(Solution Manual) Serway And Vuille's College Physics (Volume 1)

Serway,Vuille, 9

Physics

Solution Manual

9th Edition

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INSTRUCTOR ’S SOLUTIONS MANUAL FOR SERWAY AND VUILLE’S COLLEGE PHYSICS NINTH EDITION , VOLUME 1 INSTRUCTOR ’S SOLUTIONS MANUAL FOR SERWAY AND VUILLE ’S COLLEGE PHYSICS NINTH EDITION , VOLUME 1 Charles Teague Emeritus, Eastern Kentucky University Australia • Brazil • Japan • Korea • Mexico • Singapore • Spain • United Kingdom • United States 56157_00_fm_pi-viii.indd i 10/13/10 9:54:04 AM© 2011 Brooks/Cole, Cengage Learning ALL RIGHTS RESERVED. No part of this work covered by the copyright herein may be reproduced, transmit-ted, stored, or used in any form or by any means graphic, electronic, or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks, or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, with-out the prior written permission of the publisher. For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706 For permission to use material from this text or product, submit all requests online at www.cengage.com/permissions Further permissions questions can be emailed to permissionrequest@cengage.comISBN-13: 978-0-8400-6870-5ISBN-10: 0-8400-6870-0 Brooks/Cole 20 Channel Center StreetBoston, MA 02210USA Cengage Learning is a leading provider of customized learning solutions with offi ce locations around the globe, including Singapore, the United Kingdom, Australia, Mexico, Brazil, and Japan. Locate your local offi ce at: www.cengage.com/global Cengage Learning products are represented in Canada by Nelson Education, Ltd. To learn more about Brooks/Cole, visit www.cengage.com/brookscole Purchase any of our products at your local college store or at our preferred online store www.cengagebrain.com Dear Professor or Other Supplement Recipient: Cengage Learning has provided you with this product (the “Supplement”) for your review and, to the extent that you adopt the associated textbook for use in connection with your course (the “Course”), you and your students who purchase the textbook may use the Supplement as described below. Cengage Learning has established these use limita-tions in response to concerns raised by authors, professors, and other users regarding the pedagogical problems stemming from unlimited distribution of Supplements. Cengage Learning hereby grants you a nontransferable license to use the Supplement in connection with the Course, subject to the following conditions. The Supplement is for your personal, noncommercial use only and may not be reproduced, posted electronically or distributed, except that portions of the Supplement may be provided to your students IN PRINT FORM ONLY in connection with your instruction of the Course, so long as such students are advised that they may not copy or distribute any portion of the Supplement to any third party. You may not sell, license, auction, or otherwise redistribute the Supplement in any form. We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution. Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement. If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt. All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors. The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied. This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s confl ict of law rules. Thank you for your assistance in helping to safeguard the integrity of the content contained in this Supplement. We trust you fi nd the Supplement a useful teaching tool.READ IMPORTANT LICENSE INFORMATION Printed in the United States of America 1 2 3 4 5 6 7 14 13 12 11 10NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED, OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN. 56157_00_fm_pi-viii.indd ii 10/13/10 9:54:05 AMTABLE OF CONTENTS Acknowledgements v Preface vii Part 1 – Mechanics Chapter 1 – Introduction 1 Chapter 2 – Motion in One Dimension 17 Chapter 3 – Vectors and Two-Dimensional Motion 49 Chapter 4 – The Laws of Motion 89 Chapter 5 – Energy 129 Chapter 6 – Momentum and Collisions 169 Chapter 7 – Rotational Motion and the Law of Gravity 209 Chapter 8 – Rotational Equilibrium and Rotational Dynamics 243 Chapter 9 – Solids and Fluids 288 Part 2 – Thermodynamics Chapter 10 – Thermal Physics 322 Chapter 11 – Energy in Thermal Processes 344 Chapter 12 – The Laws of Thermodynamics 371 Part 3 – Vibrations and Waves Chapter 13 – Vibrations and Waves 400 Chapter 14 – Sound 426 iii 56157_00_fm_pi-viii.indd iii 10/13/10 9:54:06 AM56157_00_fm_pi-viii.indd iv 10/13/10 9:54:06 AMvACKNOWLEDGEMENTS The author would like to thank everyone who has contributed to this work. In particular, thanks go to the support staff at Cengage Learning for their excellent guidance and support in all phases of this project. Special mention goes to Physics Acquisitions Editor, Charles Hartford; Development Editor, Ed Dodd; Associate Content Project Manager, Holly Schaff ; Associate Development Editor, Brandi Kirksey; and Editorial Assistant, Brendan Killion. Susan English of Durham Technical Community College served as accuracy reviewer for this manual. Her contributions are deeply appreciated. Any remaining errors in this work are the responsibility of the author alone. I would like to acknowledge the staff of MPS Limited, a Macmillan Company for their excellent work in assembling and typing this manual and preparing diagrams and page layouts. Finally, the author would like to thank his wife, Carol, for her patience, understanding, and great support during this eff ort. v 56157_00_fm_pi-viii.indd v 10/13/10 9:54:06 AM56157_00_fm_pi-viii.indd vi 10/13/10 9:54:06 AMPREFACE This manual is written to accompany College Physics, Ninth Edition, by Raymond A. Serway and Chris Vuille. For each chapter in that text, the manual includes solutions to all end-of- chapter problems, more detailed answers to Quick Quizzes and Multiple Choice Questions than available in the main text, and answers to the even-numbered Conceptual Questions. Considerable eff ort has been made to insure that the solutions and answers given in this manual comply with the rules on signifi cant fi gures and rounding given in the Chapter 1 of the textbook. This means that intermediate answers are rounded to the proper number of signifi cant fi gures when written, and that rounded value is used in all subsequent calculations. Users should not be concerned if their answers diff er slightly in the last digit from the answers given here. Most often, this will be caused by choosing to round intermediate answers at diff erent stages of the solution. You are encouraged to keep this manual out of the hands of students as instructors in many col-leges throughout the country use this textbook, and many of them use graded problem assignments as part of the fi nal course grade. Additionally, even when the problems are not used in such a direct fashion, it is advantageous for students to struggle with some problems in order to improve their problem-solving skills. Feel free to post answers and solutions to selected questions and problems, but please preserve the manual as a whole. You may also encourage students to purchase a copy of the Student Solutions Manual with Study Guide, which provides chapter summaries as well as detailed solutions to selected problems in the main text. Attempting to keep the manual of manageable size, and recognizing that the primary users will be instructors well versed in the fi eld, answers and solutions are kept fairly brief. Answers to conceptual questions have been shortened by not off ering detailed arguments that lead to the answer. Problem solutions often omit commentary, intermediate steps, as well as initial steps that could be necessary for clear understanding by students. On occasions where selected problem solutions are to be shared with students, you may wish to supply intermediate steps and additional comments as needed. An electronic version of this manual can be obtained by requesting the Instructor’s Power Lecture CD from your local Cengage Learning Sales Representative. Contact information for your sales representative is available under the “Find Your Rep” tab found at the bottom of the page at www. academic.cengage.com. We welcome your comments on the accuracy of the solutions as presented here, as well as sugges-tions for alternative approaches. Charles Teague vii 56157_00_fm_pi-viii.indd vii 10/13/10 9:54:06 AM56157_00_fm_pi-viii.indd viii 10/13/10 9:54:06 AM1 Introduction ANSWERS TO MULTIPLE CHOICE QUESTIONS 1. Using a calculator to multiply the length by the width gives a raw answer of 67832m, but this answer must be rounded to contain the same number of signifi cant fi gures as the least accurate factor in the product. The least accurate factor is the length, which contains either 2 or 3 signifi cant fi gures, depending on whether the trailing zero is signifi cant or is being used only to locate the decimal point. Assuming the length contains 3 signifi cant fi gures, answer (c) correctly expresses the area as 67 8 1 03.× m2. However, if the length contains only 2 signifi cant fi gures, answer (d) gives the correct result as 68 1 03.× m2. 2. Both answers (d) and (e) could be physically meaningful. Answers (a), (b), and (c) must be mean- ingless since quantities can be added or subtracted only if they have the same dimensions. 3. According to Newton’s second law, Force mass acceleration =× . Thus, the units of Force must be the product of the units of mass (kg) and the units of acceleration ms2() . This yields kg m s2⋅ , which is answer (a). 4. The calculator gives an answer of 57.573 for the sum of the 4 given numbers. However, this sum must be rounded to 58 as given in answer (d) so the number of decimal places in the result is the same (zero) as the number of decimal places in the integer 15 (the term in the sum containing the smallest number of decimal places). 5. The required conversion is given by: h=()⎛ ⎝⎜⎞ ⎠⎟ 20 01 000 1 00..mmm 1.00 mcubitus 445 mm⎛ ⎛ ⎝⎜⎞ ⎠⎟=44 9. cubiti This result corresponds to answer (c). 6. The given area 1 420 ft2() contains 3 signifi cant fi gures, assuming that the trailing zero is used only to locate the decimal point. The conversion of this value to square meters is given by: A=×()⎛ ⎝⎜⎞ ⎠⎟=× 1 42 10 1 32 1032 2.. ft1.00 m 3.281 ft2mmm22=132 Note that the result contains 3 signifi cant fi gures, the same as the number of signifi cant fi gures in the least accurate factor used in the calculation. This result matches answer (b). 7. You cannot add, subtract, or equate a number apples and a number of days. Thus, the answer is yes for (a), (c), and (e). However, you can multiply or divide a number of apples and a number of days. For example, you might divide the number of apples by a number of days to fi nd the number of apples you could eat per day. In summary, the answers are (a) yes, (b) no, (c) yes, (d) no, and (e) yes. 1 56157_01_ch01_p001-016.indd 156157_01_ch01_p001-016.indd 1 10/12/10 1:26:50 PM10/12/10 1:26:50 PM2 Chapter 1 8. The given Cartesian coordinates are xy=− =50 0 1 20 0..,a n d , with the least accurate containing 3 signifi cant fi gures. Note that the specifi ed point (with xy<>00and ) is in the second quadrant. The conversion to polar coordinates is then given by: rx y=+ = − () +() =22 2250 0 1 20 0 1 30.. . tan. ... θθ==−=− = − (− y x12 00 50 024 0 24 0 and tan1) )=− °+ °= °67 3 180 113. Note that 180° was added in the last step to yield a second quadrant angle. The correct answer is therefore (b) (13.0, 113°). 9. Doing dimensional analysis on the fi rst 4 given choices yields: (a) v[] ⎡⎣⎤⎦== t2 23LT TL T (b) v[] ⎡⎣⎤⎦==−− x2 211 LT LLT (c) v222 2 3LT TLT TL T⎡⎣⎤⎦ []=()==t2 (d) v222 2LT LLT LL T⎡⎣⎤⎦ []=()==x2 Since acceleration has units of length divided by time squared, it is seen that the relation given in answer (d) is consistent with an expression yielding a value for acceleration. 10. The number of gallons of gasoline she can purchase is # gallonstotal expenditure cost per gallonE=≈33 u uros Euros LL quartquarts151 14.⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎜⎞ ⎠⎟⎟ 1 15 galgal ⎛ ⎝⎜⎜⎞ ⎠⎟⎟≈ so the correct answer is (b). 11. The situation described is shown in the drawing at the right. From this, observe that tan 2645°=h m, or h=() °= 45 26 22mmtan Thus, the correct answer is (a). 12. Note that we may write 1 365 248 0 107. × as 136 524 80 105. × . Thus, the raw answer, includ- ing the uncertainty, is x=± ×(. )136 524 80 2 105. Since the fi nal answer should contain all the digits we are sure of and one estimated digit, this result should be rounded and displayed as 137 10 1 37 1057×= × . (we are sure of the 1 and the 3, but have uncertainty about the 7). We see that this answer has three signifi cant fi gures and choice (d) is correct. ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS 2. Atomic clocks are based on the electromagnetic waves that atoms emit. Also, pulsars are highly regular astronomical clocks. 45 m h 26/H11034 45 m h 26/H11034 56157_01_ch01_p001-016.indd 256157_01_ch01_p001-016.indd 2 10/12/10 1:26:51 PM10/12/10 1:26:51 PM Introduction 3 4. (a) ~.0 5 lb 0.25 kg or ~10 kg1≈− (b) ~4l b 2k go r~ 1 0 k g0≈ (c) ~ 4000 lb 2000 kg or ~10 kg3≈ 6. Let us assume the atoms are solid spheres of diameter 10−10 m. Then, the volume of each atom is of the order of 10−30 m3. (More precisely, volume = 43 6ππrd33= .) Therefore, since 1c m 1 0 m36 3=−, the number of atoms in the 1 cm3 solid is on the order of 10 1063 0−−=1024 atoms. A more precise calculation would require knowledge of the density of the solid and the mass of each atom. However, our estimate agrees with the more precise calculation to within a factor of 10. 8. Realistically, the only lengths you might be able to verify are the length of a football fi eld and the length of a housefl y. The only time intervals subject to verifi cation would be the length of a day and the time between normal heartbeats. 10. In the metric system, units differ by powers of ten, so it’s very easy and accurate to convert from one unit to another. ANSWERS TO EVEN NUMBERED PROBLEMS 2. (a) LT2 (b) L 4. All three equations are dimensionally incorrect. 6. (a) kg m s⋅ (b) Ft = p 8. (a) 22.6 (b) 22.7 (c) 22.6 is more reliable 10. (a) 30 0 1 08.× ms (b) 2 997 9 108. × ms (c) 2 997 925 108. × ms 12. (a) 346 13mm22± (b) 66 0 1 3..mm± 14. (a) 797 (b) 1.1 (c) 17.66 16. 30 9.c m s 18. (a) 5 60 10 5 60 10 5 60 1025 7.. .×= × = × km m cm (b) 0 491 2 491 2 4 912 104.. . km m cm== × (c) 6 192 6 192 10 6 192 1035.. . kmmc m =×=× (d) 2 499 2 499 10 2 499 1035.. . kmmc m =×=× 20. 10 6.k m L 22. 92.n m s 24. 29 1 0 29 1 028..×= × mc m33 26. 25 7 1 06.× m3 56157_01_ch01_p001-016.indd 356157_01_ch01_p001-016.indd 3 10/12/10 1:26:51 PM10/12/10 1:26:51 PM4 Chapter 1 28. ∼108steps 30. ~ people with colds on any given day108 32. (a) 42 1 018 3.×−m (b) ~m1013− (c) ~ cells1016 34. (a) ∼1029prokaryotes (b) ~k g1014 (c) The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for fi xing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them! 36. 2.2 m 38. 8.1 cm40. Δsr r r r=+ − − ()12 22 12 1 22c o s θθ 42. 2.33 m44. (a) 1.50 m (b) 2.60 m 46. 8.60 m48. (a) and (b) (c) yx y x=° − () =° tan . , . tan .12 0 1 00 14 0 km (d) 14 4 1 03.× m 50. yd=⋅⋅ −tan tan tan tanθφ φθ 52. (a) 1 609.k m h (b) 88 km h (c) 16 km h 54. Assumes population of 300 million, average of 1 can week per person, and 0.5 oz per can. (a) ∼1010cans yr (b) ∼105tons yr 56. (a) 71 4 1 02.×−gal s (b) 27 0 1 043.×−ms (c) 1.03 h 58. (a) AA21 4= (b) VV21 8= 60. (a) ∼102yr (b) ∼104times 62. ∼104balls yr . Assumes 1 lost ball per hitter, 10 hitters per inning, 9 innings per game, and 81 games per year. 56157_01_ch01_p001-016.indd 456157_01_ch01_p001-016.indd 4 10/12/10 1:26:52 PM10/12/10 1:26:52 PM Introduction 5 PROBLEM SOLUTIONS 1.1 Substituting dimensions into the given equation Tg=2πC, and recognizing that 2π is a dimen- sionless constant, we have Tg[]=[] []C or TL LTTT22== = Thus, the dimensions are consistent . 1.2 (a) From x = Bt2, we fi nd that Bx t=2. Thus, B has units of [][] []Bx t==2L T2 (b) If xA f t= ()sin 2 π, then [] [ ] [ s i n ]Ax f t= ()2π But the sine of an angle is a dimensionless ratio. Therefore, [] [ ]Ax== L 1.3 (a) The units of volume, area, and height are: []V=L3, []A=L2, and []h=L We then observe that LL L32= or [VA h][ ] [ ]= Thus, the equation VA h= is dimensionally correct . (b) VR h R h A hcylinder == () = ππ22, where AR=π2 Vw h w h A hrectangular box == () = CC , where A w length width== ×C 1.4 (a) In the equation 1 22 1 2 02mmm g h vv=+ , [] []mmvv2 022 ==⎛ ⎝⎜⎞ ⎠⎟= ML TML T2 2 while mgh⎡⎣⎤⎦=⎛ ⎝⎜⎞ ⎠⎟= ML TLML T 21 2 . Thus, the equation is dimensionally incorrect . (b) In vv=+02at, [] [ ]vv==0L T but [] [ ] [ ]at a t22==⎛ ⎝⎜⎞ ⎠⎟() =L TT L22. Hence, this equation is dimensionally incorrect . (c) In the equation ma=v2, we see that [] [ ] [ ]ma m a==⎛ ⎝⎜⎞ ⎠⎟= ML TML T22, while []v22 =⎛ ⎝⎜⎞ ⎠⎟=L TL T2 2. Therefore, this equation is also dimensionally incorrect . 1.5 From the universal gravitation law, the constant G is GF r M m=2. Its units are then GFr Mm[]=[]⎡⎣⎤⎦ [] [ ]=⋅() () ⋅=⋅2kg m s m kg kgm kg s223 2 56157_01_ch01_p001-016.indd 556157_01_ch01_p001-016.indd 5 10/12/10 1:26:52 PM10/12/10 1:26:52 PM6 Chapter 1 1.6 (a) Solving KEp m=2 2 for the momentum, p, gives pm K E= ()2 where the numeral 2 is a dimensionless constant. Dimensional analysis gives the units of momentum as: pm K E[]=[] [ ] =⋅ () =⋅ = () MML T M L T MLT22 222 Therefore, in the SI system, the units of momentum are kg m s⋅() . (b) Note that the units of force are kg m s2⋅ or F[]=⋅MLT2. Then, observe that Ft p [] [ ] =⋅() ⋅= () =[] MLT T MLT2 From this, it follows that force multiplied by time is proportional to momentum: Ft p=. (See the impulse–momentum theorem in Chapter 6, Ft p⋅=ΔΔ , which says that a constant force F multiplied by a duration of time ∆t equals the change in momentum, ∆p.) 1.7 Area length width =() ×() =() ( ) = 97 2 53 5 22..mm m 1.8 (a) Computing 83() without rounding the intermediate result yields 82 2 63() = . to three signifi cant fi gures. (b) Rounding the intermediate result to three signifi cant fi gures yields 8 2 8284=→. 2.83 Then, we obtain 82 2733() =() = 2.83 . to three signifi cant fi gures. (c) The answer 22.6 is more reliable because rounding in part (b) was carried out too soon. 1.9 (a) 78 9 0 2..± has 3 significant figures with the uncertainty in the tenths position. (b) 3 788 109.× has 4 significant figures (c) 24 6 1 06.×− has 3 significant figures (d) 0 003 2 3 2 103.. =×− has 2 significant figures . The two zeros were originally included only to position the decimal. 1.10 c=×2 997 924 58 108.m s (a) Rounded to 3 signifi cant fi gures: c=×30 0 1 08.m s (b) Rounded to 5 signifi cant fi gures: c=×2 997 9 108.m s (c) Rounded to 7 signifi cant fi gures: c=×2 997 925 108.m s 1.11 Observe that the length C=56 2.c m , the width w=63 5.c m , and the height h=27 8.c m all contain 3 signifi cant fi gures. Thus, any product of these quantities should contain 3 signifi cant fi gures. (a) Cw=() () = 5 62 6 35 35 7.. .cm cm cm2 (b) Vw h=() =() () = C 35 7 2 78 99 2.. .cm cm cm23 continued on next page 56157_01_ch01_p001-016.indd 656157_01_ch01_p001-016.indd 6 10/12/10 1:26:53 PM10/12/10 1:26:53 PM Introduction 7 (c) wh=() () = 6 35 2 78 17 7.. .cm cm cm2 Vw h=() =() () = C17 7 5 62 99 5.. .cm cm cm23 (d) In the rounding process, small amounts are either added to or subtracted from an answer to satisfy the rules of signifi cant fi gures. For a given rounding, different small adjustments are made, introducing a certain amount of randomness in the last signifi cant digit of the fi nal answer. 1.12 (a) Ar== ± () =() ±() ( ) + ππ π2 221 05 02 1 05 2 1 05 02 0.. . ..mm m mm . . 22m() ⎡⎣⎤⎦ Recognize that the last term in the brackets is insignifi cant in comparison to the other two. Thus, we have A=±⎡⎣⎤⎦=±π110 4 2 346 13mm m m22 2 2. (b) Cr== ± () =± 22 1 3ππ 10.5 m 0.2 m 66.0 m m . 1.13 The least accurate dimension of the box has two signifi cant fi gures. Thus, the volume (product of the three dimensions) will contain only two signifi cant fi gures. Vw h=⋅ ⋅= () ( ) ( ) =× C 29 17 8 11 4 5 9 1033cm cm cm cm .. . 1.14 (a) The sum is rounded to 797 because 756 in the terms to be added has no positions beyond the decimal. (b) 0 003 2 356 3 3 2 10 356 3 1 140163.. . . . ×= × () ×=− must be rounded to 11. because 3.2 103×− has only two signifi cant fi gures. (c) 5.620 ×π must be rounded to 17.66 because 5.620 has only four signifi cant fi gures. 1.15 d=()⎛ ⎝⎜⎞ ⎠⎟⎛250 0005 280 1mift 1.000 mifathom 6f t⎝ ⎝⎜⎞ ⎠⎟=×21 08fathoms The answer is limited to one signifi cant fi gure because of the accuracy to which the conversion from fathoms to feet is given. 1.16 v==⎛ C t186 11 14furlongs fortnightfortnight days⎝ ⎝⎜⎜⎞ ⎠⎟⎟ ×⎛ ⎝⎜⎜⎞ ⎠⎟⎟1 86 4 1 0220 14day syds furlong .⎛ ⎛ ⎝⎜⎜⎞ ⎠⎟⎟⎛ ⎝⎜⎜⎞ ⎠⎟⎟⎛ ⎝⎜⎞ ⎠⎟3 1100 3 281ft ydcm ft . giving v=30 9.c m s 1.17 60 0 60 01..firkins firkins9g a l firkin3=⎛ ⎝⎜⎜⎞ ⎠⎟⎟. .786 L galcm Lm 10 cm33 63110 113 ⎛ ⎝⎜⎜⎞ ⎠⎟⎟⎛ ⎝⎜⎞ ⎠⎟⎛ ⎛ ⎝⎜⎞ ⎠⎟=0 204.m3 1.18 (a) C=()⎛ ⎝⎜⎞ ⎠⎟=× = 3481 60956 0 1 0 52mikm 1.000 mikm... 6 60 10 5 60 1057×= × mc m . (b) h=()⎛ ⎝⎜⎞ ⎠⎟== 110 491 2 491 612 ft.609 km 5 280 ftkm .. 2 2 4 912 104mc m =×. (c) h=()⎛ ⎝⎜⎞ ⎠⎟== 216 192 6 19 0 320 ft.609 km 5 280 ftkm .. 2 2 10 6 192 1035×= × mc m . continued on next page 56157_01_ch01_p001-016.indd 756157_01_ch01_p001-016.indd 7 10/12/10 1:26:53 PM10/12/10 1:26:53 PM8 Chapter 1 (d) d=()⎛ ⎝⎜⎞ ⎠⎟== 8012 2 499 20 f t.609 km 5 280 ft.499 km . ××= ×10 2 499 1035mc m . In (a), the answer is limited to three signifi cant fi gures because of the accuracy of the original data value, 348 miles. In (b), (c), and (d), the answers are limited to four signifi cant fi gures because of the accuracy to which the kilometers-to-feet conversion factor is given. 1.19 v=⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟ 38 01 101 1 6093 600 3..m skm mmi kms s hmi h185 0⎛ ⎝⎜⎞ ⎠⎟=. Yes, the driver is exceeding the speed limit by 10.0 mi h . 1.20 efficiency =⎛ ⎝⎜⎞ ⎠⎟ 25 01 0 6211 37...mi galkm migal 8 8610 6Lkm L⎛ ⎝⎜⎜⎞ ⎠⎟⎟= . 1.21 (a) rdiameter==⎛ ⎝⎜⎞ ⎠⎟=253 6 225 4 168 1...in cm incm (b) Ar== () =× 4 4 6 81 5 83 102 2 22ππ ..cm cm (c) Vr== () =×4 34 36 81 1 32 103 3 33ππ ..cm cm 1.22 rate=⎛ ⎝⎜⎜⎞ ⎠⎟⎟⎛ ⎝⎜⎜⎞ ⎠⎟⎟1 321 1 in dayday 24 hh 3 600 s scm 1.00 innm 10 cm2⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟=25 4 1 09.992.n m s This means that the proteins are assembled at a rate of many layers of atoms each second! 1.23 c=×⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞30 0 1 018.m s3 600 s 1hkm 10 m3⎠ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟=×167 1 1 08 mi 1.609 kmmi h . 1.24 Volume of house ft ft ft =() ( ) ( )×50 0 26 8 02 832 1... 0 0 29 1 0 29 1 01002 22−⎛ ⎝⎜⎞ ⎠⎟ =× =× ()m 1f t mmc3 3 33..m m 1mcm3 ⎛ ⎝⎜⎞ ⎠⎟=×3 829 1 0. 1.25 Volume acre ftm 3.281 ftft2 =()⎛ ⎝⎜⎞ ⎠⎟25 01 43 560.11 acrem 3.281 ft⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟⎡ ⎣⎢ ⎢⎤ ⎦⎥ ⎥=×130 8 1 02 .443m 1.26 Volume of pyramid area of base height 13=() () =1 3 1 3. .0 acres ft acre 481 ft 9.08 1027() () ⎡⎣⎤⎦() =× 43 560 f 9.08 10 ftm 1f t733 3=×()× ⎛ ⎝⎜⎞ ⎠⎟=−2 832 1022..557 106× m3 1.27 Volume of cube quart ==L31 (Where L = length of one side of the cube.) Thus, L31=( )⎛ ⎝⎜⎜⎞ ⎠⎟⎟quart1 gallon 4 quarts3.786 l iiter 1 gallon1000 cm 1 liter943 ⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟=77c m3 and L==9473cm 9.82 cm3 56157_01_ch01_p001-016.indd 856157_01_ch01_p001-016.indd 8 10/18/10 1:21:48 PM10/18/10 1:21:48 PM Introduction 9 1.28 We estimate that the length of a step for an average person is about 18 inches, or roughly 0.5 m. Then, an estimate for the number of steps required to travel a distance equal to the circumference of the Earth would be NRE== ≈Circumference Step Length Step Length2 26 π π.3 38 10 0581 06 7×()≈×m ms t e psteps. or N∼108steps 1.29. We assume an average respiration rate of about 10 breaths/minute and a typical life span of 70 years. Then, an estimate of the number of breaths an average person would take in a lifetime is n=⎛ ⎝⎜⎞ ⎠⎟()× ⎛ ⎝⎜⎜10 703 156 107breaths minyrs 1y r. ⎞ ⎞ ⎠⎟⎟⎛ ⎝⎜⎞ ⎠⎟=×141 08 min 60 sbreaths or n∼108breaths 1.30 We assume that the average person catches a cold twice a year and is sick an average of 7 days (or 1 week) each time. Thus, on average, each person is sick for 2 weeks out of each year (52 weeks). The probability that a particular person will be sick at any given time equals the percentage of time that person is sick, or probability of sicknessweeks 52 weeks==21 26 The population of the Earth is approximately 7 billion. The number of people expected to have a cold on any given day is then Number sick population probability of sickness =() (() =×()⎛ ⎝⎜⎞ ⎠⎟=× 71 01 2631 0 1 098 8or∼ 1.31 (a) Assume that a typical intestinal tract has a length of about 7 m and average diameter of 4 cm. The estimated total intestinal volume is then VAd total3 mmm ==⎛ ⎝⎜⎞ ⎠⎟=()() = CCπ π2 2 400 4 47 0 009.. The approximate volume occupied by a single bacterium is V typical length scalebacteria m ∼() =() =−− 3 6310 10118m3 If it is assumed that bacteria occupy one hundredth of the total intestinal volume, the esti- mate of the number of microorganisms in the human intestinal tract is nV V==()=×−total bacteria3 3m m100 0 009 100 1091 018.113 1410 orn∼ (b) The large value of the number of bacteria estimated to exist in the intestinal tract means that they are probably not dangerous. Intestinal bacteria help digest food and provide important nutrients. Humans and bacteria enjoy a mutually benefi cial symbiotic relationship. 1.32 (a) Vrcell mm == × () =×−− 4 34 310 1 0 42 1 036318 3ππ .. (b) Consider your body to be a cylinder having a radius of about 6 inches (or 0.15 m) and a height of about 1.5 meters. Then, its volume is VA h r hbody mm m o r == () =() ( ) = ππ2 2 301 5 15 01 1.. . ∼ ∼10 m−13 continued on next page 56157_01_ch01_p001-016.indd 956157_01_ch01_p001-016.indd 9 10/12/10 1:26:54 PM10/12/10 1:26:54 PM10 Chapter 1 (c) The estimate of the number of cells in the body is then nV V==×=×−body cellm mor01 1 42 1 026 1 03 18 316 . .. ∼ ∼1016 1.33 A reasonable guess for the diameter of a tire might be 3 ft, with a circumference ( Cr D==2ππ = distance travels per revolution) of about 9 ft. Thus, the total number of revolutions the tire might make is ntotal distance traveled distance per revolutio=n n=() ( )=×50 000 mi ft mi ft revrev, or ~5 280 931 071 10 rev7 1.34 Answers to this problem will vary, dependent on the assumptions one makes. This solution assumes that bacteria and other prokaryotes occupy approximately one ten-millionth (10−7) of the Earth’s volume, and that the density of a prokaryote, like the density of the human body, is approximately equal to that of water (10 3 kg/m3). (a) estimated number nV V==− total single prokaryote∼107(() ()−V VR lengthEarth single prokaryoteEarth∼1073 s scale ()() ( ) ()− −3763 632910 10 1010 ∼∼m m (b) m density total volume nVtotal water single pr=() () ∼ρ ookaryote3kg mm⎛ ⎝⎜⎞ ⎠⎟=⎛ ⎝⎜⎞ ⎠⎟() ( )−10 10 10 132 9 63∼0 014kg (c) The very large mass of prokaryotes implies they are important to the biosphere. They are responsible for fi xing carbon, producing oxygen, and breaking up pollutants, among many other biological roles. Humans depend on them! 1.35 The x coordinate is found as xr== ( ) = cos . cosθ25 3 5m° 2 . 0 m and the y coordinate yr== ( ) = sin . sinθ25 3 5m° 1 . 4 m 1.36 The x distance out to the fl y is 2.0 m and the y distance up to the fl y is 1.0 m. Thus, we can use the Pythagorean theorem to fi nd the distance from the origin to the fl y as mm dx y=+ = ( )+( )=222220 10.. 2.2 m 1.37 The distance from the origin to the fl y is r in polar coordinates, and this was found to be 2.2 m in Problem 36. The angle θ is the angle between r and the horizontal reference line (the x axis in this case). Thus, the angle can be found as tan.. θ== =y x1005 0m 2.0 m and θ= () =°−tan .105 0 2 7 The polar coordinates are r= = °2.2 m and 27 θ 1.38 The x distance between the two points is Δxxx=−= − − =21 30 50 80.. .cm cm cm and the y distance between them is Δyyy=−= − =21 30 40 10.. .cm cm cm . The distance between them is found from the Pythagorean theorem: dx y=+ = () +() = ΔΔ2 2 2280 10 6 5..cm cm cm =28.1 cm 1.39 Refer to the Figure given in Problem 1.40 below. The Cartesian coordinates for the two given points are: xr x11 1 2 20 0 5 00 12 9 == () °= = cos . cos . .θ mm r r yr22 1150 0 5 00 32 1 cos . cos . . sθ=() −°() = =mm iin . sin . . sinθ12 2 20 0 5 00 15 3=() °= = mm yr θ θ250 0 5 00 38 3=() −°() =− .s i n . .mm continued on next page 56157_01_ch01_p001-016.indd 1056157_01_ch01_p001-016.indd 10 10/12/10 1:26:55 PM10/12/10 1:26:55 PM Introduction 11 The distance between the two points is then: ΔΔΔsxy=() +() =−() ++()2 2 212 9 32 1 15 3 38 3.. ..mm mm2 256 9=.m 1.40 Consider the Figure shown at the right. The Cartesian coordinates for the two points are: xr xr yr11 1 2 2 2 11 1== =cos cos sinθθ θ yr22 2=sinθ The distance between the two points is the length of the hypotenuse of the shaded triangle and is given by ΔΔΔsxy x x y y=() +() =−() +−()2 2 122 122 or Δsr r r r r=+ −() +122 1222 21 2 1 2 122 cos cos cos cos sinθθ θ θ2 2 1222 21 2 1 2 122 122 θθ θ θ θ+− () =+rr r rsin sin sin cos sin θθθ θ θ θ θ1222 22 21 2 12 12 () ++() −+ rr rcos sin cos cos sin s iinθ2 () Applying the identities cos sin221 θθ+= and cos cos sin sin cosθθ θ θ θ θ12 1 2 1 2 += − () , this reduces to Δsr r r r r r=+ − + () =+12 22 12 1 2 1 2 12 222 cos cos sin sin θθ θ θ −−− () 212 1 2rrcosθθ 1.41 (a) With a=60 0.m and b being two sides of this right triangle having hypotenuse c=90 0.m , the Pythagorean theorem gives the unknown side as bc a=− = () −() =22 2290 0 60 0 m m 6.71 m.. (b) tan.θ== =a b60 0m 6.71 m0.894 (c) sin. .φ== =b c67 1 90 0m m0.746 1.42 From the diagram, cos .75 0°() =dL Thus, dL=° () =() ° () = cos 75.0 m cos 75.0 90 0. 2.33 my r1 r2(x1, y1) (x2, y2)Δy/H11005y1/H11002y2 Δx/H11005x1/H11002x2Δs xq1q2y r1 r2(x1, y1) (x2, y2)Δy/H11005y1/H11002y2 Δx/H11005x1/H11002x2Δs xq1q2 75.0/H11034L/H110059.00 m d 75.0/H11034L/H110059.00 m d 56157_01_ch01_p001-016.indd 1156157_01_ch01_p001-016.indd 11 10/12/10 1:26:55 PM10/12/10 1:26:55 PM12 Chapter 1 1.43 The circumference of the fountain is Cr=2π, so the radius is rC== =215 0 223 9ππ..mm Thus, tan ..55 023 9° () ==h rh m which gives h=() ° () = 23 9 5 50.t a n .m 34 1.m 1.44 (a) sinθ=opposite side hypotenuse so, opposite side m ° =() ( ) = 30 0 3 00.s i n . 15 0.m (b) cosθ=adjacent side hypotenuse so, adjacent side m ° =() ( ) = 30 0 3 00.c o s . 26 0.m 1.45 (a) The side opposite θ = 3.00 (b) The side adjacent to φ = 3.00 (c) cosθ==4.00 5.000.800 (d) sinφ==4.00 5.000.800 (e) tanφ==4.00 3.0013 3. 1.46 Using the diagram at the right, the Pythagorean theorem yields c=() +() = 50 0 70 022..mm 8.60 m 1.47 From the diagram given in Problem 1.46 above, it is seen that tan. .. θ==50 0 70 00 714 and θ= () =°−tan .10 714 35.5 1.48 (a) and (b) See the Figure given at the right. (c) Applying the defi nition of the tangent function to the large right triangle containing the 12.0° angle gives: yx=°tan .12 0 [1] Also, applying the defi nition of the tangent function to the smaller right triangle containing the 14.0° angle gives: y x−=°10 014 0.tan .km [2] (d) From Equation [1] above, observe that xy=°tan .12 0 Substituting this result into Equation [2] gives y y⋅° −() °=°tan . .t a n .tan .12 0 10 0 1 2014 0km 7.00 m5.00 mc q 7.00 m5.00 mc q continued on next page 56157_01_ch01_p001-016.indd 1256157_01_ch01_p001-016.indd 12 10/15/10 9:53:57 AM10/15/10 9:53:57 AM Introduction 13 Then, solving for the height of the mountain, y, yields y=() °° °− °=10 0 1 20 1 40 14 0 12 01. tan . tan . t a n. t a n..km4 44 1 44 103km m =×. 1.49 Using the sketch at the right: w 10035 0m,o r =°tan . w=() °= 100 35 0mt a n . 70.0 m 1.50 The fi gure at the right shows the situation described in the problem statement. Applying the defi nition of the tangent function to the large right triangle containing the angle θ in the Figure, one obtains yx=tanθ [1] Also, applying the defi nition of the tangent function to the small right triangle containing the angle φ gives y xd−=tanφ [2] Solving Equation [1] for x and substituting the result into Equation [2] yields y ydy yd tantantan ta θφθ −=⋅ −⋅orn ntanθφ = The last result simplifi es to yy d⋅= ⋅− ⋅⋅tan tan tan tanθφθ φ Solving for y: ydtan tan tan tanθφ θ φ− () =− ⋅ ⋅ or ydd=−⋅⋅ −=⋅⋅ −tan tan tan tantan tan tan tanθφ θφθφ φθ 1.51 (a) Given that aF m∝ , we have Fm a∝ . Therefore, the units of force are those of ma, [] [ ] [] [ ]Fm am a== = () = MLT M L T2- 2 (b) []F=⎛ ⎝⎜⎞ ⎠⎟=⋅ML TML Tso newto22nn=kg m s2⋅ 1.52 (a) 111mi hmi h1.609 km mi=⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟=1 609 .km h (b) vmaxmi hmi h1.609 km h mi h==⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟= 55 55188km h (c) Δvmaxmi hmi hmi h1.609 km h mi h=−=⎛ ⎝⎜⎞ ⎠⎟⎛65 55 101⎝ ⎝⎜⎞ ⎠⎟=16km hww 56157_01_ch01_p001-016.indd 1356157_01_ch01_p001-016.indd 13 10/15/10 9:53:59 AM10/15/10 9:53:59 AM14 Chapter 1 1.53 (a) Since 11 02mc m= , then 1 1 10 10 103 23236m m cm cm cm33 3=() =() =() = , giving mass density volumekg 1.0 cm3=( )( )=× ⎛ ⎝−10 1 03. ⎜ ⎜⎞ ⎠⎟( ) =×⎛ ⎝⎜⎞ ⎠⎟( )−10 10 1 0 103. ..m kg cmm103 33663 3cm 1mkg⎛ ⎝⎜⎞ ⎠⎟=×10 1 03. As a rough calculation, treat each of the following objects as if they were 100% water. (b) cell: mass density volumekg m3=× =⎛ ⎝⎜⎞ ⎠⎟ ×−104 305 0 1 036π. mmk g () =×−31652 1 0. (c) kidney: mass density volumekg m3=× =⎛ ⎝⎜⎞ ⎠⎟=⎛ ⎝⎜⎞ ⎠⎟ ρπ4 31033r4 4 340 1 0 02 723π..×() =−mk g (d) fl y: mass density volume density kg m3=× = () () =⎛ ⎝⎜πrh2 310⎞ ⎞ ⎠⎟ ×() ×() =×−− −π10 1 0 40 1 0 13 1 03235.. . mm k g 1.54 Assume an average of 1 can per person each week and a population of 300 million. (a) number cans yearnumber cans person weekpopu =⎛ ⎝⎜⎞ ⎠⎟ llation weeks year 1can person week31 08() ( ) ≈⎛ ⎝⎜⎞ ⎠⎟× people 52 weeks yr cans yr , or ~1010() () ≈×21 010cans yr (b) number of tons weight can number cans year 0.5=() ( ) ≈oz canlb 16 ozton 2 000 lb⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟⎡ 11 ⎣ ⎣⎢⎤ ⎦⎥×⎛ ⎝⎜⎞ ⎠⎟ ≈×21 0can yr 3 10 ton yr or ~1010 55, ton yr Assumes an average weight of 0.5 oz of aluminum per can. 1.55 The term s has dimensions of L, a has dimensions of LT−2, and t has dimensions of T. Therefore, the equation, sk a tmn= with k being dimensionless, has dimensions of LL T T2=() ()−m n or LT L T10=− mn m 2 The powers of L and T must be the same on each side of the equation. Therefore, L1 = Lm and m=1 Likewise, equating powers of T, we see that nm−=20 , or nm==22 Dimensional analysis cannot determine the value ofk, a dimensionless constant. 1.56 (a) The rate of fi lling in gallons per second is rate=⎛ ⎝⎜⎞ ⎠⎟=×− 30 0 70 01 6071 4 1 02 . ..gal minmin sg gal s continued on next page 56157_01_ch01_p001-016.indd 1456157_01_ch01_p001-016.indd 14 10/12/10 1:26:57 PM10/12/10 1:26:57 PM Introduction 15 (b) Note that 11 0 1 01 101032363 333mc m c mL cmL =() =()⎛ ⎝⎜⎞ ⎠⎟= . Thus, rate=×⎛ ⎝⎜⎞ ⎠⎟⎛−71 4 1 03 786 11 1023 3.. gal sL galm L ⎝ ⎝⎜⎞ ⎠⎟=×−27 0 1 043.m s (c) tV rate==×=×−filled 3m mss10 0 27 0 1 037 0 1 013 43 . ..h h sh3 60010 3⎛ ⎝⎜⎞ ⎠⎟=. 1.57 The volume of paint used is given by V = Ah, where A is the area covered and h is the thickness of the layer. Thus, hV A==×=× = ×− −− 37 9 1 0 25 01 52 10 152 1033 246 . ..m mmm = =152 mμ 1.58 (a) For a sphere, AR=42π. In this case, the radius of the second sphere is twice that of the fi rst, or R R212= . Hence, A AR RRRR R 2 122 1222 1212 124 42== =()=π π4 (b) For a sphere, the volume is VR=4 33π Thus, V VR RR RR R2 123 1323 1313 1343 432=() ()==()=π π8 1.59 The estimate of the total distance cars are driven each year is d=() () =× cars in use distance traveled per car 100 1 10 10 1 1064 1 2cars mi car mi () ( ) =× At a rate of 20 mi/gal, the fuel used per year would be Vd rate1 112 10 11 0 2051 0 ==×=×mi mi galgal If the rate increased to 25 mi gal , the annual fuel consumption would be Vd rate2 212 10 11 0 2541 0 ==×=×mi mi galgal and the fuel savings each year would be savings V V =−= × − × = ×1210 10 105 10 4 10 1 10 gal gal gal 1.60 (a) The amount paid per year would be annual amount 000dollars s8.64 10 s 1.00 day4 =⎛ ⎝⎜⎞ ⎠⎟× ⎛ ⎝⎜ 1⎞ ⎞ ⎠⎟⎛ ⎝⎜⎞ ⎠⎟=×365.25 days yr3.16 10dollars yr10 Therefore, it would take 10 1031 012 2 × ×=×dollars 3.16 10 dollars yryr10, , or yr ~102 (b) The circumference of the Earth at the equator is Cr== × ( )=× 2 2 6 378 10 4 007 1067ππ .. mm continued on next page 56157_01_ch01_p001-016.indd 1556157_01_ch01_p001-016.indd 15 10/12/10 1:26:57 PM10/12/10 1:26:57 PM16 Chapter 1 The length of one dollar bill is 0.155 m, so the length of ten trillion bills is C=⎛ ⎝⎜⎞ ⎠⎟ ×() =× 0 155 10 10 1 1012.m dollardollars m12. Thus, the ten trillion dollars would encircle the Earth nC==× ×=×C 11 021 012 4 m 4.007 10 mor7, ~104times 1.61 (a) 1y r 1 y r365.2 days 1y r8.64 10 s 1d a4 =()⎛ ⎝⎜⎜⎞ ⎠⎟⎟× y y3.16 10 s7 ⎛ ⎝⎜⎞ ⎠⎟=× (b) Consider a segment of the surface of the Moon which has an area of 1 m2 and a depth of 1 m. When fi lled with meteorites, each having a diameter 10−6 m, the number of meteorites along each edge of this box is n==length of an edge meteorite diameterm 1 10− −=6610 m The total number of meteorites in the fi lled box is then Nn== ()=36 3 1 810 10 At the rate of 1 meteorite per second, the time to fi ll the box is t== ()×⎛ ⎝⎜⎞ ⎠⎟=× 10 101 31 6 1 031 018 18 710ssy syr,.or ~1010yr 1.62 We will assume that, on average, 1 ball will be lost per hitter, that there will be about 10 hitters per inning, a game has 9 innings, and the team plays 81 home games per season. Our estimate of the number of game balls needed per season is then number of balls needed number lost per hitter =() nnumber hitters/game home games/year 1b a l l() ( ) = pper hitter 10hitters inning9innings ga()⎛ ⎝⎜⎞ ⎠⎟mmegames year 7300balls yea⎛ ⎝⎜⎞ ⎠⎟⎡ ⎣⎢⎤ ⎦⎥⎛ ⎝⎜⎞ ⎠⎟ =81 r rorballs year~104 1.63 The volume of the Milky Way galaxy is roughly VA tdtG==⎛ ⎝⎜⎞ ⎠⎟≈( )( )ππ2 21219 4410 10 mm o r rmVG∼1061 3 If, within the Milky Way galaxy, there is typically one neutron star in a spherical volume of radius r=×31 018m, then the galactic volume per neutron star is Vr V131 8356 3 14 34 331 0 11 0 ==× () =× ππ mm o r ∼ ∼1056 3m The order of magnitude of the number of neutron stars in the Milky Way is then nV VG= 161 3 56 310 10∼m m or n∼105neutron stars 56157_01_ch01_p001-016.indd 1656157_01_ch01_p001-016.indd 16 10/12/10 1:26:58 PM10/12/10 1:26:58 PM