 W light bulb, what is the resistance of and current passing through this light bulb? The answer to this question and similar questions is found in two equations that you learned earlier in this chapter: (~V)2
P= -R- and P=I~V
These equations describe the power dissipated by a resistor. In these equations, P is the power in watts, .::1 V is the potential difference in volts, R is the resistance in ohms, and I is the current in amperes. 620
Chapter 17
In this graphing calculator activity, you will calculate a series of tables of resistance and current versus potential difference for various values of dissipated power. By analyzing these tables, you will better understand the relationships between power, potential difference, resistance, and current. (You will also be able to answer the question about the 100 W light bulb.) Go online to HMDScience.com to find this graphing calculator activity.
ALTERNATIVE ASSESSMENT 1. Imagine that you are assisting nuclear scientists who
need to accelerate electrons between electrically charged plates. Design and sketch a piece of equipment that could accelerate electrons to 107 m /s. What should the potential difference be between the plates? How would protons move inside this device? What would you change in order to accelerate the electrons to 100 m i s?
2. Tantalum is an element widely used in electrolytic capacitors. Research tantulurn and its properties. Where on Earth is it found? In what form is it found? How expensive is it? Present your findings to the class in the form of a report, poster, or computer presentation. 3. Research an operational maglev train, such as the commercially operating train in Shangai, China, or the demonstration trains in Japan or Germany. Alternatively, research a maglev system that is under construction or being proposed for development. Investigate the cost of development, major hurdles that had to be overcome or will need to be overcome, and the advantages and disadvantages of the train. Suppose that there is a proposal for a maglev train in your area. Develop an argument for or against the proposed train, based on your research. Write a paper to convince other citizens of your position.
4. Visit an electric parts or electronic parts store or consult a print or online catalog to learn about different kinds of resistors. Find out what the different resistors look like, what they are made of, what their resistance is, how they are labeled, and what they are used for. Summarize your findings in a poster or a brochure entitled A Consumer’s Guide to Resistors. 5. The units of measurement you learned about in this chapter were named after four famous scientists: Andre-Marie Ampere, Michael Faraday, Georg Simon Ohm, and Alessandro Volta. Research their lives, works, discoveries, and contributions. Create a presentation about one of these scientists. The presentation can be in the form of a report, poster, short video, or computer presentation. 6. A thermistor is a device that changes its resistan ce as
its temperature changes. Thermistors are often used in digital thermometers. Another common temperature sensor is the thermocouple, which generates a potential difference that depends on its temperature. Many thermostats use thermistors or thermocouples to regulate temperature. Research how thermistors or thermocouples work, and how they are used in one of the applications mentioned above. Create a slideshow or a poster with the results of your research.
Chapter Review
621
MULTIPLE CHOICE
Use the following passage to answer questions 3-4.
Use the diagram below to answer questions 1-2.
A proton (q = 1.6 x 10- 19 C) moves 2.0 x 10-6 min the direction of an electric field that has a magnitude of 2.0 N/ C.
E
3. What is the change in the electrical potential energy associated with the proton? A. – 6.4 X 10-25 J B. – 4.0 X 10-6 v C. + 6.4 X 10- 25 J D. + 4.0 x 10-6 v
A-
1. What changes would take place if the electron moved from point A to point B in the uniform electric field? A. The electron’s electrical potential energy would increase; its electric potential would increase. B. The electron’s electrical potential energy would increase; its electric potential would decrease. C. The electron’s electrical potential energy would decrease; its electric potential would decrease. D. Neither the electron’s electrical potential energy nor its electric potential would change. 2. What changes would take place if the electron moved from point A to point C in the uniform electric field? F. The electron’s electrical potential energy would increase; its electric potential would increase. G. The electron’s electrical potential e nergy would increase; its electric potential would decrease. H. The electron’s electrical potential energy would decrease; its electric potential would decrease. J. Neither the electron’s electrical potential energy nor its electric potential would change.
622
Chapter 17
4. What is the potential difference between the proton’s starting point and ending point? f. – 6.4 X 10- 25 J G. – 4.0 x 10- 6 v H. + 6.4 X 10- 25 J J. + 4.0 X 10- By
5. If the negative terminal of a 12 V battery is grounded, what is the potential of the positive terminal? A. -12V
B. +ov C. +6V D.
+ 12V
6. If the area of the plates of a parallel-plate capacitor is doubled while the spacing between the plates is halved, how is the capacitance affected? F. C is doubled G. C is increased by four times H. C is decreased by¼ J. C does not change
. TEST PREP
Use the following passage to answer questions 7-8.
SHORT RESPONSE
A potential difference of 10.0 V exists across the plates of a capacitor when the charge on each plate is 40.0 µC.
13. Electrons are moving from left to right in a wire.
7. What is the capacitance of the capacitor? A. 2.00 x 10- 4 F B. 4.00 x 10- 4 F C. 2.00 x 10- 6 F D.4.00 x 1o- 6 F8. How much electrical potential energy
is stored in the capacitor? f. 2.00 X 10- 4 J G. 4.00 x 10- 4 J H. 2.00 x 10- 6 J J. 4.00 X 10- 6 J 9. How long does it take 5.0 C of charge to pass
through a given cross section of a copper wire if I = 5.0A? A. 0.20 s B. I.Os C. 5.0 s D. 25 s 10. A potential difference of 12 V produces a current of 0.40 A in a piece of copper wire. What is the resis-
tance of the wire? F. 4.8 n G. 12n H. 30D J. 36D 11. How many joules of energy are dissipated by a 50.0 W light bulb in 2.00 s? A. 25.0 J B. 50.0 J C. 100 J D. 200 J
No other charged particles are moving in the wire. In what direction is the conventional current? 14. What is drift velocity, and how does it compare with
the speed at which an electric field travels through a wire? 15. List four factors that can affect the resistance of
a wire.
EXTENDED RESPONSE 16. A parallel-plate capacitor is made of two circular plates, each of which has a diameter of 2.50 x 10- 3 m.
The plates of the capacitor are separated by a space of 1.40 X 10- 4 m. a. Assuming that the capacitor is operating in a vacuum and that the permittivity of a vacuum (80 = 8.85 x 10- 12 C2 / N• m 2 ) can be used, determine the capacitance of the capacitor. b. How much charge will be stored on each plate of the capacitor when the capacitor’s plates are connected across a potential difference of 0.12 V? c. What is the electrical potential energy stored in the capacitor when fully charged by the potential difference of 0.12 V? d. What is the potential difference between a point midway between the plates and a point that is 1.10 x 10- 4 m from one of the plates? e. If the potential difference of 0.12 V is removed from the circuit and the circuit is allowed to discharge until the charge on the plates has decreased to 70.7 percent of its fully charged value, what will the potential difference across the capacitor be?
12. How much power is needed to operate a radio that
draws 7.0 A of current when a potential difference of 115 Vis applied across it? F. 6.1 x 10- 2 W G. 2.3 x 10°w H. 1.6 X 101 W J. 8.0 x 102 w
Test Tip If at any point while taking a test you do not clearly understand the directions or the wording of a question, raise your hand and ask for help.
Standards-Based Assessment
623
Hybrid Electric Vehicles At the start of the twentieth century, electric-powered vehicles and gasoline-powered vehicles were competing for dominance in the emerging automobile industry. Electric cars were considered more reliable, and certainly quieter and less polluting, than gasoline-powered cars. However, they could go only a few miles before they needed recharging, so they were suitable only for use over short distances. As more roads were paved and as more people wanted to travel farther, electric cars were abandoned in favor of cars that burned gasoline in internal combustion engines (ICEs). Identify the Problem: Pollution and Price
As the twentieth century progressed, industry spread, and the number of cars on the road increased. The air in North America became more polluted, and people searched for ways to reduce the pollution and its harmful effects on human health. ICEs emit nitrogen oxides, carbon monoxide, and unburned hydrocarbons-all of which, along with ozone, make up a major part of urban air pollution. In addition, ICEs give off large quantities of carbon dioxide, which contributes to Earth’s greenhouse effect and increases the threat of global warming. The price of gasoline has also forced people to look for alternatives to ICEs. In the 1970s, a global energy crisis emerged as several oil-exporting countries cut off their oil exports for political and economic reasons. Oil and gas prices rose dramatically, and many people suddenly had no access to gasoline or could no longer afford it. Although the crisis subsided, worldwide economic and political instability, along with a growing awareness that global oil supplies are finite, has kept oil and gas prices high.
624
Brainstorm Solutions
In recent decades, federal and state laws have required industries and businesses-from steelmakers to dry cleaners-to limit polluting emissions. Regulations and incentives have also been put in place to increase the fuel-efficiency and reduce the emissions of passenger cars. Although overall air quality has improved as a result of these efforts, air pollution is still a serious problem, largely due to emissions from vehicles with ICEs. As the problems with air pollution and rising oil prices have become more apparent, people have started to reexamine alternatives to gasoline-powered ICEs. In the 1990s, several electric vehicles (EVs), which run solely on electricity, were developed for passenger use. While the performance of these EVs was comparable to gasoline-powered cars, they typically had driving ranges of only 80-240 km (50-150 miles) and were more expensive than gasoline-powered models. A solution was still needed that involved designing a nonpolluting car that could travel greater distances and be affordable to buy and operate.
Select a Solution
Test and Evaluate
In the mid- to late 1990s, several automakers designed and developed hybrid electric vehicles (HEVs), which use electricity in combination with a gasoline engine. HEVs are a step toward solving the problems with air pollution and the price of oil. HEVs have been more commercially successful than pure EVs. Today, practically every car manufacturer offers an HEV model.
Both series and parallel HEVs have longer driving ranges than their pure EV counterparts. Some HEVs can get as many as 50 miles per gallon in combined city/highway driving. As a result, these HEVs can travel over 600 miles in city driving without refueling. Furthermore, because the ICE charges the battery, an HEV never needs to be plugged in. As a result, the car’s owner does not have to worry about power failures or paying more for electricity.
Build a Prototype
All HEVs combine the power of a battery-driven electric motor with the power of an ICE. However, different models do this in different ways. In a series design, an electric motor powers the car directly, while the ICE serves only to power a generator that recharges the battery for the electric motor. With this design, the ICE is very efficient because it is always recharging the battery. However, series-design HEVs have less on-demand power for acceleration. In a parallel design, both the electric motor and the ICE attach directly to the drive train to power the wheels. With this design, the electric motor provides the primary power when driving in stop-and-go traffic. The ICE kicks in at higher speeds, when the ICE is more efficient. Unlike conventional gasolinepowered vehicles, parallel-design HEVs get better gas mileage and produce fewer emissions in town than they do on the highway.
Design Your Own Conduct Research
1. Go to a local car dealer and ask about hybrid electric vehicles. Do they have any HEV models available? Are they going to offer any new HEV models in the future? Do these models use a series design, a parallel design, or another type of design? Evaluate and Communicate
Redesign to Improve
Some HEVs combine these designs. For limited distances and speeds under 25 mph, the HEV can run without the engine running. However, for extended distances and speeds greater than 25 mph, the engine provides power directly and must run. In addition to maximizing the efficiency of the electric motor and the engine, many HEVs also have regenerative braking systems that recapture some of the power lost during braking and use it to recharge the battery. The result is a more efficient car that produces fewer emissions and gets better gas mileage than a comparable car powered solely by gasoline.
2. The federal government and some states offer tax deductions and other incentives for people who own HEVs or other alternative-fuel vehicles. Hold a discussion or debate on this question: “Should the government spend taxpayers’ money to subsidize the purchase of alternative-fuel vehicles that people otherwise might not buy?” Build a Prototype
3. Check the Internet for information on HEV technology. Use this information to build a model that demonstrates how the engine, battery, motor, and braking system work as a unit in an HEV.
625
SECTION 1 Objectives ►
I ► I ►
Interpret and construct circuit diagrams. Identify circuits as open or closed. Deduce the potential difference across the circuit load, given the potential difference across the battery’s terminals.
Schematic Diagrams and Circuits Key Terms schematic diagram
electric circuit
Schematic Diagrams Take a few minutes to examine the battery and light bulb in Figure 1.1 (a); then draw a diagram of each element in the photograph and its connection. How easily could your diagram be interpreted by someone else? Could the elements in your diagram be used to depict a string of decorative lights, such as those draped over the trees of the San Antonio Riverwalk?
schematic diagram a representation of a circuit that uses lines to represent wires and different symbols to represent components
A diagram that depicts the construction of an electrical apparatus is called a schematic diagram. The schematic diagram shown in Figure 1.1 (b) uses symbols to represent the bulb, battery, and wire from Figure 1.1(a). Note that these same symbols can be used to describe these elements in any electrical apparatus. This way, schematic diagrams can be read by anyone familiar with the standard set of symbols. Reading schematic diagrams allows us to determine how the parts in an electrical device are arranged. In this chapter, you will see how the arrangement of resistors in an electrical device can affect the current in and potential difference across the other elements in the device. The ability to interpret schematic diagrams for complicated electrical equipment is an essential skill for solving problems involving electricity. As shown in Figure 1.2 on the next page, each element used in a piece of electrical equipment is represented by a symbol in schematic diagrams that reflects the element’s construction or function. For example, the schematic-diagram symbol that represents an open switch resembles the open knife switch that is shown in the corresponding photograph. Note that Figure 1.2 also includes other forms of schematic-diagram symbols; these alternative symbols will not be used in this book.
A Battery and Light Bulb (a) When this battery is connected to a light bulb, the potential difference across the battery generates a current that illuminates the bulb. (b) The connections between the light bulb and battery can be represented in a schematic diagram.
628
Chapter 18
Component
Symbol used in this book
Other forms of this symbol
Wire or conductor
Explanation
• Wires that connect elements are conductors. • Because wires offer negligible resistance, they are represented by straight lines
7 Resistor or circuit load
• Resistors are shown having multiple bends, illustrating resistance to the movement of charges.
Bulb or lamp
• The multiple bends of the filament indicate that the light bulb behaves as a resistor. • The symbol for the filament of the bulb is often enclosed in a circle to emphasize the enclosure of a resistor in a bulb.
\ I ;
_0 Plug
®
• The plug symbol looks like a container for two prongs. • The emf between the two prongs of a plug is symbolized by lines of unequal length.
—;1 c-
• Differences in line length indicate a potential difference between positive and negative terminals of the battery. • The longer line represents the positive terminal of the battery.
Battery
—;1 1l1lcMultiple cells
Switch
/
• The small circles indicate the two places where the switch makes contact with the wires. Most switches work by breaking only one of the contacts, not both.
oOpen
–0——0–
Closed
Capacitor
– H-
—;1~
• The two parallel plates of a capacitor are symbolized by two parallel lines of equal length. • One curved line indicates that the capacitor can be used with only direct current sources with the polarity as shown.
Circuits and Circuit Elements
629
Electric Circuits A Complete Circuit When all electrical components are connected, charges can move freely in a circuit. The movement of charges in a circuit can be halted by opening the switch.
Think about how you get the bulb in Figure 1.3 to light up. Will the bulb stay lit if the switch is opened? Is there any way to light the bulb without connecting the wires to the battery? The filament of the light bulb acts as a resistor. When a wire connects the terminals of the battery to the light bulb, as shown in Figure 1.3, charges built up on one terminal of the battery have a path to follow to reach the opposite charges on the other terminal. Because there are charges moving through the wire, a current exists. This current causes the filament to heat up and glow. Together, the bulb, battery, switch, and wire form an electric circuit. An electric circuit is a path through which charges can flow. A schematic diagram for a circuit is sometimes called a circuit diagram.
electric circuit a set of electrical components connected such that they provide one or more complete paths for the movement of charges
Any element or group of elements in a circuit that dissipates energy is called a load. A simple circuit consists of a source of potential difference and electrical energy, such as a battery, and a load, such as a bulb or group of bulbs. Because the connecting wire and switch have negligible resistance, we will not consider these elements as part of the load. In Figure 1.3, the path from one battery terminal to the other is complete, a potential difference exists, and electrons move from one terminal to the other. In other words, there is a closed-loop path for electrons to follow. This is called a closed circuit. The switch in the circuit in Figure 1.3 must be closed in order for a steady current to exist. Without a complete path, there is no charge flow and therefore no current. This situation is an open circuit. If the switch in Figure 1.3 were open, as shown in Figure 1.2, the circuit would be open, the current would be zero, and the bulb would not light up.
Bird on a Wire Why is it possible for a bird to
be perched on a high-voltage wire without being electrocuted? (Hint: Consider the potential difference between the bird’s two feet.) Parachutist on a Wire Suppose a parachutist
lands on a high-voltage wire and grabs the wire in preparation to be rescued. W ill the parachutist
630
Chapter 18
be electrocuted? If the wire breaks, why should the parachutist let go of the wire as it falls to the ground? (Hint: First consider the potential difference between the parachutist ‘s two hands holding the wire. Then consider the potential difference between the wire and the ground.)
CFLs and LEDs he most familiar of light bulbs, incandescent bulbs, may soon be a relic of the past. Thomas Edison first invented these bulbs in 1879 and they have been in use ever since. They work by heating a small metal filament that glows and produces light. Although incandescent bulbs give off very warm and pleasant light, they are extremely inefficient. Nearly 90% of the energy they use is converted into heat and only 10% is converted into light. New federal law requires that by 2014 all bulbs be at least 30% more efficient. Two new types of bulbs look to replace incandescent bulbs.
1r
The first type of light bulb is called compact fluorescent light (or CFL for short). CFLs work by running an electrical current through a tube that contains a mixture of gases. The atoms of gas absorb energy from the electricity and emit ultraviolet light. Humans, however, cannot see ultraviolet light. What happens next is that the ultraviolet light hits the surface of the tube that has been coated with a chemical that absorbs the ultraviolet light and emits visible light.
~
electrons here release no energy, so LEDs are more energy efficient than both incandescent and CFLs. In addition, because LEDs are made of solid material, they can be very small and are very durable so they last a long time. Although both CFLs and LEDS cost considerably more than incandescent bulbs, they use much less energy to produce the same amount of light. In addition, they have a much longer life span. When both of these factors are taken into account, replacing your incandescent bulbs with CFLs or LEDs may cost more up front, but they end up saving money over the life of the bulb.
The second type of light bulb is called light-emitting diode (or LED for short). LEDs work by moving electrons and protons in a solid piece of material called a semiconductor. As the electrons move through this material they lose energy and release light. The
Short circuits can be hazardous.
u
.5
lE.r: e ~ ~
a::
s&. “‘”‘~ E
~:,
“‘©l
Without a load, such as a bulb or other resistor, the circuit contains little resistance to the movem ent of charges. This situation is called a short circuit. For example, a short circuit occurs wh en a wire is connected from one terminal of a battery to the other by a wire with little resistance. This commonly occurs when uninsulated wires connected to different terminals come into contact with each other. When short circuits occur in the wiring of your home, the increase in current can become unsafe. Most wires cannot withstand the increased current, and th ey begin to overheat. The wire’s insulation may even melt or cause a fire.
-c-
a
Circ uits and Circuit Elements
631
QuickLAB MATERIALS
• • • •
1 miniature light bulb 1 D-cell battery wires rubber band or tape
SAFETY ♦ Do not perform this lab
with any batteries or electrical devices other than those listed here. Never work with electricity near water. Be sure the floor and all work surfaces are dry.
SIMPLE CIRCUITS Connect the bulb to the battery using two w ires, using a rubber band or tape to hold the wire to the battery. Once you have gotten the bulb to light, try different arrangements to see whether there is more than one way to get the bulb to light. Can you make the bulb light using just one wire? Diagram each arrangement that you try, and note whether it produces light. Explain exactly which parts of the bulb, battery, and wire must be connected for the light bulb to produce light.
The source of potential difference and electrical energy is the circuit’s emf. Will a bulb in a circuit light up if you remove the battery? Without a potential difference, there is no charge flow and no current. The battery is necessary because the battery is the source of potential difference and electrical energy for the circuit. So, the bulb must be connected to the battery to be lit. Any device that increases the potential energy of charges circulating in a circuit is a source of emf, or electromotive force. The emf is the energy per unit charge supplied by a source of electric current. Think of such a source as a “charge pump” that forces electrons to move in a certain direction. Batteries and generators are examples of emf sources.
For conventional current, the terminal voltage is less than the emf. Look at the battery attached to the light bulb in the circuit shown in Figure 1.4. As shown in the inset, instead of behaving only like a source of emf, the battery behaves as if it contains both an emf source and a resistor. The battery’s internal resistance to current is the result of moving charges colliding with atoms inside the battery while the charges are traveling from one terminal to the other. Thus, when charges move conventionally in a battery, the potential difference across the battery’s terminals, the terminal voltage, is actually slightly less than the emf. Unless otherwise stated, any reference in this book to the potential difference across a battery should be thought of as the potential difference measured across the battery’s terminals rather than as the emf of the battery. In other words, all examples and end-of-chapter problems will disregard the internal resistance of the battery.
A Battery’s Internal Resistance
(a) A battery in a circuit behaves as if it contains both (b) an emf source and (c) an internal resistance. For simplicity’s sake, in problem solving it will be assumed that this internal resistance is insignificant.
s (c) (b)
632
Chapter 18
Small internal resistance

Potential difference across a load equals the terminal voltage. When charges move within a battery from one terminal to the other, the chemical energy of the battery is converted to the electrical potential energy of the charges. As charges move through the circuit, their , electrical potential energy is converted to other forms of energy. For instance, when the load is a resistor, the electrical potential energy of the charges is converted to the internal energy of the resistor and dissipated as thermal energy and light energy. Because energy is conserved, the energy gained and the energy lost must be equal for one complete trip around the circuit ( starting and ending at the same place). Thus, the electrical potential energy gained in the battery must equal the energy dissipated by the load. Because the potential difference is the measurement of potential energy per amount of charge, the potential increase across the battery must equal the potential decrease across the load.
~
SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Identify the types of elements in the schematic diagram illustrated in Figure 1.5 and the number of each type.
2. Using the symbols listed in Figure 1.2, draw a schematic diagram of a working circuit that contains two resistors, an emf source, and a closed switch. 3. In which of the circuits pictured below will there be no current?
4. If the potential difference across the bulb in a certain flashlight is 3.0 V, what is the potential difference across the combination of batteries used to power it?
Critical Thinking 5. In what forms is the electrical energy that is supplied to a string of decorative lights dissipated? Circuits and Circuit Elements
633
S.T.E.M. Transistors and Integr ated Circuits ou may have heard about objects called semiconductors. Semiconductors are materials that have properties between those of insulators and conductors. They play an important role in today’s world, as they are the foundation of circuits found in virtually every electronic device. Most commercial semiconductors are made primarily of either silicon or germanium. The conductive properties of semiconductors can be enhanced by adding impurities to the base material in a process called doping. Depending on how a semiconductor is doped, it can be either an n-type semiconductor or a p-type semiconductor. N-type semiconductors carry negative charges (in the form of electrons), and p-type semiconductors carry positive charges. The positive charges in a p-type semiconductor are not actually positively charged particles. They are “holes” created by the absence of electrons. The most interesting and useful properties of semiconductors emerge when more than one type of semiconductor is used in a device. One such device is a diode, which is made by placing a p-type semiconductor next to an n-type semiconductor. The junction where the two types meet is called a p-n junction. A diode has almost infinite resistance in one direction and
Motherboards, such as the one pictured above, include multiple transistors. 634
nearly zero resistance in the other direction. One useful application of diodes is the conversion of alternating current to direct current. A transistor is a device that contains three layers of semiconductors. Transistors can be either pnp transistors or npn transistors, depending on the order of the layers. A transistor is like two diodes placed back-to-back. You might think this would mean that no current exists in a transistor, as there is infinite resistance at one or the other of the p-n junctions. However, if a small voltage is applied to the middle layer of the transistor, the p-n junctions are altered in such a way that a large amount of current can be in the transistor. As a result, transistors can be used as switches, allowing a small current to turn a larger current on or off. Transistor-based switches are the building blocks of computers. A single switch turned on or off can represent a binary digit, or bit, which is always either a one or a zero. An integrated circuit is a collection of transistors, diodes, capacitors, and resistors embedded in a single piece of silicon, known as a chip. Much of the rapid progress in the computer and electronics industries in the past few decades has been a result of improvements in semiconductor technologies. These improvements allow smaller and smaller transistors and other circuit elements to be placed on chips. A typical computer motherboard, such as the one shown here, contains several integrated circuits, each one containing several million transistors.
SECTION 2
Resistors in Series or in Parallel
Objectives ►
Calculate the equivalent resistance for a circuit of resistors in series, and find the current in and potential difference across each resistor in the circuit.

Calculate the equivalent resistance for a circuit of resistors in parallel, and find the current in and potential difference across each resistor in the circuit.
Key Terms series parallel
Resistors in Series In a circuit that consists of a single bulb and a battery, the potential difference across the bulb equals the terminal voltage. The total current in the circuit can be found using the equation ~ V = IR. What happens when a second bulb is added to such a circuit, as shown in Figure 2.1? When moving through this circuit, charges that pass through one bulb must also move through the second bulb. Because all charges in the circuit must follow the same conducting path, these bulbs are said to be connected in series.
series describes two or more components of a circuit that provide a single path for current
Resistors in series carry the same current. Light-bulb filaments are resistors; thus, Figure 2.1(b) represents the two bulbs in Figure 2.1 (a) as resistors. Because charge is conserved, charges cannot build up or disappear at a point. For this reason, the amount of charge that enters one bulb in a given time interval equals the amount of charge that exits that bulb in the same amount of time. Because there is only one path for a charge to follow, the amount of charge entering and exiting the first bulb must equal the amount of charge that enters and exits the second bulb in the same time interval. Because the current is the amount of charge moving past a point per unit of time, the current in the first bulb must equal the current in the second bulb. This is true for any number of resistors arranged in series. When many
resistors are connected in series, the current in each resistor is the same.
IJ0ii;jfjj Two Bulbs in Series These two light bulbs are connected in series. Because light-bulb filaments are resistors, (a) the two bulbs in this series circuit can be represented by (b) two resistors in the schematic diagram shown on the right.
Circuits and Circuit Elements
635
The total current in a series circuit depends on how many resistors are present and on how much resistance each offers. Thus, to find the total current, first use the individual resistance values to find the total resistance of the circuit, called the equivalent resistance. Then the equivalent resistance can be used to find the current.
The equivalent resistance in a series circuit is the sum of the circuit’s resistances. As described in Section 1, the potential difference across the battery, ~ V, must equal the potential difference across the load,~ V1 + ~ Vz, where ~ V1 is the potential difference across R1 and ~ V2 is the potential difference across R2 • ~V=~V1 +~V2
According to ~ V = IR, the potential difference across each resistor is equal to the current in that resistor multiplied by the resistance. ~V= I 1R 1 + I2 R 2
Because the resistors are in series, the current in each is the same. For this reason, I 1 and I 2 can be replaced with a single variable for the current, I. ~V= I(R 1 + R2)
Equivalent Resistance for a Series Circuit (a) The two resistors in the actual circuit have the same effect on the current in the ci rcuit as (b) the equivalent resistor.
Finding a value for the equivalent resistance of the circuit is now possible. If you imagine the equivalent resistance replacing the original two resistors, as shown in Figure 2.2, you can treat the circuit as if it contains only one resistor and use ~ V = IR to relate the total potential difference, current, and equivalent resistance. ~V= I(Req)
Now set the last two equations for~ V equal to each other, and divide by the current. I
I
Req=R1 +R2
Thus, the equivalent resistance of the series combination is the sum of the individual resistances. An extension of this analysis shows that the equivalent resistance of two or more resistors connected in series can be calculated using the following equation. Resistors in Series
Req=R1 +R2+R3 ••. Equivalent resistance equals the total of individual resistances in series.
Because Req represents the sum of the individual resistances that have been connected in series, the equivalent resistance ofa series combination of resistors is always greater than any individual resistance.
636
Chapter 18
To find the total current in a series circuit, first simplify the circuit to a single equivalent resistance using the boxed equation above; then use b.. V = IR to calculate the current. I= b..V Req
Because the current in each bulb is equal to the total current, you can also use b.. V = IR to calculate the potential difference across each resistor. b.. V 1 = IR1
and
b.. V2
= IR2
The method described above can be used to find the potential difference across resistors in a series circuit containing any number of resistors.
Resistors in Series Sample Problem A A 9.0 V battery is connected to four light bulbs, as shown at right. Find the equivalent resistance for the circuit and the current in the circuit.
4.o n
d~
s.o n
2.00
0
ANALYZE
Given:
~V=9.0V
R 1 =2.0O
R2 = 4.0 0
R3
R4
= 5.0 0
= 7.0 0
Unknown:
? R eq – .
Diagram:
4.o n
I=?
s.o n
7.on
~ n 2.0
E)
PLAN
9.0V
Choose an equation or situation:
Because the resistors are connected end to end, they are in series. Thus, the equivalent resistance can be calculated with the equation for resistors in series.
Req =R1 +R2 +R3 ••· The following equation can be used to calculate the current.
~V= IReq Rearrange the equation to isolate the unknown: No rearrangement is necessary to calculate Req, but b.. V = IReq must be
rearranged to calculate current.
I = ~V Req
G·id!i,M§-►
Circuits and Circuit Elements
637
Resistors in Series
E)
SOLVE
(continued)
Substitute the values into the equation and solve: R eq
= 2.0 fl+ 4.0 fl+ 5.0 fl+ 7.0 fl
I Req = 18.0 fl I Substitute the equivalent resistance value into the equation for current.
I = ~V = 9.0V R eq 18.0 fl
II= 0.50A I 0
CHECKYOUR WORK
For resistors connected in series, the equivalent resistance should be greater than the largest resistance in the circuit.
18.0 n > 7.o n Practice 1. A 12.0 V storage battery is connected to three resistors, 6.75 n, 15.3 n, and 21.6 n, respectively. The resistors are joined in series.
a. Calculate the equivalent resistance. b. What is the current in the circuit? 2. A 4.0 n resistor, an 8.0 n resistor, and a 12.0 n resistor are connected in series with a 24.0 V battery. a. Calculate the equivalent resistance.
b. Calculate the current in the circuit. c. What is the current in each resistor? 3. Because the current in the equivalent resistor of Sample Problem A is 0.50 A, it must also be the current in each resistor of the original circuit. Find the potential difference across each resistor.
4. A series combination of two resistors, 7.25 n and 4.03 n, is connected to a 9.00 V battery. a. Calculate the equivalent resistance of the circuit and the current. b. What is the potential difference across each resistor? 5. A 7.0 n resistor is connected in series with another resistor and a 4.5 V battery. The current in the circuit is 0.60 A. Calculate the value of the unknown resistance.
6. Several light bulbs are connected in series across a 115 V source of emf. a. What is the equivalent resistance if the current in the circuit is 1.70 A?
b. If each light bulb has a resistance of 1.50 n, how many light bulbs are in the circuit?
638
Chapter 18
Series circuits require all elements to conduct. What happens to a series circuit when a single bulb burns out? Consider what a circuit diagram for a string oflights with one broken filament would look like. As the schematic diagram in Figure 2.3 shows, the broken , filament means that there is a gap in the conducting pathway used to make up the circuit. Because the circuit is no longer closed, there is no current in it and all of the bulbs go dark. Why, then, would anyone arrange resistors in series? Resistors can be placed in series with a device in order to regulate the current in that device. In the case of decorative lights, adding an additional bulb will decrease the current in each bulb. Thus, the filament of each bulb need not withstand such a high current. Another advantage to placing resistors in series is that several lesser resistances can be used to add up to a single greater resistance that is unavailable. Finally, in some cases, it is important to have a circuit that will have no current if any one of its component parts fails. This technique is used in a variety of contexts, including some burglar alarm systems.
Burned-Out Filament in a Series Circuit A burned-out filament in a bulb has the same effect as an open switch. Because this series circuit is no longer complete, there is no current in the circuit.
Resistors in Parallel As discussed above, when a single bulb in a series light set burns out, the entire string of lights goes dark because the circuit is no longer closed. What would happen if there were alternative pathways for the movement of charge, as shown in Figure 2.4? A wiring arrangement that provides alternative pathways for the movement of a charge is a parallel arrangement. The bulbs of the decorative light set shown in the schematic diagram in Figure 2.4 are arranged in parallel with each other.
parallel describes two or more components of a circuit that provide separate conducting paths for current because t he components are connected across common points or junctions
A Parallel Circuit These decorative lights are wired in parallel. Notice that in a parallel arrangement there is more than one path for current.
Circuits and Circuit Elements
639
Resistors in parallel have the same potential differences across them.
A Simple Parallel Circuit {a) This simple parallel circuit with two bulbs connected to a battery can be represented by {b) the schematic diagram shown on the right.
To explore the consequences of arranging resistors in parallel, consider the two bulbs connected to a battery in Figure 2.5(a). In this arrangement, the left side of each bulb is connected to the positive terminal of the battery, and the right side of each bulb is connected to the negative terminal. Because the sides of each bulb are connected to common points, the potential difference across each bulb is the same. If the common points are the battery’s terminals, as they are in the figure, the potential difference across each resistor is also equal to the terminal voltage of the battery. The current in each bulb, however, is not always the same.
The sum of currents in parallel resistors equals the total current. In Figure 2.5, when a certain amount of charge leaves the positive terminal and reaches the branch on the left side of the circuit, some of the charge moves through the top bulb and some moves through the bottom bulb. If one of the bulbs has less resistance, more charge moves through that bulb because the bulb offers less opposition to the flow of charges. Because charge is conserved, the sum of the currents in each bulb equals the current l delivered by the battery. This is true for all resistors in parallel. l
= 11 + 12 + 13 …
The parallel circuit shown in Figure 2.5 can be simplified to an equivalent resistance with a method similar to the one used for series circuits. To do this, first show the relationship among the currents. l
= 11 + 12
QuickLAB Cut the regular drinking straws and thin stirring straws into equal lengths. Tape them end to end in long tubes to form series combinations. Form parallel combinat ions by taping the straws together side by side.
Try several combinations of like and unlike straws. Blow through each combination of tubes, holding your fingers in front of the
640
Chapter 18
openings to compare the airflow (or current) that you achieve with each combination. Rank the combinations according to how much resistance they offer. Classify them according to the amount of current created in each. Straws in series
MATERIALS • 4 regular drinking straws • 4 stirring straws or coffee stirrers • tape
Straws in parallel
Then substitute the equivalents for current according to ~ V = IR.
~v ~v1 ~v2 -=–+-Req
R1
R2
Car Headlights How can you
tell that the headlights on a car are wired in parallel rather than in series? How would the brightness of the bulbs differ if they were wired in series across the same 12 V battery instead of in parallel?
Because the potential difference across each bulb in a parallel arrangement equals the terminal voltage(~ V = ~ V 1 = ~ V2 ), you can divide each side of the equation by~ V to get the following equation. _l_ = _!_ + _l_ Req R1 R2
An extension of this analysis shows that the equivalent resistance of two or more resistors connected in parallel can be calculated using the , following equation. Resistors in Parallel 1
1
1
Simple Circuits Sketch as
many different circuits as you can using three light bulbseach of which has t he same resistance-and a battery.
1
Req = R1 + R2 + R3 · · ·
The equivalent resistance of resistors in parallel can be calculated using a reciprocal relationship.
Notice that this equation does not give the value of the equivalent resistance directly. You must take the reciprocal of your answer to obtain the value of the equivalent resistance. Because of the reciprocal relationship, the equivalent resistance for a parallel arrangement of resistors must always be less than the smallest resistance in the group of resistors. The conclusions made about both series and parallel circuits are summarized in Figure 2.6.
Series
Parallel
schematic diagram
current
I= J1
= I 2 = 13 . . .
= same for each resistor
potential difference
equivalent resistance
~V
= ~ V1 + ~ V2 + ~ V3 . ..
I= I 1 + 12
+ 13 . ..
= sum of currents
~V
= ~ V1 + ~ V2 + ~ V3 . ..
= sum of potential differences
= same for each resistor
Req = R1 +R2 +R3 …
_1_ = _!_ + _!_
= sum of individual resistances
Req
R1
R2
+ _!_ R3
= reciprocal sum of resistances C ircuits and Circuit Elements
641
Resistors in Parallel Sample Problem B A 9.0 V battery is connected to four resistors, as shown at right. Find the equivalent resistance for the circuit and the total current in the circuit.
0
ANALYZE
Given:
2.on
~V=9.0V
R 1 =2.0fi
R 2 =4.0fi R4 Unknown:
= 7.0 f2
R eq — ?.
Diagram: 2 . 0 ! 1 ~ 4.0!1 5.0!1 7.0!1 9.0V
E)
PLAN
Choose an equation or situation: Because both sides of each resistor are connected to common points, they are in parallel. Thus, the equivalent resistance can be calculated with the equation for resistors in parallel.
1 Req
1
1
1
=R + R + R … for parallel 1
2
3
The following equation can be used to calculate the current.
~V= IReq Rearrange the equation to isolate the unknown: No rearrangement is necessary to calculate R eqi rearrange .6. V = IReq to calculate the total current delivered by the battery.
I = ~V Req
E)
SOLVE
Tips and Tricks The equation for resistors in parallel gives you the reciprocal of the equivalent resistance. Be sure to take the reciprocal of this value in the final step to find the equivalent resistance.
Substitute the values into the equation and solve:
_l_=_l_+_l_+_l_+_l_ Req 2.0 f2 4.0 f2 5.0 f2 7.0 f2 _l_ = 0.5 + 0.25 + 0.20 + 0.14 = 1.09 Req l f2 1 f2 1 f2 1 f2 1 f2 1 Req = 1.09
n
I Req = 0.917 f2 I 642
Chapter 18
G·i,i!i,\11§.►
Resistors in Parallel (continued)
—— 1 Substitute that equivalent resistance value in the equation for current.
..6. Vtat
l =–R eq
Calculator Solution The calculator answer is 9.814612868, but because the potential difference, 9.0 V, has only two significant digits, the answer is reported as 9.8 A.
9.0V o.917 n
I = 9.8A
0
CHECKYOUR WORK
For resistors connected in parallel, the equivalent resistance should be less than the smallest resistance.
o.917 n < 2.0 n
Practice 1. The potential difference across the equivalent resistance in Sample Problem B equals the potential difference across each of the individual parallel resistors. Calculate the value for the current in each resistor. 2. A length of wire is cut into five equal pieces. The five pieces are then connected in
parallel, with the resulting resistance being 2.00 D. What was the resistance of the original length of wire before it was cut up? 3. A 4.0 D resistor, an 8.0 D resistor, and a 12.0 D resistor are connected in parallel
across a 24.0 V battery.
a. What is the equivalent resistance of the circuit? b. What is the current in each resistor? 4. An 18.0 D, 9.00 D, and 6.00 D resistor are connected in parallel to an emf source.
A current of 4.00 A is in the 9.00 D resistor.
a. Calculate the equivalent resistance of the circuit. b. What is the potential difference across the source? c. Calculate the current in the other resistors.
Parallel circuits do not require all elements to conduct. What happens when a bulb burns out in a string of decorative lights that is wired in parallel? There is no current in that branch of the circuit, but each of the parallel branches provides a separate alternative pathway for current. Thus, the potential difference supplied to the other branches and the current in these branches remain the same, and the bulbs in these branches remain lit. When resistors are wired in parallel with an emf source, the potential difference across each resistor always equals the potential difference across the source. Because household circuits are arranged in parallel, a ppliance manufacturers are able to standardize their design, producing Circuits and Circuit Elements
643
, Did YOU Know?. – – – – – – – – – – – . Because the potential difference provided by a wall outlet in a home in North America is not the same as the potential difference that is standard on other continents, appliances made in North America are not always compatible with wall outlets in homes on other continents.

devices that all operate at the same potential difference. As a result, manufacturers can choose the resistance to ensure that the current will be neither too high nor too low for the internal wiring and other components that make up the device. Additionally, the equivalent resistance of several parallel resistors is less than the resistance of any of the individual resistors. Thus, a low equivalent resistance can be created with a group of resistors of higher resistances.
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Two resistors are wired in series. In another circuit, the same two resistors are wired in parallel. In which circuit is the equivalent resistance greater?
2. A 5 n, a 10 n, and a 15 0 resistor are connected in series. a. Which resistor has the most current in it? b. Which resistor has the largest potential difference across it? 3. A 5 n, a 10 n, and a 15 n resistor are connected in parallel. a. Which resistor has the most current in it? b. Which resistor has the largest potential difference across it? 4. Find the current in and potential difference across each of the resistors in the following circuits: a. a 2.0 n and a 4.0 n resistor wired in series with a 12 V source b. a 2.0 n and a 4.0 n resistor wired in parallel with a 12 V source
Interpreting Graphics 5. The brightness of a bulb depends only on the bulb’s resistance and on the potential difference across it. A bulb with a greater potential difference dissipates more power and thus is brighter. The five bulbs shown in Figure 2.7 are identical, and so are the three batteries. Rank the bulbs in order of brightness from greatest to least, indicating if any are equal. Explain your reasoning. (Disregard the resistance of the wires.)
{d) {e)
644
Chapter 18
SECTION 3
Complex Resistor Combinations
Objectives ►
Calculate the equivalent resistance for a complex circuit involving both series and parallel portions.

Calculate the current in and potential difference across individual elements within a complex circuit.
Resistors Combined Both in Parallel and in Series Series and parallel circuits are not often encountered independent of one another. Most circuits today employ both series and parallel wiring to utilize the advantages of each type. A common example of a complex circuit is the electrical wiring typical in a home. In a home, a fuse or circuit breaker is connected in series to numerous outlets, which are wired to one another in parallel. An example of a typical household circuit is shown in Figure 3.1. As a result of the outlets being wired in parallel, all the appliances operate independently; if one is switched off, any others remain on. Wiring the outlets in parallel ensures that an identical potential difference exists across any appliance. This way, appliance manufacturers can produce appliances that all use the same standard potential difference.
To prevent excessive current, a fuse or circuit breaker must be placed in series with all of the outlets. Fuses and circuit breakers open the circuit when the current becomes too high. A fuse is a small metallic strip that melts if the current exceeds a certain value. After a fuse has melted, it must be replaced. A A Household Circuit (a) When all of these circuit breaker, a more modern device, triggers a switch devices are plugged into the same household circuit, when current reaches a certain value. The switch must be (b) the result is a parallel combination of resistors in reset, rather than replaced, after the circuit overload has series with a circuit breaker. been removed. Both fuses and circuit breakers must be in series with the entire load to prevent excessive current from reaching any appliance. In fact, if all the devices in Figure 3.1 were used at once, the circuit would be overloaded. The circuit breaker would interrupt the current. Fuses and circuit breakers are carefully selected to meet the demands of a circuit. If the circuit is to carry currents as large as 30 A, an appropriate fuse or circuit breaker must be used. Because the fuse or circuit breaker is placed in series with the rest of the circuit, the current in the fuse or circuit breaker is the same as the total current in the circuit. To find this current, one must determine the equivalent resistance. When determining the equivalent resistance for a complex circuit, you must simplify the circuit into groups of series and parallel resistors and then find the equivalent resistance for each group by using the rules for finding the equivalent resistance of series and parallel resistors.
(a)
Microwave: 8.0 n Blender: 41.1 !1 Toaster: 16.9 !1
LlV = 120 V (b)
Circuits and Circuit Elements
645
PREMIUM CONTENT
~ Interactive Demo \:;I HMDScience.com
Equivalent Resistance Sample Problem C Determine the equivalent resistance of the complex circuit shown below.
0
f:)
E)
ANALYZE
PLAN
SOLVE
The best approach is to divide the circuit into groups of series and parallel resistors. This way, the methods presented in Sample Problems A and B can be used to calculate the equivalent resistance for each group. Redraw the circuit as a group of resistors along one side of the circuit. Because bends in a wire do not affect the circuit, they do not need to be represented in a schematic diagram. Redraw the circuit without the corners, keeping the arrangement of the circuit elements the same, as shown at right.
6.00
6.00
2.00
1.00
4.00
Tips and Tricks For now, disregard the emf source, and work only with the resistances.
6.00
2.00
4.0 n
Identify components in series, and calculate their equivalent resistance. Resistors in groups (a) and (b) are in series. For group (a): R eq = 3.0 0 + 6.0 0 = 9.0 0 For group (b ): R eq = 6.0 0 + 2.0 0 = 8.0 0 6.00
2.00
8.00 r:-:-:7

9.o n
2.7 n
(~
12.7 0 r:-:-:7
7 1.0 n
G.;J CS·M!i,\114-► 646
Chapter 18
Equivalent Resistance (continued) —————————1 Identify components in parallel, and calculate their equivalent resistance. Resistors in group (c) are in parallel. For group (c):
Tips and Tricks It doesn’t matter in what order the operations of simplifying the circuit are done, as long as the simpler equivalent circuits still have the same current in and potential difference across the load.
_l_=_l_ +-1-= 0.12!1 + 0.25 = 0.37 Req 8.0 !1 4.0 !1 1 1 !1 1 !1 Req=2.7!t Repeat steps 2 and 3 until the resistors in the circuit are reduced to a single equivalent resistance.
The remainder of the resistors, group (d), are in series. For group (d): Req = 9.0 n + 2.7 n + 1.0 n
I Req = 12.7 !1 I Practice 1. For each of the following sets of values, determine the equivalent
Ra
resistance for the circuit shown in Figure 3.2.
a. Ra = 25.0 n
Rb = 3.on
Re = 40.0 n
b. Ra= 12.0 n
Rb= 35.0 D
Re = 25.0 n
c. Ra = 15.0 n
Rb= 28.0 n
Re = 12.0 n
40.0V
Rb
Re Figure 3.2
2. For each of the following sets of values, determine the equivalent
Ra
resistance for the circuit shown in Figure 3.3.
a. Ra = 25.0 n Rd= 15.0 D
Rb = 3.on Re = 18.0 n
Re = 40.0 n
b. Ra = 12.0 n Rd= 50.0 D
Rb= 35.0 n Re = 45.0 n
Re = 25.0 n
25.0V Figure 3.3
Work backward to find the current in and potential difference across a part of a circuit. Now that the equivalent resistance for a complex circuit has been determined, you can work backward to find the current in and potential difference across any resistor in that circuit. In the household example, substitute potential difference and equivalent resistance in .6. V = IR to find the total current in the circuit. Because the fuse or circuit breaker is in series with the load, the current in it is equal to the total current. Once this total current is determined, .6. V = IR can again be used to find the potential difference across the fuse or circuit breaker. There is no single formula for finding the current in and potential difference across a resistor buried inside a complex circuit. Instead, .6. V = IR and the rules reviewed in Figure 3.4 must be applied to smaller pieces of the circuit until the desired values are found. Circuits and Circuit Elements
647
Series
Parallel
current
same as total
add to find total
potential difference
add to find total
same as total PREMIUM CONTENT
Current in and Potential Difference Across a Resistor
#:\
\::I
Interactive Demo HMOScience.com
I
Sample Problem D Determine the current in and potential difference across the 2.0 n resistor highlighted in the figure below.
0
E)
ANALYZE
PLAN
Tips and Tricks It is not necessary to solve for Req first and then work backward to find current in or potential difference across a particular resistor, as shown in this Sample Problem, but working through these steps keeps the mathematical operations at each step simpler.
648
Chapter 18
First determine the total circuit current by reducing the resistors to a single equivalent resistance. Then rebuild the circuit in steps, calculating the current and potential difference for the equivalent resistance of each group until the current in and potential difference across the 2.0 n resistor are known.
6.0 !1
6.0 !1
1.0 !1
4.0!1 3.0 !1
Determine the equivalent resistance of the circuit. The equivalent resistance of the circuit is 12.7 O; this value is calculated in Sample Problem C. Calculate the total current in the circuit. Substitute the potential difference and equivalent resistance in ~ V = IR, and rearrange the equation to find the current delivered by the battery.
2.0 !1
9.0V
6.00
2.00
8.00 ..-:-:7

9.00
2.70
(~
12.70 ..-:-:7
7 1.00
Determine a path from the equivalent resistance found in step I to the 2.0 n resistor. Review the path taken to find the equivalent resistance in the figure at right, and work backward through this path. The equivalent resistance for the entire circuit is the same as the equivalent resistance for group (d). The center resistor in group (d) in turn is the equivalent resistance for group ( c ). The top resistor in group (c) is the equivalent resistance for group (b ), and the right resistor in group (b) is the 2.0 n resistor.
~ G·i,i!i,\it4-►
Current in and Potential Difference Across a Resistor
E)
SOLVE
(continued)
Follow the path determined in step 3, and calculate the current in and potential difference across each equivalent resistance. Repeat this process until the desired values are found. A. Regroup, evaluate, and calculate. Replace the circuit’s equivalent resistance with group (d). The resistors in group (d) are in series; therefore, the current in each resistor is the same as the current in the equivalent resistance, which equals 0.71 A. The potential difference across the 2.7 n resistor in group (d) can be calculated using ~ V = IR. Given:
I= 0.71 A
Unknown:
.6.V= ?
R = 2.7 D,
.6. V =I= (0.71 A)(2.7 D,) = ll.9 VI
B. Regroup, evaluate, and calculate. Replace the center resistor with group (c). The resistors in group (c) are in parallel; therefore, the potential difference across each resistor is the same as the potential difference across the 2.7 n equivalent resistance, which equals 1.9 V. The current in the 8.0 n resistor in group (c) can be calculated using~ V = IR . Given:
.6.V= 1.9V R
Unknown:
I= ?
= 8.0 D,
I = .6. V = 1.9 V = I0.24AI R
son
C. Regroup, evaluate, and calculate. Replace the 8.0 n resistor with group (b). Tips and Tricks You can check each step in problems like Sample Problem D by using ~ V = IR for each resistor in a set. You can also check the sum of ~ V for series circuits and the sum of I for parallel circuits.
The resistors in group (b) are in series; therefore, the current in each resistor is the same as the current in the 8.0 n equivalent resistance, which equals 0.24 A.
I I = 0.24AI
The potential difference across the 2.0 n resistor can be calculated using ~ V = IR. Given:
I= 0.24A
Unknown:
.6.V=?
R=2.on
.6. V =IR= (0.24 A) (2.0 0) = 0.48 V l.6.V= 0.48V I
CH·i ,rn ,\it#-►
Circuits and Circuit Elements
649
Current in and Potential Difference Across a Resistor
(continued)
Practice 1. Calculate the current in and potential difference across each of the resistors shown in the schematic diagram in Figure 3.5.
R e= 4.0 D. Rd=4.0 D.
R e = 4.0 D.
14.0V Figure 3.5
Decorative Lights and Bulbs Filament
ight sets arranged in series cannot remain lit if a bulb burns out. Wiring in parallel can eliminate this problem, but each bulb must then be able to withstand 120 V. To eliminate the drawbacks of either approach, modern light sets typically contain two or three sections connected to each other in parallel, each of which contains bulbs in series. When one bulb is removed from a modern light set, half or one-third of the lights in the set go dark because the bulbs in that section are wired in series. When a bulb bums out, however, all of the other bulbs in the set remain lit. How is this possible? Modern decorative bulbs have a short loop of insulated wire, called the jumper, that is wrapped around the wires connected to the filament, as shown at right. There is no current in the insulated wire when the bulb is functioning properly. When the filament breaks, however, the current in the section is zero and the potential difference across the two wires connected to the broken filament is then 120 V. This large potential difference creates a spark
650
Chapter 18
Jumper
…—-rt:
Glass insulator
across the two wires that burns the insulation off the small loop of wire. Once that occurs, the small loop closes the circuit, and the other bulbs in the section remain lit.
~•-.,11
Because the small loop in the burned out bulb has very little resistance, the equivalent resistance of that portion of the light set decreases; its current increases. This increased current results in a slight increase in each bulb’s brightness. As more bulbs burn out, the temperature in each bulb increases and can become a fire hazard; thus, bulbs should be replaced soon after burning out.

SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Find the equivalent resistance of the complex circuit shown in Figure 3.6.
2. What is the current in the 1.5 n resistor in the complex circuit shown in Figure 3.6?
5.00
3. What is the potential difference across the 1.5 n resistor in the circuit shown in Figure 3.6?
18.0V
1.50
4. A certain strand of miniature lights contains 35 bulbs wired in series, with each bulb having a resistance of 15.0 n. What is the equivalent resistance when three such strands are connected in parallel across a potential difference of 120.0 V?
5.00
5.00
5.00
5. What is the current in and potential difference across each of the bulbs in the strands oflights described in item 4?
6. If one of the bulbs in one of the three strands of lights in item 4 goes out while the other bulbs in that strand remain lit, what is the current in and potential difference across each of the lit bulbs in that strand?
Interpreting Graphics 7.
depicts a household circuit containing several appliances and a circuit breaker attached to a 120 V source of potential difference.
Figure 3. 7
a. Is the current in the toaster equal to the current in the microwave? b. Is the potential difference across the microwave equal to the potential difference across the popcorn popper?
c. Is the current in the circuit breaker equal to the total current in all of the appliances combined? d. Determine the equivalent resistance for the circuit.
e. Determine how much current is in the toaster.
Toaster: 16.9 n Microwave: 8.0 n Popcorn popper: 10.0 n
120V
Circuits and Circuit Elements
651
Semiconductor Technician lectronic chips are used in a wide variety of devices, from toys to phones to computers. To learn more about chip making as a career, read the interview with etch process engineering technician Brad Baker, who works for Motorola. What training did you receive in order to become a semiconductor technician?
Brad Baker is creating a recipe on the plasma etch tool to test a new process.
My experience is fairly unique. My degree is in psychology. You have to have an associate’s degree in some sort of electrical or engineering field or an undergraduate degree in any field. What about semiconductor manufacturing made it more interesting than other fields?
While attending college, I worked at an airline. There was not a lot of opportunity to advance, which helped point me in other directions. Circuitry has a lot of parallels to the biological aspects of the brain, which is what I studied in school. We use the scientific method a lot. What is the nature of your work?
I work on the etch process team. Device engineers design the actual semiconductor. Our job is to figure out how to make what they have requested. It’s sort of like being a chef. Once you have experience, you know which ingredient to add. What is your favorite thing about your job?
I feel like a scientist. My company gives us the freedom to try new things and develop new processes.
What advice do you have for students who are interested in semiconductor engineering?
The field is very science oriented, so choose chemical engineering, electrical engineering, or material science as majors. Other strengths are the ability to understand and meet challenges, knowledge of trouble-shooting techniques, patience, and analytical skills. Also, everything is computer automated, so you have to know how to use computers.
Has your job changed since you started it?
Each generation of device is smaller, so we have to do more in less space. As the devices get smaller, it becomes more challenging to get a design process that is powerful enough but doesn’t etch too much or too little.
Brad Baker
SECTION 1
Schematic Diagrams and Circuits
,: ,
• Schematic diagrams use standardized symbols to summarize the contents of electric circuits.
1 ,
1
r.- ,
schematic diagram electric circuit
• A circuit is a set of electrical components connected so that they provide one or more complete paths for the movement of charges. • Any device that transforms nonelectrical energy into electrical energy, such as a battery or a generator, is a source of emf. • If the internal resistance of a battery is neglected, the emf can be considered equal to the terminal voltage, the potential difference across the source’s two terminals.
SECTION 2
Resistors in Series or in Parallel
, c,
,c, ·.-
• Resistors in series have the same current.
series
• The equivalent resistance of a set of resistors connected in series is the sum of the individual resistances.
parallel
• The sum of currents in parallel resistors equals the total current. • The equivalent resistance of a set of resistors connected in parallel is calcu lated using an inverse relationship.
SECTION 3
Complex Resistor Combinations
• Many complex circuits can be understood by isolating segments that are in series or in parallel and simplifying them to their equivalent resistances.
DIAGRAM SYMBOLS
Quantities
I
R
~v
current
resistance
potential difference
Units
A
n V
Conversions
amperes
ohms
volts
= C/s = coulombs of charge per second = VIA = volts per ampere of current ———= J/C = joules of energy per coulomb of charge
7
Wire or conductor
~
Resistor or circuit load Bulb or lamp
B_
Plug
®
Battery / direct-current emf source
. -I

Switch ————-+–
Capacitor
~ I-=-
/
o-
– H-
Problem Solving See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
Chapter Summary
653
Schematic Diagrams and Circuits
10. Which of the switches in the circuit below will complete a circuit when closed? Which will cause
a short circuit? B
REVIEWING MAIN IDEAS 1. Why are schematic diagrams useful?
C
2. Draw a circuit diagram for a circuit containing three
5.0 n resistors, a 6.0 V battery, and a switch.
A
3. The switch in the circuit shown below can be set to connect to points A, B, or C. Which of these connections will provide a complete circuit?
Resistors in Series or in Parallel
b lBO– – – – – -,
REVIEWING MAIN IDEAS 11. If four resistors in a circuit are connected in series,
4. If the batteries in a cassette recorder provide a terminal voltage of 12.0 V, what is the potential difference across the entire recorder?
which of the following is the same for the resistors in the circuit? a. potential difference across the resistors b. current in the resistors
5. In a case in which the internal resistance of a battery
is significant, which is greater? a. the terminal voltage b. the emf of the battery
CONCEPTUAL QUESTIONS
12. If four resistors in a circuit are in parallel, which of the
following is the same for the resistors in the circuit? a. potential difference across the resistors b. current in the resistors
CONCEPTUAL QUESTIONS
6. Do charges move from a source of potential differ-
ence into a load or through both the source and the load?
7. Assuming that you want to create a circuit that has current in it, why should there be no openings in the circuit?
13. A short circuit is a circuit containing a path of very
low resistance in parallel with some other part of the circuit. Discuss the effect of a short circuit on the current within the portion of the circuit that has very low resistance. 14. Fuses protect electrical devices by opening a circuit if
8. Suppose a 9 V battery is connected across a light bulb. In what form is the electrical energy supplied by the battery dissipated by the light bulb? 9. Why is it dangerous to use an electrical appliance
when you are in the bathtub?
654
Chapter 18
the current in the circuit is too high. Would a fuse work successfully if it were connected in parallel with the device that it is supposed to protect?
15. What might be an advantage of using two identical resistors in parallel that are connected in series with another identical parallel pair, as shown below, instead of using a single resistor?
~ PRACTICE PROBLEMS For problems 16-17, see Sample Problem A. 16. A length of wire is cut into five equal pieces. If each piece has a resistance of 0.15 n, what was the resistance of the original length of wire? 17. A 4.0 D resistor, an 8.0 D resistor, and a 12 D resistor are connected in series with a 24 V battery. Determine the following: a. the equivalent resistance for the circuit b. the current in the circuit
21. The technician in item 20 finds another resistor, so now there are three resistors with the same resistance. a. How many different resistances can the technician achieve? b. Express the effective resistance of each possibility in terms of R. 22. Three identical light bulbs are connected in circuit to a battery, as shown below. Compare the level of brightness of each bulb when all the bulbs are illuminated. What happens to the brightness of each bulb if the following changes are made to the circuit? a. Bulb A is removed from its socket. b. Bulb C is removed from its socket. c. A wire is connected directly between points D andE. d. A wire is connected directly between points D andF.
For problems 18-19, see Sample Problem B. 18. The resistors in item 17 are connected in parallel across a 24 V battery. Determine the following: a. the equivalent resistance for the circuit b. the current delivered by the battery
19. An 18.0 n resistor, 9.00 n resistor, and 6.00 n resistor are connected in parallel across a 12 V battery. Determine the following: a. the equivalent resistance for the circuit b. the current delivered by the battery
PRACTICE PROBLEMS For problems 23-24, see Sample Problem C.
Complex Resistor Combinations
23. Find the equivalent resistance of the circuit shown in the figure below. 30.0V
CONCEPTUAL QUESTIONS
180
20. A technician has two resistors, each of which has the same resistance, R. a. How many different resistances can the technician achieve? b. Express the effective resistance of each possibility in terms of R.
9.0 0 120
6.00
Chapter Review
655
24. Find the equivalent resistance of the circuit shown in the figure below. 1.0 o
1.00
12.0V
7.00
1.50
7.00
For problems 25-26, see Sample Problem D. 25. For the circuit shown below, determine the current in each resistor and the potential difference across each resistor.
6.00 9.00
3.00
12V
26. For the circuit shown in the figure below, determine the following:
6.00
3.00
4.00
18.0V
a. b. c. d.
the current in the 2.0 n resistor the potential dilierence across the 2.0 n resistor the potential dilierence across the 12.0 n resistor the current in the 12.0 n resistor
Mixed Review REVIEWING MAIN IDEAS 27. An 8.0 n resistor and a 6.0 n resistor are connected in series with a battery. The potential dilierence across the 6.0 n resistor is measured as 12 V. Find the potential difference across the battery.
656
Chapter 18
28. A 9.0 n resistor and a 6.0 n resistor are connected in parallel to a battery, and the current in the 9.0 resistor is found to be 0.25 A. Find the potential dilierence across the battery.
n
29. A 9.0 n resistor and a 6.0 n resistor are connected in series to a battery, and the current through the 9.0 resistor is 0.25 A. What is the potential difference across the battery?
n
30. A 9.0 n resistor and a 6.0 n resistor are connected in series with an emf source. The potential difference across the 6.0 resistor is measured with a voltmeter to be 12 V. Find the potential dilierence across the emf source.
n
31 . An 18.0 n, 9.00 n, and 6.00 n resistor are connected in series with an emf source. The current in the 9.00 n resistor is measured to be 4.00 A. a. Calculate the equivalent resistance of the three resistors in the circuit. b. Find the potential difference across the emf source. c. Find the current in the other resistors. 32. The stockroom has only 20 n and 50 n resistors. a. You need a resistance of 45 n. How can this resistance be achieved using three resistors? b. Describe two ways to achieve a resistance of35 using four resistors.
n
33. The equivalent resistance of the circuit shown below is 60.0 n. Use the diagram to determine the value ofR. R
90.00
10.00
10.00
90.00
34. Two identical parallel-wired strings of25 bulbs are connected to each other in series. If the equivalent resistance of the combination is 150.0 n and it is connected across a potential difference of 120.0 V, what is the resistance of each individual bulb?
35. The figures {a)-{e) below depict five resistance diagrams. Each individual resistance is 6.0 n. {a)
~
{b)
~
{c)¾
{d)
~
{e)
~
a. Which resistance combination has the largest equivalent resistance? b. Which resistance combination has the smallest equivalent resistance? c. Which resistance combination has an equivalent resistance of 4.0 D? d. Which resistance combination has an equivalent resistance of9.0 D?
39. A resistor with an unknown resistance is connected in parallel to a 12 n resistor. When both resistors are connected to an emf source of 12 V, the current in the unknown resistor is measured with an ammeter to be 3.0 A. What is the resistance of the unknown resistor? 40. The resistors described in item 37 are reconnected in parallel to the same 18.0 V battery. Find the current in each resistor and the potential difference across each resistor.
41. The equivalent resistance for the circuit shown below drops to one-half its original value when the switch, S, is closed. Determine the value of R. R
10.00
36. Three small lamps are connected to a 9.0 V battery, as shown below.
R 3 = 2.0 n
9.0V
90.00
42. You can obtain only four 20.0 n resistors from the stockroom. a. How can you achieve a resistance of 50.0 n under these circumstances? b. What can you do if you need a 5.0 n resistor? 43. Four resistors are connected to a battery with a terminal voltage of 12.0 V, as shown below. Determine the following:
a. What is the equivalent resistance of this circuit? b. What is the current in the battery?
30.00
so.on
c. What is the current in each bulb? d. What is the potential difference across each bulb? 37. An 18.0 n resistor and a 6.0 n resistor are connected in series to an 18.0 V battery. Find the current in and the potential difference across each resistor.
38. A 30.0 n resistor is connected in parallel to a 15.0 n resistor. These are joined in series to a 5.00 n resistor and a source with a potential difference of30.0 V. a. Draw a schematic diagram for this circuit. b. Calculate the equivalent resistance. c. Calculate the current in each resistor. d. Calculate the potential difference across each resistor.
90.00
a. the equivalent resistance for the circuit b. the current in the battery c. the current in the 30.0 n resistor d. the power dissipated by the 50.0 n resistor e. the power dissipated by the 20.0 n resistor (~V)2 (Hint: Remember that P = – R- = I~ V.)
Chapter Review
657
44. Two resistors, A and B, are connected in series to a 6.0 V battery. A voltmeter connected across resistor A measures a potential difference of 4.0 V. When the two resistors are connected in parallel across the 6.0 V battery, the current in Bis found to be 2.0 A. Find the resistances of A and B.
47. The power supplied to the circuit shown below is 4.00 W. Determine the following: 10.on
3.on
45. Draw a schematic diagram of nine 100 n resistors arranged in a series-parallel network so that the total resistance of the network is also 100 n. All nine resistors must be used. 46. For the circuit below, find the following: 5.on
3.on
3.on
10.0 n
10.0 n
4.0 n
28V
4.o n
2.0 n
3.o n
a. the equivalent resistance of the circuit b. the current in the 5.0 n resistor
a. the equivalent resistance of the circuit b. the potential difference across the battery 48. Your toaster oven and coffee maker each dissipate 1200 W of power. Can you operate both of these appliances at the same time if the 120 V line you use in your kitchen has a circuit breaker rated at 15 A? Explain. 49. An electric heater is rated at 1300 W, a toaster is rated at 1100 W, and an electric grill is rated at 1500 W. The three appliances are connected in parallel across a 120 V emf source. a. Find the current in each appliance. b. Is a 30.0 A circuit breaker sufficient in this situation? Explain.
Parallel Resistors Electric circuits are often composed of combinations of series and parallel circuits. The overall resistance of a circuit is determined by dividing the circuit into groups of series and parallel resistors and determining the equivalent resistance of each group. As you learned earlier in this chapter, the equivalent resistance of parallel resistors is given by the following equation:
_1_ = _1 + -1 + -1 + · · · Req Rl R2 R3 One interesting consequence of this equation is that the equivalent resistance for resistors in parallel will always be less than the smallest resistor in the group.
658
Chapter 18
In this graphing calculator activity, you will determine the equivalent resistance for various resistors in parallel. You will confirm that the equivalent resistance is always less than the smallest resistor, and you will relate the number of resistors and changes in resistance to the equivalent resistance. Go online to HMDScience.com to find this graphing calculator activity.
ALTERNATIVE ASSESSMENT 1. How many ways can two or more batteries be connected in a circuit with a light bulb? How will the current change depending on the arrangement? First draw diagrams of the circuits you want to test. Then identify the measurements you need to make to answer the question. If your teacher approves your plan, obtain the necessary equipment and perform the experiment. 2. Research the career of an electrical engineer or technician. Prepare materials for people interested in this career field. Include information on where people in this career field work, which tools and equipment they use, and the challenges of their field. Indicate what training is typically necessary to enter the field.
4. You and your friend want to start a business exporting small electrical appliances. You have found people willing to be your partners to distribute these appliances in Germany. Write a letter to these potential partners that describes your product line and that asks for the information you will need about the electric power, sources, consumption, and distribution in Germany. 5. Contact an electrician, builder, or contractor, and ask to see a house electrical plan. Study the diagram to identify the circuit breakers, their connections to different appliances in the home, and the limitations they impose on the circuit’s design. Find out how much current, on average, is in each appliance in the house. Draw a diagram of the house, showing which circuit breakers control which appliances. Your diagram should also keep the current in each of these appliances under the performance and safety limits.
3. The manager of an automotive repair shop has been contacted by two competing firms that are selling ammeters to be used in testing automobile electrical systems. One firm has published claims that its ammeter is better because it has high internal resistance. The other firm has published claims that its ammeter is better because it has low resistance. Write a report with your recommendation to the manager of the automotive repair shop. Include diagrams and calculations that explain how you reached your conclusion.
Chapter Review
659
MULTIPLE CHOICE 1. Which of the following is the correct term for a circuit that does not have a closed-loop path for electron flow? A. closed circuit B. dead circuit C. open circuit D. short circuit 2. Which of the following is the correct term for a circuit in which the load has been unintentionally bypassed? F. closed circuit G. dead circuit H. open circuit J. short circuit
5. Which of the following is the correct equation for the current in the resistor? A. l=IA +IB+ l e
B. IB = b.V Req
C.
[B
= 1total + IA
Use the diagram below to answer questions 6-7.
A
B
C
Use the diagram below to answer questions 3-5.
6. Which of the following is the correct equation for the equivalent resistance of the circuit? F. Req =RA+ RB+ Re G. _l_ = _l_ + _!_ + _l_ Req RA RB Re
C 3. Which of the circuit elements contribute to the load of the circuit? A. Only A B. A and B, but not C C. OnlyC D. A, B, andC 4. Which of the following is the correct equation for the equivalent resistance of the circuit?
F. Req =RA + RB G. _l_ = _!_ + _!_ Req RA RB
H. Req = Ib.V J. _l_ = _!_ + _!_ + _l_ Req RA RB Re
660
Chapter 18
H. Req = Ib.V
J. Req =RA + (JB
+
;J-l
7. Which of the following is the correct equation for the current in resistor B? A. l=IA +IB+le b.V B. IB= R eq
C.
[B
= 1total + IA
b.VB D. IB=-RB
8. Three 2.0 n resistors are connected in series to a 12 V battery. What is the potential difference across each resistor? F. 2.0V G. 4.0V H. 12V J. 36V
. TEST PREP
Use the following passage to answer questions 9-11.
EXTENDED RESPONSE
Six light bulbs are connected in parallel to a 9.0 V battery. Each bulb has a resistance of 3.0 n.
15. Using standard symbols for circuit elements, draw a diagram of a circuit that contains a battery, an open switch, and a light bulb in parallel with a resistor. Add an arrow to indicate the direction of current if the switch were closed.
9. What is the potential difference across each bulb? A. 1.5V B. 3.0V C. 9.0V D. 27V
Use the diagram below to answer questions 16-17.
10. What is the current in each bulb? F. 0.5A G. 3.0A H. 4.5A J. 18A 11. What is the total current in the circuit? A. 0.5A B. 3.0A C. 4.5A D. 18A
SHORT RESPONSE 12. Which is greater, a battery’s terminal voltage or the same battery’s emf? Explain why these two quantities are not equal. 13. Describe how a short circuit could lead to a fire. 14. Explain the advantage of wiring the bulbs in a string of decorative lights in parallel rather than in series.
1.50
12V
R=3.0fl
16. For the circuit shown, calculate the following: a. the equivalent resistance of the circuit b. the current in the light bulb. Show all your work for both calculations.
17. After a period of time, the 6.0 n resistor fails and breaks. Describe what happens to the brightness of the bulb. Support your answer. 18. Find the current in and potential difference across each of the resistors in the following circuits: a. a 4.0 n and a 12.0 n resistor wired in series with a 4.0 V source. b. a 4.0 n and a 12.0 n resistor wired in parallel with a 4.0 V source. Show all your work for each calculation. 19. Find the current in and potential difference across each of the resistors in the following circuits: a. a 150 n and a 180 n resistor wired in series with a 12Vsource. b. a 150 n and a 180 n resistor wired in parallel with a 12 V source. Show all your work for each calculation.
Test Tip Prepare yourself for taking an important test by getting plenty of sleep the night before and by eating a healthy breakfast on the day of the test.
Standards-Based Assessment
661
SECTION 1 Objectives ►
For given situations, predict whether magnets will repel or attract each other.

Describe the magnetic field around a permanent magnet.
I

Describe the orientation of Earth’s magnetic field.
Magnets and Magnetic Fields Key Terms magnetic domains
magnetic field
Magnets Most people have had experience with different kinds of magnets, such as those shown in Figure 1.1. You have probably seen a variety of magnet shapes, such as horseshoe magnets, bar magnets, and the flat magnets frequently used to attach items to a refrigerator. All types of magnets attract iron-containing objects such as paper clips and nails. In the following discussion, we will assume that the magnet has the shape of a bar. Iron objects are most strongly attracted to the ends of such a magnet. These ends are called poles; one is called the north pole, and the other is called the south pole. The names derive from the behavior of a magnet on Earth. If a bar magnet is suspended from its midpoint so that it can swing freely in a horizontal plane, it will rotate until its north pole points north and its south pole points south. In fact, a compass is just a magnetized needle that swings freely on a pivot. The list of important technological applications of magnetism is very long. For instance, large electromagnets are used to pick up heavy loads. Magnets are also used in meters, motors, generators, and loudspeakers. Magnetic tapes are routinely used in sound- and video-recording equipment, and magnetic recording material is used on computer disks. Superconducting magnets are currently being used to contain extremely high-temperature plasmas that are used in controlled nuclear fusion research. Superconducting magnets are Variety of Magnets Magnets come in a variety of shapes also used to levitate modern trains. These maglev and sizes, but like poles of two magnets always repel one another. trains are faster and provide a smoother ride than the ordinary track system because of the absence of friction between the train and the track.
Like poles repel each other, and unlike poles attract each other. The magnetic force between two magnets can be likened to the electric force between charged objects in that unlike poles of two magnets attract one another and like poles repel one another. Thus, the north pole of a magnet is attracted to the south pole of another magnet, and two north poles (or two south poles) brought close together repel each other. Electric charges differ from magnetic poles in that they can be isolated, whereas magnetic poles cannot. 664
Chapter 19
In fact, no matter how many times a permanent magnet is cut, each piece always has a north pole and a south pole. Thus, magnetic poles always occur in pairs.
Magnetic Domains The magnetic properties of many materials are explained in terms of a model in which an electron is said to spin on its axis much like a top does. (This classical description should not be taken literally. The property of electron spin can be understood only with the methods of quantum mechanics.) The spinning electron represents a charge that is in motion. As you will learn in the next section of this chapter, moving charges create magnetic fields. In atoms containing many electrons, the electrons usually pair up with their spins opposite each other causing their fields to cancel each other. For this reason, most substances, such as wood and plastic, are not magnetic. However, in materials such as iron, cobalt, and nickel, the magnetic fields produced by the electron spins do not cancel completely. Such materials are said to be ferromagnetic. In ferromagnetic materials, strong coupling occurs between neighboring atoms to form large groups of atoms whose net spins are aligned; these groups are called magnetic domains. Domains typically range in size from about 10-4 cm to 10- 1 cm. In an unmagnetized substance, the domains are randomly oriented, as shown in Figure 1.2(a). When an external magnetic field is applied, the orientation of the magnetic fields of each domain may change slightly to more closely align with the external magnetic field, or the domains that are already aligned with the external field may grow at the expense of the other domains. This alignment enhances the applied magnetic field.
Domains of Unmagnetized and Magnetized Materials When a substance is unmagnetized its domains are randomly oriented, as shown in (a). When a substance is magnetized its domains are more closely aligned, as shown in (b).
;r
~
~
I (a)
~r .1f

.1f ~
,
.1f
.1f
.1f
(b)
magnetic domain a region composed of a group of atoms whose magnetic fields are aligned in the same direction
Some materials can be made into permanent magnets. Just as two materials, such as rubber and wool, can become charged after they are rubbed together, an unmagnetized piece of iron can become a permanent magnet by being stroked with a permanent magnet. Magnetism can be induced by other means as well. For example, if a piece of unmagnetized iron is placed near a strong permanent magnet, the piece of iron will eventually become magnetized. The process can be reversed either by heating and cooling the iron or by hammering the iron, because these actions cause the magnetic domains to jiggle and lose their alignment. A magnetic piece of material is classified as magnetically hard or soft, depending on the extent to which it retains its magnetism. Soft magnetic materials, such as iron, are easily magnetized but also tend to lose their magnetism easily. In hard magnetic materials, domain alignment persists after the external magnetic field is removed; the result is a permanent magnet. In contrast, hard magnetic materials, such as cobalt and nickel, are difficult to magnetize, but once they are magnetized, they tend to retain their magnetism. In soft magnetic materials, once the external field is removed, the random motion of the particles in the material changes the orientation of the domains and the material returns to an unmagnetized state. Magnetism
665
Magnetic Fields magnetic field a region in which a magnetic force can be detected
You know that the interaction between charged objects can be described using the concept of an electric field. A similar approach can be used to describe the magnetic field that surrounds any magnetized material. As with an electric field, a magnetic field, B, is a vector quantity that has both magnitude and direction.
Magnetic field lines can be drawn with the aid of a compass. FIGURE 1.3
CONVENTIONS FOR REPRESENTING THE DIRECTION OF A MAGNETIC FIELD In the plane of the page
Into the page Out of the page

The magnetic field of a bar magnet can be explored using a compass, as illustrated in Figure 1.4. If a small, freely suspended bar magnet, such as the needle of a compass, is brought near a magnetic field, the compass needle will align with the magnetic field lines. The direction of the magnetic field, B, at any location is defined as the direction that the north pole of a compass needle points to at that location. Magnetic field lines appear to begin at the north pole of a magnet and to end at the south pole of a magnet. However, magnetic field lines have no beginning or end. Rather, they always form a closed loop. In a permanent magnet, the field lines actually continue within the magnet itself to form a closed loop. (These lines are not shown in the illustration.) This text will follow a simple convention to indicate the direction of B. An arrow will be used to show a magnetic field that is in the same plane as the page, as shown in Figure 1.3. When the field is directed into the page, we will use a series of blue crosses to represent the tails of arrows. If the field is directed out of the page, we will use a series of blue dots to represent the tips of arrows.
Magnetic flux relates to the strength of a magnetic field. Magnetic Field of a Bar Magnet The magnetic field (a) of a bar magnet can be traced with a compass (b). Note that the north poles of the compasses point in the direction of the field lines from the magnet’s north pole to its south pole.
One useful way to model magnetic field strength is to define a quantity called magnetic flux, M. It is defined as the number of field lines that cross a certain area at right angles to that area. Magnetic flux can be calculated by the following equation.
r
Magnetic Flux
I
M = AB cos 0
magnetic flux= (surface area) x ~ magnetic field component normal to the plane of surfac~
l
Now look again at Figure 1.4. Imagine two circles of the same size that are perpendicular to the axis of the magnet. One circle is located near one pole of the magnet, and the other circle is alongside the magnet. More magnetic field lines cross the circle that is near the pole of the magnet. This greater flux indicates that the magnetic field is strongest at the magnet’s poles.
666
Chapter 19
Earth has a magnetic field similar to that of a bar magnet. The north and south poles of a small bar magnet are correctly described as the “north-seeking” and “south-seeking” poles. This description means that if a magnet is used as a compass, the north pole of the magnet will seek, or point to, a location near the geographic North Pole of Earth. Because unlike poles attract, we can deduce that the geographic North Pole of Earth corresponds to the magnetic south pole and the geographic South Pole of Earth corresponds to the magnetic north pole. Note that the configuration of Earth’s magnetic field, pictured in Figure 1.5, resembles the field that would be produced if a bar magnet were buried within Earth.
If a compass needle is allowed to rotate both perpendicular to and parallel to the surface of Earth, the needle will be exactly parallel with respect to Earth’s surface only near the equator. As the compass is moved northward, the needle will rotate so that it points more toward the surface of Earth. Finally, at a point just north of Hudson Bay, in Canada, the north pole of the needle will point perpendicular to Earth’s surface. This site is considered to be the location of the magnetic south pole of Earth. It is approximately 1500 km from Earth’s geographic North Pole. Similarly, the magnetic north pole of Earth is roughly the same distance from the geographic South Pole.
Earth’s Magnetic Field Earth’s magnetic field has a configuration similar to a bar magnet’s. Note that the magnetic south pole is near the geographic North Pole and that the magnetic north pole is near the geographic South Pole. Magnetic south pole Geographic North Pole
,__ ~
‘~
Geographic South Pole
The difference between true north, which is defined by the axis of rotation of Earth, and north indicated by a compass, varies from point to point on Earth. This difference is referred to as magnetic declination. An imaginary line running roughly north-south near the center of North America currently has zero declination. Along this line, a compass will indicate true north. However, in the state of Washington, a compass aligns about 20° east of true north. To further complicate matters, geological evidence indicates that Earth’s magnetic field has changed-and even reversed-throughout Earth’s history. Although Earth has large deposits of iron ore deep beneath its surface, the high temperatures there prevent the iron from retaining permanent
Magnetic north pole
..Did YOU Know? – – – – – – : , , ‘ , : , :
By convention, the north pole of a magnet is frequently painted red . This practice comes from the longstanding use of magnets, in the form of compasses, as navigational aids. Long before global positioning system (GPS) satellites, the compass gave humans an easy way to orient themselves.
QuickLAB Stand in front of the file cab inet, and hold the compass face up and parallel to the ground. Now move t he compass from the top of the file cabinet to the bottom. Making sure that the compass is parallel to the ground, check to see if the direction of the compass needle changes as it moves from the top of the cabinet to the bottom. If the
compass needle changes direction, the file cabinet is magnetized. Can you explain what might have caused t he file cabinet to become magnetized? Remember that Earth’s magnetic field has a vertical component as well as a horizontal component. Try tracing the field around some large metal objects around your
house. Can you find an object that has been magnetized by the horizontal component of Earth’s magnetic field?
MATERIALS
• compass • metal file cabinet
Magnetism
667
magnetization. It is considered likely that the source of Earth’s magnetic field is the movement of charges in convection currents inside Earth’s liquid core. These currents occur because the temperature in Earth’s core is unevenly distributed. Charged ions circling inside the interior of Earth likely produce a magnetic field. There is also evidence that the strength of a planet’s magnetic field is linked to the planet’s rate of rotation. Jupiter rotates at a faster rate than Earth, and recent space probes indicate that Jupiter’s magnetic field is stronger than Earth’s. Conversely, Venus rotates more slowly than Earth and has a weaker magnetic field than Earth. Investigations continue into the cause of Earth’s magnetism.

SECTION 1 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. For each of the cases in the figure below, identify whether the magnets will attract or repel one another.
a. Is
NI Is
NI
b Is
NI IN
sl
c. G~
2. When you break a bar magnet in half, how many poles does each piece have?
Interpreting Graphics 3. Which of the compass-needle orientations in the figure below might correctly describe the magnet’s field at that point? (b)
8
(a)
CS) ..__I s _
(f)e
(c)
(D
(d}
CS) (S)(e)
___, NI
Critical Thinking 4. Satellite ground operators use feedback from a device called a magnetometer, which senses the direction of Earth’s magnetic field, to decide which torque coil to activate. What direction will the magnetometer read for Earth’s magnetic field when the satellite passes over Earth’s equator? 5. In order to protect other equipment, the body of a satellite must remain unmagnetized, even when the torque coils have been activated. Would hard or soft magnetic materials be best for building the rest of the satellite?
668
Chapter 19
S.T.E.M. Magnetic Resonance Imaging agnetic resonance imaging, or MRI, is an imaging technique that has been used in clinical medicine since the early 1980s. MRI allows doctors to make two-dimensional images of or three-dimensional models of parts of the human body. The use of MRI in medicine has grown rapidly. MRI produces high-resolution images that can be tailored to study different types of tissues, depending on the application. Also, MRI procedures are generally much safer than computerized axial tomography (CAT) scans, which flood the body with X rays. A typical MRI machine looks like a giant cube, 2-3 meters on each side, with a cylindrical hollow in the center to accommodate the patient as shown in the photo at right. The MRI machine uses electromagnets to create magnetic fields ranging in strength from 0.5-2.0 T. These fields are strong enough to erase credit cards and to pull pens out of pockets, even across the MRI exam room. Because resistance would cause normal electromagnets to dissipate a huge amount of heat when creating fields this strong, the electromagnets in most MRI machines contain superconducting wires that have zero resistance.
~ e!’
“‘
,::;
i::>
ill fil a:
The creation of an image with MRI depends on the behavior of atomic nuclei within a magnetic field. In a strong magnetic field, the nucleus of an atom tends to line up along the direction of the field.This behavior is particularly true for hydrogen atoms, which are the most common atoms in the body. The primary magnet in an MRI system creates a strong, uniform magnetic field centered on the part of the patient that is being examined. The field causes hydrogen nuclei in the body to line up in the direction of the field. Smaller magnets, called gradient magnets, are then turned on and off to
The imaging magnet in most MRI machines is of the superconducting type. The magnet is the most expensive component of the MRI system. create small variations, or pulses, in the overall magnetic field. Each pulse causes the hydrogen nuclei to shift away from their alignment. After the pulse, the nuclei return to alignment, and as they do so, they emit radio frequency electromagnetic waves. Scanners within the MRI machine detect these radio waves, and a computer processes the waves into images. Different types of tissues can be seen with MRI, depending on the frequency and duration of the pulses. MRI is particularly good for imaging the brain and spinal tissues and can be used to study brain function, brain tumors, multiple sclerosis, and other neurological disorders. MRI can also be used to create images of blood vessels without the surrounding tissue, which can be very useful for studying the circulatory system. The main drawbacks of MRI are that MRI systems are very expensive and that MRI cannot be used on some patients, such as those with pacemakers or certain types of metal implants.
669
SECTION 2
Magnetism from Electricity
Objectives ►
Describe the magnetic field produced by current in a straight conductor and in a solenoid.

Use the right-hand rule to determine the direction of the magnetic field in a currentcarrying wire.
Key Term solenoid
Magnetic Field of a Current-Carrying Wire Scientists in the late 1700s suspected that there was a relationship between electricity and magnetism, but no theory had been developed to guide their experiments. In 1820, Danish physicist Hans Christian Oersted devised a method to study this relationship. Following a lecture to his advanced class, Oersted demonstrated that when brought near a currentcarrying wire, a compass needle is deflected from its usual north-south orientation. He published an account of this discovery in July 1820, and his work stimulated other scientists all over Europe to repeat the experiment.
A long, straight, current-carrying wire has a cylindrical magnetic field. The experiment shown in Figure 2.1 (a) uses iron filings to show that a current-carrying conductor produces a magnetic field. In a similar experiment, several compass needles are placed in a horizontal plane near a long vertical wire, as illustrated in Figure 2.2(b). When no current is in the wire, all needles point in the same direction (that of Earth’s magnetic field). However, when the wire carries a strong, steady current, all the needles deflect in directions tangent to concentric circles around the wire. This result points out the direction of B, the magnetic field induced by the current. When the current is reversed, the needles reverse direction.
Magnetic Field of a Current-Carrying Wire {a) When the wire carries a strong
current, the alignments of the iron filings show that the magnetic field induced by the current forms concentric circles around the wire. {b) Compasses can be used to show the direction of the magnetic field induced by the wire.
“E
“‘ ti
cc
@
(a)
670
Chapter 19
(b)
The right-hand rule can be used to determine the direction of the magnetic field. These observations show that the direction of Bis consistent with a simple rule for conventional current, known as the right-hand rule: If the wire is grasped in the right hand with the thumb in the direction of the current, as shown in Figure 2.2, the four fingers will curl in the direction of B.
The Right-Hand Rule You can use the right-hand rule to find the direction of this magnetic field.
As shown in Figure 2.1 (a), the lines of B form concentric circles about the wire. By symmetry, the magnitude of B is the same everywhere on a circular path centered on the wire and lying in a plane perpendicular to the wire. Experiments show that Bis proportional to the current in the wire and inversely proportional to the distance from the wire.
f
Magnetic Field of a Current Loop The right-hand rule can also be applied to find the direction of the magnetic field of a current-carrying loop, such as the loop represented in Figure 2.3(a). Regardless of where on the loop you apply the right-hand rule, the field within the loop points in the same direction-upward. Note that the field lines of the current-carrying loop resemble those of a bar magnet, as shown in Figure 2.3(b). If a long, straight wire is bent into a coil of several closely spaced loops, as shown on the next page in Figure 2.4, the resulting device is called a solenoid.
solenoid a long, helically wound coil of insulated wire
Current-Carrying Loop
(a) The magnetic field of a current loop is similar to (b) that of a bar magnet.
QuickLAB Wind the wire around the nail, as shown below. Remove the insulation from the ends of the wire, and hold these ends against the metal terminals of the battery.
Use the compass to determine whether the nail is magnetized. Next, flip the battery so that the direction of the current is reversed. Again, bring the compass toward the same part of the nail. Can you
explain why the compass needle now points in a different direction? Bring paper clips near the nail while connected to the battery. What happens to the paper clips? How many can you pick up? MATERIALS • D-cell battery • 1 m length of insulated wire • large nail • compass • metal paper clips
(b)
Magnetism
671
Solenoids produce a strong magnetic field by combining several loops. A solenoid is important in many applications because it acts as a magnet when it carries a current. The magnetic field strength inside a solenoid increases with the current and is proportional to the number of coils per unit length. The magnetic field of a solenoid can be increased by inserting an iron rod through the center of the coil; this device is often called an electromagnet. The magnetic field that is induced in the rod adds to the magnetic field of the solenoid, often creating a powerful magnet. Figure 2.4 shows the
magnetic field lines of a solenoid. Note that the field lines inside the solenoid point in the same direction, are nearly parallel, are uniformly spaced, and are close together. This indicates that the field inside the solenoid is strong and nearly uniform. The field outside the solenoid is nonuniform and much weaker than the interior field. Solenoids are used in a wide variety of applications, from most of the appliances in your home to very high-precision medical equipment.
Magnetic Field in a Solenoid The magnetic field inside a solenoid is strong and nearly uniform. Note that the field lines resemble those of a bar magnet, so a solenoid effectively has north and south poles.
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What is the shape of the magnetic field produced by a straight currentcarrying wire? 2. Why is the magnetic field inside a solenoid stronger than the magnetic field outside?
3. If electrons behave like magnets, then why aren’t all atoms magnets?
Critical Thinking 4. In some satellites, torque coils are replaced by devices called torque rods. In torque rods, a ferromagnetic material is inserted inside the coil. Why does a torque rod have a stronger magnetic field than a torque coil?
672
Chapter 19
SECTION 3
Magnetic Force
Objectives ►
Given the force on a charge in a magnetic field, determine the strength of the magnetic field.

Use the right-hand rule to find the direction of the force on a charge moving through a magnetic field.

Determine the magnitude and direction of the force on a wire carrying current in a magnetic field.
Charged Particles in a Magnetic Field Although experiments show that a constant magnetic field does not exert a net force on a stationary charged particle, charges moving through a magnetic field do experience a magnetic force. This force has its maximum value when the charge moves perpendicular to the magnetic field, decreases in value at other angles, and becomes zero when the particle moves along the field lines. To keep the math simple in this book, we will limit our discussion to situations in which charges move parallel or perpendicular to the magnetic field lines.
A charge moving through a magnetic field experiences a force. Recall that the electric field at a point in space is defined as the electric force per unit charge acting on some test charge placed at that point. In a similar manner, we can describe the properties of the magnetic field, B, in terms of the magnetic force exerted on a test charge at a given point. Our test object is assumed to be a positive charge, q, moving with velocity v perpendicular to B. It has been found experimentally that the strength of the magnetic force on the particle moving perpendicular to the field is equal to the product of the magnitude of the charge, q, the magnitude of the velocity, v, and the strength of the external magnetic field, B, as shown by the following relationship. F magnetic = qvB
This expression can be rearranged as follows: Magnitude of a Magnetic Field
Fmagnetic B=—-
qv
magnetic force on a charged particle magnetic f i e l d = – – – – – – – – – – – – – – – (magnitude of charge)(speed of charge)
If the force is in newtons, the charge is in coulombs, and the speed is in meters per second, the unit of magnetic field strength is the tesla (T). Thus, if a 1 C charge moving at 1 m / s perpendicular to a magnetic field experiences a magnetic force of 1 N, the magnitude of the magnetic field is equal to 1 T. Most magnetic fields are much smaller than 1 T. We can express the units of the magnetic field as follows: T=
N
N
V•s
C•m/ s
A•m
m2
Conventional laboratory magnets can produce magnetic fields up to about 1.5 T. Superconducting magnets that can generate magnetic fields as great as 30 T have been constructed. For comparison, Earth’s magnetic field near its surface is about 50 µT (5 x 10- 5 T). Magnetism
673
Alternative RightHand Rule Use this alternative right-hand rule to find the direction of the magnetic force on a positive charge.
An alternative right-hand rule can be used to find the direction of the magnetic force. Experiments show that the direction of the magnetic force is always perpendicular to both the velocity, v, and the magnetic field, B. To determine the direction of the force, use the right-hand rule. As before, place your fingers in the direction ofB with your thumb pointing in the direction of v, as illustrated in Figure 3.1. The magnetic force, F magnetic’ on a positive charge is directed out of the palm of your hand.
B
t V
If the charge is negative rather than positive, the force is directed opposite that shown in Figure 3.1. That is, if q is negative, simply use the right-hand rule to find the direction of Fmagnetic for positive q, and then reverse this direction for the negative charge.
Auroras very so often, the sky in far north and far south latitudes lights up with a spectacular natural lights show. These phenomena, called aurora borealis in the Northern Hemisphere and aurora australias in the Southern Hemisphere, are due to the interaction between charged particles and the Earth’s magnetic field. The sun constantly emits charged particles, protons and electrons,
674
Chapter 19
which eventually make their way to Earth. Once they reach Earth they move through its magnetic field. This in turn produces a force that causes the charges to accelerate and move toward the poles. The charges, guided along the Earth’s magnetic field, spiral toward the lower atmosphere. They eventually collide with atoms of nitrogen and oxygen. These atoms, in turn, get excited by the collision and emit light, ranging from brilliant reds to sparkling greens. The color of these lights depends on the atom being excited and its altitude. Aurora are most often seen near the poles because Earth’s magnetic field lines are most concentrated there, and because the field lines are at the correct height to produce these seemingly magical interactions.
PREMIUM CONTENT
~ Interactive Demo \:,:/ HMOScience.com
Sample Problem A A proton moving east experiences a force of 8.8 x 10- 19 N upward due to the Earth’s magnetic field. At this location, the field has a magnitude of 5.5 x 10- 5 T to the north. Find the speed of the particle.
0
ANALYZE
Given:
q = l.60 x 10- 19 C Fmagnetic
Unknown:
E)
SOLVE
B = 5.5 x 10-s T
= 8.8 X 10-19 N
v=?
Use the definition of magnetic field strength. Rearrange to solve for v.
B
= F magnetic qv Fmagnetic
V =—-
qB
Tips and Tricks The directions given can be used to verify the right-hand rule. Imagine standing at this location and facing north. Turn the palm of your right hand upward (the direction of the force) with your thumb pointing east (the direction of the velocity). If your palm and thumb point in these directions, your fingers point directly north in the direction of the magnetic field, as they should.
V=
8.8 X 10-19 N (1.60 x 10-19 C) (5.5 x 10-s T)
Iv= l.0 x 10
5
mis
= 1.0 X
10sm/s
I
Practice 1. A proton moves perpendicularly to a magnetic field that has a magnitude of 4.20 x 10- 2 T. What is the speed of the particle if the magnitude of the magnetic force on it is 2.40 x 10- 14 N? 2. If an electron in an electron beam experiences a downward force of2.0 x 10- 14 N while traveling in a magnetic field of 8.3 x 10- 2 T west, what is the direction and magnitude of the velocity? 3. A uniform 1.5 T magnetic field points north. If an electron moves vertically downward (toward the ground) with a speed of2.5 x 107 mis through this field, what force (magnitude and direction) will act on it?
M agnetism
675
A charge moving through a magnetic field follows a circular path. Charge Moving Through a Uniform Magnetic Field When the velocity, v, of a charged particle is perpendicular to a uniform magnetic field, the particle moves in a circle whose plane is perpendicular to B.
‘ Fmagnetic
I
I
x:x

I
\
+ q I
\ I
\
X
‘,X ”
X
……..
X/ X
_____ …. ; .,
Consider a positively charged particle moving in a uniform magnetic field. Suppose the direction of the particle’s initial velocity is exactly perpendicular to the field, as in Figure 3.2. Application of the right-hand rule for the charge q shows that the direction of the magnetic force, F magnetic’ at the charge’s location is to the left. Furthermore, application of the right-hand rule at any point shows that the magnetic force is always directed toward the center of the circular path. Therefore, the magnetic force is, in effect, a force that maintains circular motion and changes only the direction of v, not its magnitude. Now consider a charged particle traveling with its initial velocity at some angle to a uniform magnetic field. A component of the particle’s initial velocity is parallel to the magnetic field. This parallel part is not affected by the magnetic field, and that part of the motion will remain the same. The perpendicular part results in a circular motion, as described above. The particle will follow a helical path, like the red stripes on a candy cane, whose axis is parallel to the magnetic field.
X
Magnetic Force on a Current-Carrying Conductor Recall that current consists of many charged particles in motion. If a force is exerted on a single charged particle when the particle moves through a magnetic field, it should be no surprise that a current-carrying wire also experiences a force when it is placed in a magnetic field. The resultant force on the wire is the sum of the individual magnetic forces on the charged particles. The force on the particles is transmitted to the bulk of the wire through collisions with the atoms making up the wire.
Force on a Current-Carrying Wire in a Magnetic Field A current-carrying conductor in a magnetic field experiences a force that is perpendicular to the direction of the current.
X X
X
Consider a straight segment of wire of length ecarrying current, I, in a uniform external magnetic field, B, as in Figure 3.3. When the current and magnetic field are perpendicular, the magnitude of the total magnetic force on the wire is given by the following relationship. Force on a Current-Carrying Conductor Perpendicular to a Magnetic Field F magnetic
X
:1
X
= BIE
magnitude of magnetic force = (magnitude of magnetic field) (current)(Iength of conductor within B)
e
Fmagnetic
X
X
X
X
676
Chapter 19
B
X
The direction of the magnetic force on a wire can be obtained by using the right-hand rule. However, in this case, you must place your thumb in the direction of the current rather than in the direction of the velocity, v. In Figure 3.3, the direction of the magnetic force on the wire is to the left. When the current is either in the direction of the field or opposite the direction of the field, the magnetic force on the wire is zero.
Two parallel conducting wires exert a force on one another. Because a current in a conductor creates its own magnetic field, it is easy to understand that two current-carrying wires placed close together exert magnetic forces on each other. When the two conductors are parallel to each other, the direction of the magnetic field created by one is perpendicular to the direction of the current of the other, and vice versa. In this way, a force of F magnetic = Elf acts on each wire, where B is the magnitude of the magnetic field created by the other wire. Consider the two long, straight, parallel wires shown in Figure 3.4. When the current in each is in the same direction, the two wires attract one another. Confirm this by using the right-hand rule. Point your thumb in the direction of current in one wire, and point your fingers in the direction of the field produced by the other wire. By doing this, you find that the direction of the force (pointing out from the palm of your hand) is toward the other wire. When the currents in each wire are in opposite directions, the wires repel one another.
Force Between Parallel Conducting Wires Two parallel wires, each carrying a steady current, exert magnetic forces on each other. The force is (a) attractive if the currents have the same direction and (b) repulsive if the two currents have opposite directions.
Loudspeakers use magnetic force to produce sound. The loudspeakers in most sound systems use a magnetic force acting on a current-carrying wire in a magnetic field to produce sound waves. One speaker design, shown in Figure 3.5, consists of a coil of wire, a flexible paper cone attached to the coil that acts as the speaker, and a permanent magnet. In a speaker system, a sound signal is converted to a varying electric signal by the microphone. This electrical signal is amplified and sent to the loudspeaker. At the loudspeaker, this varying electrical current causes a varying magnetic force on the coil. This alternating force on the coil results in vibrations of the attached cone, which produce variations in the density of the air in front of it. In this way, an electric signal is converted to a sound wave that closely resembles the sound wave produced by the source.
Loudspeaker In a loudspeaker, when the direction and magnitude of the current in the coil of wire change, the paper cone attached to the coil moves, producing sound waves.
Paper cone
N
Magnetism
677
PREMIUM CONTENT
~ Interactive
Force on a Current-Carrying Conductor
\.::I
Demo
HMDScience .com
Sample Problem B A wire 36 m long carries a current of 22 A from east to west. If the magnetic force on the wire due to Earth’s magnetic field is downward ( toward Earth) and has a magnitude of 4.0 x 10-2 N, find the magnitude and direction of the magnetic field at this location.
0
ANALYZE
Given:
E=36m
l=22A
F magnetic= 4.0 X 10-2 N
Unknown:
E)
SOLVE
Use the equation for the force on a current-carrying conductor perpendicular to a magnetic field.
Fmagnetic = Bie Rearrange to solve for B.
B = Fmagnetic = 4.0 X 10- 2 N = Is.o If (22 A)(36 m) ·
X
10- s T I
·
Using the right-hand rule to find the direction ofB, face north with your thumb pointing to the west (in the direction of the current) and the palm of your hand down (in the direction of the force). Your fingers point north. Thus, Earth’s magnetic field is from south to north.
Practice 1. A 6.0 m wire carries a current of 7.0 A toward the +x direction. A magnetic force of 7 .0 x 1o- 6 N acts on the wire in the -y direction. Find the magnitude and direction of the magnetic field producing the force. 2. A wire 1.0 m long experiences a magnetic force of 0.50 N due to a perpendicular uniform magnetic field. If the wire carries a current of 10.0 A, what is the magnitude of the magnetic field?
3. The magnetic force on a straight 0.15 m segment of wire carrying a current of 4.5 A is 1.0 N. What is the magnitude of the component of the magnetic field that is perpendicular to the wire?
4. The magnetic force acting on a wire that is perpendicular to a 1.5 T uniform magnetic field is 4.4 N. If the current in the wire is 5.0 A, what is the length of the wire that is inside the magnetic field?
678
Chapter 19

Galvanometers A galvanometer is a device used in the construction of both ammeters and voltmeters. Its operation is based on the fact that a torque acts on a current loop in the presence of a magnetic field. Figure 3.6 shows a simplified arrangement of the main components of a galvanometer. It consists of a coil of wire wrapped around a soft iron core mounted so that it is free to pivot in the magnetic field provided by the permanent magnet. The torque experienced by the coil is proportional to the current in the coil. This means that the larger the current, the greater the torque and the more the coil will rotate before the spring tightens enough to stop the movement. Hence, the amount of deflection of the needle is proportional to the current in the coil. When there is no current in the coil, the spring returns the needle to zero. Once the instrument is properly calibrated, it can be used in conjunction with other circuit elements as an ammeter (to measure currents) or as a voltmeter (to measure potential differences).
A Galvanometer In a galvanometer, when current enters the coil, which is in a magnetic field, the magnetic force causes the coil to twist.
Spring
SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. A particle with a charge of 0.030 C experiences a magnetic force of 1.5 N while moving at right angles to a uniform magnetic field. If the speed of the charge is 620 m/ s, what is the magnitude of the magnetic field the particle passes through?
2. An electron moving north encounters a uniform magnetic field. If the magnetic field points east, what is the direction of the magnetic force on the electron? 3. A straight segment of wire has a length of 25 cm and carries a current of 5.0 A. If the wire is perpendicular to a magnetic field of 0.60 T, then what is the magnitude of the magnetic force on this segment of the wire?
4. Two parallel wires have charges moving in the same direction. Is the force between them attractive or repulsive?
Interpreting Graphics 5. Find the direction of the magnetic force on the current-carrying wire in Figure 3.7.
sl
sl I
Magnetism
679
SECTION 1
Magnets and Magnetic Fields
1 1 , 1 , , ·.1
• Like magnetic poles repel, and unlike poles attract. • A magnetic domain is a group of atoms whose magnetic fields are aligned.
magnetic domain magnetic field
• The direction of any magnetic field is defined as the direction the north pole of a magnet would point if placed in the field. The magnetic field of a magnet points from the north pole of the magnet to the south pole. • The magnetic north pole of Earth corresponds to the geographic South Pole, and the magnetic south pole corresponds to the geographic North Pole.
SECTION 2
Magnetism from Electricity
, c,
• A magnetic field exists around any current-carrying wire; the direction of the magnetic field follows a circular path around the wire.
1c, ,.1
solenoid
• The magnetic field created by a solenoid or coil is similar to the magnetic field of a permanent magnet.
SECTION 3
Magnetic Force
• The direction of the force on a positive charge moving through a magnetic field can be found by using the alternate right-hand rule.
• A current-carrying wi re in an external magnetic field undergoes a magnetic force. The direction of the magnetic force on the wire can be found by using the alternate right-hand rule. • Two parallel current-carrying wires exert on one another forces that are equal in magnitude and opposite in direction. If the currents are in the same direction, the two wires attract one another. If the currents are in opposite directions, the wires repel one another.
DIAGRAM SYMBOLS
Magnetic field vector B
magnetic field
Fmagnetic magnetic force
e
length of conductor in field
T tesla N newtons
N Cem/s kg•m s2
N A•m
– ——
Magnetic field pointing into the page Magnetic field pointing out of the page
m meters
t )(

♦——————–I
Problem Solving See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
680
Chapter 19
Magnets and Magnetic Fields REVIEWING MAIN IDEAS
10. You have two iron bars and a ball of string in your
possession; one iron bar is magnetized, and one iron bar is not. How can you determine which iron bar is magnetized?
1. What is the minimum number of poles for a magnet? 11. Why does a very strong magnet attract both poles of a
2. When you break a magnet in half, how many poles
weak magnet?
does each piece have? 12. A magnet attracts a piece of iron. The iron can then
3. The north pole of a magnet is attracted to the geographic North Pole ofEarth, yet like poles repel. Can you explain this?
4. Which way would a compass needle point if you were at the magnetic north pole? 5. What is a magnetic domain? 6. Why are iron atoms so strongly affected by magnetic fields? 7. When a magnetized steel needle is strongly heated in
a Bunsen burner flame, it becomes demagnetized. Explain why. 8. If an unmagnetized piece of iron is attracted to one pole of a magnet, will it be repelled by the opposite pole?
CONCEPTUAL QUESTIONS 9. In the figure below, two permanent magnets with holes bored through their centers are placed one over the other. Because the poles of the upper magnet are the reverse of those of the lower, the upper magnet levitates above the lower magnet. If the upper magnet were displaced slightly, either up or down, what would be the resulting motion? Explain. What would happen if the upper magnet were inverted?
attract another piece of iron. Explain, on the basis of alignment of domains, what happens in each piece of iron. 13. When a small magnet is repeatedly dropped, it
becomes demagnetized. Explain what happens to the magnet at the atomic level.
Magnetism from Electricity REVIEWING MAIN IDEAS 14. A conductor carrying a current is arranged so that electrons flow in one segment from east to west. If a
compass is held over this segment of the wire, in what direction is the needle deflected? (Hint: Recall that current is defined as the motion of positive charges.) 15. What factors does the strength of the magnetic field
of a solenoid depend on?
CONCEPTUAL QUESTIONS 16. A solenoid with ends marked A and Bis suspended
by a thread so that the core can rotate in the horizontal plane. A current is maintained in the coil so that the electrons move clockwise when viewed from end A toward end B. How will the coil align itself in Earth’s magnetic field? 17. Is it possible to orient a current-carrying loop of wire
in a uniform magnetic field so that the loop will not tend to rotate?
Chapter Review
681
18. If a solenoid were suspended by a string so that it could rotate freely, could it be used as a compass when it carried a direct current? Could it also be used if the current were alternating in direction?
26. For each situation below, use the movement of the positively charged particle and the direction of the magnetic force acting on it to find the direction of the magnetic field. (a)
Magnetic Force
l r (b)
REVIEWING MAIN IDEAS
V
19. Two charged particles are projected into a region where there is a magnetic field perpendicular to their velocities. If the particles are deflected in opposite directions, what can you say about them? 20. Suppose an electron is chasing a proton up this page when suddenly a magnetic field pointing into the page is applied. What would happen to the particles?
21. Why does the picture on a television screen become distorted when a magnet is brought near the screen? 22. A proton moving horizontally enters a region where there is a uniform magnetic field perpendicular to the proton’s velocity, as shown below. Describe the proton’s subsequent motion. How would an electron behave under the same circumstances?
X X 0–;;-+ X V X X
X X X X X
X ,C Bin
X X X
23. Explain why two parallel wires carrying currents in opposite directions repel each other.
24. Can a stationary magnetic field set a resting electron in motion? Explain. 25. At a given instant, a proton moves in the positive x direction in a region where there is a magnetic field in the negative z direction. What is the direction of the magnetic force? Does the proton continue to move along the x -axis? Explain.
682
Chapter 19
(c) F
—0 Vin
0
CONCEPTUAL QUESTIONS 27. A stream of electrons is projected horizontally to the right. A straight conductor carrying a current is supported parallel to and above the electron stream. a. What is the effect on the electron stream if the current in the conductor is left to right? b. What is the effect if the current is reversed? 28. If the conductor in item 27 is replaced by a magnet with a downward magnetic field, what is the effect on the electron stream?
29. Two wires carrying equal but opposite currents are twisted together in the construction of a circuit. Why does this technique reduce stray magnetic fields?
PRACTICE PROBLEMS For problems 30-31, see Sample Problem A.
30. A duck flying due east passes over Atlanta, where the magnetic field ofEarth is 5.0 x 10- 5 T directed north. The duck has a positive charge of 4.0 x 10- 8 C. If the magnetic force acting on the duck is 3.0 x 10- 11 N upward, what is the duck’s velocity? 31. A proton moves eastward in the plane of Earth’s magnetic equator, where Earth’s magnetic field points north and has a magnitude of 5.0 x 10- 5 T. What velocity must the proton have for the magnetic force to just cancel the gravitational force? For problems 32-33, see Sample Problem B. 32. A wire carries a 10.0 A current at an angle 90.0° from the direction of a magnetic field. If the magnitude of the magnetic force on a 5.00 m length of the wire is 15.0 N, what is the strength of the m agnetic field?
33. A thin 1.00 m long copper rod in a uniform magnetic field has a mass of 50.0 g. When the rod carries a current of 0.245 A, it floats in the magnetic field. What is the field strength of the magnetic field?
Mixed Review REVIEWING MAIN IDEAS 34. A proton moves at 2.50 x 106 m/s horizontally at a right angle to a magnetic field. a. What is the strength of the magnetic field required to exactly balance the weight of the proton and keep it moving horizontally? b. Should the direction of the magnetic field be in a horizontal or a vertical plane? 35. Find the direction of the force on a proton moving through each magnetic field in the four figures below.
(a)
(b)
V
x+x XX X
V
.111
B♦♦♦
41. In the figure below, a 15 cm length of conducting wire that is free to move is held in place between two thin conducting wires. All the wires are in a magnetic field. When a 5.0 A current is in the wire, as shown in the figure, the wire segment moves upward at a constant velocity. Assuming the wire slides without friction on the two vertical conductors and has a mass of 0.15 kg, find the magnitude and direction of the minimum magnetic field that is required to move the wire.
-+ -+
x 8 in
xxxx
(c)
:r~ V
40. A proton travels with a speed of 3.0 x 106 m /s at an angle of 37° west of north. A magnetic field of 0.30 T points to the north. Determine the following: a. the magnitude of the magnetic force on the proton b. the direction of the magnetic force on the proton c. the proton’s acceleration as it moves through the magnetic field (Hint: The magnetic force experienced by the proton in the magnetic field is proportional to the component of the proton’s velocity that is perpendicular to the magnetic field.)
(d)
Bout
V
•••• •••• •••• ••••
36. Find the direction of the force on an electron moving through each magnetic field in the four figures in item 35 above.
37. In the four figures in item 35, assume that in each case the velocity vector shown is replaced with a wire carrying a current in the direction of the velocity vector. Find the direction of the magnetic force acting on each wire. 38. A proton moves at a speed of2.0 x 107 m / s at right angles to a magnetic field with a magnitude of 0.10 T. Find the magnitude of the acceleration of the proton. 39. A proton moves perpendicularly to a uniform m agnetic field, B, with a speed of 1.0 x 107 m /s and
15cm
l
5.0A
~ 5.0A
t
5.0A
42. A current, I= 15 A, is directed along the positive x-axis and perpendicular to a uniform magnetic field. The conductor experiences a magnetic force per unit length of0.12 Nim in the negative y direction. Calculate the magnitude and direction of the magnetic field in the region through which the current passes. 43. A proton moving perpendicular to a magnetic field of strength 3.5 mT experiences a force due to the field of 4.5 x 10- 2 1 N. Calculate the following: a. the speed of the proton b. the kinetic energy of the proton Recall that a proton has a charge of 1.60 x 10- 19 C and a mass of 1.67 x 10- 27 kg.
experiences an acceleration of2.0 x 1013 m / s2 in the positive x direction when its velocity is in the positive z direction. Determine the magnitude and direction of the field.
Chapter Review
683
44. A singly charged positive ion that has a mass of 6.68 x 10- 27 kg moves clockwise with a speed of 1.00 x 104 m / s. The positively charged ion moves in a circular path that has a radius of 3.00 cm. Find the direction and strength of the uniform magnetic field through which the charge is moving. (Hint: The magnetic force exerted on the positive ion is the centripetal force, and the speed given for the positive ion is its tangential speed.)
46. Calculate the force on an electron in each of the following situations: a. moving at 2.0 percent the speed oflight and perpendicular to a 3.0 T magnetic field b. 3.0 x 10- 6 m from a proton c. in Earth’s gravitational field at the surface of Earth Use the following: qe = -1.6 x 10- 19 C; me= 9.1 X 10- 3 1 kg; qp = 1.6 X 10- 19 C; 2 2 8 9 C = 3.0 X 10 m / s; kc = 9.0 X 10 N•m / C
45. What speed would a proton need to achieve in order to circle Earth 1000.0 km above the magnetic equator? Assume that Earth’s magnetic field is everywhere perpendicular to the path of the proton and that Earth’s magnetic field h as an intensity of 4.00 x 10- 8 T. (Hint: The magnetic force exerted on the proton is equal to the centripetal force, and the speed needed by the proton is its tangential speed. Remember that the radius of the circular orbit should also include the radius of Earth. Ignore relativistic effects.)
Solenoids A solenoid consists of a long, helically wound coil of insulated wire. When it carries a current, a solenoid acts as a magnet. The magnetic field strength (B) increases linearly with the current (J) and with the number of coils per unit length. Because there is a direct relation between B and I, the following equation applies to any solenoid:
B = al + b In this equation, the parameters a and bare different for different solenoids. The a and b parameters can be determined if the magnetic field strength of the solenoid is
684
Chapter 19
known at two different currents. Once you determine a and b, you can predict the magnetic field strength of a solenoid for various currents. The graphing calculator program that accompanies this activity uses this procedure. You will be given the magnetic field and current data for various solenoids. You will then use this information and the program to predict the magnetic field strength of each solenoid. Go online to HMDScience.com to find this graphing calculator activity.
ALTERNATIVE ASSESSMENT 1. During a field investigation with your class, you find a roundish chunk of metal that attracts iron objects. Design a procedure to determine whether the object is magnetic and, if so, to locate its poles. Describe the limitations of your method. What materials would you need? How would you draw your conclusions? List all the possible results you can anticipate and the conclusions you could draw from each result.
2. Imagine you have been hired by a manufacturer interested in making kitchen magnets. The manufacturer wants you to determine how to combine several magnets to get a very strong magnet. He also wants to know what protective material to use to cover the magnets. Develop a method for measuring the strength of different magnets by recording the maximum number of paper clips they can hold under various conditions. First open a paper clip to use as a hook. Test the strength of different magnets and combinations of magnets by holding up the magnet, placing the open clip on the magnet, and hooking the rest of the paper clips so that they hang below the magnet. Examine the effect oflayering different materials between the magnet and the clips. Organize your data in tables and graphs to present your conclusions.
3. Research phenomena related to one of the following topics, and prepare a report or presentation with pictures and data. a. How does Earth’s magnetic field vary with latitude, with longitude, with the distance from Earth, and in time? b. How do people who rely on compasses account for these differences in Earth’s magnetic field? c. What is the Van Allen belt? d. How do solar flares occur? e. How do solar flares affect Earth?
4. Obtain old buzzers, bells, telephone receivers, speakers, motors from power or kitchen tools, and so on to take apart. Identify the mechanical and electromagnetic components. Examine their connections. How do they produce magnetic fields? Work in a cooperative group to describe and organize your findings about several devices for a display entitled “Anatomy of Electromagnetic Devices:’ 5. Magnetic force was first described by the ancient Greeks, who mined a magnetic mineral called magnetite. Magnetite was used in early experiments on magnetic force. Research the historical development of the concept of magnetic force. Describe the work of Peregrinus, William Gilbert, Oersted, Faraday, and other scientists.
Chapter Review
685
MULTIPLE CHOICE 1. Which of the following statements best describes the domains in unmagnetized iron? A. There are no domains. B. There are domains, but the domains are smaller than in magnetized iron. C. There are domains, but the domains are oriented randomly. D. There are domains, but the domains are not magnetized.
4. How can you increase the strength of a magnetic field inside a solenoid? F. increase the number of coils per unit length G. increase the current H. place an iron rod inside the solenoid J. all of the above Use the diagram below to answer questions 5-6.
X X X X X
2. Which of the following statements is most correct? F. The north pole of a freely rotating magnet points north because the magnetic pole near the geographic North Pole is like the north pole of a magnet. G. The north pole of a freely rotating magnet points north because the magnetic pole near the geographic North Pole is like the south pole of a magnet. H. The north pole of a freely rotating magnet points south because the magnetic pole near the geographic South Pole is like the north pole of a magnet. J. The north pole of a freely rotating magnet points south because the magnetic pole near the geographic South Pole is like the south pole of a magnet.
3. If you are standing at Earth’s magnetic north pole and holding a bar magnet that is free to rotate in three dimensions, which direction will the south pole of the magnet point? A. straight up B. straight down C. parallel to the ground, toward the north D. parallel to the ground, toward the south
686
Chapter 19
X X X X X
X X X X X
X X X X X
X X XBin X X
J;
5. How will the electron move once it passes into the magnetic field? A. It will curve to the right and then continue moving in a straight line to the right. B. It will curve to the left and then continue moving in a straight line to the left. C. It will move in a clockwise circle. D. It will move in a counterclockwise circle. 6. What will be the magnitude of the force on the electron once it passes into the magnetic field?
F. qvB G. – qvB H qv ‘ B
J. Blt
7. An alpha particle (q = 3.2 x 10- 19 C) moves at a speed of2.5 x 106 m/ s perpendicular to a magnetic field of strength 2.0 x 10- 4 T. What is the magnitude of the magnetic force on the particle? A. 1.6 x 10- 16 N B. – 1.6 x 10- 16 N C. 4.0 x 10- 9 N D. zero
. TEST PREP
Use the passage below to answer questions 8-9.
A wire 25 cm long carries a 12 A current from east to west. Earth’s magnetic field at the wire’s location has a magnitude of 4.8 x 10- 5 T and is directed from south to north. 8. What is the magnitude of the magnetic force on
the wire? F. 2.3 X 10- 5 N G. 1.4 x 10- 4 N H. 2.3 x 10- 3 N J. 1.4 X 10- 2 N 9. What is the direction of the magnetic force on the wire? A. north B. south C. up, away from Earth D. down, toward Earth
11. What is the direction of the force on wire 2 as a result of B1? A. to the left B. to the right C. into the page D. out of the page 12. What is the magnitude of the magnetic force on wire 2?
F. B/1t1 G. B//2 H. B/2 t 2 J. B/l2
SHORT RESPONSE 13. Sketch the magnetic field lines around a bar magnet. 14. Describe how to use the right-hand rule to deter-
Use the diagram below to answer questions 10-12.
mine the direction of a magnetic field around a current-carrying wire. 15. Draw a diagram showing the path of a positively charged particle moving in the plane of a piece of paper if a uniform magnetic field is coming out of the page.
EXTENDED RESPONSE
Wire 1 carries current 11 and creates magnetic field B1 . Wire 2 carries current 12 and creates magnetic field B2 . 10. What is the direction of the magnetic field B 1 at the location of wire 2? F. to the left G. to the right H. into the page J. out of the page
16. A proton (q = 1.6 x 10- 19 C; m = 1.7 x 10- 27 kg) is in a uniform 0.25 T magnetic field. The proton moves in a clockwise circle with a tangential speed of2.8 x 105 m / s. a. What is the direction of the magnetic field? Explain how you determined this. b. What is the radius of the circle? Show your work.
Test Tip If you are asked to write out an answer, to show your calculations, or to draw a diagram, be sure to write clearly, to show all steps of your work, and to add clear labels to your diagrams. You may receive some credit for using the right approach to a problem, even if you do not arrive at the correct final answer.
Standards-Based Assessment
687
Can Cell Phones Cause Cancer? Cell phones transfer messages by sending and receiving electromagnetic waves. The electromagnetic spectrum includes low-energy waves, such as radio waves, and highenergy waves, such as X rays and gamma rays. High-energy electromagnetic waves are ionizing, which means they have enough energy to remove an electron from its orbit. Ionizing electromagnetic radiation can damage living tissue and cause DNA mutations, which is why exposure to X rays should be limited. Cell phones use radio frequencies (RFs) ranging from about 800 to 900 MHz. These nonionizing waves do not alter the molecular structure of living tissue. Though they can cause the atoms in a molecule to vibrate, they do not have enough energy to remove electrons from their orbits. At high enough levels, however, nonionizing radiation can cause biological damage by heating living tissue. But the amount of heat that a cell phone’s radiation generates is very small and is much smaller than the energy generated in a microwave oven. Identify the Problem: Effects of Nonionizing Radiation The effects of nonionizing radiation on the human body are not fully known. Several studies have been conducted to determine whether there is a link between cell phone use and brain cancer. Scientists conducting these studies have attempted to determine whether the risk of brain cancer is greater for cell phone users than for nonusers. Even if a link is found, it is not necessarily a cause-and-effect link. In other words, even if cell phone users do have a higher risk of cancer, cell phone use is not necessarily the cause. Other issues complicate research into the issue. Cell phones were not widely available until the 1990s, and brain tumors develop over many years. Therefore, long-range studies are required to assess the effects of nonionizing radiation from cell phones. One such study is now underway. Conduct Research The International Cohort Study on Mobile Phone Users (COSMOS) aims to conduct long-term health monitoring of a large group of
688
Ce/I-phone usage has become increasingly popular among children, adolescents, and adults. Little is known, however, about its long-term health effects.
people to determine if there are any health issues linked to prolonged exposure to radio frequency energy from cell phone use. The COSMOS study will follow approximately 250 000 cell phone users in Europe for over 25 years. MOBI-KIDS is another international study investigating the relationship between exposure to radio frequency energy from communication technologies, including cell phones, and brain cancer in young people. This study, which involves 13 countries, began in 201 Oand will continue for 5 years. lnterphone is still another international study designed to determine whether cell phones increase the risk of head and neck cancer. In this study, scientists compared cell phone usage for more than 5000 people with brain tumors and a similar number of healthy controls. Results of this study did not conclusively show that cell phones caused brain cancer. However, the study did suggest that people who used cell phones an average of more than a half hour per day, every day, for over 1oyears had a slight increase in brain cancer. But the scientists cautioned that the increase was not significant enough to determine a relationship between heavy use of cell phones and brain cancer. However, a December 2010 report questioned the findings of the lnterphone study. In this latest report, scientists examined the findings of the earlier report to include a wider age group
and redefine how users were classified. Based on their results, the scientists concluded that there was a significant link between heavy cell phone use and brain cancer. Select a Solution While researchers continue their investigation of nonionizing radiation, concerned cell phone users can take measures to limit their exposure to RFs. Exposure depends on a number of factors, including the amount of time spent using the phone, the amount of cell phone traffic in the area, and the distance between the antenna and the user’s head. One way to reduce exposure is to minimize the time spent on cellular calls. Another option is to use a hands-free device that puts the antenna farther from the head. Additionally, pregnant women should avoid carrying a cell phone next to their abdomen. Because children have smaller
and thinner skulls, they should limit cell phone use. Some scientists also warn against using a cell phone in areas with a weak signal, because the phones emit more radiation during those times. Finally, researchers caution to not go to sleep with a cell phone turned on and placed next to your bedside or under your pillow.
Design Your Own Conduct Research
Radio towers, such as the one in this image, send out radio signals that are picked up by cell phones and translated into sounds and images.
As cell phones have grown in popularity, concerns have arisen not only about brain cancer, but also about the safety of driving while using a cell phone. Several countries and many states in the United States have banned the use of cell phones while driving. Conduct research to find out about studies conducted on this issue. Is it hazardous? Should laws be passed in all states that prevent the use of a cell phone while driving in a school zone? Write a paper summarizing your findings. Test and Evaluate Cell phone makers are now required to report the specific absorption rate (SAR), the amount of RF energy absorbed by the user. The maximum allowed SAR is 1.6 watts per kilogram. Find the SAR of several top models of phones. If you own a cell phone, see if you can determine the SAR of your phone. Communicate Use the Internet to research one of the recent studies done on cell phone use and brain cancer. Write a short report describing the study, including the subjects and control group, the method of obtaining data, and the conclusions reached by the researchers. Share your report with the class.
689
SECTION 1
Electricitv from Magnetism
Objectives

Recognize that relative motion between a conductor and a magnetic field induces an emf in the conductor.

Describe how the change in the number of magnetic field lines through a circuit loop affects the magnitude and direction of the induced electric current.

Key Term electromagnetic induction
Electromagnetic Induction Recall that when you were studying circuits, you were asked if it was possible to produce an electric current using only wires and no battery. So far, all electric circuits that you have studied have used a battery or an electrical power supply to create a potential difference within a circuit. The electric field associated with that potential difference causes charges to move through the circuit and to create a current.
Apply Lenz’s law and Faraday’s law of induction to solve problems involving induced emf and current.
It is also possible to induce a current in a circuit without the use of a
battery or an electrical power supply. You have learned that a current in a circuit is the source of a magnetic field. Conversely, a current results when a closed electric circuit moves with respect to a magnetic field, as shown in Figure 1.1. The process of inducing a current in a circuit by changing the magnetic field that passes through the circuit is called electromagnetic induction.
electromagnetic induction the process of creat ing a current in a circuit loop by changing the magnetic flux in the loop
Consider a closed circuit consisting of only a resistor that is in the vicinity of a magnet. There is no battery to supply a current. If neither the magnet nor the circuit is moving with respect to the other, no current will be present in the circuit. But, if the circuit moves toward or away from the magnet or the magnet moves toward or away from the circuit, a current is induced. As long as there is relative motion between the two, a current is created in the circuit.
The separation of charges by the magnetic force induces an emf. It may seem strange that there can be an induced emf and a correspond-
ing induced current without a battery or similar source of electrical
Electromagnetic Induction When the circuit loop crosses the Iines of the magnetic field, a current is induced in the circuit, as indicated by the movement of the galvanometer needle.
692
Chapter 20
Galvanometer
I
0
+
energy. Recall from that a moving charge can be deflected by a magnetic field. This deflection can be used to explain how an emf occurs in a wire that moves through a magnetic field. Consider a conducting wire pulled through a magnetic field, as shown on the left in Figure 1.2. You learned when studying magnetism that charged particles moving with a velocity at an angle to the magnetic field will experience a magnetic force. According to the right-hand rule, this force will be perpendicular to both the magnetic field and the motion of the charges. For free positive charges in the wire, the force is directed downward along the wire. For negative charges, the force is upward. This effect is equivalent to replacing the segment of wire and the magnetic field with a battery that has a potential difference, or emf, between its terminals, as shown on the right in Figure 1.2. As long as the conducting wire moves through the magnetic field, the emf will be maintained. The polarity of the induced emf depends on the direction in which the wire is moved through the magnetic field. For instance, if the wire in Figure 1.2 is moved to the right, the right-hand rule predicts that the negative charges will be pushed upward. If the wire is moved to the left, the negative charges will be pushed downward. The magnitude of the induced emf depends on the velocity with which the wire is moving through the magnetic field, on the length of the wire in the field, and on the strength of the magnetic field.
Potential Difference in a Wire The separation of positive and negative moving charges by the magnetic force creates a potential difference (emO between the ends of the conductor.
1 •• •• •• •• e ve •• @ee •• —–+-
I I

=+ –
..L I
B ( out of page)
The angle between a magnetic field and a circuit affects induction. One way to induce an emf in a closed loop of wire is to move all or part of the loop into or out of a constant magnetic field. No emf is induced if the loop is static and the magnetic field is constant. The magnitude of the induced emf and current depend partly on how the loop is oriented to the magnetic field, as shown in Figure 1.3. The induced current is largest if the plane of the loop is perpendicular to the magnetic field, as in (a); it is smaller if the plane is tilted into the field, as in (b); and it is zero if the plane is parallel to the field, as in (c). The role that the orientation of the loop plays in inducing the current can be explained by the force that the magnetic field exerts on the charges in the moving loop. Only the component of the magnetic field perpendicular to
Orientation of a Loop in a Magnetic Field These three loops of wire are moving out of a region that has a constant magnetic field. The induced emf and current are largest when the plane of the loop is perpendicular to the magnetic field (a), smaller when the plane of the loop is tilted (b), and zero when the plane of the loop and the magnetic field are parallel (c).
V
(a)
(b)
•••
(c)
• • • c!=>.• • •••
Electromagnetic Induction
693
both the plane and the motion of the loop exerts a magnetic force on the charges in the loop. If the area of the loop is moved parallel to the magnetic field, there is no magnetic field component perpendicular to the plane of the loop and therefore no induced emf to move the charges around the circuit.
Change in the number of magnetic field lines induces a current. ..Did YOU Know?. – – – – – – – – – – – , In 1996, the space shuttle Columbia attempted to use a 20.7 km conducting tether to study Earth’s magnetic field in space. The plan was to drag the tether through the magnetic field, inducing an emf in the tether. The magnitude of the emf would directly vary with the strength of the magnetic field. Unfortunately, the tether broke before it was fully extended, so the experiment was abandoned.
: : :
:
So far, you have learned that moving a circuit loop into or out of a magnetic field can induce an emf and a current in the circuit. Changing the size of the loop or the strength of the magnetic field also will induce an emf in the circuit. One way to predict whether a current will be induced in a given situation is to consider how many magnetic field lines cut through the loop. For example, moving the circuit into the magnetic field causes some lines to move into the loop. Changing the size of the circuit loop or rotating the loop changes the number of field lines passing through the loop, as does changing the magnetic field’s strength or direction. Figure 1.4 summarizes these three ways of inducing a current.
Characteristics of Induced Current Suppose a bar magnet is pushed into a coil of wire. As the magnet moves into the coil, the strength of the magnetic field within the coil increases, and a current is induced in the circuit. This induced current in turn produces its own magnetic field, whose direction can be found by using the right-hand rule. If you were to apply this rule for several cases, you would notice that the induced magnetic field direction depends on the change in the applied field.
Description
Before
After
~• •.
• v• •
Circuit is moved into or out of magnetic field (either circuit or magnet moving). I
Circuit is rotated in the magnetic field (angle between area of circuit and magnetic field changes).
-~B
694
Chapter 20
B
..-~-
• • • •~ .• • ·• •0· . • __ , , -. ••• • ••• •• ·.r-;-i.•. •• • • ••• • •
B
Intensity and/or direction of magnetic field is varied.
Ll •• • •• • I
B
I
Magnet Moving Toward Coil When a bar magnet is moved toward a coil, the induced magnetic field is similar to the field of a bar magnet with the orientation shown.
::::::::
Falling Magnet A bar magnet
N
~ “Induced current
Magnetic field from induced current
is dropped toward the floor, on which lies a large ring of conducting metal. The magnet’s length-and thus the poles of the magnet- is parallel to the direction of motion. Disregarding air resistance, does the magnet fall toward the ring with the constant acceleration of a freely falling body? Explain your answer.

v –
Approaching magnetic field
Induction in a Bracelet
As the magnet approaches, the magnetic field passing through the coil increases in strength. The induced current in the coil is in a direction that produces a magnetic field that opposes the increasing strength of the approaching field. So, the induced magnetic field is in the opposite direction of the increasing magnetic field. The induced magnetic field is similar to the field of a bar magnet that is oriented as shown in Figure 1.5. The coil and the approaching magnet create a pair of forces that repel each other.
Suppose you are wearing a bracelet that is an unbroken ring of copper. If you walk briskly into a strong magnetic field while wearing the bracelet, how would you hold your wrist with respect to the magnetic field in order to avoid inducing a current in the bracelet?
If the magnet is moved away from the coil, the magnetic field passing through the coil decreases in strength. Again, the current induced in the coil produces a magnetic field that opposes the decreasing strength of the receding field. This means that the magnetic field that the coil sets up is in the same direction as the receding magnetic field. The induced magnetic field is similar to the field of a bar magnet oriented as shown in Figure 1.6. In this case the coil and magnet attract each other.
Magnet Moving Away from Coil When a bar magnet is moved away from a coil, the induced magnetic field is similar to the field of a bar magnet with the orientation shown. Wire
s
~
Induced current
:::::::
V
Magoet;, field fi’om
Rei.;og
induced current
magnetic field
Electromagnetic Induc t ion
695
The rule for finding the direction of the induced current is called Lenz’s law and is expressed as follows: The magnetic field of the induced current is in a direction to produce a field that opposes the change causing it.
Note that the field of the induced current does not oppose the applied field but rather the change in the applied field. If the applied field changes, the induced field tends to keep the total field strength constant.
Magnetic Field of a Conducting Loop at an Angle The angle 0 is defined as the angle between the magnetic field and the normal to the plane of the loop. B cos 0 equals the strength of the magnetic field perpendicular to the plane of the loop. B cos 0 B Normal to plane of loop
Loop
Faraday’s law of induction predicts the magnitude of the induced emf. Lenz’s law allows you to determine the direction of an induced current in a circuit. Lenz’s law does not provide information on the magnitude of the induced current or the induced emf. To calculate the magnitude of the induced emf, you must use Faraday’s law of magnetic induction. For a single loop of a circuit, this may be expressed as follows: 6_q>M
emf=–6.t Recall from the chapter on magnetism that the magnetic flux, M’ can be written as AB cos 0. This equation means that a change with time of any of the three variables-applied magnetic field strength, B ; circuit area, A; or angle of orientation, 0-can give rise to an induced emf. The term B cos 0 represents the component of the magnetic field perpendicular to the plane of the loop. The angle 0 is measured between the applied magnetic field and the normal to the plane of the loop, as indicated in Figure 1.7. The minus sign in front of the equation is included to indicate the polarity of the induced emf. The sign indicates that the induced magnetic field opposes the change in the applied magnetic field as stated by Lenz’s law. If a circuit contains a number, N, of tightly wound loops, the average induced emf is simply N times the induced emf for a single loop. The equation thus takes the general form of Faraday’s law of magnetic induction.
Faraday’s Law of Magnetic Induction
~.Nflt The negative sign signifies that N decreases with time; that is, flN is negative. The quantity>. is called the decay constant. The value of>. for any isotope indicates the rate at which that isotope decays. Isotopes with a large decay constant decay quickly, and those with a small decay constant decay slowly. The number of decays per unit time, -flNI flt, is called the decay rate, or activity, of the sample. Note that the activity of a sample equals the decay constant times the number of radioactive nuclei in the sample, as follows: activity= -flN = >.N flt The SI unit of activity is the becquerel (Bq). One becquerel is equal to 1 decay/ s. The curie (Ci), which was the original unit of activity, is the approximate activity of 1 g of radium. One curie is equal to 3. 7 x 1010 Bq.
Half-life measures how long it takes half a sample to decay. Another quantity that characterizes radioactive decay is the half-life, written as T 112. The half-life of a substance is the time it takes for half of the radioactive nuclei in a sample to decay. The half-life of any substance is inversely proportional to the decay constant of the substance.
Decay Series Suppose a radioactive parent substance with a very long half-life has a daughter with a very short half-life. Describe what happens to a freshly purified sample of the parent substance. Probability of Decay ‘T he more probable the decay, the shorter the half-life.” Explain this statement.
half-life the time needed for half of the original nuclei of a sample of a radioactive substance to undergo radioactive decay
Decay of Radium The radioactive 2
nucleus ~~Ra (radium-226) has a half-life of about 1.6 x 103 years. Although the solar system is approximately 5 billion years old, we still find this radium nucleus in nature. Explain how this is possible.
0
• · •
·•

Subatomic Physics
785
Substances with large decay constants have short half-lives. The relationship between half-life and decay constant is given in the equation below. A derivation of this equation is beyond the scope of this book, but it involves the natural logarithm of 2. Because ln 2 = 0.693, this factor occurs in the final equation.
Half-Life
_ 0.693 T 112- A half-life=
0 693 •
decay constant
Consider a sample that begins with N radioactive nuclei. By definition, after one half-life, ½N radioactive nuclei remain. After two half-lives, half of these will have decayed, so ¼N radioactive nuclei remain. After three half-lives, ½Nwill remain, and so on.
Measuring Nuclear Decay Sample Problem C The half-life of the radioactive radium (226Ra) nucleus is 5.0 x 1010 s. A sample contains 3.0 x 1016 nuclei. What is the decay constant for this decay? How many radium nuclei, in curies, will decay per second?
0
E)
ANALYZE
PLAN
1010 s
Given:
T112 = 5.0
X
N = 3.0
Unknown:
,,\ = ?
activity = ? Ci
X
1016
Choose an equation or situation:
To find the decay constant, use the equation for half-life.
_ 0.693
T 112 –
,,\
The number of nuclei that decay per second is given by the equation for the activity of a sample.
activity = ,,\N Rearrange the equation to isolate the unknown:
The first equation must be rearranged to isolate the decay constant, .\.
,a.,,rn ,M&- ► 786
Chapter 22
Measuring Nuclear Decay
E)
(continued)
Substitute the values into the equations and solve:
SOLVE
.,\ = 0.693 = Tl/2 Tips and Tricks Always pay attention to units. Here, the activity is divided by the conversion factor 3.7 x 1010 s- 1/Ci to convert the answer from becquerels to curies, as specified in the problem statement.
0
CHECK YOUR WORK
0.693 10 5.0 X 10 s
I.,\ = 1.4 x 10-11 s-1 I . .
‘\ N
act1V1ty = /\
1 s- )
16 X 10 )
(1.4 X 10-ll (3.0 = – – – – – – 10 – -1 – – – 3.7 X 10 s- /Ci
5 I activity = 1.1 x 10- Ci I
Because the half-life is on the order of 1010 s, the decay constant, which approximately equals 0.7 divided by the half-life, should equal a little less than 10- 10 s- 1. Thus, 1.4 x 10- 11 s- 1 is a reasonable answer for the decay constant.
Practice 2
1. The half-life of J!Po is 164 µs. A polonium-214 sample contains 2.0 x 106 nuclei.
What is the decay constant for the decay? How many polonium nuclei, in curies, will decay per second? 2
2. The half-life of JjBi is 19.7 min. A bismuth-214 sample contains 2.0 x 109 nuclei.
What is the decay constant for the decay? How many bismuth nuclei, in curies, will decay per second? 1
3. The half-life of JJ1 is 8.07 days. Calculate the decay constant for this isotope. What is the activity in Ci for a sample that contains 2.5 x 1010 iodine-131 nuclei? 4. Suppose that you start with 1.00 x 10- 3 g of a pure radioactive substance and
determine 2.0 h later that only 0.25 x 10- 3 g of the substance is left undecayed. What is the half-life of this substance?
ilRn)
2
5. Radon-222 ( is a radioactive gas with a half-life of3.82 days. A gas sample contains 4.0 x 108 radon atoms initially. a. Estimate how many radon atoms will remain after 12 days.
b. Estimate how many radon nuclei will have decayed by this time.
Subatomic Physics
787
Half-Life of Carbon-14 The radioactive isotope carbon-14 has a half-life of 5715 years. In each successive 5715-year period, half the remaining carbon-14 nuclei decay to nitrogen-14. 0
14
C atoms
O
14
N atoms
A decay curve is a plot of the number of radioactive parent nuclei remaining in a sample as a function of time. A typical decay curve for a radioactive sample is shown in Figure 2.7. After each half-life, half the remaining parent nuclei have decayed. This is represented in the circles to the right of the decay curve. The blue spheres are the parent nuclei (carbon-14), and the red spheres are daughter nuclei (nitrogen-14). Notice that the total number of nuclei remains constant, while the number of carbon atoms continually decreases over time. For example, the initial sample contains 8 carbon-14 atoms. After one half-life, there are 4 carbon-14 atoms and 4 nitrogen-14 atoms. By the next half-life, the number of carbon-14 atoms is reduced to 2, and the process continues. As the number of carbon-14 atoms decreases, the number ofnitrogen-14 atoms increases.
Time
Ti;2
3Ti;2
Living organisms have a constant ratio of carbon-14 to carbon-12 because they continuously exchange carbon dioxide with their surroundings. When an organism dies, this ratio changes due to the decay of carbon-14. Measuring the ratio between carbon-14, which decays as shown in Figure 2.7, and carbon-12, which does not decay, provides an approximate date as to when the organism was alive.
SECTION 2 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. Explain the main differences between alpha, beta, and gamma decays.
2. Complete the following radioactive decay formulas: 2 a. J~Th – t ? + iHe
fB
1
+ je + D 1 c. ? – t iHe + t8Nd b.
-t?
3. A radioactive sample consists of 5.3 x 105 nuclei. There is one decay every4.2h. a. What is the decay constant for the sample? b. What is the half-life for the sample?
Critical Thinking 4. The 14 C content decreases after the death of a living system with a half-life of 5715 years. If an archaeologist finds an ancient fire pit containing partially consumed firewood and if the 14C content of the wood is only 12.5 percent that of an equal carbon sample from a present-day tree, what is the age of the ancient site?
788
Chapter 22
Nuclear Reactions Fission and Fusion Any process that involves a change in the nucleus of an atom is called a nuclear reaction. Nuclear reactions include fission, in which a nucleus splits into two or more nuclei, and fusion, in which two or more nuclei combine.
Stable nuclei can be converted to unstable nuclei. When a nucleus is bombarded with energetic particles, it may capture a particle, such as a neutron. As a result, the nucleus will no longer be stable and will disintegrate. For example, protons can be released when alpha particles collide with nitrogen atoms, as follows:
According to this expression, an alpha particle (tHe) strikes a nitrogen 1 nucleus ( *N) and produces an unknown product nucleus (X) and a proton ( H). By balancing atomic numbers and mass numbers, we can conclude that the unknown product has a mass number of 17 and an atomic number of 8. Because the element with an atomic number of 8 is oxygen, the product can be written symbolically as 1~ 0, and the reaction can be written as follows:
t
4He + 14N —-+ 1aF —-+ 17O + 1H 2 7 9 8 1 This nuclear reaction starts with two stable isotopes-helium and nitrogen-that form an unstable intermediate nucleus, fluorine. The intermediate nucleus then disintegrates into two different stable isotopes, hydrogen and oxygen. This reaction, which was the first nuclear reaction to be observed, was detected by Rutherford in 1919.
Heavy nuclei can undergo nuclear fission. Nuclear fission occurs when a heavy nucleus splits into two lighter nuclei. For fission to occur naturally, the nucleus must release energy. This means that the nucleons in the daughter nuclei must be more tightly bound and therefore have less mass than the nucleons in the parent nucleus. This decrease in mass per nucleon appears as released energy when fission occurs, often in forms such as photons or kinetic energy of the fission products. Because fission produces lighter nuclei, the binding energy per nucleon must increase with decreasing atomic number. Figure 3.1 shows that this is possible only for atoms in which A > 58. Thus, fission occurs naturally only for heavy atoms.
Binding Energy per Nucleon Light nuclei are very loosely bound. The binding energy of heavy nuclei is roughly the same for all nuclei. Binding Energy versus Mass Number Region of greatest stability
i
9.0
u = C
6.0
~ 8.0 C Q a, 7.0

a. 5.0 a,
>,
4.0 e’ a, C a, 3.0 en C
:a C iii
2.0 1.0 0 50 150 250 Mass number (A)
Subatomic Physics
789
One example of this process is the fission ofuranium-235. First, the nucleus is bombarded with neutrons. When the nucleus absorbs a neutron, it becomes unstable and decays. The fission of 235U can be represented as follows:
6n + i~u —-+ i~u* —-+ X + Y + neutrons 2
2
The isotope i~u* is an intermediate state that lasts only for about 10- 12 s before splitting into X and Y. Many combinations ofX and Y are possible. In the fission of uranium, about 90 different daughter nuclei can be formed. The process also results in the production of about two or three neutrons per fission event. 2
A typical reaction of this type is as follows:
in+ ~~u—-+ 2
1
tiBa
+ ~iKr + 3 6n
To estimate the energy released in a typical fission process, note that the binding energy per nucleon is about 7.6 MeV for heavy nuclei (those having a mass number of approximately 240) and about 8.5 MeV for nuclei of intermediate mass (see Figure 3.1 ) . The amount of energy released in a fission event is the difference in these binding energies (8.5 MeV – 7.6 MeV, or about 0.9 MeV per nucleon). Assuming a total of240 nucleons, this is about 220 MeV. This is a very large amount of energy relative to the energy released in typical chemical reactions. For example, the energy released in burning one molecule of the octane used in gasoline engines is about one hundred-millionth the energy released in a single fission event.
Neutrons released in fission can trigger a chain reaction.
A Nuclear Chain Reaction A nuclear chain
In
0
reaction can be initiated by the capture of a neutron.
In
0
I n 235u 0 92 93l(r 36
In
0
235u 92 I40Ba 56
In
0
I44cs 55
790
Chapter 22
I40Ba 56
235u 92
When 235U undergoes fission, an average of about 2.5 neutrons are emitted per event. The released neutrons can be captured by other nuclei, making these nuclei unstable. This triggers additional fission events, which lead to the possibility of a chain reaction, as shown in Figure 3.2. Calculations show that if the chain reaction is not controlled-that is, if it does not proceed slowly- it could result in the release of an enormous amount of energy and a violent explosion. If the energy in 1 kg of 235U were released, it would equal the energy released by the detonation of about 20 000 tons of TNT. This is the principle behind the first nuclear bomb, shown in Figure 3.3, which was essentially an uncontrolled fission reaction.
A nuclear reactor is a system designed to maintain a controlled, self-sustained chain reaction. Such a system was first achieved with uranium as the fuel in 1942 by Enrico Fermi, at the University of Chicago. Primarily, it is the uranium-235 isotope that releases energy through nuclear fission. Uranium from ore typically contains only about 0. 7 percent of 235U, with the remaining 99.3 percent being the 238U isotope. Because uranium-238 tends to absorb neutrons without fissioning, reactor fuels must be processed to increase the proportion of 235U so that the reaction can sustain itself. This process is called enrichment.
Atomic Bomb The first nuclear fission bomb, often called the atomic bomb, was tested in New Mexico in 1945.
At this time, all nuclear reactors operate through fission. One difficulty associated with fission reactors is the safe disposal of radioactive materials when the core is replaced. Transportation of reactor fuel and reactor wastes poses safety risks. As with all energy sources, the risks must be weighed against the benefits and the availability of the energy source.
Light nuclei can undergo nuclear fusion. Nuclear fusion occurs when two light nuclei combine to form a heavier nucleus. As with fission, the product of a fusion event must have a greater binding energy than the original nuclei for energy to be released in the reaction. Because fusion reactions produce heavier nuclei, the binding energy per nucleon must increase as atomic number increases. As shown in Figure 3.1 , this is possible only for atoms with A< 58. Hence,fusion occurs naturally only for light atoms. One example of this process is the fusion reactions that occur in stars. All stars generate energy through fusion. About 90 percent of the stars, including our sun, fuse hydrogen and possibly helium. Some other stars fuse helium or other heavier elements. The proton-proton cycle is a series of three nuclear-fusion reactions that are believed to be stages in the liberation of energy in our sun and other stars rich in hydrogen. In the proton-proton cycle, four protons combine to form an alpha particle and two positrons, releasing 25 MeV of energy in the process. The first two steps in this cycle are as follows:
‘.Did YOU Know?
This is followed by either of the following processes:
, What has been called the atomic bomb since 1945 is actually a tremendous , nuclear fission reaction. Likewise, the so-called hydrogen bomb is an uncontrolled nuclear fusion reaction in ‘ which hydrogen nuclei merge to form helium nuclei.
The released energy is carried primarily by gamma rays, positrons, and neutrinos. These energy-liberating fusion reactions are called thermonuclearfusion reactions. The hydrogen (fusion) bomb, first detonated in 1952, is an example of an uncontrolled thermonuclear fusion reaction.
Subatomic Physics
791
Fusion reactors are being developed. The enormous amount of energy released in fusion reactions suggests the possibility of harnessing this energy for useful purposes on Earth. Efforts are under way to create controlled thermonuclear reactions in the form of a fusion reactor. Because of the ready availability of its fuel sourcewater-controlled fusion is often called the ultimate energy source. For example, if deuterium (iH) were used as the fuel, 0.16 g of deuterium could be extracted from just 1 L of water at a cost of about one cent. Such rates would make the fuel costs of even an inefficient reactor almost insignificant. An additional advantage of fusion reactors is that few radioactive byproducts are formed. The proton-proton cycle shows that the end product of the fusion of hydrogen nuclei is safe, nonradioactive helium. Unfortunately, a thermonuclear reactor that can deliver a net power output for an extended time is not yet a reality. Many difficulties must be resolved before a successful device is constructed. For example, the energy released in a gas undergoing nuclear fusion depends on the number of fusion reactions that can occur in a given amount of time. This varies with the density of the gas because collisions are more frequent in a denser gas. It also depends on the amount of time the gas is confined. In addition, the Coulomb repulsion force between two charged nuclei must be overcome before they can fuse. The fundamental challenge is to give the nuclei enough kinetic energy to overcome this repulsive force. This can be accomplished by heating the fuel to extremely high temperatures (about 108 K, or about 10 times greater than the interior temperature of the sun). Such high temperatures are difficult and expensive to obtain in a laboratory or a power plant.

SECTION 3 FORMATIVE ASSESSMENT Reviewing Main Ideas 1. What are the similarities and differences between fission and fusion?
2. Explain how nuclear reactors utilize chain reactions. 3. What is enrichment? Why is enrichment necessary when uranium is used as a reactor fuel?
4. A fission reaction leads to the formation of 141 Ba and 92Kr when 235U absorbs a neutron. a. How is this reaction expressed symbolically? b. How many neutrons are released in this reaction? 5. What are some advantages to fusion reactors (as opposed to fission reactors)? What are some difficulties in the development of a fusion reactor?
Critical Thinking 6. Why would a fusion reactor produce less radioactive waste material than a fission reactor does? 792
Chapter 22
Particle Physics The Particle View of Nature Particle physics seeks to discover the ultimate structure of matter: elementary particles. Elementary particles, which are the fundamental units that compose matter, do not appear to be divisible and have neither size nor structure.
Many new particles have been produced in accelerators. Until 1932, scientists thought protons and electrons were elementary particles because these particles were stable. However, beginning in 1945, experiments at particle accelerators, such as the Stanford Linear Accelerator shown in Figure 4.1, have demonstrated that new particles are often formed in high-energy collisions between known particles. These new particles tend to be very unstable and have very short half-lives, ranging from 10-6 s to 10- 23 s. So far, more than 300 new particles have been catalogued.
Particle Accelerator The Stanford Linear Accelerator, in California, creates high-energy particle collisions that provide evidence of new particles.
There are four fundamental interactions in nature. The key to understanding the properties of elementary particles is to be able to describe the interactions between them. All particles in nature are subject to four fundamental interactions: strong, electromagnetic, weak, and gravitational. The strong interaction is responsible for the binding of neutrons and protons into nuclei, as we have seen. This interaction, which represents the “glue” that holds the nucleons together, is the strongest of all the fundamental interactions. It is very short-ranged and is negligible for separations greater than about 10- 15 m (the approximate size of a nucleus). The electromagnetic interaction, which is about 10-2 times the strength of the strong interaction at nuclear distances, is responsible for the attraction of unlike charges and the repulsion of like charges. This interaction is responsible for the binding of atoms and molecules. It is a long-range interaction that decreases in strength as the inverse square of the separation between interacting particles, as described in the chapter “Electric Forces and Fields:’ The weak interaction is a short-range nuclear interaction that is involved in beta decay. Its strength is only about 10- 13 times that of the strong interaction. However, because the strength of an interaction depends on the distance through which it acts, the relative strengths of two interactions differ depending on what separation distance is used. The strength of the weak interaction, for example, is sometimes cited to be as large as 1o- 6 times that of the strong interaction. Keep in mind that these relative strengths are merely estimates and they depend on the assumed separation distance. Subatomic Physics
793
Feynman Diagram of Electrons Exchanging a Photon In particle physics, the electromagnetic interaction is modeled as an exchange of photons. The wavy red line represents a photon, and the blue lines represent electrons.
Finally, the gravitational interaction is a long-range interaction with a strength of only about 10- 38 times that of the strong interaction. Although this familiar interaction is what holds the planets, stars, and galaxies together, its effect on elementary particles is negligible. The gravitational interaction is the weakest of all four fundamental interactions.
A force can be thought of as mediated by an exchange of particles. Notice that in this section, we have referred to a force as an interaction. This is because in particle physics the interaction of matter is usually described not in terms of forces but in terms of the exchange of special particles. In the case of the electromagnetic interaction, the particles are photons. Thus, it is said that the electromagnetic force is mediated by photons. Figure 4.2 shows how two
electrons might repel each other through the exchange of a photon. Because momentum is conserved, the electron emitting a photon changes direction slightly. As the photon is absorbed, the other electron’s direction must also change. The net effect is that the two electrons change direction and move away from each other.
a,
E i=
Likewise, the strong interaction is mediated by particles called gluons, the weak interaction is mediated by particles called the Wand Z bosons, and the gravitational interaction is mediated by gravitons. All of these except gravitons have been detected. The four fundamental interactions of nature and their mediating field particles are summarized in Figure 4.3.
Interaction (force)
Relative strength
Range of force
Mediating field particle
strong
1
::::: 1 fm
gluon
electromagnetic
10- 2
proportional to 1/r 2
photon
weak
10-13
< 10-3 fm
w± and Z bosons
gravitational
10-38
proportional to 1/r 2
graviton
Classification of Particles Did YOU Know?. – – – – – – – – – – – , The interaction of charged particles by the exchange of photons is described by a theory called quantum , electrodynamics, or QED.
794
Chapter 22
All particles other than the mediating field particles can be classified into two broad categories: leptons and hadrons. The difference between the two is whether they interact through the strong interaction. Leptons are a group of particles that participate in the weak, gravitational, and electromagnetic interactions but not in the strong interaction. Hadrons are particles that interact through all four fundamental interactions, including the strong interaction.
Leptons are thought to be elementary particles. Electrons and neutrinos are both leptons. Like all leptons, they have no measurable size or internal structure and do not seem to break down into smaller units. Because of this, leptons appear to be truly elementary. The number of known leptons is small. Currently, scientists believe there are only six leptons: the electron, the muon, the tau, and a neutrino associated with each. Each of these six leptons also has an antiparticle.
Hadrons include mesons and baryons.
Particle Subdivisions Leptons appear to be elementary, while hadrons consist of smaller particles called quarks. As a result, hadrons can be further subdivided into baryons and mesons, based on their internal composition. Classification of Particles
Hadrons, the strongly interacting particles, can be further divided into two classes: mesons and baryons. Originally, mesons and baryons were classified according to their masses. Baryons were heavier than mesons, and both were heavier than leptons. However, this distinction no longer holds. Today, mesons and baryons are distinguished by their internal structure. All mesons are unstable. Because of this, they are not constituents of normal, everyday matter. Baryons have masses equal to or greater than the proton mass. The most common examples of baryons are protons and neutrons, which are constituents of normal, everyday matter. A summary of this classification of particles is given in Figure 4.4.
Matter
Hadrons
Baryons
Leptons
Mesons
Hadrons are thought to be made of quarks. Particle-collision experiments involving hadrons seem to involve many short-lived particles, implying that hadrons are made up of more-fundamental particles. Furthermore, there are numerous hadrons, and many of them are known to decay into other hadrons. These facts strongly suggest that hadrons, unlike leptons, cannot be truly elementary. In 1963, Murray Gell-Mann and George Zweig independently proposed that hadrons have a more elementary substructure. According to their model, all hadrons are composed of two or three fundamental particles, which came to be called quarks. In the original model, there were three types of quarks, designated by the symbols u, d, and s. These were given the arbitrary names up, down, and sideways (now more commonly referred to as strange). Associated with each quark is an antiquark of opposite charge. The difference between mesons and baryons is due to the number of quarks that compose them. The compositions of all hadrons known when Gell-Mann and Zweig presented their models could be completely specified by three simple rules, which are summarized in Figure 4.5. Later evidence from collision experiments encouraged theorists to propose the existence of three more quarks, now known as charm, top, and bottom. These six quarks seem to fit together in pairs: up and down, charm and stran ge, and top and bottom. All quarks have a charge associated with them. The charge of a hadron is equal to the sum of the charges of its constituent quarks and is either zero or a multiple of e, the fundamental unit of charge. This implies that quarks have a very unusual property-fractional electric charge. In other words, the charge of the electron is no longer thought to be the smallest
Particle
Composition
meson
one quark and one antiquark
baryon
three quarks
antibaryon
three antiquarks
Subatomic Physics
795
Did YOU Know?. – – – – – – – – – – – , Murray Gell-Mann borrowed the word , quark from the passage “Three quarks for Muster Mark” in James Joyce’s book Finnegans Wake.
possible nonzero charge that a particle can have. The charges for all six quarks that have been discovered and their corresponding antiquarks are summarized in Figure 4.6.
Quark
Charge
interacts with a meson. Can a baryon be produced in such an
Charge
up (u)
+Ie
u
2 –e
down (d)
1 –e
d
+le
3
3
3
3
charm (c)
+Ie
c
2 –e
strange (s)
1 –e
s
+le
3 3
Particle-Antiparticle Interactions An antibaryon
Antiquark
3 3
top (t)
+Ie
t
2 –e
bottom (b)
1 –e
b
+le
3 3
interaction? Explain.
3
3
Strong and Weak Interactions Two protons in a
nucleus interact via the strong interaction. Are they also subject to the weak interaction?
Figure 4.7 represents the
quark compositions of several hadrons, both baryons and mesons. Just two of the quarks, u and d, are needed to construct the hadrons encountered in ordinary matter (protons and neutrons). The other quarks are needed only to construct rare forms of matter that are typically found only in high-energy situations, such as particle collisions. The charges of the quarks that make up each hadron in Figure 4.7 add up to zero or a multiple of e. For example, the proton contains three quarks (u, u, and d) having charges of +fe, +fe, and The total charge of the proton is +e, as you would expect. Likewise, the total charge of quarks in a neutron is zero ( +¾e, and
-½e.
-½e,
-½e).
You may be wondering whether such discoveries will ever end. At present, physicists believe that six quarks and six leptons (and their antiparticles) are the fundamental particles.
Quarks Baryons contain three quarks, while mesons contain a quark and an antiquark. The baryons represented are a proton (p +) and a neutron (n). The mesons shown are a pion (1r +) and a kaon (K- ), both rare particles.
796
Chapter 22
Baryons
p+
Mesons
n
n+
K-
00 G O
Despite many extensive efforts, no isolated quark has ever been observed. Physicists now believe that quarks are permanently confined inside ordinary particles by the strong force. This force is often called the colorforce for quarks. Of course, quarks are not really colored. Color is merely a name given to the property of quarks that allows them to attract one another and form composite particles. The attractive force between nucleons is a byproduct of the strong force between quarks.
. Did YOU Know?. – – – – – – – – – – – , The word atom is from the Greek word atomos, meaning “indivisible.” At one time, atoms were thought to be the indivisible constituents of matter; that is, they were regarded as elementary particles. Today, quarks and leptons are considered to be elementary particles.
The Standard Model The current model used in particle physics to understand matter is called the standard model. This model was developed over many years by a variety of people. Although the details of the standard model are complex, the model’s essential elements can be summarized by using Figure 4.8.
According to the standard model, the strong force is mediated by gluons. This force holds quarks together to form composite particles, such as protons, neutrons, and mesons. Leptons participate only in the electromagnetic, gravitational, and weak interactions. The combination of composite particles, such as protons and neutrons, with leptons, such as electrons, makes the constituents of all matter, which are atoms.
The Standard Model This schematic diagram summarizes the main elements of the standard model, including the fundamental forces, the mediating field particles, and the constituents of matter.
Matter and energy
~~
Forces
Constituents
Strong
Gluon
Quarks
u Electromagnetic
——-Weak
Photon
d
C
I
µ
e
Graviton
I
Leptons
Wand Z bosons
Gravity
s
t b
Ve
I

I v:
The standard model can help explain the early universe. Particle physics helps us understand the evolution of the universe. Ifwe extrapolate our knowledge of the history of the universe, we find that time itself goes back only about 13 billion to 15 billion years. At that time, the universe was inconceivably dense. In the brief instant after this singular moment, the universe expanded rapidly in an event called the big bang. Immediately afterward, there were such extremes in the density of matter and energy that all four fundamental interactions of physics operated in a single, unified way. The temperatures and energy present reduced everything into an undifferentiated “quark soup:’ Subatomic Physics
797
Evolution of the Universe from the Big Bang The four fundamental interactions of nature were indistinguishable during the early moments of the big bang.
I
QoaO.N
activity or decay rate
Bq Ci
becquerel curie
r,12
half-life
s
seconds
e)
———
Problem Solving See Appendix D: Equations for a summary of the equations introduced in this chapter. If you need more problem-solving practice, see Appendix I: Additional Problems.
Chapter Summary
803
11. The figure below shows the steps by which i~u decays to 2 g~Pb. Draw this diagram, and enter the correct isotope symbol in each square. 2
The Nucleus REVIEWING MAIN IDEAS 1
1. How many protons are there in the nucleus ~~Au? How many neutrons? How many electrons are there in the neutral atom?
a 235u 92
+/3~
2. What are isotopes?
….
a
3. What holds the nucleons in a nucleus together? ,a-♦
CONCEPTUAL QUESTIONS
a
,a·
2gJPb
4. Is it possible to accurately predict an atom’s mass from its atomic number? Explain.
12. What factors make fusion difficult to achieve?
5. What would happen if the binding energy of a nucleus was zero?
CONCEPTUAL QUESTIONS
6. Why do heavier elements require more neutrons to maintain stability?
13. If a film is kept in a box, alpha particles from a radioactive source outside the box cannot expose the film, but beta particles can. Explain.
PRACTICE PROBLEMS For problems 7-9, see Sample Problem A and refer to Appendix H. 1
7. Calculate the total binding energy of ~C. 8. Calculate the total binding energy of tritium ( yH) and helium-3 (~He). 9. Calculate the average binding energy per nucleon of i~Mg and ~~Rb.
Nuclear Decay and Reactions REVIEWING MAIN IDEAS 10. Explain the main differences between alpha, beta, and gamma emissions.
804
Chapter 22
14. An alpha particle has twice the charge of a beta particle. Why does the beta particle deflect more when both pass between electrically charged plates, assuming they both have the same speed?
15. Suppose you have a single atom of a radioactive material whose half-life is one year. Can you be certain that the nucleus will have decayed after two years? Explain. 16. Why is carbon dating unable to provide accurate estimates of very old materials? 17. A free neutron undergoes beta decay with a half-life of about 15 min. Can a free proton undergo a similar decay? (Hint: Compare the masses of the proton and the neutron.) 18. Is it possible for a 1 ~C (12.000 000 u) nucleus to spontaneously decay into three alpha particles? Explain.
19. Why is the temperature required for deuteriumtritium fusion lower than that needed for deuteriumdeuterium fusion? (Hint: Consider the Coulomb repulsion and nuclear attraction for each case.)
PRACTICE PROBLEMS For problems 20-21, see Sample Problem B. 20. Determine the product of the following reaction: ~Li + ~He -+ ? +
in
21. Complete the following nuclear reactions: 1 1 a. ? + ~N -+ ~H + ;0
b. ~Li + ~ H -+ ~He + ?
28. What are each of the baryons in item 27 called? 29. How many quarks or antiquarks are there in the following particles? a. a baryon b. an antibaryon c. ameson d. an antimeson
CONCEPTUAL QUESTIONS 30. Compare a neutrino with a photon. 31. Consider the statement, “All mesons are hadrons, but not all hadrons are mesons:’ Is this statement true? Explain.
For problems 22-24, see Sample Problem C.
22. A radioactive sample contains 1.67 x 10 11 atoms 1 of ~~Ag (half-life= 2.42 min) at some instant. Calculate the decay constant and the activity of the sample in mCi. 23. How long will it take a sample of polonium-210 with a half-life of 140 days to decay to one-sixteenth its original strength? 24. The amount of carbon-14 (1!C) in a wooden artifact is measured to be 6.25 percent the amount in a fresh sample of wood from th e same region. The half-life of carbon-14 is 5715 years. Assuming the sam e amount of carbon-14 was initially present in the artifact, determine the age of the artifact.
Particle Physics REVIEWING MAIN IDEAS 25. Describe the properties of quarks. 26. What is the electric charge of the particles with the following quark compositions?
a. udd b. uud
c. ud 27. What is the electric charge of the baryons with the following quark compositions?
a. uud b. udd
Mixed Review REVIEWING MAIN IDEAS 32. Complete the following nuclear reaction: gAl + -+ ? + i~P?
iHe
33. Consider the hydrogen atom to be a sphere with a radius equal to the Bohr radius, 0.53 x 10- 10 m, and calculate the approximate value of the ratio of atomic density to nuclear density. 34. Certain stars are thought to collapse at the end of their lives, combining their protons and electrons to form a neutron star. Such a star could be thought of as a giant atomic nucleus. If a star with a m ass equal to that of the sun (1.99 x 1030 kg) were to collapse into neutrons, what would be the radius of the star?
35. Calculate the difference in binding energy for the 1 1 two nuclei and ~N.
io
36. A piece of charcoal known to be approximately 25 000 years old contains 7.96 x 1010 C-14 atoms. a. Determine the number of decays per minute expected from this sample. (The half-life of C-14 is 5715 years.) b. If the radioactive background in the counter without a sample is 20.0 counts per minute and we assume 100.0 percent efficiency in counting, explain why 25 000 is close to the limit of dating with this technique.
Chapter Review
805
1
37. Natural gold has only one stable isotope, ~~Au. If gold is bombarded with slow neutrons, 13- particles are emitted. a. Write the appropriate reaction equation. b. Calculate the maximum energy of the emitted beta particles.
41. An all-electric home uses about 2.0 x 103 kW•h of electrical energy per month. How many 235U atoms would be required to provide this house with its energy needs for one year? Assume 100.0 percent conversion efficiency and 208 MeVreleased per fission.
38. 1\vo ways 235U can undergo fission when bombarded with a neutron are described below. In each case, neutrons are also released. Find the number of neutrons released in each of the following: a. 140Xe and 94 Sr released as fission fragments b. 132Sn and 101Mo released as fission fragments
42. When 180 is struck by a proton, 18 F and another particle are produced. What is the other particle?
39. When a ~Li nucleus is struck by a proton, an alpha particle and a product nucleus are released. What is the product nucleus? 1
40. Suppose ~B is struck by an alpha particle, releasing a proton and a product nucleus in the reaction. What is the product nucleus?
43. When a star has exhausted its hydrogen fuel, it may fuse other nuclear fuels, such as helium. At temperatures above 1.0 x 108 K, helium fusion can occur. a. 1\vo alpha particles fuse to produce a nucleus, A, and a gamma ray. What is nucleus A? b. Nucleus A absorbs an alpha particle to produce a nucleus, B, and a gamma ray. What is nucleus B? 44. A sample of a radioactive isotope is measured to have an activity of 240.0 mCi. If the sample has a half-life of 14 days, how many nuclei of the isotope are there at this time?
Nuclear Decay In nuclear decay, a radioactive substance is transformed into another substance that may or may not be radioactive. The amount of radioactive material remaining is given by the following equation: m = mr13- >.t
In this nuclear decay equation, m0 is the initial mass and ,\ is the decay constant. As you learned earlier in this chapter, the decay constant is related to the half-life by the following equation: _ 0.693 T112 ,\
806
Chapter 22
One of the interesting aspects of nuclear decay is that radioactive substances have a wide range of half-lives- from femtoseconds to billions of years. And all of these radioactive substances obey both of these equations. In this graphing calculator activity, the calculator will use these equations to make graphs of the amount of remaining mass versus time. By analyzing these graphs, you will be able to make predictions about radioactive substances that have various initial masses and various half-lives. Go online to HMDScience.com to find this graphing calculator activity.
45. At some instant of time, the activity of a sample of radioactive material is 5.0 µCi. If the sample contains 1.0 x 109 radioactive nuclei, what is the half-life of the material?
47. If the average energy released in a fission event is 208 MeV, find the total number of fission events required to provide enough energy to keep a 100.0 W light bulb burning for 1.0 h.
46. It has been estimated that Earth has 9.1 x 10 11 kg of natural uranium that can be economically mined. Of this total, 0. 70 percent is 235U. If all the world’s energy needs (7.0 x 10 12 J/s) were supplied by 235U fission, how long would this supply last? Assume that 208 MeV of energy is released per fission event and that the mass of 235U is about 3.9 x 10- 25 kg.
48. How many atoms of 235U must undergo fission to operate a 1.0 x 103 MW power plant for one day if the conversion efficiency is 30.0 percent? Assume 208 MeVreleased per fission event.
ALTERNATIVE ASSESSMENT 1. You are designing a nuclear power plant for a space station to be established on Mars. Material A is radioactive and has a half-life of two years. Material B is also radioactive and has a half-life of one year. Atoms of material B have half the mass of atoms of material A. Discuss the benefits and drawbacks involved with each of these fuels. 2. Design a questionnaire to investigate what people in your community know about nuclear power and how they feel about it. Give the questionnaire to your classmates for their comments, and if your teacher approves, conduct a study with people in your community. Present your results in the form of a class presentation and discussion. 3. Investigate careers in nuclear medicine. Interview people who work with radiation or with isotopic tracers in a hospital. Find out what kind of patients they treat or test and the technology they use. What training is necessary for this type of career?
4. Research the lives and careers of female nuclear physicists such as Marie Curie, Lise Meitner, Ida Tacke Noddack, and Maria Goeppert-Mayer. Create a presentation about one of these scientists. The presentation can be in the form of a report, poster, short video, or computer presentation.
5. Research how radioactive decay is used to date archaeological remains and fossils. What nuclear reactions are involved in the carbon-14 dating technique? What assumptions are made when the carbon-14 dating technique is used? What time scale is the carbon-14 technique suitable for? Is the carbon-14 technique appropriate to determine the age of a painting suspected to be 375 years old? Summarize your findings in a brochure or poster for visitors to a science museum. 6. Research the problem of nuclear waste in the United States. How much is there? What kinds of radioactive waste are there? Where are they produced? What are the costs and hazards associated with different techniques for disposal of radioactive waste? How do other countries deal with the problem? Choose the disposal option you think is most appropriate, and write a position paper. Include information about all options and the reasons for your choice.
7. Some modern physicists have developed string theory in an attempt to unify the four fundamental forces. Conduct research to learn about this theory. What are the main principles of string theory? Why do some scientists oppose it? Share your results with the class in a short lecture presentation.
Chapter Review
807
MULTIPLE CHOICE
5. A nuclear reaction of major historical note took
1. Which of the following statements correctly 1
describes a nucleus with the symbol tc? A. It is the nucleus of a cobalt atom with eight protons and six neutrons. B. It is the nucleus of a carbon atom with eight protons and six neutrons. C. It is the nucleus of a carbon atom with six protons and eight neutrons. D. It is the nucleus of a carbon atom with six protons and fourteen neutrons. 2. One unified mass unit (u) is equivalent to a mass of 1.66 x 10- 27 kg. What is the equivalent rest energy in joules? f. 8.27 X 10- 45 J G. 4.98 x 10- 19 J H. 1.49 x 10- lO J J. 9.31 X 108 J
3. What kind of force holds protons and neutrons together in a nucleus? A. electric force B. gravitational force C. binding force D. strong force
4. What type of nuclear decay most often produces the greatest mass loss? F. alpha decay G. beta decay H. gamma decay J. All of the above produce the same mass loss.
place in 1932, when a beryllium target was bombarded with alpha particles. Analysis of the experiment showed that the following reaction took 1 place: 1He + :Be —. iC + X. What is X in this reaction?
A. oe l B. – 0IP
C.
ln 0
D. ~p
6. What fraction of a radioactive sample has decayed after two half-lives have elapsed? 1
F. 4 1
G. 2 H

l4
J. The whole sample has decayed. 7. A sample of organic material is found to contain 18 g of carbon-14. Based on samples of pottery
found at a dig, investigators believe the material is about 23 000 years old. The half-life of carbon-14 is 5715 years. Estimate what percentage of the material’s carbon-14 has decayed. A. 4.0% B. 25% C. 75% D. 94% 8. The half-life of radium-228 is 5.76 years. At some instant, a sample contains 2.0 x 109 nuclei.
Calculate the decay constant and the activity of the sample. f. A = 3.81 X 10- 9 s- 1; activity= 2.1 x 10- 10 Ci G. ,\ = 3.81 x 10- 9 s- 1; activity = 7.8 Ci H. ,\ = 0.120 s- 1; activity= 6.5 x 10- 3 Ci J. ,\ = 2.6 x 108 s- 1; activity = 1.4 x 107 Ci
808
Chapter 22
. TEST PREP
9. What must be true in order for a nuclear reaction to happen naturally? A. The nucleus must release energy in the reaction. B. The binding energy per nucleon must decrease in the reaction. C. The binding energy per nucleon must increase in the reaction. D. There must be an input of energy to cause the reaction. 10. Which is the weakest of the four fundamental interactions? F. electromagnetic G. gravitational H. strong J. weak 11. Which of the following choices does not correctly match a fundamental interaction with its mediating particles? A. strong: gluons B. electromagnetic: electrons C. weak: Wand Z bosons D. gravitational: gravitons 12. What is the charge of a baryon containing one up quark (u) and two down quarks (d)?
F. G. H. J.
– 1
241
16. Smoke detectors use the isotope Am in their operation. The half-life of Am is 432 years. If the smoke detector is improperly discarded in a landfill, estimate how long its activity will take to decrease to a relatively safe level of 0.1 percent of its original activity. (Hint: The estimation process that you should use notes that the activity decreases to 50 percent in one half-life, to 25 percent in two halflives, and so on.)
EXTENDED RESPONSE 17. Iron-56 (~i Fe) has an atomic mass of 55.934 940 u . The atomic mass of hydrogen is 1.007 825 u, and m n = 1.008 665 u . Show your work for the following calculations: a. Find the mass defect in the iron-56 nucleus. b. Calculate the binding energy in the iron-56 nucleus. c. How much energy would be needed to dissociate all the particles in an iron-56 nucleus? 18. Use the table below to calculate the energy released 2 in the alpha decay of ~~U. Show your work.
Nucleus
Mass
o
238 LJ 92
238.050 784 LI
+l +2
234 Th 90
234.043 593 LI
4 2 He
4.002 602 LI
SHORT RESPONSE 13. Suppose it could be shown that the ratio of carbon-14 to carbon-12 in living organisms was much greater thousands of years ago than it is today. How would this affect the ages we assign to ancient samples of once-living matter? 14. A fission reactor produces energy to drive a generator. Describe briefly how this energy is produced. 15. Balance the following nuclear reaction:
in + ? – 1He+ ~Li Test Tip If you finish a test early, go back and check your work before turning in the test.
Standards-Based Assessment
809
Nuclear Waste What Can We Do with Nuclear Waste?
As radioactive isotopes decay, nuclear waste emits all common forms of radioactivity-alpha particles, beta particles, and gamma radiation. When this radiation penetrates living cells, it knocks electrons away from atoms, causing them to become electrically charged ions. As a result, vital biological molecules break apart or form abnormal chemical bonds with other molecules. Often, a cell can repair this damage, but if too many molecules are disrupted, the cell will die. The ionizing radiation can also damage a cell’s genetic material (DNA and RNA), causing the cell to divide again and again, out of control. This condition is called cancer. Identify the Need: Safe Storage
Because of these hazards, nuclear waste must be sealed and stored until the radioactive isotopes in the waste decay to the point at which radiation reaches a safe level. Nuclear wastes include low-level wastes from the nuclear medicine departments at hospitals, where radioactive isotopes are used to diagnose and treat diseases. The greatest disposal problem involves high-level waste, or HLW. Some kinds of HLW will require safe storage for at least 1O000 years. Brainstorm Solutions
Much HLW consists of used fuel rods from reactors at nuclear power plants. About a third of these rods are replaced every year or two because their supply of fissionable uranium-235 becomes depleted, or spent. When nuclear power plants in the United States began operating in 1957, engineers had planned to reprocess spent fuel to reclaim fissionable isotopes of uranium to make new fuel rods. But people feared that the plutonium byproduct made available by reprocessing might be used to build bombs, so that plan was abandoned. Since that time, HLW has continued to accumulate at power plant sites in “temporary” storage facilities that are now nearly full. When there is no more storage space, plants will have to cease operation. Consequently, states and utility companies are demanding that the federal government 810
honor the Nuclear Waste Policy Act of 1982 in which the federal government agreed to provide permanent storage sites. Any site used for disposal of HLW must be far away from population centers and likely to remain geologically stable for thousands of years. One possible type of location is certain deep spots of the oceans, where, some scientists claim, the seabed is geologically stable, as well as devoid of life. Sealed stainless-steel canisters of waste could be packed into rocket-shaped carriers that would bury themselves deep into sediments when they hit the ocean bottom. Opponents say that the canisters have not been proven safe, and that if they leaked, the radioactivity could kill off photosynthetic marine algae that replenish much of the world’s oxygen. Proponents claim that the ocean bottom already contains many radioactive minerals and that the radioactivity from all HLWs in existence would not harm marine algae. Some people have proposed salt domes as a nuclear waste storage site. A salt dome is a geologic formation of salt covered with a cap made of rock. Salt domes form as a consequence of the relative buoyancy of salt when it is buried beneath other types of sediment. The salt flows upward to form salt domes, sheets, pillars, and other structures. People have tapped these salt domes to obtain halite, commonly known as table salt. People have also used these salt domes to store nuclear wastes. For example, Germany began storing nuclear wastes in a mine contained within a salt dome in the 1960s. However, in 201 OGermany decided to start removing thousands of barrels of these wastes after the salt dome was determined to be unstable. Redesign to Improve
Scientists in the United States have considered still other proposals as well, but since the Nuclear Waste Policy Act of 1982 most of the attention had focused on the development of a disposal site beneath Yucca Mountain in Nevada. The design of this site includes sloping shafts that lead to a 570-hectare (1400-acre) storage area 300 meters deep in the mountain’s interior.
The planned nuclear waste storage facility at Yucca Mountain, in Nevada, has been on hold for decades.
However, in March 2010, the U.S. Department of Energy stated that Yucca Mountain was no longer being considered as a site for nuclear waste storage. Instead, the federal government established a presidential advisory panel to investigate alternative solutions to the problem of nuclear waste storage. This panel began hearings to solicit input from the public, including elected officials. Communicate In January 2011, Senator Lindsey Graham of South Carolina was one of several officials and business leaders who communicated their opinions to the presidential advisory panel. Senator Graham recommended that the government restart the development of Yucca Mountain as a nuclear waste repository. He claimed that the government’s decision meant that there will be nowhere to send radioactive waste from the Savannah River Site, the former nuclear weapons complex on the Georgia-South Carolina border. As a result, abandoning Yucca Mountain would turn the Savannah River Site into a permanent nuclear waste dump. Not surprisingly, Senator Harry Reid of Nevada has supported the government’s decision to shut down Yucca Mountain. He claims that Nevada’s residents do not want all the country’s HLWs to be stored in their “backyard.” The battle to find a solution to the problem of nuclear waste storage still continues.
Design Your Own Conduct Research Some people support the construction of “integrated spent-fuel recycling facilities” for the United States. Do research to determine how such facilities would operate to reprocess the radioactive materials in spent fuel rods. Include information on countries where these facilities are currently in operation. Brainstorm Solutions Recycling spent nuclear fuel is one approach to reducing nuclear waste, but it does not eliminate it. With a partner or in a small group, brainstorm multiple ideas for safely storing nuclear waste. Because you are brainstorming, write down each idea without stopping to evaluate it. Select a Solution Now review the ideas you came up with for safely storing nuclear waste. Eliminate ideas that, on further thought, are not safe or practical. For example, sending waste into space on a rocket may sound like a good idea, but if a rocket blew up it would scatter waste over a wide area. Choose one that might be worthwhile to investigate.
811
PHYSICS AND ITS WORLD 1953 n>-..
= 2dsin0
Rosalind Franklin, a crystallographer and chemist, produces an x-ray image of DNA.
1957
1969
1975
The Soviet Union launches Sputnik, the world’s first sateIIite, into space.
The Woodstock Music Festival takes places on a farm in New York. Woodstock proves to be the beginning of the end of the hippie counterculture movement of the 1960s.
After over a decade of war, Saigon falls to the NorthVietnamese army and “5 €. }3 United States troops leave Vietnam, -~ z ending the Vietnam War. >-‘ r————.,-~ ~ 13 a:
@
1954 In the landmark case Brown versus Board of Education, the United States Supreme Court unanimously rules that “separate educational facilities are inherently unequal,” paving the way for school desegregation and the civil rights movement.
812
1975
1968 uud udd Scientists at the Stanford Linear Accelerator run experiments that provide the first direct evidence of the existence of quarks.
1969 On July 20 the United States sends Apollo 11 to the moon. Neil Armstrong and Buzz Aldrin become the first two humans to walk on the moon.
In her observations of spiral galaxies, astronomer Vera Rubin notices that stars on the outer edge of the galaxy move at the same speed as those closer to the center of the galaxy. Her discovery lends support to the theory of dark matter.
-.::-
g
1981
1989-1990
NASA launches the first shuttle, Columbia, into space. After a two-day mission, Columbia returns to Earth.
Scientists at the European Organization for Nuclear Research (CERN), in an effort to facilitate the sharing of information, establish the World Wide Web.
1971-1985
Throughout the 1970s, scientists such as Michael Green theorize that even the smallest parts of matter, quarks, are made up of smaller units. They call this the superstring theory.
1997
>–z
2001
Scientist Nicolas Gisin sends two photons in opposite directions down a wire. When the photons are 7 miles apart, they encounter two paths. Although the photons are unable to communicate, the paths taken by both match.
In a deadly terrorist attack on U.S. soil, two airplanes hit the World Trade Center in New York; one hits the Pentagon in Washington, D.C.; one crashes in a field in Pennsylvania. Thousands of people are killed.
1986
1994
2003
A failure at the Chernobyl nuclear power plant triggers a huge explosion, sending radioactive material not only into the local environment, but also into the atmosphere.
Apartheid ends in South Africa after nearly 50 years of forced segregation in which the minority white population denied basic rights to the majority black population. Former political prisoner Nelson Mandela becomes president.
In her lab, Deborah S. Jin discovers a sixth form of matter called fermionic condensates. This matter forms when fermions are cooled to very low temperatures close to Okelvin.
813
Mathematical Review Scientific Notation Positive exponents Many quantities that scientists deal with often have very large or very small values. For example, the speed of light is about 300 000 000 m l s, and the ink required to make the dot over an i in this textbook has a mass of about 0.000 000 001 kg. Obviously, it is cumbersome to work with numbers such as these. We avoid this problem by using a method based on powers of the number 10.
10° = 1 101 = 10 2
10 = 10 3 10 = 10
X
10 = 100
X
10
X
10 = 1000
104 = 10
X
10
X
10
X
10 = 10 000
105 = 10
X
10
X
10
X
10
X
10 = 100 000
The number of zeros determines the power to which 10 is raised, or the exponent of 10. For example, the speed oflight, 300 000 000 m / s, can be expressed as 3 x 108 m / s. In this case, the exponent of 10 is 8. Negative exponents For numbers less than one, we note the following:
10-1 = 110 = 0.1 10-
2
= lO
! lO = 0.01
10-3 = 10
X
110
X
10- 4 = 10
X
10
! 10
10
1 10
lO-S = 10
X
X
10
= 0.001 X
10
= 0.0001
X
10
X
10
= O.OOO Ol
The value of the negative exponent equals the number of places the decimal point must be moved to be to the right of the first nonzero digit (in these cases, the digit 1). Numbers that are expressed as a number between 1 and 10 multiplied by a power of 10 are said to be in scientific notation. For example, 5 943 000 000 is 5.943 x 109 when expressed in scientific notation, and 0.000 0832 is 8.32 x 10- 5 when expressed in scientific notation.
R2
Appendix A
Multiplication and division in scientific notation When numbers
expressed in scientific notation are being multiplied, the following general rule is very useful: 10n X 10m = 10(n+m)
Note that n and m can be any numbers; they are not necessarily integers. For example, 102 x 105 = 107, and 10114 x 10112 = 10314. The rule also 8 3 applies to negative exponents. For example, 10 x 10- = 10- 5 . When dividing numbers expressed in scientific notation, note the following: 10n = 10n X 10-m = 1o(n- m)
10m
For example,
103 102
= 10(3- 2) = 101.
Fractions The rules for multiplying, dividing, adding, and subtracting fractions are summarized in Figure 1, where a, b, c, and d are four numbers.
Operation
Rule
Example
Multiplication
(%) (1) = :~
(2 ) (4 )
(:)
(;) (:)
Division
(~)
Addition and Subtraction
ad be
Q.+_£= ad± be b -d bd
3 5
2 3
(2)(4) 8 = (3)(5) = 15
(2)(5) (3)(4)
4 5
10 12
5
6
(2)(5) – (3)(4) (3)(5)
2 15
Powers Rules of exponents When powers of a given quantity, x, are multiplied,
the rule used for scientific notation applies: (x n)(x m)
= x(n+m)
For example, (x 2)(x 4 )
= xl
2
Hl = x 6 .
When dividing the powers of a given quantity, note the following: n
~
xm
= X(n- m)
For example, x X
8
= x(s- 2) = x6. 2
Mathematical Review
R3
-•

A power that is a fraction, such as½, corresponds to a root as follows: xll n
=’1x
For example, 4 113 calculations.)
=V4 = 1.5874. (A scientific calculator is useful for such
Finally, any quantity, xn, that is raised to the mth power is as follows:
(xn)m = xnm ‘ For example, (x2 ) 3
= x( 2)(3 ) = x!’.
The basic rules of exponents are summarized in Figure 2.
x1 x” x”‘
=X
= x(n-m)
Algebra Solving for unknowns When algebraic operations are performed, the laws of arithmetic apply. Symbols such as x, y, and z are usually used to represent quantities that are not specified. Such unspecified quantities are called unknowns.
First, consider the following equation: Bx= 32
If we wish to solve for x, we can divide each side of the equation by the
same factor without disturbing the equality. In this case, if we divide both sides by 8, we have the following: Bx

8
32

8
X=4 Next, consider the following equation:
x +2 = 8 In this type of expression, we can add or subtract the same quantity from each side. If we subtract 2 from each side, we get the following:
x+2 – 2=8 – 2 x= 6
In general, if x + a = b, then x = b – a.
R4
Appendix A
Now, consider the following equation:
If we multiply each side by 5, we are left with x isolated on the left and a
value of 45 on the right.
(s)
(¾) = (9)(s)
x=45 In all cases, whatever operation is performed on the left side of the equation must also be performed on the right side.
Factoring Some useful formulas for factoring an equation are given in Figure 3. As an example of a common factor, consider the equation Sx + Sy+ Sz = 0. This equation can be expressed as S(x + y + z) = 0. The expression a 2 + 2ab + b2, which is an example of a perfect square, is equivalent to the expression (a+ b )2. For example, if a= 2 and b = 3, then 22 + (2)(2)(3) + 32 = (2 + 3)2, or (4 + 12 + 9) = 52 = 25. Finally, for an example of the difference of two squares, let a= 6 and b = 3. In this case, (62 – 32 ) = (6 + 3)(6 – 3), or (36 – 9) = (9)(3) = 27.
ax+ ay + az = a(x + y + z)
common factor perfect square difference of two squares
Quadratic Equations The general form of a quadratic equation is as follows: ax2 + bx + c= 0 In this equation, xis the unknown quantity and a, b, and care numerical factors known as coefficients. This equation has two roots, given by the following: X=
2 – 4ac b±Vb —–2a
If b2 ~ 4ac, the value inside the square-root symbol will be positive or zero and the roots will be real. If b2 < 4ac, the value inside the square-root
symbol will be negative and the roots will be imaginary numbers. In problems in this physics book, imaginary roots should not occur.
Mathematical Review
R5
-•

I
Find the solutions for the equation x 2
SOLVE
+ 5x + 4 = 0.
The given equation can be expressed as (l)x 2 + (5)x + (4) = 0. In other words, a = 1, b = 5, and c = 4. The two roots of this equation can be found by substituting these values into the quadratic equation, as follows:
b±Vb2
– 4ac X=—–2a
– 5 ±y52 – (4)(1)(4)
(2)(1)
– 5±\/9 2
-5 +3 2
-5 + 3 -5 – 3 The two roots are x = = – 1 and x = = 4 2 2 Ix = -1 and x = -4 I
We can evaluate these answers by substituting th em into the given equation an d verifying that the result is zero. x2+5x+4 = 0 For x = -1, (-1)2 + 5(-1) + 4 = 1 – 5 + 4 = 0. For x = -4, ( -4)2 + 5(-4) + 4 = 16 – 20 + 4 = 0.
1 ….
r.:•••••Jr::.l
I Factor the equation 2×2 – 3x SOLVE
4 = 0.
Th e given equation can be expressed as (2)x2 + (- 3x) + (- 4) = 0. Thus, a = 2, b = -3, and c = -4. Substitute these values into the quadratic equation to factor the given equation. X
V
-b ±V b 2 – 4ac 3 ± (-3) 2 – ( 4)(2)( -4) 3 ± v’4l 3 ± 6.403 – —— – ———- — – —2a (2)(2) 4 4
3 4 3 3 3 The two roots are x = + :Ao = 2.351 and x = – ~ o = – 0.851. Ix= 2.351 and x = – 0.851 I
Again, evaluate these answers by substituting them into the given equation. 2×2-3x-4=0 for X = 2.351, 2(2.351) 2

3(2.351) – 4 = 11.054 – 7.053 – 4::,:: 0.
For x = – 0.851, 2( – 0.851)2 – 3( – 0.851) – 4 = 1.448 + 2.553 – 4::,:: 0.
R6
Appendix A
Linear Equations A linear equation has the following general form: y=ax+b In this equation, a and b are constants. This equation is called linear because the graph of y versus xis a straight line, as shown in Figure 4. The constant b, called the intercept, represents the value of y where the straight line intersects the y-axis. The constant a is equal to the slope of the straight ‘ line and is also equal to the tangent of the angle that the line makes with the x-axis ( 0). If any two points on the straight line are specified by the coordinates (x1, y 1) and (x2, y 2), as in Figure 4, then the slope of the straight line can be expressed as follows: slope =
Y2-Y1 X2 – X1
y
Lil X
~y
~x
For example, if the two points shown in Figure 4 are (2, 4) and (6, 9), then the slope of the line is as follows : 4 9 slope = ( – ) = ~ (6 – 2)
4
Note that a and b can have either positive or negative values. If a > 0, the straight line has a positive slope, as in Figure 4. If a < 0, the straight line has a negative slope. Furthermore, if b > 0, they intercept is positive (above the x-axis), while if b < 0, they intercept is negative (below the x-axis). Figure 5 gives an example of each of these four possible cases, which are summarized in Figure 6.
Constants
Slope
y intercept
a> 0, b > 0
posit ive slope
positive y intercept
a> 0, b < 0
posit ive slope
negative y intercept
a< 0, b > 0
negative slope
positive y intercept
a< 0, b < 0
negative slope
negative y intercept
Solving Simultaneous Linear Equations Consider the following equations: 3x + 5y= 15
This equation has two unknowns, x and y. Such an equation does not have a unique solution. That is, (x = 0, y = 3), and (x = 5, y = 0), and (x = 2, y = ¾)are all solutions to this equation. If a problem has two unknowns, a unique solution is possible only if there are two independent equations. In general, if a problem has nunknowns, its solution requires n independent equations. There are three basic methods that can be used to solve simultaneous equations. Each of these methods is discussed below, and an example is given for each.
Mathematical Review
R7
-•

First method: substitution One way to solve two simultaneous equations
involving two unknowns, x and y, is to solve one of the equations for one of the unknown values in terms of the other unknown value. In other words, either solve one equation for x in terms of y or solve one equation for yin terms of x. Once you have an expression for either x or y, substitute this expression into the other original equation. At this point, the equation has only one unknown quantity. This unknown can be found through algebraic manipulations and then can be used to determine the other unknown.
I
Solve the following two simultaneous equations:
1. Sx
+ y = -8.
2. 2x- 2y = 4 SOLVE
First solve for either x or yin one of the equations. We’ll begin by solving equation 2 for x. 2. 2x-2y= 4
2x= 4 + 2y 4+ 2y X=–=2+y 2
Next, we substitute this equation for x into equation 1 and solve for y. 1. 5x+ y= -8 5(2 + y) + y = -8 10 + 5y+ y= – 8 6y= -18 ly= – 3 1
To find x, substitute this value for y into the equation for x derived from equation 2. X= 2 + y= 2+- 3
B There is always more than one way to solve simultaneous equations by substitution. In this example, we first solved equation 2 for x. However, we could have begun by solving equation 2 for y or equation 1 for x or y. Any of these processes would result in the same answer.
R8
Appendix A
Second method: canceling one term Simultaneous equations can also be solved by multiplying both sides of one of the equations by a value that will make either the x value or they value in that equation equal to and opposite the corresponding value in the second equation. When the two equations are added together, that unknown value drops out and only one of the unknown values remains. This unknown can be found through algebraic manipulations and then can be used to determine the other unknown.
Solve the following two simultaneous equations: 1. 3x+y=-6
2. – 4x – 2y = 6 SOLVE
First, multiply each term of one of the equations by a factor that will make either the x or they values cancel when the two equations are added together. In this case, we can multiply each term in equation 1 by the factor 2. The positive 2y in equation 1 will then cancel the negative 2y in equation 2. 1. 3x+ y= -6 (2)(3x) + (2)(y) = -(2)(6) 6x+ 2y = – 12
Next, add the two equations together and solve for x. 2. -4x-2y= 6 1. 6x+ 2y= -12 2x=-6
Then, substitute this value of x into either equation to find y. 1. 3x+ y = -6 y = -6 – 3x = -6 – (3)( -3) = -6 + 9
IY=
3
I
In this example, we multiplied both sides of equation 1 by 2 so that they terms would cancel when the two equations were added together. As with substitution, this is only one of many possible ways to solve the equations. For example, we could have multiplied both sides of equation 2 by ¾ so that the x terms would cancel when the two equations were added together.
Mathematical Review
R9
Third method: graphing the equations Two linear equations with two unknowns can also be solved by a graphical method. If the straight lines corresponding to the two equations are plotted in a conventional coordinate system, the intersection of the two lines represents the solution.
Solve the following two simultaneous equations:
1.x-y=2 2. X – 2y = -1 SOLVE
These two equations are plotted in Figure 7. To plot an equation, rewrite the equation in the form y = ax + b, where a is the slope and b is they intercept. In this example, the equations can be rewritten as follows:
y=x-2 y
y
6 5
4 3
2 1
x- y= 2
= __!_x + __!_
2 2 Once one point of a line is known, any other point on that line can be found with the slope of the line. For example, the slope of the first line is 1, and we know that (0, -2) is a point on this line. Ifwe choose the point x = 2, we have (2, y 2). The coordinate y2 can be found as follows: y2 -y1 y2 – (-2) slope = = 1 Xz – x1 2-0 Y2= 0
Connecting the two known coordinates, (0, -2) and (2, 0), results in a graph of the line. The second line can be plotted with the same method. As shown in Figure 7, the intersection of the two lines has the coordinates x = 5, y = 3. This intersection represents the solution to the equations. You should check this solution using either of the analytical techniques discussed above.
Logarithms Suppose that a quantity, x, is expressed as a power of another quantity, a.
x = aY The number a is called the base number. The logarithm of x with respect to the base, a, is equal to the exponent to which a must be raised in order to satisfy the expression x = aY. y
= logax
Conversely, the antilogarithm of y is the number x. x
= antilogaY
Common and natural bases In practice, the two bases most often used are base 10, called the common logarithm base, and base e = 2.718 … , called the natural logarithm base. When common logarithms are used, y and x are related as follows: y
R10
Appendix A
= log10 x, or x = IoY
When natural logarithms are used, the symbol ln is used to signify that the logarithm has a base of e; in other words, log~ = ln x. y
= ln x, or x = eY
For example, log10 52 = 1.716, so antilog 10 1.716 = 101.716 = 52. Likewise, ln 52 = 3.951, so antiln 3.951 = e3· 951 = 52. Note that you can convert between base 10 and base e with the equality ln x = (2.302 585)log10 x. Some useful properties oflogarithms are summarized in Figure 8.
Example
Rule
log (ab) log
= log a + log b
(f) = log a –
log b
log (an) = n log a
= log 2 + log 5
log (2)(5) log
1=
log 73
log 3 – log 4
= 3 log 7
In e = 1
e5 = 5
In ffl-= a
In
In(:) = – Ina
1 In – = – In 8 8
Conversions Between Fractions, Decimals, and Percentages The rules for converting numbers from fractions to decimals and percentages and from percentages to decimals are summarized in Figure 9.
Conversion
Rule
Example
Fraction to decimal
divide numerator by denominator
~=069
Fraction to percentage
convert to decimal, then multiply by 100%
Percentage to decimal
move decimal point two places to the left, and remove the percent sign
45
1~
.
= (0.69)(100%) = 69%
69% = 0.69
Mathematical Review
R11
-•

Geometry Figure 10 provides equations for
the area and volume of several geometrical shapes used throughout this text.
Geometrical shape
Useful equations
area= lw perimeter = 2(1
+ w)
rectangle
area= 1rr2 circumference = 2m-
circle
area=
ibh
b triangle
surface area = volume =
4’7rr2
~ 1rr3
sphere
surface area = 27rr2 + 21rrl volume = ‘lrr2l
cylinder
surface area = 2(lh + lw + hw) volume = lwh
h~ I rectangular box
R12
Appendix A
Trigonometry and the Pythagorean Theorem The portion of mathematics that is based on the relationships between the sides and angles of triangles is called trigonometry. Many of the concepts of this branch of mathematics are of great importance in the study of physics. To review some of the basic concepts of trigonometry, consider the right triangle shown in Figure 11, where side a is opposite the angle 0, side b is adjacent to the angle 0, and side c is the hypotenuse of the triangle (the side opposite the right angle). The most common trigonometry functions are summarized in Figure 12, using Figure 11 as an example.
sin 0 = 1
B=t
cos tan 0= g_ b
a 90° b sine (sin)
sin 0 = side opposite 0 hypotenuse
cosine (cos)
cos 0
= side adjacent to 0 = !!_
tangent (tan)
tan
=
inverse sine (sin- 1)
. 1 (side 0) = sin. 1 (a) – – opposite —0= sinhypotenuse c
inverse cosine (cos- 1)
adjacent to 0)_(b) 0_ _(sidehypotenuse
inverse tangent (tan – 1)
0 = tan- 1
0
= !:!:… c
hypotenuse
c
side opposite 0 = !:!:… side adjacent to 0 b
-COS 1 – – – – – – – -COS 1 c (
side opposite 0 ) = tan – 1 (!:!:…) side adjacent to 0 b
When 0 = 30°, for example, the ratio of a to c is always 0.50. In other words, sin 30° = 0.50. Sine, cosine, and tangent are quantities without units because each represents the ratio of two lengths. Furthermore, note the following trigonometry identity: side opposite 0 sin 0 cos 0
hypotenuse side adjacent to 0 hypotenuse
=
side opposite 0 side adjacent to 0
= tan 0
Some additional trigonometry identities are as follows: sin2 0 + cos2 0 = 1 sin 0 = cos(90° – 0) cos 0 = sin(90° – 0)
Mathematical Review
R13
-•

Determining an unknown side The first three functions given in Figure 12 can be used to determine any unknown side of a right triangle when one side and one of the non-right angles are known. For example, if 0 = 30° and a = 1.0 m, the other two sides of the triangle can be found as follows:
sin0=~
c-_ a __ I.Om – sin 0 – sin 300
I c= 2.0ml tan0=: b-_a_ _ I.Om – tan 0 – tan 30° lb=l.7ml Determining an unknown angle In some cases, you might know the value of the sine, cosine, or tangent of an angle and need to know the value of the angle itself. The inverse sine, cosine, and tangent functions given in Figure 12 can be used for this purpose. For example, in Figure 11, suppose you know that side a = 1.0 m and side c = 2.0 m . To find the angle 0, you could use the inverse sine function, sin- 1, as follows: 0 = sin- 1 (!!:_) = sin- 1 (I.Om)= sin- 1(0.50) C 2.Qm
, Converting from degrees to radians The two most common units used to measure angles are degrees and radians. A full circle is represented by 360 degrees (360°) or by 2-rr radians (2-rr rad). As such, the following conversions can be used:
[angle (
0 ) ]
=
!
1 0
[angle (rad)]
[angle (rad)] = ; [angle ( 1 0
b
Appendix A
)]
Pythagorean theorem Another useful equation when working with right triangles is the Pythagorean theorem. If a and b are the two legs of a right triangle and c is the hypotenuse, as in Figure 13, the Pythagorean theorem can be expressed as follows:
c2 = a2 + b2
R14
0
In other words, the square of the hypotenuse of a right triangle equals the sum of the squares of the other two legs of the triangle. The Pythagorean theorem is useful when two sides of a right triangle are known but the third side is not. For example, if c = 2.0 m and a = 1.0 m, you could find busing the Pythagorean theorem as follows: b = Vc2 – a b
2
= V4.0 m 2 –
= ..j(2.0 m) 1.0 m 2
2

(1.0 m)
2
= V3.0 m 2
Law of sines and law of cosines The law of sines may be used to find angles of any general triangle. The law of cosines is used for calculating one side of a triangle when the angle opposite and the other two sides are known. If a, b, and care the three sides of the triangle and 0a’ 0b, and 0care the three angles opposite those sides, as shown in Figure 14, the following relationships hold true: C b a
c2 = a 2 + b 2 – 2ab cos 0C
Accuracy in Laboratory Calculations Absolute error Some laboratory experiments, such as one that measures free-fall acceleration, may involve finding a value that is already known. In this type of experiment, the accuracy of your measurements can be determined by comparing your results with the accepted value. The absolute value of the difference between your experimental or calculated result and the accepted value is called the absolute error. Thus, absolute error can be found with the following equation:
absolute error = Iexperimental – accepted I Be sure not to confuse accuracy with precision. The accuracy of a measurement refers to how close that measurement is to the accepted value for the quantity being measured. Precision depends on the instruments used to measure a quantity. A meterstick that includes millimeters, for example, will give a more precise result than a meterstick whose smallest unit of measure is a centimeter. Thus, a measurement of9.61 m / s2 for free-fall acceleration is more precise than a measurement of9.8 m /s2, but 9.8 m /s2 is more accurate than 9.61 m /s2 • Relative error Note that a measurement that has a relatively large absolute error may be more accurate than a measurement that has a smaller absolute error if the first m easurement involved much larger quantities. For this reason, the percentage error, or relative error, is often more meaningful than the absolute error. The relative error of a measured value can be found with the following equation: . (experimental – accepted) d relative error = accepte Mathematical Review
R15
-•

In other words, the relative error is the difference between the experimental value and the accepted value divided by the accepted value. Because relative error takes the size of the quantity being measured into account, the accuracy of two different measurements can be compared by comparing their relative errors. For example, consider two laboratory experiments in which you are determining values that are fairly well known. In the first, you determine that free-fall acceleration at Earth’s surface is 10.31 mls2 . In the second, you find that the speed of sound in air at 25°C is 355 mls. The accepted values for these quantities are 9.81 mls2 and 346 mis, respectively. Now we’ll find the absolute and relative errors for each experiment.
For the first experiment, the absolute and relative errors can be calculated as follows : absolute error= Iexperimental – accepted I= I 10.31 mls2 – 9.81 ml s2 I
I absolute error= 0.50 mls2 I . (experimental – accepted) (10.31 ml s2 – 9.81 m/s2 ) relative error = d = accepte 9.81 mi s2
I relative error= 0.051 = 5.1% I
For the second experiment, the absolute and relative errors can be calculated as follows: absolute error= Iexperimental – accepted I= I 355 mi s – 346 mi s I
I absolute error= 9 mi s
J
(experimental – accepted) (355 mis – 346 mi s) re1at1ve error = d = I accepte 346 m s .
I relative error = 0.026 = 2.6% I \.
Note that the absolute error is less in the first experiment, while the relative error is less in the second experiment. The absolute error is less in the first experiment because typical values for free-fall acceleration are much smaller than typical values for the speed of sound in air. The relative errors take this difference into account. Thus, comparing the relative errors shows that the speed of sound is measured with greater accuracy than is the free-fall acceleration.
R16
Appendix A
The Scientific Process Science and Its Scope Science is a specific way of looking at and understanding the world around us. The scope of science encompasses a search for understanding natural and physical phenomena. For example, biologists explore how living things function in their environment. Geologists examine how Earth’s structures and materials have changed over time. Chemists investigate the nature of matter and the changes it undergoes. Physicists search for an understanding of the interactions between matter and energy. Often, the areas that scientists investigate overlap. As a result, there are biochemists who study the chemistry of living things, geophysicists who investigate the physical properties of Earth, and physical chemists who apply physical laws to chemical reactions. Moreover, the scope of science is not limited to investigating phenomena on Earth. In fact, the scope of science extends throughout the universe.
Science and Its Limitations Science is limited to investigating phenomena that can be examined usefully in a scientific way. Some questions are outside the realm of science because they deal with phenomena that are not scientifically testable. In other words, scientists must be able to use scientific processes in their search for an answer or a solution. For example, they may need to design a controlled experiment, analyze the results in a logical way, or develop scientific models to explain data. As a result, the scope of science does not extend to issues of morals, values, or the supernatural. In effect, the scope of science is limited to answering the question “how;’ not “why;’ Sometimes technology is a limitation for scientists. For example, scientists who studied space were once limited to observing only what they could see with their eyes. In the 1600s, Galileo used his telescope to observe things the eye could not see, such as the large moons of Jupiter. Since Galileo, scientists have developed instruments that have allowed them to see even farther and fainter objects. They have even put telescopes into space, such as the Hubble Space Telescope, and sent probes to the edges of our solar system. However, there is still a limit to what our current technology can detect. One of the deepest mysteries in space involves dark energy, which scientists think is responsible for the expansion of the universe. To detect this dark energy, scientists will need to build a space telescope that is able to make the large number and specific types of observations that are needed.
Science and Its Methods All scientists use certain processes in their search for explanations as to how the natural world operates. These processes include making observations, asking questions, forming a reasonable answer, evaluating the validity of the explanation, and communicating the results. Taken together, the processes constitute the scientific method. The scientific method is not a series of exact steps, but rather a strategy for drawing sound conclusions. The scientific method also includes procedures that a scientist follows, such as conducting an experiment in a laboratory or using a computer to analyze data. A scientist chooses the procedures to use depending on the nature of the investigation.
The Scientific Process
R17
There is no one correct scientific procedure. One scientist might use a field study to investigate a geologic formation, while another might do a chemical analysis on a rock sample. Another scientist might develop an experimental procedure to determine how to slow down the division of cells. Still another scientist might use a computer to create a model of a molecule. Sometimes different groups come to the same scientific conclusions through two different approaches. For example, in 1964, Arno Penzias and Robert Wilson were using a supersensitive antenna in their research laboratory to pick up certain radio waves coming from space. However, there was a background noise that they were picking up everywhere they pointed the antenna. Penzias and Wilson could not eliminate this steady noise. They checked their equipment and found nothing unusual. The scientists even cleaned the antenna, but the noise still persisted. They concluded that this radiation must actually be coming from space. While these two scientists were working with their antenna, another team of scientists just 60 km away at Princeton University was about to start a search for the same cosmic radiation that Penzias and Wilson had discovered. The members of the Princeton team had reasoned that when the universe was formed, a tremendous blast of radiation must have been released into space. The Princeton team had been planning to make observations designed to find and measure this radiation. When another scientist became aware of the coincidence, he put Penzias and Wilson in touch with the Princeton team. Penzias and Wilson had observed the radiation the Princeton team had predicted.
Science and Its Investigators Sometimes, scientists investigating the same phenomenon might interpret the results quite differently. One scientist might have one explanation, while the other scientist has a completely different explanation. One example is the behavior of light. Some scientists explained the behavior oflight in terms of waves. Others explained the same behavior of light in terms of particles. Today, scientists recognize that light has a dual nature-its behavior resembles that of both waves and particles. In this case, both explanations were logically consistent. Moreover, these explanations were tested by other scientists who confirmed the results.
Science and Its Evidence Any explanation proposed by a scientist must abide by the rules of evidence. The data must support the conclusion. If they don’t, then the explanation must be modified or even discarded. For example, observational evidence of the orbit of Mercury could not be explained by Newton’s law of gravity. Albert Einstein proposed a new way of thinking about gravity that explained the change in Mercury’s orbit. When scientists were able to directly observe some of the results that Einstein predicted, they accepted his theory of general relativity. Scientists also expect that all results can be replicated by other scientists working under the same conditions. For instance, in 1989, two scientists reported that they had performed “cold fusion:’ In effect, the scientists claimed to have carried out nuclear fusion at room temperature in a container on a countertop. People were at first hopeful that this discovery would lead to cheap and plentiful energy sources. However, a group of scientists organized that same year by the U.S. Department of Energy found no evidence to support “cold fusion:’ Other groups were unable to obtain the same results of the original experimenters. If no one can replicate a scientific result, then that result is usually not accepted as valid. R18
Appendix B
Science and Its Theories Scientists hope to arrive at a conclusion about the phenomenon they investigate. They start by making observations and asking questions. Then they suggest a reasonable explanation for what they observe. This explanation is known as a hypothesis. A hypothesis is a rational explanation of a single event or phenomenon based upon what is observed, but which has not been proven. A hypothesis usually develops from observations about the natural world. Sometimes the results of observations are expected, but sometimes they are not. Unexpected results can lead to new hypotheses. For example, in 1820, a Danish scientist named Hans Christian Oersted discovered a relationship between magnetism and electricity. While working with equipment for a lecture demonstration, Oersted placed a compass near a wire connected to an apparatus that generated an electrical current. Oersted noticed that the needle on the compass jumped and pointed toward the wire. He formed a hypothesis that electricity and magnetism were related. A hypothesis is a testable explanation. One way to test a hypothesis is by carrying out experiments that test the predictions made by a hypothesis. Oersted conducted further experiments and found that he could control the direction in which the compass needle pointed by moving the wire. His new observations supported his hypothesis. A good scientist recognizes that there is a chance that a test can fail to support the hypothesis. If Oersted’s compass needle had jumped for some other reason, he would have gotten different results. When this occurs, the scientist needs to rethink the hypothesis and construct another explanation for the event or phenomenon. Unlike a hypothesis, a theory is a well-established and highly reliable explanation accepted by scientists. A theory is an explanation of a set of related observations or events based upon proven hypotheses, and it is verified multiple times by different groups of scientists. A theory may also develop from a collection of hypotheses that have been tested and verified. For example, during the nineteenth century, various scientists developed hypotheses to account for observations that linked electricity and magnetism. In 1873, James Maxwell published his book Treatise on Electricity and Magnetism. The theory of electromagnetism is now a well-established part of science.
Science and Its Laws In science, a law is a descriptive statement that reliably predicts events under certain conditions. A law can sometimes be expressed in terms of a single mathematical equation. Some scientific laws include Newton’s laws of motion, the laws of thermodynamics, the ideal gas laws, and the laws of conservation of mass and energy. It is important to know the conditions under which a law is valid before using it. Laws can be valid over a wide range of circumstances, or over a very limited range of circumstances. A law is not the same as a theory. A law describes what is observed in nature under certain conditions. A theory is a system of ideas that explains many related observations and is supported by a large body of evidence acquired through scientific investigation. For example, Newton’s law of gravitation predicts the size of the gravitational force between masses. It says nothing of what causes this force. Einstein’s theory of gravity, however, explains that motion due to gravity is due to the bending of space-time caused by mass. Laws and theories do, however, share certain features- both are supported by observational evidence, both are widely accepted by scientists, and both may need to be modified or abandoned if conflicting evidence is discovered. They are both tools that help scientists to answer questions about the world around them. The Scientific Process
R19
Symbols Diagram Symbols MECHANICS Symbol
Meaning
displacement vector, displacement component
WAVES AND ELECTROMAGNETISM Symbol
Meaning
V
ray (light or sound) positive charge
velocity vector, velocity component
negative charge acceleration vector force vector, force component
electric field lines
momentum vector gravitational field vector
:::====>
angle marking
electric field vector electric current
rotational motion magnetic field lines
THERMODYNAMICS Symbol
Meaning
energy transferred as heat energy transferred as work
C
R20
Appendix C
cycle or process
)(

magnetic field vector (into page, out of page)
MATHEMATICAL SYMBOLS Symbol
Ll
0
Meaning
Symbol
(Greek delta) change in some quantity
::s
less than or equal to
(Greek sigma) sum of quantities
ex
is proportional to
(Greek theta) any angle
::::::
is approximately equal to
equal to
>
8 X 1024 y < 1 .25 X 1021 y 107 y
X
> 2.36
X
1021 y
10.5 y
La
Praseodymium
1015 y
2 X 106 y 30 y
Lanthanum
59
X
1.6
57

9.3
55 y
51


462 d
1.05
X
1011 y
> 5 x 1016 y
Table of Isotopes and Atomic Masses
R49
z
Element
60
Neodymium
Symbol
Nd
Average Atomic Mass (u) 144.24

Mass Number (*indicates radioactivity) A
Atomic Mass (u)
Percent Abundance
142 143 144* 145 146
141.907 718 142.909 809 143.910 082 144.912 568 145.913 113
27.13 12.18 23.80 8.30 17.19
145* 146*
144.912 745 145.914 968
61
Promet hium
Pm
62
Samarium
Sm
150.36
147* 148* 149* 150 152 154
146.914 894 147.914 819 148.917 180 149.917 273 151.919728 153.922 206
15.0 11.3 13.8 7.4 26.7 22.7
63
Europium
Eu
151 .96
151 152* 153
150.919 846 151 .921740 152.921 226
47.8



T,12
2.3x 1015 y
17.7 y 5.5 y 1.06 X 1011 y 7 X 1015 y >2 x 1015 y
13.5 y 52.2
64
Gadolinium
Gd
157.25
155 156 157 158 160
154.922 618 155.922 119 156.923 957 157.924 099 159.927 050
14.80 20.47 15.65 24.84 21.86
65
Terbi um
Tb
158.9253
159
158.925 345
100
66
Dysprosium
Dy
162.5
161 162 163 164
160.926 930 161.926796 162.928 729 163.929 172
18.9 25.5 24.9 28.2
67
Holmium
Ho
164.9303
165
164.930 316
100
68
Erbium
Er
167.26
166 167 168 170
165.930 292 166.932 047 167.932 369 169.935 462
33.6 22.95 27.8 14.9
69
Thulium
Tm
168.9342
169 171*
168.934 213 170.936 428
100

Half-life (if radioactive)




1.92 y
70
Ytterbium
Yb
173.04
171 172 173 174 176
170.936 171.936 172.938 173.938 175.942
324 379 209 861 564
14.3 21.9 16.12 31.8 12.7
71
Lutetium
Lu
174.967
175 176*
174.940 772 175.942 679
97.41 2.59

3.78
X
1010y

72
Hafnium
Hf
178.49
177 178 179 180
176.943 218 177.943 697 178.945 813 179.946 547
18.606 27.297 13.029 35.100
73
Tantalum
Ta
180.9479
181
180.947 993
99.988
74
Tungsten
w
183.85
182 183 184 186
181.948202 182.950 221 183.950 929 185.954 358
26.3 14.28 30.7 28.6
75
Rhenium
Re
186.207
185 187*
184.952 951 186.955 746
37.40 62.60


R50
Appendix H
4.4
X
1010 y
Average Atomic Mass (u)
Mass Number (*indicates radioactivity) A
Atomic Mass (u)
Percent Abundance
z
Element
Symbol
76
Osmium
Os
190.2
188 189 190 192
187.955832 188.958 139 189.958 439 191.961 468
13.3 16.1 26.4 41 .0
77
Iridium
Ir
192.2
191 193
190.960 585 192.962 916
37.3 62.7
78
Platinum
Pt
195.08
194 195 196
193.962 655 194.964 765 195.964 926
32.9 33.8 25.3
79
Gold
Au
196.9665
197
196.966 543
100
80
Mercury
Hg
200.59
198 199 200 201 202
197.966 743 198.968 253 199.968 299 200.970 276 201.970 617
9.97 16.87 23.10 13.10 29.86
81
Thallium
Tl
204.383
203 204* 205 208*
202.972 320 203.073 839 204.974 400 207.981 992
29.524
206 207 208 212*
205.974 440 206.974 871 207.976 627 211.991 872
24.1 22.1 52.4
209 212*
208.980 374 211.991 259
100


Half-life (if radioactive) 7 112


82
Lead
Pb
207.2
83
Bismuth
Bi

70.476 3.053 m
10.64 h 60.6 m
84
Polonium
Po
209* 212* 216*
208.982 405 211.988 842 216.001 889
102 y 0.30 µs 0.145 s
85
Astatine
At
218* 219*
218.008 685 219.011 294
1.6 s 0.9m
86
Radon
Rn
220* 222*
220.011 369 222.017 571
55.6 s 3.823 d
87
Francium
Fr
223*
223.019 733
22 m
88
Radium
Ra
224* 226* 228*
224.020 187 226.025 402 228.031 064
3.66 d 1.6 X 103 Y 5.75 y
89
Actinium
Ac
227* 228*
227.027 701 228.028 716
18.72 y 1.913 y
90
Thorium
Th
232* 234*
232.038 051 234.043 593
91
Protactinium
Pa
231 * 234*
231.035 880 234.043 300
92
Uranium
u
234* 235* 238*
234.040 946 235.043 924 238.050 784


208.9803
3.78 y



100
1.40 X 1010 y 24.1 d 32.760 y 6.7 h
0.0055 0.720 99.2745
2.46 7.04 4.47
X X X
105 y 108 y 1Q9 y
93
Neptunium
Np
236* 237*
236.046 560 237.048 168
1.15 x 1o5 y 2.14 X 106 y
94
Plutonium
Pu
239* 244*
239.052 157 244.064 200
2.412 X 105 y 8.1 x 107 y

Table of Isotopes and Atomic Masses
R51
Additional Problems The Science ol Physics 1. Mt. Waialeale in Hawaii gets 1.168 x 103 cm ofrainfall per year. Express this quantity in meters. 2. An acre is equal to about 4.0469 x 103 m 2 . Express this area in square kilometers.
3. A group drinks about 6.4 x 104 cm3 of water per person per year. Express this in cubic meters. 4. The largest ston e jar on the Plain ofJars in Laos has a mass of 6.0 x 103 kg. Express this mass in milligrams. 5. Half of a sample of the radioactive isotope beryllium-8 decays in 6.7 x 10- 17 s. Express this time in picoseconds.
Motion in One Dimension 6. The fastest airplane is the Lockheed SR-71. If an SR-71
flies 15.0 km west in 15.3 s, what is its average velocity in kilometers per hour? 7. Except for a 22.0 min rest stop, Emily d rives with a
constant velocity of 89.5 km/h, north. How long does the trip take if Emily’s average velocity is 77 .8 km/ h, north? 8. A spaceship accelerates uniformly for 1220 km. How much time is required for the spaceship to increase its speed from 11.1 km/ s to 11.7 km/ s? 9. A polar bear initially running at 4.0 m l s accelerates uniformly for 18 s. If the bear travels 135 min this time, what is its maximum speed? 10. A walrus accelerates from 7.0 km/ h to 34.5 km/ h over a distance of 95 m. What is the magnitude of the walrus’s acceleration? 11 . A snail can move about 4.0 min 5.0 min. What is the average speed of the snail? 12. A crate is accelerated at 0.035 m/ s2 for 28.0 s along a conveyor belt. If the crate’s initial speed is 0.76 m / s, what is its final speed?
17. A jet slows uniformly from 153 km/ h to 0 km/ h over 42.0 m. What is the jet’s acceleration? 18. A softball thrown straight up at 17.5 mi s is caught 3.60 s later. How high does the ball rise? 19. A child, starting from rest, sleds down a snow-covered slope in 5.50 s. If the child’s final speed is 14.0 mis, what the length of the slope?
20. A sky diver opens her parachute and drifts down for 34.0 s with a constant velocity of 6.50 m / s. What is the sky diver’s displacement? 21. In a race, a tortoise runs at 10.0 cm/ sand a hare runs at 200.0 cm/ s. Both start at the same time, but the hare stops to rest for 2.00 min. The tortoise wins by 20.0 cm. At what time does the tortoise cross the finish line?
22. What is the length of the race in problem 21? 23. The cable pulling an elevator upward at 12.5 m / s breaks. How long does it take for the elevator to come to rest?
24. A disk is uniformly accelerated from rest for 0.910 s over 7.19 km. What is its final speed? 25. A tiger accelerates 3.0 m / s2 for 4.1 s to reach a final speed of 55.0 km/ h. What was its initial speed in kilometers per hour? 26. A shark accelerates uniformly from 2.8 km/ h to 32.0 km/ h in 1.5 s. How large is its acceleration?
27. The 1903 Wright flyer was accelerated at 4.88 m / s2 along a track that was 18.3 m long. How long did it take to accelerate the flyer from rest? 28. A drag racer starts at rest and reaches a speed of 386.0 km/ h with an average acceleration of 16.5 m / s2 • How long does this acceleration take?
29. A hummingbird accelerates at – 9.20 m/ s2 such that its velocity changes from + 50.0 km/h to 0 km/ h. What is its displacement?
13. A person throws a ball vertically and catches it after 5.10 s. What is the ball’s initial velocity?
30. A train backs up from an initial velocity of – 4.0 m / s and an average acceleration of – 0.27 m / s2 • What is the train’s velocity after 17 s?
14. A bicyclist accelerates – 0.870 m/ s2 during a 3.80 s interval. What is the change in the velocity of the bicyclist and bicycle?
31. A cross-country skier skiing with an initial velocity of +4.42 m i s slows uniformly at – 0. 75 m/ s2 • How long does it take the skier to stop?
15. A hockey puck slides 55.0 m in 1.25 s with a uniform acceleration. If the puck’s final speed is 43.2 m/s, what was its initial speed?
R52
16. A small rocket launched from rest travels 12.4 m upward in 2.0 s. What is the rocket’s net acceleration?
Appendix I
32. What is the skier’s displacement in problem 31?
47. Find the displacement direction in problem 46.
33. A speedboat uniformly increases its speed from 25 m / s west to 35 ml s west. How long does it take the boat to travel 250 m west?
48. A train travels 478 km southwest along a straight stretch. If the train is displaced south by 42 km, what is the train’s displacement to the west?
34. A ship accelerates at – 7.6 x 10- 2 m/ s2 so that it comes to rest at the dock 255 m away in 82.0 s. What is the ship’s initial speed?
49. Find the displacement direction in problem 48.
35. A student skates downhill with an average acceleration of0.85 m / s 2• Her initial speed is 4.5 m / s, and her final speed is 10.8 m is. How long does she take to skate down the hill? 36. A wrench dropped from a tall building is caught in a safety net when the wrench has a velocity of – 49.5 m/ s. How far did it fall? 37. A rocket sled comes to a complete stop from a speed of 320 km/ h in 0.18 s. What is the sled’s average acceleration? 38. A racehorse uniformly accelerates 7.56 m / s2, reaching its final speed after running 19.0 m. If the horse starts at rest, what is its final speed?
39. An arrow is shot upward at a speed of 85.1 m is. How long does the archer have to move from the launching spot before the arrow returns to Earth? 40. A handball strikes a wall with a forward speed of 13.7 m /s and bounces back with a speed of 11.5 m/s. If the ball changes velocity in 0.021 s, what is the handball’s average acceleration? 41. A ball accelerates at 6.1 m / s2 from 1.8 m /s to 9.4 m / s.
How far does the ball travel?
42. A small sandbag is dropped from rest from a hovering hot-air balloon. After 2.0 s, what is the sandbag’s displacement below the balloon? 43. A hippopotamus accelerates at 0.678 m /s2 until it reaches a speed of 8.33 m /s. If the hippopotamus runs 46.3 m, what was its initial speed? 44. A ball is hit upward with a speed of7.5 m /s. How long does the ball take to reach maximum height? 45. A surface probe on the planet Mercury falls 17.6 m downward from a ledge. If free-fall acceleration near Mercury is -3.70 m /s 2, what is the probe’s velocity wh en it reaches the ground?
Two-Dimensional Motion and Vectors 46. A plane moves 599 m northeast along a runway. If the northern component of this displacement is 89 m, how large is the eastern component?
50. A ship’s total displacement is 7400 km at 26° south of west. If the ship sails 3200 km south, what is the western component of its journey? 51. The distance from an observer on a plain to the top of a
nearby mountain is 5.3 km at 8.4° above the horizontal. How tall is the mountain?
52. A skyrocket travels 113 m at an angle of 82.4° with respect to the ground and toward the south. What is the rocket’s horizontal displacement? 53. A hot-air balloon descends with a velocity of 55 km/ h at an angle of 37° below the horizontal. What is the vertical velocity of the balloon?
:t>
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
0 C”‘

(‘t)
3 CJ)
54. A stretch of road extends 55 km at 37° north of east, then continues for 66 km due east. What is a driver’s resultant displacement along this road? 55. A driver travels 4.1 km west, 17.3 km north, and finally 1.2 km at an angle of 24.6° west of north. What is the driver’s displacement? 56. A tornado picks up a car and hurls it horizontally 125 m
with a speed of 90.0 m /s. How long does it take the car to reach the ground?
57. A squirrel knocks a nut horizontally at a speed of 10.0 cm/ s. If the nut lands at a horizontal distance of 18.6 cm, how high up is the squirrel? 58. A flare is fired at an angle of 35° to the ground at an initial speed of 250 m l s. How long does it take for the flare to reach its maximum altitude? 59. A football kicked with an initial speed of23.l m /s reaches a maximum height of 16.9 m. At what angle was the ball kicked?
60. A bird flies north at 58.0 km/ h relative to the wind. The wind is blowing at 55.0 km/ h south relative to Earth. How long will it take the bird to fly 1.4 km relative to Earth? 61. A racecar moving at 286 km/ h is 0. 750 km behind a car
moving at 252 km/ h. How long will it take the faster car to catch up to the slower car? 62. A helicopter flies 165 m horizontally and then moves downward to land 45 m below. What is the helicopter’s resultant displacement?
63. A toy parachute floats 13.0 m downward. If the parachute travels 9.0 m horizontally, what is the resultant displacement?
Additional Problems
R53
64. A billiard ball travels 2. 7 m at an angle of 13° with respect to the long side of the table. What are the components of the ball’s displacement? 65. A golf ball has a velocity of 1.20 m/s at 14.0° east of
north. What are the velocity components?
79. What is the range of an arrow shot horizontally at
85.3 m / s from 1.50 m above the ground? 80. A drop of water in a fountain takes 0.50 s to travel 1.5 m
horizontally. The water is projected upward at an angle of 33°. What is the drop’s initial speed?
66. A tiger leaps with an initial velocity of 55.0 km/ h at an angle of 13.0° with respect to the horizontal. What are the components of the tiger’s velocity?
81. A golf ball is hit up a 41.0° ramp to travel 4.46 m horizontally and 0.35 m below the edge of the ramp. What is the ball’s initial speed?
67. A tramway extends 3.88 km up a mountain from a
82. A flare is fired with a velocity of 87 km/h west from a car
station 0.8 km above sea level. If the horizontal displacement is 3.45 km, how far above sea level is the mountain peak? 68. A bullet travels 850 m, ricochets, and moves another
640 mat an angle of 36° from its previous forward motion. What is the bullet’s resultant displacement?
69. A bird flies 46 km at 15° south of east, then 22 km at 13° east of south, and finally 14 km at 14° west of south. What is the bird’s displacement? 70. A ball is kicked with a horizontal speed of9.37 m / s
off the top of a mountain. The ball moves 85.0 m horizontally before hitting the ground. How tall is the mountain?
71 . A ball is kicked with a horizontal speed of 1.50 m / s from a height of 2.50 x 102 m . What is its horizontal displacement when it hits the ground?
traveling 145 km/ h north. With respect to Earth, what is the flare’s resultant displacement 0.45 s after being launched? 83. A sailboat travels south at 12.0 km/h with respect to the
water against a current 15.0° south of east at 4.0 km/ h. What is the boat’s velocity?
Forces and the Laws ol Motion 84. A boat exerts a 9.5 x 104 N force 15.0° north of west on a
barge. Another exerts a 7.5 x 104 N force north. What direction is the barge moved?
85. A shopper exerts a force on a cart of 76 N at an angle of 40.0° below the horizontal. How much force pushes the cart in the forward direction? 86. How much force pushes the cart in problem 85 against
72. What is the velocity of the ball in problem 71 when it
the floor?
reaches the ground? 87. What are the magnitudes of the largest and smallest net 73. A shingle slides off a roof at a speed of 2.0 m / s and an
angle of 30.0° below the horizontal. How long does it take the shingle to fall 45 m?
forces that can be produced by combining a force of 6.0 Nanda force of 8.0 N? 88. A buoyant force of790 N lifts a 214 kg sinking boat.
74. A ball is thrown with an initial speed of 10.0 mi s and an
angle of37.0° above the horizontal. What are the vertical and horizontal components of the ball’s displacement after 2.5 s? 75. A rocket moves north at 55.0 km/ h with respect to the air. It encounters a wind from 17.0° north of west at
40.0 km/ h with respect to Earth. What is the rocket’s velocity with respect to Earth? 76. How far to the north and west does the rocket in
problem 75 travel after 15.0 min? 77. A cable car travels 2.00 x 102 m on level ground, then
3.00 x 102 m at an incline of 3.0°, and then 2.00 x 102 m at an incline of8.8°. What is the final displacem ent of the cable car? 78. A hurricane moves 790 km at 18° north of west, then
due west for 150 km, then north for 470 km, and finally 15° east of north for 240 km. What is the hurricane’s resultant displacement?
R54
Appendix I
What is the boat’s net acceleration? 89. A house is lifted by a net force of 2850 N and moves
from rest to an upward speed of 15 cm/ sin 5.0 s. What is the mass of the house?
90. An 8.0 kg bag is lifted 20.0 cm in 0.50 s. If it is initially at rest, what is the net force on the bag? 91. A 90.0 kg skier glides at constant speed down a 17.0° slope. Find the frictional force on the skier. 92. A snowboarder slides down a 5.0° slope at a constant
speed. What is the coefficient of kinetic friction between the snow and the board?
93. A 2.00 kg block is in equilibrium on a 36.0° incline. What is the normal force on the block? 94. A 1.8 x 103 kg car is parked on a hill on a 15.0° incline. A 1.25 x 104 N frictional force holds the car in place. Find the coefficient of static friction.
95. The coefficient of kinetic friction between a jar slid across a table and the table is 0.20. What is the magnitude of the jar’s acceleration? 96. A force of 5.0 N to the left causes a 1.35 kg book to have a net acceleration of 0. 76 ml s2 to the left. What is the frictional force on the book? 97. A child pulls a toy by exerting a force of 15.0 Nat an angle of 55.0° with respect to the floor. What are the components of the force? 98. A car is pulled by three forces: 600.0 N to the north, 750.0 N to the east, and 675 N at 30.0° south of east. What direction does the car move? 99. Suppose a catcher exerts a force of -65.0 N to stop a baseball with a mass of 0.145 kg. What is the ball’s net acceleration as it is being caught?
100. A 2.0 kg fish pulled upward by a fisherman rises 1.9 m in 2.4 s, starting from rest. What is the net force on the fish during this interval? 101. An 18.0 N force pulls a cart against a 15.0 N frictional force. The speed of the cart increases 1.0 mis every 5.0 s. What is the cart’s mass? 102. A 47 kg sled carries a 33 kg load. The coefficient of kinetic friction between the sled and snow is 0.075. What is th e magnitude of the frictional force on the sled as it moves up a hill with a 15° incline? 103. Ice blocks slide with an acceleration of 1.22 m/s2 down a chute at an angle of 12.0° below the horizontal. What is the coefficient of kinetic friction between the ice and chute? 104. A 1760 N force pulls a 266 kg load up a 17° incline. What is the coefficient of static friction between the load and the incline?
109. A traffic signal is supported by two cables, each of which makes an angle of 40.0° with the vertical. If each cable can exert a maximum force of7.50 x 102 N, what is the largest weight they can support? 110. A certain cable of an elevator is designed to exert a force of 4.5 x 104 N. If the maximum acceleration that a loaded car can withstand is 3.5 m/s2 , what is the combined mass of the car and its contents? 111. A frictional force of 2400 N keeps a crate of machine parts from sliding down a ramp with an incline of 30.0°. The coefficient of static friction between the box and the ramp is 0.20. What is the normal force of the ramp on the box? 112. Find the mass of the crate in problem ll l. 2
113. A 5.1 x 10 kg bundle of bricks is pulled up a ramp at an incline of 14° to a construction site. The force needed to move the bricks up the ramp is 4.1 x 103 N. What is the coefficient of static friction between the bricks and the ramp?
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
0 C”‘

(‘t)
3 CJ)
Work and Energy 114. If 2.13 x 106 J of work must be done on a roller-coaster car to move it 3.00 x 102 m, how large is the net force acting on the car?
115. A force of715 N is applied to a roller-coaster car to push it horizontally. If 2. 72 x 104 J of work is done on the car, how far has it been pushed? 116. In 0.181 s, through a distance of8.05 m, a test pilot’s speed decreases from 88.9 m/s to 0 m /s. If the pilot’s mass is 70.0 kg, how much work is done against his body?
105. A 4.26 x 107 N force pulls a ship at a constant speed along a dry dock. Th e coefficient of kinetic friction between the ship and dry dock is 0.25. Find the normal force exerted on the ship.
117. What is the kinetic energy of a disk with a mass of 0.20 g and a speed of 15.8 km/ s?
106. If the incline of the dry dock in problem 105 is 10.0°, what is the ship’s mass?
119. A golf ball with a mass of 47.0 g has a kinetic energy of 1433 J. What is the ball’s speed?
107. A 65.0 kg skier is pulled up an 18.0° slope by a force of 2.50 x 102 N. If the net acceleration uphill is 0.44 m / s2 , what is the frictional force between the skis and the snow?
120. A turtle, swimming at 9.78 m / s, has a kinetic energy of 6.08 x 104 J. What is the turtle’s mass?
108. Four forces are acting on a hot-air balloon: F 1 = 2280.0 N up, F2 = 2250.0 N down, F 3 = 85.0 N west, and F4 = 12.0 N east. What is the direction of the net external force on the balloon?
:t>
118. A9.00 x 102 kgwalrusis swimmingataspeedof 35.0 km/ h. What is its kinetic energy?
121. A 50.0 kg parachutist is falling at a speed of 47.00 m / s when her parachute opens. Her speed upon landing is 5.00 mis. How much work is done by the air to reduce the parachutist’s speed? 122. An llOO kg car accelerates from 48.0 km/ h to 59.0 km/ h over 100.0 m. What was the magnitude of the net force acting on it?
Additional Problems
R55
123. What is the gravitational potential energy of a 64.0 kg person at 5334 m above sea level? 124. A spring has a force constant of 550 N/ m. What is the elastic potential energy stored in the spring when the spring is compressed 1.2 cm? 125. What is the kinetic energy of a 0.500 g raindrop that falls 0.250 km? Ignore air resistance. 126. A 50.0 g projectile is fired upward at 3.00 x 102 mis and lands at 89.0 m/ s. How much mechanical energy is lost to air resistance? 127. How long does it take for 4.5 x 106 J of work to be done by a 380.3 kW engine? 128. A ship’s engine has a power output of 13.0 MW. How much work can it do in 15.0 min? 129. A catcher picks up a baseball from the ground with a net upward force of 7.25 x 10-2 N so that 4.35 x 10-2 J of net work is done. How far is the ball lifted? 3
130. A crane does 1.31 x 10 J of net work when lifting cement 76.2 m. How large is the net force doing this work? 131. A girl exerts a force of35.0 Nat an angle of20.0° to the horizontal to move a wagon 15.0 m along a level path. What is the net work done on it if a frictional force of 24.0 N is present? 132. The Queen Mary had a mass of 7.5 x 107 kg and a top cruising speed of 57 km/ h. What was the kinetic energy of the ship at that speed? 133. How fast is a 55.0 kg sky diver falling when her kinetic energy is 7.81 x 104 J? 134. A hockey puck with an initial speed of8.0 mis coasts 45 m to a stop. If the force of friction on the puck is 0.12 N, what is the puck’s mass?
104 kg jet travel
135. How far does a 1.30 x if it is slowed from 2.40 x 102 km/h to 0 km/h by an acceleration of – 30.8 m/s2 ? 136. An automobile is raised 7.0 m, resulting in an increase in gravitational potential energy of 6.6 x 104 J. What is the automobile’s mass?
R56
139. A ball falls 3.0 m down a vertical pipe, the end of which bends horizontally. How fast does the ball leave the pipe if no energy is lost to friction? 140. A spacecraft’s engines do 1.4 x 1013 J of work in 8.5 min. What is the power output of these engines? 141. A runner exerts a force of334 N against the ground while using 2100 W of power. How long does it take him to run a distance of50.0 m? 142. A high-speed boat has four 300.0 kW motors. How much work is done in 25 s by the motors? 143. A 92 N force pushes an 18 kg box of books, initially at rest, 7.6 m across a floor. The coefficient of kinetic friction between the floor and the box is 0.35. What is the final kinetic energy of the box of books?
144. A guardrail can be bent by 5.00 cm and then restore its shape. What is its force constant if struck by a car with 1.09 x 104 J of kinetic energy? 145. A 25.0 kg trunk strikes the ground with a speed of 12.5 m/s. If no energy is lost from air resistance, what is the height from which the trunk fell? 146. Sliding a 5.0 kg ston e up a frictionless ramp with a 25.0° incline increases its gravitational potential energy by 2.4 x 102 J. How long is the ramp? 147. A constant 4.00 x 102 N force moves a 2.00 x 102 kg iceboat 0.90 km. Frictional force is negligible, and the boat starts at rest. Find the boat’s final speed. 148. A 50.0 kg circus clown jumps from a platform into a net 1.00 m above the ground. The net is stretched 0.65 m and has a force constant of 3.4 x 104 Ni m. What is the height of the platform?
Momentum and Collisions 149. If a 50.0 kg cheetah, initially at rest, runs 274 m north in 8.65 s, what is its momentum? 150. If a 1.46 x 105 kg whale has a momentum of9.73 x 105 kg• m/s to the south, what is its velocity? 151 . A star has a momentum of8.62 x 1036 kg• m / s and a speed of255 km/ s. What is its mass?
137. A spring in a pogo stick has a force constant of 1.5 x 104 Ni m. How far is the spring compressed when its elastic potential energy is 120 J?
152. A 5.00 g projectile has a velocity of 255 m / s right. Find the force to stop this projectile in 1.45 s.
138. A 100.0 g arrow is pulled back 30.0 cm against a bowstring. The bowstring’s force constant is 1250 N/ m. At what speed will the arrow leave the bow?
153. How long does it take a 0.17 kg hockey puck to decrease its speed by 9.0 m /s if the coefficient of kinetic friction is 0.050?
Appendix I
154. A 705 kg racecar driven by a 65 kg driver moves with a velocity of 382 km/h right. Find the force to bring the car and driver to a stop in 12.0 s.
169. A 1.1 x 103 kg walrus starts swimming east from rest and reaches a velocity of9.7 mi s in 19 s. What is the net force acting on the walrus?
155. Find the stopping distance in problem 154.
170. A 12.0 kg wagon at rest is pulled by a 15.0 N force at an angle of 20.0° above the horizontal. If an 11.0 N frictional force resists the forward force, how long will the wagon take to reach a speed of 4.50 mis?
156. A 50.0 g shell fired from a 3.00 kg rifle has a speed of 400.0 m / s. With what velocity does the rifle recoil in the opposite direction? 157. A twig at rest in a pond moves with a speed of 0.40 cm/ s opposite a 2.5 g snail, which has a speed of 1.2 cm/ s. What is the mass of the twig? 158. A 25.0 kg sled holding a 42.0 kg child has a speed of 3.50 m / s. They collide with and pick up a snowman, initially at rest. The resulting speed of the snowman, sled, and child is 2.90 mis. What is the snowman’s mass? 159. An 8500 kg railway car moves right at 4.5 m / s, and a 9800 kg railway car moves left at 3.9 m / s. The cars collide and stick together. What is the final velocity of the system?
171. A 42 g meteoroid moving forward at 7.82 x 103 m/ s collides with a spacecraft. What force is needed to stop the meteoroid in 1.0 x 10-6 s? 172. A 455 kg polar bear slides for 12.2 s across the ice. If the coefficient of kinetic friction between the bear and the ice is 0.071, what is the change in the bear’s momentum as it comes to a stop? 173. How far does the bear in problem 172 slide?
160. What is the change in kinetic energy for the two railway cars in problem 159? 161. A 55 g clay ball moving at 1.5 mis collides with a 55 g clay ball at rest. By what percentage does the kinetic energy change after the inelastic collision?
176. A 1.36 x 104 kg barge is loaded with 8.4 x 103 kg of coal. What was the unloaded barge’s speed if the loaded barge has a speed of 1.3 m / s?
162. A 45 g golf ball collides elastically with an identical ball at rest and stops. If the second ball’s final speed is 3.0 mi s, what was the first ball’s initial speed?
177. A 1292 kg automobile moves east at 88.0 km/ h. If all forces remain constant, what is the car’s velocity if its mass is reduced to 1255 kg?
163. A 5.00 x 102 kg racehorse gallops with a momentum of 8.22 x 103 kg•m/ s to the west. What is the horse’s velocity?
178. A 68 kg student steps into a 68 kg boat at rest, causing both to move west at a speed of 0.85 m / s. What was the student’s initial velocity?
164. A 3.0 x 107 kg ship collides elastically with a 2.5 x 107 kg ship moving north at 4.0 km/ h. After the collision, the first ship moves north at 3.1 km/ h and the second ship moves south at 6.9 km/ h. Find the unknown velocity.
179. A 1400 kg automobile, heading north at 45 km/h, collides inelastically with a 2500 kg truck traveling east at 33 km/ h. What is the vehicles’ final velocity?
166. A bird with a speed of 50.0 km/h has a momentum of magnitude of 0.278 kg•m/ s. What is the bird’s mass? 167. A 75 N force pulls a child and sled initially at rest down a snowy hill. If the combined mass of the sled and child is 55 kg, what is their speed after 7.5 s? 168. A student exerts a net force of – 1.5 N over a period of 0.25 s to bring a falling 60.0 g egg to a stop. What is the egg’s initial speed?
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
0 C”‘

(‘t)
3 CJ)
174. How long will it take a -1.26 x 104 N force to stop a 2.30 x 103 kg truck moving at a speed of 22.2 m / s? 175. A 63 kg skater at rest catches a sandbag moving north at 5.4 m / s. The skater and bag then move north at 1.5 m/ s. Find the sandbag’s mass.
165. A high-speed train has a mass of7.10 x 105 kg and moves at a speed of270.0 km/ h. What is the magnitude of the train’s momentum?
:t>
180. An artist throws 1.3 kg of paint onto a 4.5 kg canvas at rest. The paint-covered canvas slides backward at 0.83 m / s. What is the change in the kinetic energy of the paint and canvas? 181. Find the change in kinetic energy if a 0.650 kg fish leaping to the right at 15.0 m / s collides inelastically with a 0.950 kg fish leaping to the left at 13.5 m / s. 182. A 10.0 kg cart moving at 6.0 mi s hits a 2.5 kg cart moving at 3.0 mis in the opposite direction. Find the carts’ final speed after an inelastic collision. 183. A ball, thrown right 6.00 m / s, hits a 1.25 kg panel at rest, then bounces back at 4.90 m / s. The panel moves right at 1.09 mi s. Find the ball’s m ass.
Additional Problems
R57
184. A 2150 kg car, moving east at 10.0 m / s, collides and joins with a 3250 kg car. The cars move east together at 5.22 m / s. What is the 3250 kg car’s initial velocity?
197. A 2.05 x 108 kg asteroid has an orbit with a 7378 km radius. The centripetal force on the asteroid is 3.00 x 109 N. Find the asteroid’s tangential speed.
185. Find the change in kinetic energy in problem 184.
198. Find the gravitational force between a 0.500 kg mass and a 2.50 x 1012 kg mountain that is 10.0 km away.
186. A 15.0 g toy car moving to the right at 20.0 cm/ s collides elastically with a 20.0 g toy car moving left at 30.0 cm/ s. The 15.0 g car then moves left at 37.1 cm/ s. Find the 20.0 g car’s final velocity.
187. A remora swimming right at 5.0 m/ s attaches to a 150.0 kg shark moving left at 7.00 m / s. Both move left at 6.25 m / s. Find the remora’s mass. 188. A 6.5 x 1012 kg comet, moving at 420 m / s, catches up to and collides inelastically with a 1.50 x 1013 kg comet moving at 250 m / s. Find the change in the comets’ kinetic energy. 189. A 7.00 kg ball moves east at 2.00 m / s, collides with a 7.00 kg ball at rest, and then moves 30.0° north of east at 1.73 mis. What is the second ball’s final velocity? 190. A 2.0 kg block moving at 8.0 m/ s on a frictionless surface collides elastically with a block at rest. The first block moves in the same direction at 2.0 m / s. What is the second block’s mass?
Circular Motion and Gravitation 191. A pebble that is 3.81 m from the eye ofa tornado has a tangential speed of 124 m / s. What is the magnitude of the pebble’s centripetal acceleration?
R58
199. The gravitational force between Ganymede and Jupiter is 1.636 x 1022 N. Jupiter’s mass is 1.90 x 1027 kg, and the distance between the two bodies is 1.071 x 106 km. What is Ganymede’s mass? 200. At the sun’s surface, the gravitational force on 1.00 kg is 274 N. The sun’s mass is 1.99 x 1030 kg. If the sun is assumed spherical, what is the sun’s radius? 201. At the surface of a red giant star, the gravitational force on 1.00 kg is only 2.19 x 10-3 N. Ifits mass equals 3.98 x 1031 kg, what is the star’s radius? 202. Uranus has a mass of 8.6 x 1025 kg. The mean distance between the centers of the planet and its moon Miranda is 1.3 x 105 km. If the orbit is circular, what is Miranda’s period in hours? 203. What is the tangential speed in problem 202? 204. The rod connected halfway along the 0.660 m radius of a wheel exerts a 2.27 x 105 N force. How large is the maximum torque? 205. A golfer exerts a torque of 0.46 N•m on a golf club. If the club exerts a force of 0.53 Non a stationary golf ball, what is the length ofthe club? 206. What is the orbital radius of the Martian moon Deimos if it orbits 6.42 x 1023 kg Mars in 30.3 h?
192. A racecar speeds along a curve with a tangential speed of75.0 mi s. The centripetal acceleration on the car is 22.0 m / s 2 . Find the radius of the curve.
207. A 4.00 x 102 N•m torque is produced applying a force 1.60 m from the fulcrum and at an angle of 80.0° to the lever. How large is the force?
193. A subject in a large centrifuge has a radius of8.9 m and a centripetal acceleration of20g (g = 9.81 m / s2). What is the tangential speed of the subject?
208. A customer 11 m from the center of a revolving restaurant has a speed of 1.92 x 10-2 mi s. How large a centripetal acceleration acts on the customer?
194. A 1250 kg automobile with a tangential speed of 48.0 km/ h follows a circular road that has a radius of 35.0 m. How large is the centripetal force?
209. A toy train on a circular track has a tangential speed of 0.35 mi s and a centripetal acceleration of0.29 m / s2. What is the radius of the track?
195. A rock in a sling is 0.40 m from the axis of rotation and has a tangential speed of6.0 m /s. What is the rock’s mass if the centripetal force is 8.00 x 102 N?
210. A person against the inner wall of a hollow cylinder with a 150 m radius feels a centripetal acceleration of 9.81 m/s2 • Find the cylinder’s tangential speed.
196. A 7.55 x 1013 kg comet orbits the sun with a speed of 0.173 km/ s. If the centripetal force on the comet is 505 N, how far is it from the sun?
211 . The tangential speed of 0.20 kg toy carts is 5.6 m / s when they are 0.25 m from a turning shaft. How large is the centripetal force on the carts?
Appendix I
212. A 1250 kg car on a curve with a 35.0 m radius has a centripetal force from friction and gravity of 8.07 x 103 N. What is the car’s tangential speed?
228. A block of ebony with a volume of 2.5 x 10- 3 m 3 is placed in fresh water. If the apparent weight of the block is 7.4 N, what is the density of ebony?
213. Two wrestlers, 2.50 x 10-2 m apart, exert a 2.77 x 10- 3 N gravitational force on each other. One has a mass of 157 kg. What is the other’s mass?
229. One piston of a hydraulic lift holds 1.40 x 103 kg. The other holds an ice block (p = 917 kgl m 3 ) that is 0.076 m thick. Find the first piston’s area.
105
104
214. A 1.81 x kg blue whale is 1.5 m from a 2.04 x kg whale shark. What is the gravitational force between them? 105
215. Triton’s orbit around Neptune has a radius of3.56 x km. Neptune’s mass is 1.03 x 1026 kg. What is Triton’s period?
4
230. A hydraulic-lift piston raises a 4.45 x 10 N weight by 448 m. How large is the force on the other piston if it is pushed 8.00 m downward? glcm 3
231. A platinum flute with a density of21.5 is submerged in fresh water. If its apparent weight is 40.2 N, what is the flute’s mass?
:t>
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
0 C”‘

216. Find the tangential speed in problem 215.
(‘t)
217. A moon orbits a 1.0 x 1026 kg planet in 365 days. What is the radius ofthe moon’s orbit?
Heat
3 CJ)
218. What force is required to produce a 1.4 N•m torque wh en applied to a door at a 60.0° angle and 0.40 m from the hinge?
232. Surface temperature on Mercury ranges from 463 K during the day to 93 Kat night. Express this temperature range in degrees Celsius.
219. What is the maximum torque that the force in problem 218 can exert?
233. Solve problem 233 for degrees Fahrenheit.
220. A worker hanging 65.0° from the vane of a windmill exerts an 8.25 x 103 N•m torque. If the worker weighs 587 N, what is the vane’s length?
Fluid Mechanics 221. A cube of volume 1.00 m 3 floats in gasoline, which has a density of 675 kgl m 3 • How large a buoyant force acts on the cube? 222. A cube 10.0 cm on each side has a density of 2.053 x 104 kgl m 3 • Its apparent weigh t in fresh water is 192 N. Find the buoyant force. 223. A 1.47 x 106 kg steel hull has a base that is 2.50 x 103 m 2 in area. If it is placed in sea water (p = 1.025 x 103 kgl m 3), how deep does the hull sink? 224. What size force will open a door of area 1.54 m 2 if the net pressure on the door is 1.013 x 103 Pa? 225. Gas at a pressure of 1.50 x 106 Pa exerts a force of 1.22 x 104 N on the upper surface of a piston. What is the piston’s upper surface area? 226. In a barometer, the mercury column’s weight equals the force from air pressure on the mercury’s surface. Mercury’s density is 13.6 x 103 kgl m 3 . What is the air’s pressure if the column is 760 mm high?
234. The temperature in Fort Assiniboine, Montana, went from – 5°F to + 37°F on January 19, 1892. Calculate this change in temperature in kelvins. 235. An acorn falls 9.5 m, absorbing 0.85 of its initial potential energy. If 1200 Jl kg will raise the acorn’s temperature 1.0°c, what is its temperature increase? 236. A bicyclist on level ground brakes from 13.4 mi s to O mis. What is the cyclist’s and bicycle’s mass if the increase in internal energy is 5836 J? 237. A 61.4 kg roller skater on level ground brakes from 20.5 mis to Omi s. What is the total change in the internal energy of the system? 238. A 0.225 kg tin can (cP = 2.2 x 103 Jlkg• °C) is cooled in water, to which it transfers 3.9 x 104 J of energy. By how much does the can’s temperature change? 239. What mass of bismuth (cp = 121 Jl kg•°C) increases temperature by 5.0°C when 25 J are added by heat? 240. Placing a 0.250 kg pot in 1.00 kg of water raises the water’s temperature 1.00°c. The pot’s temperature drops l 7.5°C. Find the pot’s specific heat capacity. 241. Lavas at Kilauea in Hawaii have temperatures of 2192°F. Express this quantity in degrees Celsius. 242. The present temperature of the background radiation in the universe is 2.7 K. What is this temperature in degrees Celsius?
227. A cube of osmium with a volume of 166 cm3 is placed in fresh water. The cube’s apparent weight is 35.0 N. What is the density of osmium?
Additional Problems
R59
243. The human body cannot survive at a temperature of 42°C for very long. Express this quantity in kelvins.
258. Find the efficiency of an engine that receives 571 J as heat and loses 463 J as heat per cycle.
244. Two sticks rubbed together gain 2.15 x 104 J from kinetic energy and lose 33 percent of it to the air. How much does the sticks’ internal energy change?
259. A 5.4 x 10- 4 m 3 increase in steam’s volume does 1.3 J of work on a piston. What is the pressure?
245. A stone falls 561.7 m. When the stone lands, the internal energy of the ground and the stone increases by 105 J. What is the stone’s mass? 246. A 2.5 kg block of ice at 0.0°C slows on a level floor from 5.7 m is to O m is. If3.3 x 105 J cause 1.0 kg of ice to melt, how much of the ice melts? 247. Placing a 3.0 kg skillet in 5.0 kg of water raises the water’s temperature 2.25°C and lowers the skillet’s temperature 29.6°C. Find the skillet’s specific heat. 248. Air has a specific heat of 1.0 x 103 Jlkg• °C. If air’s temperature increases 55°C when 45 x 106 J are added to it by heat, what is the air’s mass? 249. A 0.23 kg tantalum part has a specific heat capacity of 140 Jlkg•°C. By how much does the part’s temperature change if it gives up 3.0 x 104 J as heat?
Thermodynamics 250. A volume of air increases 0.227 m 3 at a net pressure of 2.07 x 107 Pa. How much work is done on the air? 251 . The air in a hot-air balloon does 3.29 x 106 J of work, increasing the balloon’s volume by 2190 m 3 • What is the net pressure in the balloon? 252. Filling a fire extinguisher with nitrogen gas at a net pressure of25.0 kPa requires 472.5 J of work on the gas. Find the change in the gas’s volume. 253. The internal energy of air in a closed car rises 873 J. How much heat energy is transferred to the air? 254. A system’s initial internal energy increases from 39 J to 163 J. If 114 J of heat are added to the system, how much work is done on the system? 255. A gas does 623 J of work on its surroundings when 867 J are added to the gas as heat. What is the change in the internal energy of the gas? 256. An engine with an efficiency of 0.29 takes in 693 J as heat. How much work does the engine do? 257. An engine with an efficiency of0.19 does 998 J of work. How much energy is taken in by heat?
R60
Appendix I
260. A pressure of 655 kPa does 393 J of work inflating a bike tire. Find the change in volume. 261. An engine’s internal energy changes from 8093 J to 2.0920 x 104 J. If 6932 J are added as heat, how much work is done on or by the system? 262. Steam expands from a geyser to do 192 kJ of work. If the system’s internal energy increases by 786 kJ, how much energy is transferred as heat? 263. If 632 kJ are added to a boiler and 102 kJ of work are done as steam escapes from a safety valve, what is the net change in the system’s internal energy? 264. A power plant with an efficiency of 0.35 percent requires 7.37 x 108 J of energy as heat. How much work is done by the power plant? 265. An engine with an efficiency of 0.11 does 1150 J of work. How much energy is taken in as heat? 266. A test engine performs 128 J of work and receives 581 J of energy as heat. What is the engine’s efficiency?
Vibrations and Waves 267. A scale with a spring constant of 420 N i m is compressed 4.3 cm. What is the spring force? 268. A 669 N weight attached to a giant spring stretches it 6.5 cm. What is the spring constant? 269. An archer applies a force of 52 N on a bowstring with a spring constant of 490 Ni m. What is the bowstring’s displacement? 270. On Mercury, a pendulum 1.14 m long would have a 3.55 s period. Calculate ag for Mercury. 271 . Find the length of a pendulum that oscillates with a frequency of 2.5 Hz.
272. Calculate the period of a 6.200 m long pendulum in Oslo, Norway, where ag = 9.819 m l s2 . 273. Find the pendulum’s frequency in problem 272. 274. A 24 kg child jumps on a trampoline with a spring constant of364 Nim. What is the oscillation period? 275. A 32 N weight oscillates with a 0.42 s period when on a spring scale. Find the spring constant.
276. Find the mass of a ball that oscillates at a period of 0.079 son a spring with a constant of 63 Nim. 277. A dolphin hears a 280 kHz sound with a wavelength of 0.51 cm. What is the wave’s speed? 278. If a sound wave with a frequency of 20.0 Hz has a speed of331 m is, what is its wavelength? 104
m i s and a 279. A sound wave has a speed of2.42 x wavelength of 1.1 m. Find the wave’s frequency. 280. An elastic string with a spring constant of 65 NIm is stretched 15 cm and released. What is the spring force exerted by the string? 281 . The spring in a seat compresses 7.2 cm under a 620 N weight. What is the spring constant? 282. A 3.0 kg mass is hung from a spring with a spring constant of36 Ni m. Find the displacement. 283. Calculate the period of a 2.500 m long pendulum in Quito, Ecuador, where a = 9.780 m l s2 .
295. A 1.53 m long pipe that is closed on one end has a seventh harmonic frequency of 466.2 Hz. What is the speed ofthe waves in the pipe? 296. A pipe open at both ends has a fundamental frequency of 125 Hz. If the pipe is 1.32 m long, what is the speed of the waves in the pipe? 10-3
W. At what 297. Traffic has a power output of 1.57 x distance is the intensity 5.20 x 10- 3 Wlm2? 298. If a mosquito’s buzzing has an intensity of 9.3 x 10-8 Wl m 2 at a distance of0.21 m, how much sound power does the mosquito generate? 299. A note from a flute (a pipe with a closed end) has a first harmonic of 392.0 Hz. How long is the flute if the sound’s speed is 331 m is? 300. An organ pipe open at both ends has a first harmonic of 370.0 Hz when the speed of sound is 331 mis. What is the length of this pipe?
:t>
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
0 C”‘

(‘t)
3 CJ)
8
284. How long is a pendulum with a frequency of 0.50 Hz? 285. A tractor seat supported by a spring with a spring constant of 2.03 x 103 Nim oscillates at a frequency of 0.79 Hz. What is the mass on the spring? 286. An 87 N tree branch oscillates with a period of 0.64 s. What is the branch’s spring constant? 287. What is the oscillation period for an 8.2 kg baby in a seat that has a spring constant of221 Nim? 288. An organ creates a sound with a speed of331 mis and a wavelength of 10.6 m. Find the frequency. 289. What is the speed of an earthquake S-wave with a 2.3 x 104 m wavelength and a 0.065 Hz frequency?
Sound 290. What is the distance from a sound with 5.88 x 10- 5 W power if its intensity is 3.9 x 10-6 Wlm 2? 291 . Sound waves from a stereo have a power output of 3.5 Wat 0.50 m. What is the sound’s intensity? 292. What is a vacuum cleaner’s power output if the sound’s intensity 1.5 m away is 4.5 x 10- 4 Wl m 2 ?
Light and Rellection 301 . A 7.6270 x 108 Hz radio wave has a wavelength of 39.296 cm. What is this wave’s speed? 302. An X ray’s wavelength is 3.2 nm. Using the speed of light in a vacuum, calculate the frequency of the X ray. 303. What is the wavelength of ultraviolet light with a frequency of 9.5 x 10 14 Hz? 304. A concave mirror has a focal length of 17 cm. Where must a 2. 7 cm tall coin be placed for its image to appear 23 cm in front of the mirror’s surface? 305. How tall is the coin’s image in problem 304? 306. A concave mirror’s focal length is 9.50 cm. A 3.0 cm tall pin appears to be 15.5 cm in front of the mirror. How far from the mirror is the pin? 307. How tall is the pin’s image in problem 306? 308. A convex mirror’s magnification is 0.11. Suppose you are 1.75 m tall. How tall is your image? 309. How far in front of the m irror in problem 308 are you if your image is 42 cm behind the mirror?
293. Waves travel at 499 m is on a 0.850 m long cello string. Find the string’s fundamental frequency.
310. A mirror’s focal length is – 12 cm. What is the object distance if an image forms 9.00 cm behind the surface of the mirror?
294. A mandolin string’s first harmonic is 392 Hz. How long is the string if the wave speed on it is 329 m l s?
311. What is the magnification in problem 310? 312. A metal bowl is like a concave spherical mirror. You are 35 cm in front of the bowl and see an image at 42 cm. What is the bowl’s focal length?
Additional Problems
R61
313. For problem 312, find the bowl’s radius of curvature. 314. A concave spherical mirror on a dressing table has a focal length of 60.0 cm. If someone sits 35.0 cm in front ofit, where is the image? 315. What is the magnification in problem 314? 316. An image appears 5.2 cm behind the surface of a convex mirror when the object is 17 cm in front of the mirror. What is the mirror’s focal length?
329. How fast does microwave radiation that has a frequency of 1. 173 06 x 1011 Hz and a wavelength of 2.5556 mm travel? 330. Suppose the microwaves in your microwave oven have a frequency of 2.5 x 10 10 Hz. What is the wavelength of these microwaves?
317. If the object in problem 316 is 3.2 cm tall, how tall is its image?
331. You place an electric heater 3.00 min front of a concave spherical mirror that has a focal length of 30.0 cm. Where would your hand feel warmest?
318. In order for someone to observe an object, the wavelength of the light must be smaller than the object. Th e Bohr radius of a hydrogen atom is 5.291 770 x 10- 11 m. What is the lowest frequency that can be used to locate a hydrogen atom?
332. You see an image of your hand as you reach for a doorknob with a focal length of 6.3 cm. How far from the doorknob is your hand when the image appears at 5.1 cm behind the doorknob?
319. Meteorologists use Doppler radar to watch the movement of storms. If a weather station uses electromagnetic waves with a frequency of2.85 x 109 Hz, what is the wavelength of the radiation?
333. What is the magnification of the image in problem 332?
Refraction
320. PCS cellular phones have antennas that use radio frequencies from 1800 to 2000 MHz. What range of wavelengths corresponds to these frequencies?
334. A ray of light in air enters an amethyst crystal (n = 1.553). If the angle ofrefraction is 35°, what is the angle of incidence?
321. Suppose you have a mirror with a focal length of 32.0 cm. Where would you place your right hand so that you appear to be shaking hands with yourself?
335. Light passes from air at an angle of incidence of 59.2° into a nephrite jade vase (n = 1.61). Determine the angle of refraction in the jade.
322. A car’s headlamp is made of a light bulb in front of a concave spherical mirror. If the bulb is 5.0 cm in front of the mirror, what is the radius of the mirror?
336. Light entering a pearl travels at a speed of 1.97 x 108 m/s. What is the pearl’s index of refraction?
323. Suppose you are 19 cm in front of the bell of your friend’s trumpet and you see your image at 14 cm. If the trumpet’s bell is a concave mirror, what would be its focal length? 324. A soup ladle is like a spherical convex mirror with a focal length of 27 cm. If you are 43 cm in front of the ladle, where does the image appear? 325. What is the magnification in problem 324? 326. Just after you dry a spoon, you look into the convex part of the spoon. If the spoon has a focal length of – 8.2 cm and you are 18 cm in front of the spoon, where does the image appear? 327. The base of a lamp is made of a convex spherical mirror with a focal length of – 39 cm. Where does the image appear when you are 16 cm from the base?
R62
328. Consider the lamp and location in problem 327. If your nose is 6.0 cm long, how long does the image appear?
Appendix I
337. An object in front of a diverging lens of focal length 13.0 cm forms an image with a magnification of +5.00. How far from the lens is the object placed? 338. An object with a height of 18 cm is placed in front of a converging lens. The image height is -9.0 cm. What is the magnification of the lens? 339. If the focal length of the lens in problem 338 is 6.0 cm, how far in front of the lens is the object? 340. Where does the image appear in problem 339? 341. The critical angle for light traveling from a green tourmaline gemstone into air is 37.8°. What is tourmaline’s index of refraction? 342. Find the critical angle for light traveling from ruby (n = 1.766) into air. 343. Find the critical angle for light traveling from emerald (n =1.576) into air.
344. Malachite has two indices ofrefraction: n 1 = 1.91 and n2 = 1.66. A ray of light in air enters malachite at an incident angle of35.2°. Calculate both of the angles of refraction. 345. A ray oflight in air enters a serpentine figurine (n = 1.555). If the angle ofrefraction is 33°, what is the angle of incidence? 346. The critical angle for light traveling from an aquamarine gemstone into air is 39.18°. What is the index of refraction for aquamarine? 347. A 15 cm tall object is placed 44 cm in front of a diverging lens. A virtual image appears 14 cm in front of the lens. What is the lens’s focal length? 348. What is the image height in problem 347? 349. A lighthouse converging lens has a focal length of 4 m. What is the image distance for an object placed 4 min front of the lens? 350. What is the magnification in problem 349? 351 . Light moves from olivine (n = 1.670) into onyx. If the critical angle for olivine is 62.85°, what is the index of refraction for onyx? 352. When light in air enters an opal mounted on a ring, the light travels at a speed of 2.07 x 108 m/ s. What is opal’s index of refraction? 353. When light in air enters albite, it travels at a velocity of 1.95 x 108 m/s. What is albite’s index ofrefraction? 354. A searchlight is constructed by placing a 500 W bulb 0.5 m in front of a converging lens. The focal length of the lens is 0.5 m. What is the image distance? 355. A microscope slide is placed in front of a converging lens with a focal length of 3.6 cm. The lens forms a real image of the slide 15.2 cm behind the lens. How far is the lens from the slide? 356. Where must an object be placed to form an image 12 cm in front of a diverging lens with a focal length of44cm? 357. The critical angle for light traveling from almandine garnet into air ranges from 33.1° to 35.3°. Calculate the range of almandine garnet’s index of refraction. 358. Light moves from a clear andalusite (n = 1.64) crystal into ivory. If the critical angle for andalusite is 69.9°, what is the index of refraction for ivory?
360. Light passing through two slits with a separation of 8.04 x 10-6 m forms a third bright fringe 13.1° from the center. Find the wavelength. 361 . Two slits are separated by 0.0220 cm. Find the angle at which a first-order bright fringe is observed for light with a wavelength of527 nm. 362. For 546.1 nm light, the first-order maximum for a diffraction grating forms at 75.76°. How many lines per centimeter are on the grating?
:t>
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
363. Infrared light passes through a diffraction grating of 3600 lines/cm. Th e angle of the third-order maximum is 76.54°. What is the wavelength?
0 C”‘
364. A diffraction grating with 1950 lines/ cm is used to examine light with a wavelength of 497.3 nm. Find the angle of the first-order maximum.
(‘t)
3 CJ)
365. At what an gle does the second-order maximum in problem 364 appear? 366. Light passes through two slits separated by 3.92 x 10-6 m to form a second-order bright fringe at an angle of 13.1 °. What is the light’s wavelength?
367. Light with a wavelength of 430.8 nm shines on two slits that are 0.163 mm apart. What is the angle at which a second dark fringe is observed? 368. Light of wavelength 656.3 nm passes through two slits. The fourth-order dark fringe is 0.548° from the central maximum . Find the slit separation.
369. The first-order maximum for light with a wavelength of 447.1 nm is found at 40.25°. How many lines per centimeter does the grating have? 370. Light through a diffraction grating of9550 lines/ cm forms a second-order maximum at 54.58°. What is the wavelength of the light?
Electric Forces and Fields 371 . Charges of – 5.3 µC and +5.3 µCare separated by 4.2 cm. Find the electric force between them.
372. A dog’s fur is combed, and the comb gains a charge of 8.0 nC. Find the electric force between the fur and comb when they are 2.0 cm apart. 373. Two equal charges are separated by 6.5 x 10- 11 m . If the magnitude of the electric force between the charges is 9.92 x 10- 4 N, what is the value of q?
Interference and Diffraction 359. Light with a 587 .5 nm wavelength passes through two slits. A second-order bright fringe forms 0.130° from the center. Find the slit separation.
Additional Problems
R63
374. Two point charges of -13.0 µC and -16.0 µC exert repulsive forces on each other of 12.5 N. What is the distance between the two charges? 375. Three equal point charges of 4.00 nC lie 4.00 m apart on a line. Calculate the magnitude and direction of the net force on the middle charge. 376. A proton is at each corner of a square with sides 1.52 x 10-9 m long. Calculate the resultant force vector on the proton at the upper right corner. 377. Three 2.0 nC charges are located at coordinates (0 m, 0 m), (1.0 m, 0 m), and (1.0 m, 2.0 m). Find the resultant force on the first charge.
390. What electric charge experiences an 8.42 x 10- 9 N electric force in an electric field of 1663 N/ C? 391. Two 3.00 µC charges lie 2.00 m apart on the x-axis. Find the resultant electric field vector at a point 0.250 m on the y-axis, above the charge on the left. 392. Two electrons are 2.00 x 10- 10 m and 3.00 x 10- 10 m, respectively, from a point. Where with respect to that point must a proton be placed so that the resultant electric field strength is zero?
378. Charges of 7.2 nC and 6. 7 nC are 32 cm apart. Find the equilibrium position for a – 3.0 nC charge.
393. A -7.0 C charge is in equilibrium with a 49 C charge 18 m to the right and an unknown charge 25 m to the right. What is the unknown charge?
379. A -12.0 µC charge is between two 6.0 µC charges, 5.0 cm away from each. What electric force keeps the central charge in equilibrium?
394. Suppose two pions are separated by 8.3 x 10- 10 m . If the magnitude of the electric force between the charges is 3.34 x 10- 10 N, what is the value of q?
380. A 9.0 N/ C electric field is directed along the x-axis. Find the electric force vector on a -6.0 C charge.
395. Suppose two muons having equal but opposite charge are separated by 6.4 x 1o-8 m . If the magnitude of the electric force between the charges is 5.62 x 10- 14 N, what is the value of q?
381. What charge experiences an electric force of 6.43 x 10-9 Nin an electric field of 4.0 x 103 N/ C? 382. A 5.00 µC charge is 0.500 m above a 15.0 µC charge. Calculate the electric field at a point 1.00 m above the 15.0 mC charge. 383. Two static point charges of99.9 µC and 33.3 µC exert repulsive forces on each other of87.3 N. What is the distance between the two charges? 384. Two particles are separated by 9.30 x 10- 11 m. If the magnitude of the electric force between the charges is 2.66 x 10-8 N, what is the value of q? 385. A -23.4 nC charge is 0.500 m below a 4.65 nC charge and 1.00 m below a 0.299 nC charge. Find the resultant force vector on the – 23.4 nC charge. 386. Three point charges are on the corners of a triangle: q 1 = – 9.00 nC is at the origin; q2 = – 8.00 nC is at x = 2.00 m ; and q3 = 7.00 nC is aty = 3.00 m. Find the magnitude and direction of the resultant force on qr 387. Charges of – 2.50 nC and -7.50 nC are 20.0 cm apart. Find a 5.0 nC charge’s equilibrium position.
388. A – 4.6 C charge is in equilibrium with a – 2.3 C charge 2.0 m to the right, and an unknown charge 4.0 m to the right. What is the unknown charge?
R64
389. Find the electric force vector on a 5.0 nC charge in a 1500 N/C electric field directed along the y-axis.
Appendix I
396. Consider four electrons at the corners of a square. Each side of the square is 3.02 x 10-s m . Find the magnitude and direction of the resultant force on q3 if it is at the origin. 397. A charge of 5.5 nC and a charge of 11 nC are separated by 88 cm. Find the equilibrium position for a -22 nC charge. 398. Three charges are on the y-axis. At the origin is a charge, q 1 = 72 C; an unknown charge, q2 , is at y = 15 mm. A third charge, q 3 = – 8.0 C, is placed at y = – 9.0 mm so that it is in electrostatic equilibrium with q 1 and q2 . What is the charge on q2 ?
Electrical Energy and Current 399. A helium-filled balloon with a 14.5 nC charge rises 290 m above Earth’s surface. By how much does the electrical potential energy change if Earth’s electric field is – 105 N/C? 400. A charged airplane rises 7.3 km in a 3.4 x 105 N/C electric field. The electrical potential energy changes by – 1.39 x 1011 J. What is the charge on the plane?
401. Earth’s radius is 6.4 x 106 m. What is Earth’s capacitance if it is regarded as a conducting sphere? 402. A 0.50 pF capacitor is connected across a 1.5 V battery. How much charge can this capacitor store? 403. A 76 C charge passes through a wire’s cross-sectional area in 19 s. Find the current in the wire. 404. The current in a telephone is 1.4 A. How long does 98 C of charge take to pass a point in the wire? 405. What is a television’s total resistance if it is plugged into a 120 V outlet and carries 0.75 A of current? 406. A motor with a resistance of 12.2 n is plugged into a 120.0 V outlet. What is the current in the motor? 407. The potential difference across a motor with a 0.30 n resistance is 720 V. How much power is used? 408. What is a microwave oven’s resistance if it uses 1750 W of p ower at a voltage of 120.0 V? 409. A 64 nC charge moves 0.95 m with an electrical potential energy change of – 3.88 x 10- 5 J. What is the electric field strength?
420. The potential difference across an electric eel is 650 V. How much current would an electric eel deliver to a body with a resistance of 1.0 x 102 D? 421. If a garbage-disposal motor has a resistance of 25.0 n and carries a current of 4.66 A, what is the potential difference across the motor’s terminals? 422. A medium-sized oscillating fan draws 545 mA of current when the potential difference across its motor is 120 V. How large is the fan’s resistance? 4
416. A 114 µC charge passes through a gold wire’s crosssectional area in 0.36 s. What is the current? 417. If the current in a blender is 7.8 A, how long do 56 C of charge take to pass a point in the circuit? 418. A computer uses 3.0 A in 2.0 min. How much charge passes a point in the circuit in this time? 419. A battery-powered lantern has a resistance of 6.4 n. What potential difference is provided by the battery if the total current is 0.75 A?
:l
-“tJ Ql
1111111:
(‘t)
425. A laser uses 6.0 x 10 13 W of power. What is the potential difference across the laser’s circuit if the current in the circuit is 8.0 x 106 A?
3 CJ)
426. A blender with a 75 n resistance uses 350 W of power. What is the current in the blender’s circuit?
411. A 5.0 x 10-5 F polyester capacitor stores 6.0 x 10-4 C. Find the potential difference across the capacitor.
415. A photocopier uses 9.3 A in 15 s. How much charge passes a point in the copier’s circuit in this time?
-· 0
424. A computer with a resistance of 91.0 n uses 230.0 W of power. Find the current in the computer.
427. A theater has 25 surround-sound speakers wired in series. Each speaker has a resistance of 12.0 n. What is the equivalent resistance?
414. A television receiver contains a 14 µF capacitor charged across a potential difference of 1.5 x 104 V. How much charge does this capacitor store?
-· …..
0 C”‘
Circuits and Circuit Elements
413. The area of the plates in a 4550 pF parallel-plate capacitor is 6.4 x 10-3 m 2. Find the plate separation.
0. 0..
423. A generator produces a 2.5 x 10 V potential difference across power lines that carry 20.0 A of current. How much power is generated?
410. A -14 nC charge travels through a 156 N/ C electric field with a change of 2.1 x 10- 6 J in the electrical potential energy. How far does the charge travel?
412. Some ceramic capacitors can store 3 x 10-2 C with a potential difference of 30 kV across them. What is the capacitance of such a capacitor?
:t>
428. In case of an emergency, a corridor on an airplane has 57 lights wired in series. Each light bulb h as a resistance of2.00 n. Find the equivalent resistance. 429. Four resistors with resistances of 39 n, 82 n, 12 n, and 42 n are connected in parallel across a 3.0 V potential difference. Find the equivalent resistance. 430. Four resistors with resistances of33 n, 39 n, 47 n, and 68 n are connected in parallel across a 1.5 V potential difference. Find the equivalent resistance. 431. A 16 n resistor is connected in series with another resistor across a 12 V battery. The current in the circuit is 0.42 A. Find the unknown resistance. 432. A 24 n resistor is connected in series with another resistor across a 3.0 V battery. The current in the circuit is 62 mA. Find the unknown resistance. 433. A 3.3 n resistor and another resistor are conn ected in parallel across a 3.0 V battery. The current in the circuit is 1.41 A. Find the unknown resistance.
Additional Problems
R65
434. A 56 n resistor and another resistor are connected in parallel across a 12 V battery. The current in the circuit is 3.21 A. Find the unknown resistance.
445. For the figure above, what is the current in the 3.0 n resistors? 8.on
2.on
435. Three bulbs with resistances of 56 n, 82 n, and 24 n are wired in series. If the voltage across the circuit is 9.0 V, what is the current in the circuit?
24.0V
436. Three bulbs with resistances of 96 n, 48 n, and 29 n are wired in series. What is the current through the bulbs if the voltage across them is 115 V? 437. A refrigerator (R 1 = 75 !1) wired in parallel with an oven (R2 = 91 !1) is plugged into a 120 V outlet. What is the current in the circuit of each appliance? 438. A computer (R1 = 82 n) and printer (R2 = 24 n) are wired in parallel across a 120 V potential difference. Find the current in each machine’s circuit. 5.on
3.on
5.o n
446. For the figure above, calculate the equivalent resistance of the circuit. 447. For the figure above, what is the total current in the circuit? 448. For the figure above, what is the current in either of the 8.0 n resistors?
Magnetism
1.5n
449. A proton moves at right angles to a magnetic field of 0.8 T. If the proton’s speed is 3.0 x 107 m/s, how large is the magnetic force exerted on the proton?
12.0V 439. For the figure above, what is the equivalent resistance of the circuit? 440. For the figure above, find the current in the circuit. 441 . For the figure above, what is the potential difference across the 6.0 n resistor? 442. For the figure above, what is the current through the 6.0 n resistor?
3.on
8.on
15.0V
443. For the figure above, calculate the equivalent resistance of the circuit. 444. For the figure above, what is the total current in the circuit?
450. A weak magnetic field exerts a 1.9 x 10-22 N force on an electron moving 3.9 x 106 m /s perpendicular to the field. What is the magnetic field strength? 451. A 5.0 x 10-s T magnetic field exerts a 6.1 x 10- 17 N force on a 1.60 x 10- 19 C charge, which moves at a right angle to the field. What is the charge’s speed? 452. A 14 A current passes through a 2 m wire. A 3.6 x 10-4 T magnetic field is at right angles to the wire. What is the magnetic force on the wire? 453. A 1.0 m printer cable is perpendicular to a 1.3 x 10-4 T magnetic field. What current must the cable carry to experience a 9.1 x 10-5 N magnetic force? 454. A wire perpendicular to a 4.6 x 10-4 T magnetic field experiences a 2.9 x 10- 3 N magnetic force. How long is the wire if it carries a 10.0 A current? 455. A 12 m wire carries a 12 A current. What magnetic field causes a 7.3 x 10-2 N magnetic force to act on the wire when it is perpendicular to the field? 456. A magnetic force of 3.7 x 10- 13 N is exerted on an electron moving at 7.8 x 106 m /s perpendicular to a sunspot. How large is the sunspot’s magnetic field?
R66
Appendix I
457. An electron moves with a speed of2.2 x 106 mis at right angles through a 1.1 x 10-2 T magnetic field. How large is the magnetic force on the electron?
471. A step-up transformer converts a 4.9 kV voltage to 49 kV. If the secondary (output) coil has 480 turns, how many turns does the primary have?
458. A pulsar’s magnetic field is 1 x 10-8 T. How fast does an electron move perpendicular to this field so that a 3.2 x 10- 22 N magnetic force acts on the charge?
472. A 320-turn coil rotates from 0° to 90.0° in a 0.046 T magnetic field in 0.25 s, which induces an average emf of 4.0 V. What is the area of the coil?
459. A levitation device designed to suspend 75 kg uses 10.0 m of wire and a 4.8 x 10-4 T magnetic field, perpendicular to the wire. What current is needed? 103
460. A power line carries 1.5 x A for 15 km. Earth’s magnetic field is 2.3 x 1o- 5 T at a 45° angle to the power line. What is the magnetic force on the line?
10-5
m2
473. A 180-turn coil with a 5.0 x area is in a magnetic field that decreases by 5.2 x 10- 4 Tin 1.9 x 10- 5 s. What is the induced current if the coil’s resistance is 1.0 x 102 W? 474. A generator provides a maximum ac current of 1.2 A and a maximum output emf of 211 V. Calculate therms potential difference. 475. Calculate therms current for problem 474.
Electromagnetic Induction 461. A coil with 540 turns and a 0.016 m 2 area is rotated exactly from 0° to 90.0° in 0.050 s. How strong must a magnetic field be to induce an emf of 3.0 V? 462. A 550-turn coil with an area of 5.0 x 10-5 m 2 is in a magnetic field that decreases by 2.5 x 10-4 Tin 2.1 x 10-5 s. What is the induced emf in the coil? 463. A 246-turn coil has a 0.40 m 2 area in a magnetic field that increases from 0.237 T to 0.320 T. What time interval is needed to induce an emf of – 9.1 V? 464. A 9.5 V emf is induced in a coil that rotates from 0.0° to 90.0° in a 1.25 x 10-2 T magnetic field for 25 m s. The coil’s area is 250 cm2 . How many turns of wire are in the coil? 465. A generator provides an rms emf of 320 V across 100 n. What is the maximum emf? 466. Find the rms current in the circuit in problem 465. 467. Some wind turbines can provide an rms current of 1.3 A. What is the maximum ac current? 468. A transformer has 1400 turns on the primary and 140 turns on the secondary. What is the voltage across the primary if secondary voltage is 6.9 kV? 469. A transformer has 140 turns on the primary and 840 turns on the secondary. What is the voltage across the secondary if the primary voltage is 5.6 kV? 470. A step-down transformer converts a 3.6 kV voltage to 1.8 kV. If the primary (input) coil has 58 turns, how many turns does the secondary have?
476. A generator can provide a maximum output emf of 170 V. Calculate the rms potential difference.
:t>
0. 0..
-· …..
-· 0 :l
-“tJ Ql
1111111:
0 C”‘

(‘t)
3 CJ)
477. A step-down transformer converts 240 V across the primary to 5.0 V across the secondary. What is the step-down ratio (N1:N 2 )?
Atomic Physics 478. Determine the energy of a photon of green light with a wavelength of 527 nm. 479. Calculate the de Broglie wavelength of an electron with a velocity of2.19 x 106 m / s. 480. Calculate the frequency of ultraviolet (UV) light having a photon energy of20.7 eV. 481. X-ray radiation can have an energy of 12.4 MeV. To what wavelength does this correspond? 482. Light of wavelength 240 nm shines on a potassium surface. Potassium has a work function of 2.3 eV. What is the maximum kinetic energy of the photoelectrons? 483. Manganese has a work function of 4.1 eV. What is the wavelength of the photon that will just have the threshold energy for manganese? 484. What is the speed of a proton with a de Broglie wavelength of2.64 x 10- 14 m? 485. A cheetah can run as fast as 28 m / s. If the cheetah has a de Broglie wavelength of 8.97 x 10-37 m, what is the cheetah’s mass? 486. What is the energy of a photon of blue light with a wavelength of 430.8 nm?
Additional Problems
R67
487. Calculate the frequency of infrared (IR) light with a photon energy of 1.78 eV. 488. Calculate the wavelength of a radio wave that has a photon energy of 3.1 x 10-6 eV. 489. Light of frequency 6.5 x 10 14 Hz illuminates a lithium surface. The ejected photoelectrons are found to have a maximum kinetic energy of 0.20 eV. Find the threshold frequency of this metal. 490. Light of wavelength 519 nm shines on a rubidium surface. Rubidium has a work function of 2.16 eV. What is the maximum kinetic energy of the photoelectrons? 491. The smallest known virus moves across a Petri dish at 5.6 x 10-6 mis. If the de Broglie wavelength of the virus is 2.96 x 10- 8 m, what is the virus’s mass? 492. The threshold frequency of platinum is 1.36 x 1015 Hz. What is the work function of platinum? 493. The ship Queen Elizabeth II h as a mass of 7.6 x 107 kg. Calculate the de Broglie wavelength if this ship sails at 35m/s. 494. Cobalt has a work function of 5.0 eV. What is the wavelength of the photon that will just have the threshold energy for cobalt? 495. Light of frequency 9.89 x 1014 Hz illuminates a calcium surface. The ejected photoelectrons are found to have a maximum kinetic energy of 0.90 eV. Find the threshold frequency of this metal. 496. What is the speed of a neutron with a de Broglie wavelength of 5.6 x 10- 14 m?
Subatomic Physics 497. Calculate the binding energy of
I~ K .
498. Determine the difference in the binding energy 63 Cu of 107 Ag and 47 29 .
R68
Appendix I
499. Find the mass defect of ~~ Ni. 500. Complete this radioactive-decay formula: 2Jf Po—+?+
iHe.
501. Complete this radioactive-decay formula: l~N—+? + j e + v. 502. Complete this radioactive-decay formula: 147 Sm—+ 143Nd +? 62 60 · 503. A 3.29 x 10-3 g sample of a pure radioactive substance is found after 30.0 s to have only 8.22 x 10- 4 g left undecayed. What is the half-life of the substance?
i~Cr
is 21.6 h. A chromium-48 sample 504. The half-life of contains 6.5 x 106 nuclei. Calculate the activity of the sample in mCi. 505. How long will it take a sample oflead-212 (which has a half-life of 10.64 h) to decay to one-eighth its original strength?
gg Sn.
506. Compute the binding energy of 1
507. Calculate the difference in the binding energy of Ile and 1~o. 508. What is the mass defect of ~6Zn? 509. Complete this radioactive-decay formula: ?. —+ 13lxe + – 01 e + v. 54 510. Complete this radioactive-decay formula: 160w _ I56m + 1 74 72 . 511 . Complete this radioactive-decay formula : 1 ? —+ R~Te +
iHe.
512. A 4.14 x 10-4 g sample of a pure radioactive substance is found after 1.25 days to have only 2.07 x 10-4 g left undecayed. What is the substance’s half-life? 513. How long will it take a sample of cadmium-109 with a half-life of 462 days to decay to one-fourth its original strength? 514. The half-life of ~gFe is 2.7 years. What is the decay constant for the isotope?
The Science ol Physics PRACTICE A 1. 5 X 10- 5 m 3. a. 1 x 10- 8 m b. 1 x 10- 5 m m C. 1 X 10- 2 µm 5. 1.440 x 103 kg
PRACTICE B 1. 2.2 s 3. 5.4s 5. a. l.4m/ s b. 3.1 m / s
PRACTICEC
1. 21 m 3. 9. 1 s
PRACTICED 1 REVIEW 11. a. 2 x 102 mm b. 7.8 x 103 s C. 1.6 X 107 µg d. 7.5 x 104 cm 8. 6.75 X 10- 4 g f. 4.62 x 10- 2 cm g. 9.7 m / s 13. 1.08 X 109 km 19. a. 3
b. 4 c. 3 d. 2 21. 228.8 cm 23. b, c 29. 4 x 108 breaths 31. 5.4 X 108 s 33. 2 x 103 balls 35. 7 x 102 tuners 37. a. 22 cm ; 38 cm 2 b. 29.2 cm ; 67.9 cm 2 39. 9.818 X 10- 2 m 41. The ark (6 x 104 m 3) was about 100 time s as large as a typical h ou se (6 x 102 m 3). 43. 1.0 x 103 kg 45. a. 0.618 g/ cm 3 b. 4.57 x 1016 m 2
Motion in One Dimension PRACTICE A
1. 9.8 m / s; 29 m 3. – 7.5 m / s; 19 m
PRACTICE E 1. +2.5lm/ s 3. a. 16 m / s b. 7.0 s 5. + 2.3 m / s2
PRACTICE F 1. a. – 42 m / s b. 11 s 3. a. 8.0 m / s b. 1.63 s
2 REVIEW 1. 5.0 m ; + 5.0 m 3. t 1 : negative; t2 : positive; t3 : positive; ti n egative; t5 : zero 7. 10.1 km to the east 9. a. +70.0m b. + 140.0 m c. + 14 m / s d. +28 m / s 11. 0.2 km west of th e flagpole 17. 0.0 m / s2; + 1.36 m / s2; + 0.680 m / s2 19. 110m 21. a. – 15 m / s b. – 38 m 23. 17.5 m 25. 0.99 m / s 31. 3.94 s
33. 1.51 h 35. a. 2.00 m in b. l.00 min c. 2.00 min 37. 931 m 39. – 26 m / s; 31 m 41. 1.6 s 43. 5 s; 85 s; +60 m / s 45. – 1.5 x 103 m / s2 47. a. 3.40 s b. – 9.2 m / s c. – 31.4 m / s; – 33 m / s 49. a. 4.6 s after stock car starts b. 38 m c. +17 m / s (stockcar), +21 m / s (race car) 51. 4.44 m / s
Two-Dimensional Motion and Vectors PRACTICE A 1. a. 23 km b. 17 km to th e east 3. 15.7 m at 22° to the side of downfield
PRACTICE B 1. 95 km/ h 3. 21 m / s, 5.7 m / s
PRACTICE C 1. 49 m at 7 .3° to the right of downfield 3. 13.0 m at 57° n orth of east
PRACTICED 1. 0.66 m /s 3. 7.6 m / s
PRACTICE E
1. yes, ~ y
=-
2.3 m
3. 2.0 s; 4.8 m
1. 2.0 km to the east 3. 680 m to the n orth 5. 0.43 h
Selected Answers
R69
PRACTICE F 1. 0 m / s 3. 3.90 m is at (4.0 x 101) north of east 0
3 REVIEW 7. a. 5.20 mat 60.0° above the positive x -axis b. 3.00 mat 30.0° below the positive x-axis c. 3.00 mat 150° counterclockwise from the positive x-axis d. 5.20 mat 60.0° below the positive x -axis 9. 15.3 mat 58.4° south of east 19. if the vector is oriented at 45° from the axes 21. a. 5 blocks at 53° north of east b. 13 blocks 23. 61.8 mat 76.0° S ofE (or S of W), 25.0 mat 53.1° S ofE (or S ofW) 25. 2.81 km east, 1.31 km north 31. 45.1 m is 33. 11 m 35. a. clears the goal by 1 m b. falling 37. 80 m; 210 m 41. a. 70 m is east b. 20 m / s 43. a. 10.1 mi s at 8.53° east of north b. 48.8m 45. 7.5 min 47. a. 41.7 m / s b. 3.81 s c. vy,f = – 13.5 m / s, vx,f = 34.2 m / s, v1 = 36.7 mi s 49. 10.5 m is 51. a. 2.66m/s b. 0.64 m 53. 157 km
R70
Selected Answers
55. a. 32.5 m b. 1.78 s 57. a. 57.7km/ hat60.0° westof the vertical b. 28.8 km/ h straight down 59. 18 m; 7.9 m 61. 6.19 m / s downfield
Forces and the Laws ol Motion PRACTICE B 1. Fx = 60.6 N; FY = 35.0 N 3. 557 N at 35.7° west of north
29. 35. 37. 39. 41. 43. 45.
47.
49. 51. 53.
51 N 0.70, 0.60 0.816 1.0 m / s 2 13 N down the incline 64Nupward a. 0.25 m / s2 forward b. 18m c. 3.0 mi s a. 2 s b. The box will never move. The force exerted is not enough to overcome friction. – 1.2 m / s2; 0.12 a. 2690 N forward b. 699 N forward 13 N, 13 N, 0 N, -26 N
PRACTICE C 1. 2.2 m / s2 forward 3. 4.50 m / s 2 to the east 5. 14N
Work and Energy PRACTICE A
PRACTICED 1. 0.23 3. a. 8.7 X 102 N, 6.7 X 102 N b. 1.1 x 102 N, 84N 2 C. 1 X 103 N, 5 X 10 N d. SN, 2N
PRACTICE E 1. 2. 7 m / s2 in the positive x direction 3. a. 0.061 b. 3.61 m / s2 down the ramp
4 REVIEW 11. a. Fl (220N)andF2(114 N) both point right; F1 (220 N) points left, and F2 (114 N) points right. b. first situation: 220 N to the right, 114 N to the right; second situation: 220 N to the left, 114 N to the right 21. 55 N to the right
1. 1.50 X 107 J 3. 1.6 X 103
PRACTICE B 1. 1.7 x 102 m / s 3. the bullet with the greater mass; 2 to 1 5. 1.6 x 103 kg
PRACTICE C
1. 7.8m 3. 5.lm
PRACTICED 1. 3.3 J 3. a. 785 J b. 105 J c. 0.00 J
PRACTICE E 1. 20.7m/s 3. 14.1 m / s 5. 0.18 m
PRACTICE F 1. 66kW 3. 2.61 x 108 s (8.27 years) 5, a, 7.50 X 104 J b. 2.50 x 104 W
PRACTICE C 1. 5.33 s; 53.3 m to the west 3. a. 1.22 x 104 N to the east b. 53.3 m to the west
PRACTICED 5 REVIEW 7. 53 J, -53 J 9. 47.5 J 19. 7.6 X 104 J 21. 2.0 X 101 m 23. a. 5400 J, o J; 5400 J b. o J, – 5400 J; 5400 J c. 2700 J, -2700 J; 5400 J 33. 12.0 mis 35. 17.2 s 37. a. 0.633 J b. 0.633 J c. 2.43 mi s d. 0.422 J, 0.211 J 39. 5.0 m 41 . 2.5 m 45. a. 61 J b. – 45 J c. 0 J 47. a. 28.0 mis b. 30.0 m above the ground 49. 0.107 51. a. 66 J b. 2.3 m is c. 66 J d. – 16 J
Momentum and Collisions PRACTICE A 1. 2.5 x 103 kg•mls to the south 3. 46 mis to the east
PRACTICE B 1. 3.8 x 102 N to the left 3. 16 kg•mls to the south
1. 1.90 mis 3. a. 12.0 m is b. 9.6 mis PRACTICE E 1. 3.8 mis to the south 3. 4.25 mi s to the north 5. a. 3.0kg b. 5.32 m is
37. a. 0.0 kg•mls b. 1.1 kg•ml s upward 39. 23 mi s 41. 4.0 X 102 N 43. 2.36 X 10- 2 m 45. 0.413 47. -22 cmls, 22 cmls 49. a. 9.9 mis downward b. 1.8 x 103 N upward
Circular Motion and Gravitation PRACTICE A
PRACTICE F 1. a. 0.43 mis to the west b. 17 J 3. a. 4.6 mis to the south b. 3.9 x 103 J
1. 2.5 mis 3. l.5ml s2
PRACTICE B 1. 29.6 kg 3. 40.0N
PRACTICE G 1. a. 22.5 cmls to the right b. KE;= 6.2 x 10- 4 J = KE! 3. a. 8.0 mi s to the right b. KE; = 1.3 x 102 J = KE!
PRACTICE C 1. 0.692m 3. a. 651 N b. 246N c. 38.5 N
6 REVIEW 11. a. 8.35 x 10- 21 kg•mls
13.
23. 29. 31. 33. 35.
upward b. 4.88 kg•ml s to the right c. 7.50 x 102 kg•mls to the southwest d. 1.78 x 1029 kg•ml s forward 18 N 0.037 mi s to the south 3.00 mis a. 0.81 mis to the east b. 1.4 X 103 J 4.0 mi s 42.0 mis toward second base
PRACTICED 1. Earth: 7.69 x 103 m is, 5.51 x 103 s; Jupiter: 4.20 x 104 mis, 1.08 x 104 s; moon: 1.53 x 103 mis, 8.63 X 103 S
PRACTICE E 1. 0.75N•m 3. 133 N
7 REVIEW 9. 2.7 mi s 11. 62 kg 19. 1.0 x 10-
10
m (0.10 nm)
Selected Answers
R71
27. vt = 1630 m i s; T = 5.78 X 105 s 29. Ju piter (m = 1.9 x 1027 kg) 33. F2 37. 26N• m 39. 12 m i s 41. 220 N 43. 1800 N• m 45. 2.0 X 102 N 47. 72% 49. a. 2.25 days b. 1.60 x 104 m i s 51. a. 6300 Nern b. 550N 53. 6620 N; no (Fe = 7880 N)
Heat
Fluid Mechanics
9 REVIEW
PRACTICE A
1. a. 3.57 x 103 kgl m 3 b. 6.4 x 102 kgl m 3 3. 9.4
X 103
N
PRACTICE B 1. a. b. 3. a. b.
1.48 X 103 N 1.88 x 105 Pa 1.2 x 103 Pa 6.0 x 10- 2 N
8 REVIEW 9. 15. 21. 23. 25. 27.
29. 31. 33. 35.
37.
R72
2.1 X 103 kgl m 3 6.28 N 1.01 X 10 11 N 6.11 x 10- 1 kg 17N,31 N a. 1.0 x 103 kgl m 3 b. 3.5 x 102 Pa C. 2.1 X 103 Pa 1.7 X 10- 2 m 0.605 m 6.3 m a. 0.48 m l s 2 b. 4.0 s 1.7 X 10- 3 ffi
Selected Answers
PRACTICE C
1. 0.1504 3. a. 0.247 b. 4.9 x 104 J
PRACTICE A 1. -89.22°C, 183.93 K 3. 37.0°C, 39°C 5. -195.81 °C, – 320.5°F
PRACTICE B 1. 755 J 3. 0.96 J
PRACTICE C 1. 47°c 3. 390 J/kg• °C
9. 57.8°C, 331.0 K 25. a. 2.9 J b. It goes into the air, the ground, and the hammer. 31. 25.0°C 33. a. TR = Tp + 459.7, or TF = TR – 459.7 5 9 b. T = gTR orTR= T 5 3 35. a. Trn = zTc+50,or
5. 755 J
10 REVIEW 3. b, c, d , e 9. 1.08 x 10 3 J; done by th e gas 15. a. none (Q, W, and b.U > 0) b. b. U < 0, Q < 0 for refrigerator interior (W = 0) c. b.U0) 17. a. 1. 7 x 106 J, to the rod b. 3.3 x 102 J; by the rod c. 1. 7 x 106 J; it increases 27. 0.32 29. a. 188 J b. 1.400 x 103 J
Vibrations and Waves
1
Tc =
2
3
(TTH-50)
b. – 360° TH 37. 330 g 39. 5.7 x 103 J/min
= 95 Jl s
PRACTICE A 1. a. 15 Ni m b. less stiff 3. 2.7 X 103 Ni m
PRACTICE B 1. 1.4 X 102 m 3. 3.6m
Thermodynamics PRACTICE A 105 J b. – 4.8 x 105 J 3. 3.3x l 02 J
1. a. 6.4
X
PRACTICE B 1. 33 J 3. 1.00 X 104 J 5. 1.74 X 108 J
PRACTICE C 102 Ni m 3. 39.7 Ni m 5. a. 1. 7 s, 0.59 Hz b. 0. 14 s, 7.1 Hz c. 1.6 s, 0.62 Hz
1. 2.1
X
PRACTICED 1. 0.081 m :s ,\ :s 12 m 3. 4.74 X 1014 Hz
11 REVIEW 9. 11. 19. 21. 27. 35. 39. 43. 45. 47. 49. 51.
580N/m
43. a, 5.0 X 104 W b. 2.8 x 10- 3 w
4A
9.7m a. 0.57 s b. l.8Hz 1/3 s; 3 Hz 0.0333m a. 0.0 cm b. 48 cm a, b, and d (,\ = 0.5L, L, and 2L, respectively) 1.7 N 446m 9.70 m/s2 9:48A.M.
Light and Rellection PRACTICE A 10- 13 m 3. 85.7 m-10.1 m; The wavelengths are shorter than those of the AM radio band. 5. 5.4 X 10 14 Hz
1. 1.0
X
PRACTICE B
= 10.0 cm: no image (infinite q); p = 5.00 cm:
PRACTICE A 1. a. 8.0 X 10- 4 W/m2 b. 1.6 x 10- 3 W/ m 2 C, 6.4 X 10- 3 W/m2 3. 2.3 X lQ- 5 W 5. 4.8m
PRACTICE B 1. 440 Hz 3. a. 82.1 Hz b. 115 Hz
c. 144 Hz 12 REVIEW 23. 7.96 x 10- 2 w ; m 2 25. a. 4.0m b. 2.0m c. 1.3 m d. I.Om 29. 3Hz 35. 3.0 X 103 Hz 37. 5 beats per second 39. 0.20s 41. Lclosed = 1. 5 (Lopen)
Retraction PRACTICE A 1. 18.5° 3. 1.47
PRACTICE B 1. p
Sound
51. p = 11.3 cm 55. R = -25.0 cm 57. concave, R = 48.1 cm; M = 2.00; virtual
q = -10.0 cm, M = 2.00;
virtual, upright image 3. R = 1.00 x 102 cm; M = 2.00; virtual image
PRACTICE C 1. p
= 46.0 cm; M = 0.500;
virtual, upright image; h = 3.40cm 3. p = 45 cm; h = 17 cm; M = 0.41; virtual, upright image 5. q = – 1.31 cm; M = 0.125; virtual, upright image
13 REVIEW 7. 3.00 x 108 m / s 11. 1 X 10- 6 m 13. 9.1 x 10- 3 m (9.1 mm) 21. 1.2 mis; The image moves toward the mirror’s surface. 35. q = 26 cm; real, inverted; M=-2.0 47. inverted; p = 6.1 cm;f = 2.6
cm; real 49. q2 = 6.7 cm; real; M 1 = – 0.57, M 2 = -0.27; inverted
1. 20.0 cm, M = – 1.00; real, inverted image 3. -6.67 cm, M = 0.333; virtual, upright image
PRACTICE C 1. 42.8° 3. 49.8°
14 REVIEW 11. 13. 23. 25. 37.
39. 41. 43. 45.
47. 49. 51. 53.
26° 30.0°, 19.5°, 19.5°, 30.0° yes, because nice > nair 3.40; upright a. 31.3° b. 44.2° c. 49.8° 1.31 1.62; carbon disulfide 7.50 cm a. 6.00 cm b. A diverging lens cannot form an image larger than the object. a. 3.01 cm b. 2.05 cm blue: 47.8°, red: 48.2° 48.8° 4.54 m
55. 190 f
Selected Answers
R73
57. a. 24.7°
b. It will pass through the bottom surface because 0i< 0c (0c =41.8°). 1.38 58.0m a. 4.83 cm b. The lens must be moved 0.12cm. 1.90 cm
59. 61. 63.
65.
PRACTICE B 1. 4 7 N, along the negative x-axis; 157 N, along the positive x-axis; 11.0 x 101 N, along the negative x-axis
PRACTICE C 1. x =0.62m 3. 5.07m
PRACTICED
Interference and Diffraction PRACTICE A 1. 5.1 x 10- 7 m = 5.1 x 102 nm 3. 0.125° PRACTICE B 1. 0.02°, 0.04°, 0.11 ° 3. 11 5. 6.62 x 103 lines/ cm
15 REVIEW 5. 0 would decrease because ,\ is 9. 11 . 19. 21. 29. 31.
shorter in water. 630nm 160 µm 3.22° a. 10.09°, 13.71 °, 14.77° b. 20.51 °1 28.30°1 30.66° 432.0nm 1.93 x 10- 3 mm= 3 >-.; a maximum
Electric Forces and Fields PRACTICE A 1. 230 N (attractive) 3. 0.393 m
R74
Selected Answers
1. 1.66 x 105 N/ C, 81.1° above the positive x-axis 3. a. 3.2 x 10- 15 N, along the negative x-axis b. 3.2 x 10- 15 N, along the positive x-axis
16 REVIEW 15. 3.50 X 103 N 17. 91 N (repulsive) 19. 1.48 x 10- 7 N, along the +x direction 21. 18 cm from the 3.5 nC charge 33. 5.7 x 103 N/ C, 75° above the positive x-axis 35. a. 5.7 x 10- 27 N, in a direction opposite E b. 3.6 x 10- 8 N/ C 37. a. 2.0 x 107 N/C, along the positive x-axis b. 4.0 x 101 N 41. 7.2 x 10- 9 C 6 43. v electron = 4.4 X 10 m / s; vpro1on = 2.4 x 103 m /s 45. 5.4 X 10- 14 N 47. 2.0 x 10- 6 C 49. 32.5 m 51. a. 5.3 x 1017 m / s 2 b. 8.5 x 10- 4 m c. 2.9 x 1014 m /s2 53. a. positive b. 5.3 x 10- 7 C 55. a. 1.3 x 104 N/C b. 4.2 x 106 m /s
Electrical Energy and Current PRACTICE A 1. 6.4 3. 2.3
1010-
X X
19
C
16
J
PRACTICE B 1. a. b. 3. a. b.
4.80 x 10- 5 C 4.50 x 10- 6 J 9.00V 5.0 X 10- 12 C
PRACTICE C 1. 4.00 X 102 S 3. 6.00 X 102 S 5. a. 2.6 X 10- 3 A b. 1.6 x 1017 electrons C. 5.1 X 10- 3 A
PRACTICED 1. 0.43A 3. a. 2.5A b. 6.0A 5. 46!1
PRACTICE E 1. 14!1 3. 1.5V 5. 5.00
X
102 A
17 REVIEW 9. – 4.2 X 105 V 19. 0.22 J 23. V avg > > Vdrift 33. a. 3.5 min b. 1.2 x 1022 electrons 41. 3.4A 49. 3.6 X 106 J 51. the 75 Wbulb 53. 2.0 X 10 16 J
55. 93 n 57. 3.000 m; 2.00 x 10- 7 C 59. 4.0 x 103 V/m 61. a, 4.11 X 10- lS J b. 2.22 x 106 m / s 63. a. 1.13 x 105 V/ m b. 1.81 X 10- 14 N C. 4.39 X 10- 17 J 65. 0.545 m, – 1.20 m 67. a. 7.2 X 10- l 3 J b. 2.9 x 107 m / s 69. a. 3.0 X 10- 3 A b. 1.1 x 1018 electrons/m in 71. a. 32V b. 0.16V 73. 1.ox105 w 75. 3.2 X 105 J 77. 13.5 h 79. 2.2 X 10- 5 V
Circuits and Circuit Elements PRACTICE A 1. a. 43.6 n b. 0.275A 3. 1.0 V, 2.0 V, 2.5 V, 3.5 V
5.
o.5n
PRACTICE B 1. 4.5 A, 2.2 A, 1.8 A, 1.3 A 3. a. 2.2 n b. 6.0 A, 3.0 A, 2.00 A
PRACTICE C 1. a. 27.8 n b. 26.6 n
c. 23.4 n
PRACTICED Ra: 0.50 A, 2.5 V Rb: 0.50 A, 3.5 V RC: 1.5 A, 6.0 V
Rd: 1.0 A, 4.0 V
Re: 1.0 A, 4.0 V Rf 2.0 A, 4.0 V
Magnetism PRACTICE A
18 REVIEW
1. 3.57 x 106 mis
17. a. 24 n b. LOA 19. a. 2.99 n b. 4.0A 21. a. seven combinations R R 2R 3R b. R, 2R,3R, , , ,
3. 6.0 x 10-
2 3 3 2
23. 15 n 25. 3.0 !1: 1.8 A, 5.4 V 6.0 !1: 1.1 A, 6.5 V 9.0 !1: 0.72 A, 6.5 V
27. 28V 29. 3.8V 31. a. 33.o n b. 132V c. 4.00 A, 4.00 A 33. 10.0 n 35. a. a b. C c. d d. e 37. 18.0 !1: 0. 750 A, 13.5 V 6.0 !1: 0.750 A, 4.5 V 39. 4.o 41. 13.96 n 43. a. 62.4 b. 0.192A c. 0.102A d. 0.520W e. 0.737W 47. a. 5.1 b. 4.5V 49. a. 11 A (heater), 9.2 A (toaster), 12 A (grill) b. The total current is 32.2 A, so the 30.0 A circuit breaker will open th e circuit if these appliances are all on.
n
n
n
12
N west
PRACTICE B 1. 1.7 x 10- 7 T in + zdirection 3. 1.5 T
19 REVIEW 31. 2.1 x 10- 3 m /s 33. 2.00T 39. 2. 1 x 10- 2 T, in the negative y direction 41. 2.0 T, out of the page 43. a. 8.0 mis b. 5.4 X 10- 26 J 45. 2.82 x 107 m /s
Electromagnetic Induction PRACTICE A 1. 0.30V 3. 0.14 V
PRACTICE B 1. 4.8 A; 6.8 A, 170V 3. a. 7.42A b. 14.8 5. a. 1.10 x 102 v b. 2. lA
n
PRACTICE C 1. 55 turns 3. 25 turns 5. 147V
20 REVIEW 11. 0.12A 27. a. 2.4 X 102 V b. 2.0A
Selected Answers
R75
29. a. 8.34A b. 119V 35. 221 V 37. a. a step-down transformer b. 1.2 x 103 V 43. 790 turns 45. a. a step-up transformer b. 44OV 47. 171:1 49. 3OOV
Subatomic Physics
APPENDIX I
PRACTICE A
ADDITIONAL PROBLEMS
1. 160.65 MeV; 342.05 MeV 3. 7.933MeV
PRACTICE B 1. 12c 6 3. 14c 6 63 63 0 5. 28 Ni–+ 29 Cu + – 1e +
Atomic Physics
v
1. 2.OHz 3. 1.2 X 1015 Hz
PRACTICE C
PRACTICE B 1. 4.83 X 1014 Hz 3. 2.36eV
22 REVIEW 1. 79; 118; 79 7. 92.162MeV 9. 8.2607 MeV/ nucleon; 8.6974
PRACTICE C 1. 4.56 x 1014 Hz; line 4 3. 1.61 X 1015 Hz 5. E6 to E2 ; line 1
—————PRACTICED 1. 39.9 m / s 3. 8.84 x 10- 27 m / s 5. 1.0 X 10- 15 kg 21 REVIEW 11. 4.8 X 1017 Hz 13. 1.2 X 1015 Hz 23. a. 2.46 x 1015 Hz b. 2.92 x 1015 Hz C. 3.09 X 1015 Hz d. 3.16 x 1015 Hz 33. 1.4 x 107 m / s 35. 2.00 eV 37. 0.80 eV
R76
Selected Answers
15. 17.
1. 4.23 x 103 s- 1, 0.23 Ci 3. 9.94 X 10- 7 s- 1, 6.7 x 10- 7 Ci 5. a. about 5.0 x 107 atoms b. about 3.5 x 108 atoms
PRACTICE A
1. 3. 5. 7. 9. 11. 13.
MeV/ nucleon 21. a. 24He
b. ~He 23. 560 days 27. a. -e b. o 33. 1.2 X 10- 14 35. 3.53MeV + 37. a. 01n + 197 79Au –+ 198Hg 80 0 – 1e + v b. 7.885MeV 39. 32He 41. 2.6 x 1021 atoms 43. a. : Be b. iic 45. 3.8 X 103 S 47. 1.1 x 1016 fission events
19. 21. 23. 25. 27. 29. 31. 33. 35. 37. 39. 41. 43. 45. 47. 49. 51. 53. 55. 57. 59. 61. 63. 65. 67.
11.68 m 6.4 X 10- 2 m 3 6.7 X 10- 5 ps 2.80 h = 2 h, 48 min 4.0 x 101 km/h 48 m/h +25.0 m/s = 25.0 m/s, upward 44.8 m/s – 21.5 m / s 2 = 21.5 m / s 2, backward 38.5m 126 s 1.27 s 11 km/ h 2.74s 10.5 m, forward 5.9 s 8.3 s 7.4 s -490 m / s2 = 490 m / s2, backward 17.3 s 7.Om 2.6m/ s – 11.4 m / s = 11.4 m / s, downward 8.5° north of east 5.0° south of west 77Om -33 km/ h = 33 km/ h, downward 18.9 km, 76° north of west 17.Om 52.0° 79 s 15.8 m, 55° below the horizontal 0.290 m / s, east; 1.16 m / s, north 2.6km
69. 71. 73. 75.
66 km, 46° south of east 10.7m 3.0 s 76.9 km/h, 60.1 ° west of north 77. 7.0 x 102 m, 3.8° above the horizontal 79. 47.2 m 81. 6.36 m/s 83. 13.6 km/h, 73° south of east 85. 58N 87. 14.0 N; 2.0 N 89. 9.5 x 104 kg 91. 258 N, up the slope 93. 15.9 N 95. 2.0 m/s 2 97. Fx = 8.60 N; FY = 12.3 N 99. -448 m /s2 = 448 m/s2 , backward 101.15 kg 103.0.085 105.1.7 x 108 N 107. 24N, downhill 109.1.150 X 103 N 111. 1.2 x 104 N 113.0.60 115. 38.0 m 117. 2.5 X 104 J 119. 247 m /s 121. -5.46 X 104 J 123.3.35 X 106 J 125.1.23 J 127.12 s 129.0.600m 131.133 J 133. 53.3 mis 135. 72.2m 137.0.13 m = 13 cm 139. 7.7 m /s 141. 8.0 s 143. 230 J 145. 7.96m 147. 6.0 x 101 m /s
3
149.1.58 x 10 kg•m/s, north 151. 3.38 x 1031 kg 153.18 s 155. 637 m, to the right 157. 7.5 g 159.0.0m/ s 161. -5.0 x 101 percent 163.16.4 m/s, west 165. 5.33 x 107 kg•m/s 167. 1.0 x 101 m /s 169. 560 N, east 171. -3.3 x 108 N = 3.3 x 108 N, backward 173.52 m 175.24kg 177. 90.6 km/ h, east 179. 26 km/h, 37° north of east 181.-157J 183. 0.125 kg 185. -4.1 X 104 J 187.9.8 kg 189. 1.0 mis, 60° south of east 191.4.04 x 103 m / s2 193. 42 m / s 195.8.9kg 197. 1.04 x 104 m / s = 10.4 km/ s 199.1.48 x 1023 kg 201. 1.10 X 1012 m 203. 6.6 x 103 m /s = 6.6 km/s 205.0.87 m 207. 254N 209. 0.42 m = 42 cm 211. 25 N 213.165 kg 215. 5.09 x 105 s = 141 h 217. 5.5 X 109 m = 5.5 X 106 km 219.1.6 N•m 221. 6.62 x 103 N 223.0.574m 225. 8.13 X 10- 3 m 2 4 3 227. 2.25 x 10 kg/m
1
2
229. 2.0 X 10 m 231.4.30 kg 233. 374°F to -292°F 235. 6.6 X 10- 2°C 237.1.29 X 104 J 239.4.1 x 10- 2 kg 241.1.200 X 103°C 243.315 K 245.1.91 x 10- 2 kg= 19.1 g 247.530 J/ kge°C 249.-930°C 251. 1.50 x 103 Pa = 1.50 kPa 253.873 J 255. 244 J 257. 5.3 X 103 J 259.2.4 x 103 Pa= 2.4kPa 261. 5895 J 263. 5.30 X 102 kJ = 5.30 X 105 J 265. 1.0 X 104 J 267. -18N 269.-0.11 m = -11 cm 271. 4.0 x 10- 2 m = 4.0 cm 273. 0.2003 Hz 275. 730 Ni m 277.1.4 x 103 m /s 279. 2.2 X 104 Hz 281.8.6 X 103 N/m 283. 3.177s 285.82kg 287.1.2 s 289.1.5 x 103 m /s 291.1.1 W/m2 293. 294Hz 295.408 m / s 297.0.155m 299.0.211 m = 21.1 cm 301. 2.9971 x 108 m /s 303. 3.2 x 10- 7 m = 320 nm 305. -0.96 cm 307. -1.9 cm 309. 3.8 m 311.0.25 313.38 cm
Selected Answers
R77
315. 2.40 317. 0.98 cm 319. 10.5 cm 321. 64.0 cm in front of the mirror 323. 8.3 cm 325.0.40 327.-11 cm 329. 2.9979 x 108 m / s 331 .33.3 cm 333. 0.19 335.32.2° 337. – 10.4 cm 339.18cm 341. 1.63 343. 39.38° 345. 58° 347. -21 cm
349. oo 351 .1.486 353.1.54 355. 4.8 cm 357. 1.73 to 1.83 359. 5.18 x 10- 4 m = 0.518 mm 361. 0.137° 363. 9.0 x 10- 7 m = 9.0 x 102 nm 365.11.2° 367. 0.227° 369.1.445 x 104 lines/ cm 371.140 N attractive 373. 2.2 x 10- 17 C 375. 0.00N 377.4.0 x 10- 8 N, 9.3° below the n egative x-axis
R78
Selected Answers
379. 260 N from eith er ch arge 381.1.6 x 10- 12 C 383.0.585 m = 58.5 cm 385. 3.97 x 10- 6 N, upward 387. 0.073 m = 7.3 cm 389. 7.5 x 10- 6 N, along the +y-axis 391. 4.40 x 105 N/C, 89.1° ab ove the -x-axis 393.-7.4C 395.1.6 x 10- 19 C 397. 36 cm 399. 4.4 X 10- 4 J 401. 7.1 X 10- 4 F 403.4.0A 405.160 .Q 407.1.7 x 106 W = 1.7 MW 409. 6.4 x 102 N/ C 411. 12 V 413. 1.2 x 10- 5 m 415.1.4 x 102 C 417. 7.2 s 419. 4.8V 421. 116 V 423. 5.0 x 105 W = 0.50 MW 425. 7.5 x 106 V 427. 3.00 X 102 .Q 429. 6.0 .Q 431.13 .Q 433.6.0 .Q 435. 0.056A = 56 mA 437. 1.6 A (refrigerator); 1.3 A (oven ) 439.12.6 .Q 441. 2.6V 443. 9.4 .Q
445. l.6A 447. l.45A 449.4 X 10- 12 N 451. 7.6 x 106 m i s 453.0.70 A 455. 5.1 x 10- 4 T 457. 3.9 X 10- 15 N 459. 1.5 X 105 A 461. 1.7 X 10- 2 T 463.0.90 s 465. 450V 467. l.8A 469. 3.4 x 104 V = 34 kV 471. 48 turns 473. 2.5 x 10- 3 A = 2.5 mA 475.0.85A 477. 48:1 479. 3.32 X 10- 10 m 481. 1.00 X 10- 13 m 483. 3.0 X 10- 7 m 485. 26kg 487. 4.30 x 1014 Hz 489. 6.0 X 10 14 Hz 491. 4.0 x 10- 2 1 kg 493. 2.5 X 10- 43 m 495. 7.72 x 10 14 Hz 497. 333.73 MeV 499. 0.543 705 u 1 501. go 503.15.0 s 505. 31.92 h 507. 35.46 MeV
509.
1
JjI
1
511. ~lxe 513. 924 days
absorption spectrum a diagram or graph that indicates the wavelengths of radiant energy that a substance absorbs acceleration the rate at wh ich velocity changes over time; an object accelerates if its speed, direction, or both change accuracy a description of how close a m easurement is to the correct or accepted value of the quantity measured adiabatic process a thermodynamic process in which no energy is transferred to or from the system as heat alternating current an electric current that changes direction at regular intervals amplitude the maximum displacement from equilibrium angle of incidence the angle between a ray that strikes a surface and the line p erpendicular to that surface at the point of contact angle ofreflection the angle form ed by the line perpendicular to a surface and the direction in which a reflected ray moves angular acceleration the time rate of change of angular velocity, usually expressed in radians per second per second angular displacement th e angle through which a point, line, or body is rotated in a specified d irection and about a specified axis angular momentum for a rotating object, the product of the object’s moment of inertia and angular velocity about the same axis angular velocity the rate at which a body rotates about an axis, usually expressed in radians per second antinode a point in a standing wave, h alfway between two n odes, at which the largest displacement occurs average velocity the total displacement divided by the time interval during which the displacement occurred
back emf the emf induced in a motor’s coil that tends to reduce the current in the coil of the motor beat the periodic variation in the amplitude of a wave that is the superposition of two waves of slightly different frequencies binding energy the energy released when unbound nucleons come togeth er to form a stable nucleus, which is equivalent to the en ergy required to break the nucleus into individual nucleons blackbody radiation the radiation emitted by a blackbody, which is a perfect radiator and absorber and emits radiation based only on its temperature buoyant force the upward force exerted by a liquid on an object immersed in or floating on the liquid
compression the region of a longitudinal wave in which the density and pressure are at a maximum Compton shift an increase in the wavelength of the photon scattered by an electron relative to the wavelength of the incident photon concave spherical mirror a mirror whose reflecting surface is an inward-curved segment of a sphere constructive interference a superposition of two or more waves in which individual displacements on the same side of the equilibrium position are added together to form the resultant wave controlled experiment an experiment that tests only one factor at a time by using a comparison of a control group with an experimental group convex spherical mirror a mirror whose reflecting surface is an outward-curved segment of a sphere crest the highest point above the equilibrium position
calorimetry an experimental procedure used to m easure the en ergy transferred from one substance to another as h eat
critical angle the minimum an gle of incidence for which total internal reflection occurs
capacitance the ability of a conductor to store energy in the form of electrically separated charges
cyclic process a thermodynamic process in which a system returns to the same conditions under which it started
center of mass the point in a body at which all the mass of the body can be considered to be concentrated when an alyzing translational motion centripetal acceleration the acceleration directed toward the center of a circular path chromatic aberration the focusing of different colors of light at different distances behind a len s coefficient of friction the ratio of the magnitude of the force of friction between two objects in contact to the magnitude of the normal force with which the objects press again st each other coherence the correlation between the phases of two or more waves components of a vector the projections of a vector along the axes of a coordinate system
decibel a dimen sionless unit that describes the ratio of two intensities of sound; th e threshold of h earing is commonly used as the reference intensity destructive interference a superposition of two or more waves in which individual displacements on opposite sides of the equilibrium position are added together to form the resultant wave diffraction a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge dispersion the process of separatin g polychromatic light into its component wavelengths displacement the change in position of an object
Glossary
R79
doping the addition of an impurity element to a semiconductor Doppler effect an observed change in frequency when there is relative motion between the source of waves and an observer
environment the combination of conditions and influences outside a system that affect th e behavior of the system
drift velocity the net velocity of a charge carrier m oving in an electric field
equilibrium in physics, the state in which the net force on an object is zero
————0
excited state a state in which an atom has more energy than it does at its ground state
elastic collision a collision in which the total momentum and total kinetic energy remain constant elastic potential energy the energy available for use when an elastic body returns to its origin al configuration electric circuit a set of electrical components connected such that they provide one or more complete paths for the movement of charges electric current the rate at which charges pass through a given area electric field a region where an electric force on a test charge can be detected electric potential the work that must be performed again st electric forces to move a charge from a reference point to the point in question divided by the charge electrical conductor a material in which charges can move freely electrical insulator a material in which charges can not move freely electrical potential energy potential energy associated with a charge due to its position in an electric field electromagnetic induction the process of creating a current in a circuit by a changing m agn etic field electromagnetic radiation the transfer of energy associated with an electric and magnetic field; it varies periodically and travels at the speed of light electromagnetic wave a wave that consists of oscillating electric and magnetic fields, which radiate outward from the source at the speed oflight emission spectrum a diagram or graph that indicates the wavelengths of radiant energy that a substance emits
R80
entropy a measure of the randomness or disorder of a system
Glossary
————-•
————
fluid a nonsolid state of matter in which the atoms or molecules are free to move p ast each other, as in a gas or liquid force an action exerted on an object which may chan ge the object’s state of rest or motion; force has magnitude and direction frame of reference a system for specifying the precise location of objects in space and time free fall the m otion of a body when only the force due to gravity is acting on the body frequency the number of cycles or vibrations per unit of time; also the number of waves produced per unit of time fundamental frequency the lowest frequen cy of vibration of a standing wave
————-0 ———–generator a machine that converts mechanical en ergy into electrical energy gravitational force the mutual force of attraction between particles of matter gravitational potential energy the potential energy stored in the gravitational fields of interacting bodies ground state the lowest energy state of a quantized system
0 half-life the time needed for half of the original nuclei of a sample of a radioactive substance to undergo radioactive decay harmonic series a series of frequencies that includes the fundamental frequency and integral m ultiples of the fundamental frequency heat the energy transferred between objects because of a difference in their temperatures; energy is always transferred from higher-temperature objects to lower-temperature objects until thermal equilibrium is reached hole an energy level that is not occupied by an electron in a solid hypothesis an explanation that is based on prior scientific research or observations and that can be tested
————-0 ———–ideal fluid a fluid that has no internal friction or viscosity and is incompressible impulse the product of the force and the time over which the force acts on an object index of refraction the ratio of the speed of light in a vacuum to the speed oflight in a given transparent medium induction th e process of charging a conductor by bringing it near another charged object and grounding the conductor inertia the tendency of an object to resist being moved or, if the object is moving, to resist a change in speed or direction instantaneous velocity the velocity of an object at some instant or at a specific point in the object’s path intensity the rate at which energy flows through a unit area perpendicular to the direction of wave motion internal energy the energy of a substance due to both the random motions of its particles and to the potential energy that results from the distances and alignments between the particles
isothermal process a thermodynamic process that takes place at constant temperature
mechanical energy the sum of kinetic energy and all forms of potential energy
isotope an atom that has the same number of protons (or the same atomic number) as other atoms of the sam e element do but that has a different number of neutrons (an d thus a different atomic mass)
mechanical wave a wave that requires a medium through which to travel
isovolumetric process a thermodynamic process that takes place at constant volume so that no work is done on or by th e system
model a pattern, plan, representation, or description designed to show the structure or workings of an object, system, or concept
————-0 ———– -kinetic energy the energy of an object that is due to the object’s m otion kinetic friction the force that opposes the movement of two surfaces that are in contact and are sliding over each other
————-•
————
medium a physical environment through which a disturbance can travel
moment of inertia the tendency of a body that is rotating about a fixed axis to resist a change in this rotating motion momentum a quantity defined as the product of the mass and velocity of an object mutual inductance the ability of one circuit to indu ce an emf in a nearby circuit in the presence of a changing current
laser a device that produces coherent light of only one wavelength latent heat the energy per unit mass that is transferred during a phase change of a substance lens a transparent object that refracts light waves such that they converge or diverge to create an image lever arm the perpendicular distance from the axis of rotation to a line drawn along the direction of the force
linear polarization the alignment of electrom agnetic waves in such a way that the vibration s of the electric fields in each of the waves are parallel to each oth er longitudinal wave a wave wh ose par ticles vibrate parallel to the direction the wave is travelin g
————•
————
magnetic domain a region composed of a group of atoms whose magnetic fields are aligned in th e same direction magnetic field a region wh ere a magnetic force can b e detected mass density the concentration of matter of an object, measured as the mass per unit volume of a substance
net force a single force whose external effects on a rigid body are th e same as the effects of several actual forces acting on th e body node a point in a standing wave that maintains zero displacement normal force a force that acts on a surface in a direction perpen dicular to the surface
order number th e number assigned to interference fringes relative to the cen tral bright fringe
————-0 ————parallel describes two or more components of a circuit that provide separate conducting paths for current because the components are connected across common p oints or junctions path difference th e difference in th e distance traveled by two beams wh en they a re scattered in the sam e direction from different points perfectly inelastic collision a collision in which two objects stick together after colliding
period the time that it takes a com plete cycle or wave oscillation to occur phase change the physical change of a substance from one state (solid, liquid, or gas) to another at constant temperature and pressure photoelectric effect the emission of electrons from a material when light of certain frequencies shin es on the surface of the material photon a unit or quantum of light; a particle of electromagnetic radiation that has zero mass and carries a quantum of energy pitch a measure of how h igh or low a sound is perceived to be, depending on the frequency of the sound wave potential difference the work that must be performed against electric forces to move a charge between the two points in question divided by the charge potential energy the energy associated with an object because of the position, shape, or condition of the object power a quantity that measures the rate at which work is done or energy is transformed precision the degree of exactness of a m easurement pressure the magnitude of the force on a surface per unit area projectile motion the curved path that an object follows when thrown, launched, or otherwise projected near the surface of Earth
radian an angle whose arc length is equal to the radius of the circle, which is approximately equal to 57 .3° rarefaction th e region of a longitudinal wave in which the density and pressure are at a minimum real image an image that is formed by the intersection of light rays; a real image can be projected on a screen reflection the turning back of an electromagnetic wave at a surface refraction the bending of a wavefront as the wavefront passes between two substances in which the speed of the wave differs
Glossary
R81
resistance the opposition presented to electric current by a material or device resolving power the ability of an optical instrument to form separate images of two objects that are close together
strong force the interaction that binds nucleons together in a nucleus
resonance a phenomenon that occurs when the frequency of a force applied to a system matches the natural frequency of vibration of the system, resulting in a large amplitude of vibration
superconductor a material whose resistance is zero at a certain critical temperature, which varies with each material
resultant a vector that represents th e sum of two or more vectors
system a set of particles or interacting components considered to be a distinct physical entity for the purpose of study
rms current the value of alternating current that gives the same h eating effect that the corresponding value of direct current does rotational kinetic energy the energy of an object that is due to the object’s rotational motion
———— 0 ———–scalar a physical quantity that has magnitude but no direction schematic diagram a representation of a circuit that uses lines to represent wires and different symbols to represent components series describes two or more components of a circuit that provide a single path for current significant figures those digits in a measurement that are known with certainty plus the first digit that is uncertain simple harmonic motion vibration about an equilibrium position in which a restoring force is proportional to the displacement from equilibrium solenoid a long, helically wound coil of insulated wire specific heat capacity the quantity of heat required to raise a unit mass of homogen eous material 1 Kor 1•c in a specified way given constant pressure and volume spring constant the energy available for use when a deformed elastic object returns to its original configuration standing wave a wave pattern that results when two waves of the same frequency, wavelength, and amplitude travel in opposite directions and interfere
R82
static friction the force that resists the initiation of sliding motion between two surfaces that are in contact and at rest
Glossary
————-0 ———–tangential acceleration the acceleration of an object that is tangent to the object’s circular path tangential speed the speed of an object that is tangent to the object’s circular path temperature a measure of the average kinetic energy of the particles in an object thermal equilibrium the state in which two bodies in physical contact with each other have identical temperatures timbre the musical quality of a tone resulting from the combination of harmonics present at different intensities torque a quantity that measures the ability of a force to rotate an object around some axis total internal reflection the complete reflection that takes place within a substance when the angle of incidence of light striking the surface boundary is less than the critical angle transformer a device that increases or decreases the emf of alternating current transistor a semiconductor device that can amplify current and that is used in amplifiers, oscillators, and switches transverse wave a wave whose particles vibrate perpendicularly to the direction the wave is traveling trough the lowest point below the equilibrium position
(!) ultraviolet catastrophe the failed prediction of classical physics that the energy radiated by a blackbody at extremely short wavelengths is extremely large and that the total energy radiated is infinite uncertainty principle the principle that states that it is impossible to simultaneously determine a particle’s position and momentum with infinite accuracy
————-•
————
vector a physical quantity that has both magnitude and a direction virtual image an image from which light rays appear to diverge, even though they are not actually focused there; a virtual image cannot be projected on a screen
————-e
————-
wavelength the distance between two adjacent similar points of a wave, such as from crest to crest or from trough to trough weight a measure of the gravitational force exerted on an object; its value can ch ange with the location of th e object in the universe work the product of the component of a force along the direction of displacement and the magnitude of the displacement work function the minimum energy needed to remove an electron from a metal atom work-kinetic energy theorem the net work done by all the forces acting on an object is equal to the change in the object’s kinetic energy
Page references followed by f refer to image figures. Page references followed by t refer to tables.
e
aberration, 452, 463, 463!, 505,505f absolute pressure, 278-279 absolute zero, 302 absorption spectrum, 746- 747, 747!, 748, 749 ac. See alternating current acceleration, 44-54; angular, 64-65, 64!, 65t, 253, 256-257, 257t; average, 44- 45; centripetal, 224-226, 225!, 253, 253.f. constant, 47- 54, 47!, 48!, 54t (see also free fall; free-fall acceleration); of electric charges, 716, 745; force and, 118, 118!, 123, 124, 128- 129, 128.f. ine rtia and, 123, 124; o f m ass-spring system, 364-366, 371t; negative, 46-47, 461, 47t; of pendulum, 370, 373-374, 374.f. of reference frame, 258- 259, 258.f. ta ngential, 226, 253; total rotational, 253, 253.f. units of, 44 acceleration due to gravity. See free-fall acceleration accelerators, particle, 176, 177!, 596, 793, 793!, 799, 801 accuracy,16-17, 16.f. in laboratory calculations, Rl5- Rl6; uncertainty principle and, 757- 758 action-reaction pair, 131 adhesion, 135, 135f adiabatic process, 337, 337!, 3401 air conditioning, 320, 354 airplane, lift force on, 282, 282f air resistance: as friction, 140; Galileo’s experiments and, 8, 9, 21; projectile motion and, 94-95, 94.f. terminal velocity and, 60 algebra, review of, R4-Rl0, RSI, R71 alpha decay, 779, 780, 782, 783,784 alpha (a) particles, 744- 745, 744!, 779-780, 779t, 789, 791 alternating current (ac), 604- 605, 605.f. generators o f, 702- 703, 702!, 703.f. supplied to motor, 704 alternating-current (ac) circuits, 707-710, 707!, 708!, 708t; transformers in, 711-713, 711f
ammeters,679, 710 ampere (A), 594 amplitude: of simple harmonic motion, 372, 373,373t,374,376;ofa wave, 380, 380!, 384 analog signals, 536 Anderson, Carl, 800 angle of incidence: for reflection, 448, 448.f. for refractio n, 482, 482!, 486 angle of reflection, 448, 448f angle of refraction, 482, 482!, 486 angles: critical, 500-501, 500.f. determining an unknown angle, 86-87, Rl4, Rl5, Rl5.f. radian measure for, 62- 63, 62!, Rl 4 angular acceleration, 64- 65, 64!, 65t, 253, 256- 257, 257t angular displaceme nt, 63, 63!, 64, 65, 65t angular kinematics, 62-65, 62!, 63!, 641, 65t angular momentum, 257, 257!, 2571 angular speed, 64, 252-253, 252!, 257, 257t angular velocity, 64-65, 64!, 651 antimatter, 800-801 antineutrinos, 781-782 antinodes, 389, 389!, 418-421, 4 19t, 420!, 42If antiparticles, 795, 796, 796t, 798,800,801 apparent weight, 271, 272, 273,274 apparent weightlessness, 242-243,242! Archimedes’ principle, 272, 272!, 273, 279 arc length, 62, 62!, 63 areas, of geometrical shapes, Rl21 Aristarchus, 238 Aristotle, 52 atmosphere (atm), 276 atmospheric pressure, 276, 278,279,285,285f atmospheric refraction, 503, 503f atomic bomb, 79lf atomic mass unit, unified, 773 atomic number (Z), 772-773, 772!, 772t, 773f atomic spectra. See spectra, atomic atoms (see also Bohr model; electrons; elements; nucleus; spectra, atomic): early models of, 744-745, 744!, 745.f. electric ch a rges of particles in, 549, 550, 550t; e nergy of, 299, 299!, 2991; images of, with STM, 611, 611.f. in a laser, 534-535, 534.f. table o f
masses, R46-R51; thermal conduction by, 308; wave function and, 758-759, 758f aurora borealis, 674, 732-733,749 axis ofrotation, 62, 63, 64, 65, 224, 225, 245, 245!, 246, 246!, 255
8
back emf, 704, 704f band theory, 760-761, 760!, 76If bar codes, 537 Bardeen, John, 613 barometer, 285, 285f baryons, 795, 7951,796, 796!, 796t batteries, 628-633, 628f,630f, 632.f. chem ical energy in, 586, 604, 604!, 605; direct current generated by, 605; potential diffe re nce of, 582, 582!, 586, 632-633; in schematic diagra m