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INSTRUCTOR SOLUTIONS MANUAL(vol 1 & 2ch 01-33)

INSTRUCTORSOLUTIONS MANUALVOLUMES 1 & 2VOLUME 1DOUGLAS C. GIANCOLI’SPHYSICSPRINCIPLES WITHAPPLICATIONS7TH EDITIONBOB DAVISTAYLOR UNIVERSITYJ. ERIK HENDRICKSONUNIVERSITY OF WISCONSIN – EAU CLAIRESan Francisco Boston New YorkCape Town Hong Kong London Madrid Mexico CityMontreal Munich Paris Singapore Sydney Tokyo Toronto

President, Science, Business and Technology: Paul CoreyPublisher: Jim SmithExecutive Development Editor: Karen KarlinProject Manager: Elisa MandelbaumMarketing Manager: Will MooreSenior Managing Editor: Corinne BensonManaging Development Editor: Cathy MurphyProduction Service: PreMedia Global, Inc.ISBN 10: 0-321-74768-2ISBN 13: 978-0-321-74768-6Copyright 2014, 2009, 1998 Pearson Education, Inc 1301 Sansome St., San Francisco, CA 94111. All rightsreserved. Manufactured in the United States of America. This publication is protected by Copyright andpermission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrievalsystem, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, orlikewise. To obtain permission(s) to use material from this work, please submit a written request to PearsonEducation, Inc., Permissions Department, 1900 E. Lake Ave., Glenview, IL 60025. For information regardingpermissions, call (847) 486-2635. Many of the designations used by manufacturers and sellers to distinguishtheir products are claimed as trademarks. Where those designations appear in this book, and the publisher wasaware of a trademark claim, the designations have been printed in initial caps or all caps.www.pearsonhighered.com

CONTENTSChapter 1Chapter 2Chapter 3Chapter 4Chapter 5Chapter 6Chapter 7Chapter 8Chapter 9Chapter 10Chapter 11Chapter 12Chapter 13Chapter 14Chapter 15Chapter 16Chapter 17Chapter 18Chapter 19Chapter 20Chapter 21Chapter 22Chapter 23Chapter 24Chapter 25Chapter 26Chapter 27Chapter 28Chapter 29Chapter 30Chapter 31Chapter 32Chapter 33PREFACE ----------------------- ivIntroduction, Measurement, Estimating ------------------------------------- 1-1Describing Motion: Kinematics in One Dimension ------------------------ 2-1Kinematics in Two Dimensions; Vectors------------------------------------ 3-1Dynamics: Newton’s Laws of Motion --------------------------------------- 4-1Circular Motion; Gravitation -------------------------------------------------- 5-1Work and Energy ------------- 6-1Linear Momentum ------------ 7-1Rotational Motion ------------ 8-1Static Equilibrium; Elasticity and Fracture---------------------------------- 9-1Fluids --------------------------10-1Oscillations and Waves ature and Kinetic Theory ------------------------------14-1The Laws of -----------15-1Electric Charge and Electric -1Electric ----------------------17-1Electric Currents -------------18-1DC Circuits -------------------19-1Magnetism --------------------20-1Electromagnetic Induction and Faraday’s Law ----------------------------21-1Electromagnetic Waves -----22-1Light: Geometric Optics ----23-1The Wave Nature of -------24-1Optical ---------------------25-1The Special Theory of Relativity rly Quantum Theory and Models of the Atom--------------------------27-1Quantum Mechanics of --28-1Molecules and --------------29-1Nuclear Physics and Radioactivity ear Energy; Effects and Uses of Radiation ---------------------------31-1Elementary ------------------32-1Astrophysics and Cosmology -1 Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

PREFACEThis Instructor’s Solutions Manual provides answers and worked-out solutions to all end of chapterquestions and problems from chapters 1 – 15 of Physics: Principles with Applications, 7th Edition, byDouglas C. Giancoli. At the end of the manual are grids that correlate the 6th edition questions and problemsto the 7th edition questions and problems.We formulated the solutions so that they are, in most cases, useful both for the student and the instructor.Accordingly, some solutions may seem to have more algebra than necessary for the instructor. Other solutionsmay seem to take bigger steps than a student would normally take: e.g. simply quoting the solutions from aquadratic equation instead of explicitly solving for them. There has been an emphasis on algebraic solutions,with the substitution of values given as a very last step in most cases. We feel that this helps to keep thephysics of the problem foremost in the solution, rather than the numeric evaluation.Much effort has been put into having clear problem statements, reasonable values, pedagogically soundsolutions, and accurate answers/solutions for all of the questions and problems. Working with us was a teamof five additional solvers – Karim Diff (Santa Fe College), Thomas Hemmick (Stony Brook University),Lauren Novatne (Reedley College), Michael Ottinger (Missouri Western State University), and TrinaVanAusdal (Salt Lake Community College). Between the seven solvers we had four complete solutions forevery question and problem. From those solutions we uncovered questions about the wording of the problems,style of the possible solutions, reasonableness of the values and framework of the questions and problems, andthen consulted with one another and Doug Giancoli until we reached what we feel is both a good statementand a good solution for each question and problem in the text.Many people have been involved in the production of this manual. We especially thank Doug Giancoli forhis helpful conversations. Karen Karlin at Prentice Hall has been helpful, encouraging, and patient as we haveturned our thoughts into a manual. Michael Ottinger provided solutions for every chapter, and helped inthe preparation of the final solutions for some of the questions and problems. And the solutions fromKarim Diff, Thomas Hemmick, Lauren Novatne, and Trina VanAusdal were often thought-provoking andalways appreciated.Even with all the assistance we have had, the final responsibility for the content of this manual is ours. Wewould appreciate being notified via e-mail of any errors that are discovered. We hope that you will find thispresentation of answers and solutions useful.Bob Davis ([email protected])Upland, INJ. Erik Hendrickson ([email protected])Eau Claire, WI Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

INTRODUCTION, MEASUREMENT, ESTIMATING1Responses to Questions1.(a)A particular person’s foot. Merits: reproducible. Drawbacks: not accessible to the general public;not invariable (size changes with age, time of day, etc.); not indestructible.(b) Any person’s foot. Merits: accessible. Drawbacks: not reproducible (different people havedifferent size feet); not invariable (size changes with age, time of day, etc.); not indestructible.Neither of these options would make a good standard.2.The distance in miles is given to one significant figure, and the distance in kilometers is given to fivesignificant figures! The value in kilometers indicates more precision than really exists or than ismeaningful. The last digit represents a distance on the same order of magnitude as a car’s length!The sign should perhaps read “7.0 mi (11 km),” where each value has the same number ofsignificant figures, or “7 mi (11 km),” where each value has about the same % uncertainty.3.The number of digits you present in your answer should represent the precision with which youknow a measurement; it says very little about the accuracy of the measurement. For example, if youmeasure the length of a table to great precision, but with a measuring instrument that is notcalibrated correctly, you will not measure accurately. Accuracy is a measure of how close ameasurement is to the true value.4.If you measure the length of an object, and you report that it is “4,” you haven’t given enoughinformation for your answer to be useful. There is a large difference between an object that is4 meters long and one that is 4 feet long. Units are necessary to give meaning to a numerical answer.5.You should report a result of 8.32 cm. Your measurement had three significant figures. When youmultiply by 2, you are really multiplying by the integer 2, which is an exact value. The number ofsignificant figures is determined by the measurement.6.The correct number of significant figures is three: sin 30.0 0.500.7.Useful assumptions include the population of the city, the fraction of people who own cars, theaverage number of visits to a mechanic that each car makes in a year, the average number of weeks amechanic works in a year, and the average number of cars each mechanic can see in a week. Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.1-1

1-2Chapter 1(a)(b)There are about 800,000 people in San Francisco, as estimated in 2009 by the U.S. CensusBureau. Assume that half of them have cars. If each of these 400,000 cars needs servicing twicea year, then there are 800,000 visits to mechanics in a year. If mechanics typically work50 weeks a year, then about 16,000 cars would need to be seen each week. Assume that onaverage, a mechanic can work on 4 cars per day, or 20 cars a week. The final estimate, then, is800 car mechanics in San Francisco.Answers will vary.Responses to MisConceptual Questions1.(d)One common misconception, as indicated by answers (b) and (c), is that digital measurementsare inherently very accurate. A digital scale is only as accurate as the last digit that it displays.2.(a)The total number of digits present does not determine the accuracy, as the leading zeros in (c)and (d) are only placeholders. Rewriting the measurements in scientific notation shows that (d)has two-digit accuracy, (b) and (c) have three-digit accuracy, and (a) has four-digit accuracy.Note that since the period is shown, the zeros to the right of the numbers are significant.3.(b)The leading zeros are not significant. Rewriting this number in scientific notation shows that itonly has two significant digits.4.(b)When you add or subtract numbers, the final answer should contain no more decimal places thanthe number with the fewest decimal places. Since 25.2 has one decimal place, the answer mustbe rounded to one decimal place, or to 26.6.5.(b)The word “accuracy” is commonly misused by beginning students. If a student repeats ameasurement multiple times and obtains the same answer each time, it is often assumed to beaccurate. In fact, students are frequently given an “ideal” number of times to repeat theexperiment for “accuracy.” However, systematic errors may cause each measurement to beinaccurate. A poorly working instrument may also limit the accuracy of your measurement.6.(d)This addresses misconceptions about squared units and about which factor should be in thenumerator of the conversion. This error can be avoided when students treat the units as algebraicsymbols that must be cancelled out.7.(e)When making estimates, students frequently believe that their answers are more significant thanthey actually are. This question helps the student realize what an order-of-magnitude estimationis NOT supposed to accomplish.8.(d)This addresses the fact that the generic unit symbol, like [L], does not indicate a specificsystem of units.Solutions to Problems1.(a)2143 significant figures(b)81.604 significant figures(c)7.033 significant figures Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Introduction, Measurement, Estimating2.3.4.(d)0.031 significant figure(e)0.00862 significant figures(f)32364 significant figures(g)87002 significant figures(a)1.156 1.156 100(b)21.8 2.18 101(c)0.0068 6.8 10 3(d)328.65 3.2865 102(e)0.219 2.19 10 1(f)444 4.44 102(a)8.69 104 86,900(b)9.1 103 9100(c)8.8 10 1 0.88(d)4.76 102 476(e)3.62 10 5 0.0000362(a)14 billion years 1.4 1010 years(b) 3.156 107(1.4 1010 yr) 1 yr 1-3s 17 4.4 10 s 0.25 m 100% 4.6%5.48 m5.% uncertainty 6.(a)% uncertainty 0.2 s 100% 3.636% 4%5.5 s(b)% uncertainty 0.2 s 100% 0.3636% 0.4%55 s(c)The time of 5.5 minutes is 330 seconds.% uncertainty 0.2 s 100% 0.0606% 0.06%330 s Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-47.Chapter 1To add values with significant figures, adjust all values to be added so that their exponents are all thesame.(9.2 103 s) (8.3 104 s) (0.008 106 s) (9.2 103 s) (83 103 s) (8 103 s) (9.2 83 8) 103 s 100.2 103 s 1.00 105 sWhen you add, keep the least accurate value, so keep to the “ones” place in the last set of parentheses.8.When you multiply, the result should have as many digits as the number with the least number ofsignificant digits used in the calculation.(3.079 102 m)(0.068 10 1 m) 2.094 m 2 2.1 m 29.The uncertainty is taken to be 0.01 m.% uncertainty 10.0.01 m 21.57 m 2 100% 0.637% 1%To find the approximate uncertainty in the volume, calculate the volume for the minimum radius andthe volume for the maximum radius. Subtract the extreme volumes. The uncertainty in the volumeis then half of this variation in volume.3Vspecified 34 π rspecified 43 π (0.84 m)3 2.483 m33Vmin 34 π rmin 34 π (0.80 m)3 2.145 m33Vmax 34 π rmax 43 π (0.88 m)3 2.855 m3ΔV 12 (Vmax Vmin ) 12 (2.855 m3 2.145 m3 ) 0.355 m3The percent uncertainty is11.ΔVVspecified 0.355 m32.483 m3 100 14.3 14% .To find the approximate uncertainty in the area, calculate the area for the specified radius, theminimum radius, and the maximum radius. Subtract the extreme areas. The uncertainty in the areais then half this variation in area. The uncertainty in the radius is assumed to be 0.1 104 cm.2Aspecified π rspecified π (3.1 104 cm) 2 3.019 109 cm 22Amin π rmin π (3.0 104 cm) 2 2.827 109 cm 22Amax π rmax π (3.2 104 cm) 2 3.217 109 cm 2ΔA 12 ( Amax Amin ) 12 (3.217 109 cm 2 2.827 109 cm 2 ) 0.195 109 cm 2Thus the area should be quoted as A (3.0 0.2) 109 cm 2 .12.(a)286.6 mm286.6 10 3 m0.2866 m(b)85 μ V85 10 6 V0.000085 V(c)760 mg760 10 6 kg0.00076 kg (if last zero is not significant) Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Introduction, Measurement, Estimating(d)62.1 ps62.1 10 12 s0.0000000000621 s(e)22.5 nm22.5 10 9 m0.0000000225 m(f)2.50 gigavolts2.50 109 volts2,500,000,000 volts1-5Note that in part (f ) in particular, the correct number of significant digits cannot be determined whenyou write the number in this format.13.(a)1 106 volts1 megavolt 1 MV(b)2 10 6 meters2 micrometers 2 μ m(c)6 103 days6 kilodays 6 kdays(d)18 102 bucks18 hectobucks 18 hbucks or 1.8 kilobucks(e)7 10 7 seconds700 nanoseconds 700 ns or 0.7 μ s14. 1.000 104 m 2 3.281 ft 2 1 acre1 hectare (1 hectare) 1 hectare 1 m 4.356 104 ft 2 15.(a)93 million miles (93 106 miles)(1610 m/1 mile) 1.5 1011 m(b)1.5 1011 m (1.5 1011 m)(1 km/103 m) 1.5 108 km16. 2.471 acres To add values with significant figures, adjust all values to be added so that their units are all the same.1.80 m 142.5 cm 5.34 105 μ m 1.80 m 1.425 m 0.534 m 3.759 m 3.76 mWhen you add, the final result is to be no more accurate than the least accurate number used. In thiscase, that is the first measurement, which is accurate to the hundredths place when expressed in meters.17.18.(a)1.0 10 10 m (1.0 10 10 m )(39.37 in/1 m ) 3.9 10 9 in(b) 1 m 1 atom8(1.0 cm) 1.0 10 atoms 10100cmm 1.0 10(a) 0.621 mi 0.621 mi/h.(1 km/h) 0.621 mi/h, so the conversion factor is1 km/h 1 km (b) 3.28 ft 3.28 ft/s.(1 m/s) 3.28 ft/s, so the conversion factor is1 m/s 1m (c) 1000 m 1 h 0.278 m/s.(1 km/h) 0.278 m/s, so the conversion factor is1 km/h1km3600s Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-6Chapter 1Note that if more significant figures were used in the original factors, such as 0.6214 miles perkilometer, more significant figures could have been included in the answers.19.(a)Find the distance by multiplying the speed by the time.1.00 ly (2.998 108 m/s)(3.156 107 s) 9.462 1015 m 9.46 1015 m(b)Do a unit conversion from ly to AU. 9.462 1015 m 1 AU4(1.00 ly) 6.31 10 AU11 1.00ly1.5010m 20.One mile is 1609 m, according to the unit conversions in the front of the textbook. Thus it is 109 mlonger than a 1500-m race. The percentage difference is calculated here.109 m 100% 7.3%1500 m21.Since the meter is longer than the yard, the soccer field is longer than the football field.1.094 yd 100.0 yd 9.4 yd1msoccer football 100.0 m soccer football 100.0 m 100.0 yd 1m 8.6 m1.094 ydSince the soccer field is 109.4 yd compared with the 100.0-yd football field, the soccer fieldis 9.4% longer than the football field.22.23.(a)# of seconds in 1.00 yr: 3.156 1071.00 yr (1.00 yr) 1 yr s 7 3.16 10 s (b)# of nanoseconds in 1.00 yr: 3.156 1071.00 yr (1.00 yr) 1 yr s 1 109 ns 16 3.16 10 ns1s (c)# of years in 1.00 s: 1 yr1.00 s (1.00 s ) 3.156 107 (a) 10 15 kg 1 proton or neutron 12 10 protons or neutrons 27 1bacterium10kg (b) 1 proton or neutron 10 17 kg10 10 protons or neutrons 271DNAmolecule10kg (c) 102 kg 1 proton or neutron 29 10 protons or neutrons 271human10kg (d) 1041 kg 1 proton or neutron 68 10 protons or neutrons 271Galaxy10kg 8 3.17 10 yrs Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Introduction, Measurement, Estimating24.1-7The radius of the ball can be found from the circumference (represented by “c” in the equationsbelow), and then the volume can be found from the radius. Finally, the mass is found from the volumeof the baseball multiplied by the density ( ρ mass/volume) of a nucleon.cball 2π rball rball cball c 3 34 π ball ; Vball 34 π rball2π 2π m nucleon Vball 4 3 3 π rnucleon mmball Vball ρ nucleon Vball nucleon Vnucleon 4π32 mnucleon cball 2π 4π 1 d 3 2 nucleon(3 mnucleon Vball 4 1 π d nucleon 3 2( 3 )3 3 0.23 m cball m 27kg) (10nucleon 3 π (10 15 m) π d nucleon ) 3.9 1014 kg 4 1014 kg25.26.(a)2800 2.8 103 1 103 103(b)86.30 103 8.630 104 10 104 105(c)0.0076 7.6 10 3 10 10 3 10 2(d)15.0 108 1.5 109 1 109 109The textbook is approximately 25 cm deep and 5 cm wide. With books on both sides of a shelf, theshelf would need to be about 50 cm deep. If the aisle is 1.5 m wide, then about 1/4 of the floor space iscovered by shelving. The number of books on a single shelf level is then 1 book41 (3500 m 2 ) 7.0 10 books. With 8 shelves of books, the total number of books4(0.25m)(0.05m) stored is as follows:books 45 7.0 10 ( 8 shelves ) 6 10 booksshelflevel 27.The distance across the U.S. is about 3000 miles.(3000 mi)(1 km/0.621 mi)(1 h/10 km) 500 hOf course, it would take more time on the clock for a runner to run across the U.S. The runnerobviously could not run for 500 hours non-stop. If he or she could run for 5 hours a day, then it wouldtake about 100 days to cross the country.28.A commonly accepted measure is that a person should drink eight 8-oz. glasses of water each day.That is about 2 quarts, or 2 liters of water per day. Approximate the lifetime as 70 years.(70 yr)(365 d/1 yr)(2 L/1 d) 5 104 L29.An NCAA-regulation football field is 360 feet long (including the end zones) and 160 feet wide,which is about 110 meters by 50 meters, or 5500 m 2 . We assume the mower has a cutting width of Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-8Chapter 10.5 meters and that a person mowing can walk at about 4.5 km/h, which is about 3 mi/h. Thus thedistance to be walked is as follows:d area5500 m 2 11000 m 11 kmwidth0.5 mAt a speed of 4.5 km/h, it will take about 11 km 30.1h 2.5 h to mow the field.4.5 kmThere are about 3 108 people in the U.S. Assume that half of them have cars, that they drive anaverage of 12,000 miles per year, and that their cars get an average of 20 miles per gallon of gasoline. 1 automobile 12,000 mi/auto 1 gallon 11(3 108 people) 1 10 gal/yr2people1yr20mi 31.In estimating the number of dentists, the assumptions and estimates needed are: the population of the city the number of patients that a dentist sees in a day the number of days that a dentist works in a year the number of times that each person visits the dentist each yearWe estimate that a dentist can see 10 patients a day, that a dentist works 225 days a year, and that eachperson visits the dentist twice per year.(a) For San Francisco, the population as of 2010 was about 800,000 (according to the U.S. CensusBureau). The number of dentists is found by the following calculation: 2 visits/yr 1 yr1 dentist(8 105 people) 700 dentists 1 person 225 workdays 10 visits/workday (b)For Marion, Indiana, the population is about 30,000. The number of dentists is found by acalculation similar to that in part (a), and would be about 30 dentists . There are about 40dentists (of all types, including oral surgeons and orthodontists) listed in the 2012 Yellow Pages.32.Consider the diagram shown (not to scale). The balloon is a distance h 200 mabove the surface of the Earth, and the tangent line from the balloon height to thesurface of the Earth indicates the location of the horizon, a distance d away fromthe balloon. Use the Pythagorean theorem.drhr(r h)2 r 2 d 2 r 2 2rh h2 r 2 d 22rh h2 d 2 d 2rh h2d 2(6.4 106 m)(200 m) ( 200 m ) 5.1 104 m 5 104 m ( 80 mi)233.At 1,000 per day, you would earn 30,000 in the 30 days. With the other pay method, you wouldget 0.01(2t 1 ) on the tth day. On the first day, you get 0.01(21 1 ) 0.01. On the second day,you get 0.01(22 1 ) 0.02. On the third day, you get 0.01(23 1 ) 0.04. On the 30th day, youget 0.01(230 1 ) 5.4 106 , which is over 5 million dollars. Get paid by the second method. Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

Introduction, Measurement, Estimating34.1-9In the figure in the textbook, the distance d is perpendicular to the radius that is drawn approximatelyvertically. Thus there is a right triangle, with legs of d and R, and a hypotenuse of R h. SincehR, h 22 Rh.d 2 R 2 ( R h) 2 R 2 2 Rh h 2 d 2 2 Rh h 2 d 2 2 Rh R d22h(4400 m) 2 6.5 106 m2(1.5 m)A better measurement gives R 6.38 106 m.35.For you to see the Sun “disappear,” your line of sightto the top of the Sun must be tangent to the Earth’ssurface. Initially, you are lying down at point A, andyou see the first sunset. Then you stand up, elevatingyour eyes by the height h 130 cm. While you stand,your line of sight is tangent to the Earth’s surface atpoint B, so that is the direction to the second sunset.The angle θ is the angle through which the Sunappears to move relative to the Earth during the timeto be measured. The distance d is the distance fromyour eyes when standing to point B.hdTo 1st sunsetAθTo 2nd sunsetBRRθEarth centerUse the Pythagorean theorem for the following relationship:d 2 R 2 ( R h) 2 R 2 2 Rh h 2 d 2 2 Rh h 2The distance h is much smaller than the distance R, so h 2 2 Rh which leads to d 2 2 Rh. We alsohave from the same triangle that d /R tan θ , so d R tan θ . Combining these two relationships gives2h.d 2 2 Rh R 2 tan 2 θ , so R tan 2 θThe angle θ can be found from the height change and the radius of the Earth. The elapsed timebetween the two sightings can then be found from the angle, because we know that a full revolutiontakes 24 hours.R 2h2tan θθ360o t sec3600 s24 h 1h θt 360o36. θ tan 1Density units 2h2(1.3 m) tan 1 (3.66 10 2 )oR6.38 106 m 3600 s (3.66 10 2 )o 24h 1 h 360o 3600 s 24 h 8.8 s1h mass units M volume units L3 Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

1-1037.Chapter 1(a)For the equation υ At 3 Bt , the units of At 3 must be the same as the units of υ . So the unitsof A must be the same as the units of υ /t 3 , which would be L /T 4 . Also, the units of Bt must bethe same as the units of υ . So the units of B must be the same as the units of υ /t , which wouldbe L /T 2 .38.(b)For A, the SI units would be m/s 4 , and for B, the SI units would be m/s 2 .(a)The quantity υ t 2 has units of (m/s)(s 2 ) mis, which do not match with the units of meters for x.The quantity 2at has units (m/s 2 )(s) m/s, which also do not match with the units of meters for x.Thus this equation cannot be correct .(b)The quantity υ0t has units of (m/s)(s) m, and1 at 22has units of (m/s 2 )(s 2 ) m. Thus, sinceeach term has units of meters, this equation can be correct .(c)The quantity υ0t has units of (m/s)(s) m, and 2at 2 has units of (m/s 2 )(s 2 ) m. Thus, sinceeach term has units of meters, this equation can be correct .39.Using the units on each of the fundamental constants (c, G, and h), we find the dimensions of thePlanck length. We use the values given for the fundamental constants to find the value of the Plancklength.P P Gh c3Ghc3 [ L3 /MT 2 ][ ML2 /T ][ L /T ]3 L3 L2T 3 M L5 2 3 L [ L ]3 3 L MT L (6.67 10 11 m3 /kg is 2 )(6.63 10 34 kgim 2 /s)8(3.00 10 m/s)3 4.05 10 35 mThus the order of magnitude is 10 35 m .40.41.The percentage accuracy is2m7 100% 1 10 5 % . The distance of 20,000,000 m needs to be2 10 mdistinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distancemeasurements.Multiply the number of chips per wafer by the number of wafers that can be made from a cylinder.We assume the number of chips per wafer is more accurate than 1 significant figure.chips 1 wafer 250 mm 5 chips 3.3 10 400 wafer 0.300 mm 1 cylinder cylinder 42.Assume that the alveoli are spherical and that the volume of a typical human lung is about 2 liters,which is 0.002 m3 . The diameter can be found from the volume of a sphere,4 π r3.3 Copyright 2014 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in wr

Instructor’s Solutions Manual. provides answers and worked-out solutions to all end of chapter questions and problems from chapters 1 – 15 of . Physics: Principles with Applications, 7th Edition, by Douglas C. Giancoli. At the end of the manual are grids that correlate the 6th edition questions and problems to the 7th edition questions and .File Size: 9MBPage Count: 917Explore furtherGIANCOLI PHYSICS 7TH EDITION ANSWERS PDFthetoolcompany.netGIANCOLI PHYSICS 7TH EDITION ANSWERS PDFprotzonbeer.comGiancoli physics 7th edition answers pdf by . - Issuuissuu.comINSTRUCTOR'S SOLUTIONS MANUAL PDF: Giancoli, Physics .groups.google.comGiancoli 7th and 6th Edition physics solutionswww.giancolianswers.comRecommended to you b