Solution Manual for Physics: Principles with Applications, 7th Edition

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DOUGLAS C. GIANCOLI’SPHYSICSPRINCIPLES WITHAPPLICATIONS7THEDITIONBOB DAVISTAYLOR UNIVERSITYJ. ERIK HENDRICKSONUNIVERSITY OF WISCONSIN – EAU CLAIRE

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CONTENTSPREFACE ------------------------------------------------------------------------- ivChapter 1Introduction, Measurement, Estimating ------------------------------------- 1-1Chapter 2Describing Motion: Kinematics in One Dimension ------------------------ 2-1Chapter 3Kinematics in Two Dimensions; Vectors------------------------------------ 3-1Chapter 4Dynamics: Newton’s Laws of Motion --------------------------------------- 4-1Chapter 5Circular Motion; Gravitation -------------------------------------------------- 5-1Chapter 6Work and Energy --------------------------------------------------------------- 6-1Chapter 7Linear Momentum -------------------------------------------------------------- 7-1Chapter 8Rotational Motion -------------------------------------------------------------- 8-1Chapter 9Static Equilibrium; Elasticity and Fracture---------------------------------- 9-1Chapter 10Fluids ----------------------------------------------------------------------------10-1Chapter 11Oscillations and Waves -------------------------------------------------------11-1Chapter 12Sound----------------------------------------------------------------------------12-1Chapter 13Temperature and Kinetic Theory --------------------------------------------13-1Chapter 14Heat------------------------------------------------------------------------------14-1Chapter 15The Laws of Thermodynamics-----------------------------------------------15-1Chapter 16Electric Charge and Electric Field-------------------------------------------16-1Chapter 17Electric Potential---------------------------------------------------------------17-1Chapter 18Electric Currents ---------------------------------------------------------------18-1Chapter 19DC Circuits ---------------------------------------------------------------------19-1Chapter 20Magnetism ----------------------------------------------------------------------20-1Chapter 21Electromagnetic Induction and Faraday’s Law ----------------------------21-1Chapter 22Electromagnetic Waves -------------------------------------------------------22-1Chapter 23Light: Geometric Optics ------------------------------------------------------23-1Chapter 24The Wave Nature of Light----------------------------------------------------24-1Chapter 25Optical Instruments------------------------------------------------------------25-1Chapter 26The Special Theory of Relativity --------------------------------------------26-1Chapter 27Early Quantum Theory and Models of the Atom--------------------------27-1Chapter 28Quantum Mechanics of Atoms-----------------------------------------------28-1Chapter 29Molecules and Solids----------------------------------------------------------29-1Chapter 30Nuclear Physics and Radioactivity ------------------------------------------30-1Chapter 31Nuclear Energy; Effects and Uses of Radiation ---------------------------31-1Chapter 32Elementary Particles-----------------------------------------------------------32-1Chapter 33Astrophysics and Cosmology ------------------------------------------------33-1

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1-1Responses 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 30.00.500.° =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.INTRODUCTION,MEASUREMENT,ESTIMATING1

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1-2Chapter 1(a)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.(b)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.

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Introduction, Measurement, Estimating1-3(d)0 031 significant figure.(e)0 00862 significant figures.(f)32364 significant figures(g)87002 significant figures2.(a)01 1561.15610.=×(b)121.82.1810=×(c)30.00686.810−=×(d)2328.653.286510=×(e)10.2192.1910−=×(f)24444.4410=×3.(a)48.691086,900×=(b)39.1109100×=(c)18.8100.88−×=(d)24.7610476×=(e)53.62100.0000362−×=4.(a)1014 billion years1.410years=×(b)710173.15610s(1.410yr)4.410s1 yr⎛⎞××=×⎜⎟⎜⎟⎝⎠5.0.25 m% uncertainty100%4.6%5.48 m=×=6.(a)0.2 s% uncertainty100%3.636%4%5.5 s=×=≈(b)0.2 s% uncertainty100%3636%0.4%55 s=×= 0.≈(c)The time of 5.5 minutes is 330 seconds.0.2 s% uncertainty100%0.0606%0.06%330 s=×=≈

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1-4Chapter 17.To add values with significant figures, adjust all values to be added so that their exponents are all thesame.346333335(9.210s)(8.310s)(0.00810s)(9.210s)(8310s)(810s)(9.2838)10s100.210s1.0010s×+×+×=×+×+×=++×=×=×When 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.2122(3.07910m)(0.06810m)2.094 m2.1 m−××=≈9.The uncertainty is taken to be 0.01 m.220.01 m% uncertainty100%0.637%1%1.57 m=×=≈10.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.33344specifiedspecified33(0.84 m)2.483 mVrππ===33344minmin3333344maxmax33(0.80 m)2.145 m(0.88 m)2.855 mVrVrππππ======33311maxmin22()(2.855 m2.145 m )0.355 mVVVΔ=−=−=The percent uncertainty is33specified0.355 m10014.314% .2.483 mVVΔ=×=≈11.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 be40.110cm.×24292specifiedspecified(3.110cm)3.01910cmArππ==×=×24292minmin(3.010cm)2.82710cmArππ==×=×24292maxmax(3.210cm)3.21710cmArππ==×=×92929211maxmin22()(3.21710cm2.82710cm )0.19510cmAAAΔ=−=×−×=×Thus the area should be quoted as92(3.00.2)10cm.A=±×12.(a)286.6 mm3286.610m−×0.2866 m(b)85Vμ68510V−×0.000085 V(c)760 mg676010kg−×0.00076 kg(if last zero is not significant)

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Introduction, Measurement, Estimating1-5(d)62.1 ps1262.110s−×0.0000000000621 s(e)22.5 nm922.510m−×0.0000000225 m(f)2.50 gigavolts92.5010volts×2,500,000,000 voltsNote that in part (f) in particular, the correct number of significant digits cannot be determined whenyou write the number in this format.13.(a)6110volts×1 megavolt1 MV=(b)6210meters−×2 micrometers2mμ=(c)3610days×6 kilodays6 kdays=(d)21810bucks×18 hectobucks18 hbucks=or 1.8 kilobucks(e)7710seconds−×700 nanoseconds700 ns or 0.7sμ=14.242421.00010m3.281 ft1 acre1 hectare(1 hectare)2.471 acres1 hectare1 m4.35610ft⎛⎞⎛⎞⎛⎞×==⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟×⎝⎠⎝⎠⎝⎠15.(a)61193 million miles(9310miles)(1610 m/1 mile)1.510m=×=×(b)1111381.510m(1.510m)(1 km/10m)1.510km×=×=×16.To add values with significant figures, adjust all values to be added so that their units are all the same.51.80 m142.5 cm5.3410m1.80 m1.425 m0.534 m3.759 m3.76 mμ++×=++==When 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.(a)101091.010m1.010m39.37 in/1 m3.910in−−−×= (×)() =×(b)8101 m1 atom(1.0 cm)1.010atoms100 cm1.010m−⎛⎞⎛⎞=×⎜⎟⎜⎟ ⎜⎟×⎝⎠ ⎝⎠18.(a)0.621 mi(1 km/h)0.621 mi/h,1 km⎛⎞=⎜⎟⎝⎠so the conversion factor is0.621 mi/h .1 km/h(b)3.28 ft(1 m/s)3.28 ft/s,1 m⎛⎞=⎜⎟⎝⎠so the conversion factor is3.28 ft/s .1 m/s(c)1000 m1 h(1 km/h)0.278 m/s,1 km3600 s⎛⎞⎛⎞=⎜⎟⎜⎟⎝⎠⎝⎠so the conversion factor is0.278 m/s .1 km/h

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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.8715151.00 ly(2.99810m/s)(3.15610s)9.46210m9.4610m=××=×≈×(b)Do a unit conversion from ly to AU.154119.46210m1 AU(1.00 ly)6.3110AU1.00 ly1.5010m⎛⎞⎛⎞×=×⎜⎟⎜⎟⎜⎟⎜⎟×⎝⎠⎝⎠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 m100%7.3%1500 m×=21.Since the meter is longer than the yard, the soccer field is longer than the football field.soccerfootball1.094 yd100.0 m100.0 yd9.4 yd1 m−=×−=AAsoccerfootball1 m100.0 m100.0 yd8.6 m1.094 yd−=−×=AASince the soccer field is 109.4 yd compared with the 100.0-yd football field, the soccer fieldis9.4%longer than the football field.22.(a)# of seconds in 1.00 yr:773.15610s1.00 yr(1.00 yr)3.1610s1 yr⎛⎞×==×⎜⎟⎜⎟⎝⎠(b)# of nanoseconds in 1.00 yr:79163.15610s110ns1.00 yr(1.00 yr)3.1610ns1 yr1 s⎛⎞ ⎛⎞××==×⎜⎟ ⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠(c)# of years in 1.00 s:871 yr1.00 s1.00 s3.1710yr3.15610s−⎛⎞= ()=×⎜⎟⎜⎟×⎝⎠23.(a)15122710kg1 proton or neutron10protons or neutrons1 bacterium10kg−−⎛⎞⎛⎞=⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠(b)17102710kg1 proton or neutron10protons or neutrons1 DNA molecule10kg−−⎛⎞ ⎛⎞=⎜⎟ ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠(c)2292710kg1 proton or neutron10protons or neutrons1 human10kg−⎛⎞ ⎛⎞=⎜⎟ ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠(d)41682710kg1 proton or neutron10protons or neutrons1 Galaxy10kg−⎛⎞ ⎛⎞=⎜⎟ ⎜⎟⎜⎟⎜⎟ ⎝⎠⎝⎠

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Introduction, Measurement, Estimating1-724.The 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(mass/volume)ρ=of a nucleon.()33ballball44ballballballballball33nucleonnucleonnucleonballballnucleonballballball33441nucleonnucleon3nucleon322balln432;222cccrrVrmmmmVVVVVrdcmπππππρππππ⎛⎞=→===⎜⎟⎝⎠⎛⎞⎛⎞⎛⎞⎜⎟⎜⎟====⎜⎟⎜⎟⎜⎟⎝⎠⎜⎟⎝⎠⎝⎠⎛⎞=⎜⎟⎝⎠()3327ucleonballnucleon31541nucleonnucleon3214140.23 m(10kg)(10m)3.910kg410kgcmddπππ−−⎛⎞⎛⎞⎛⎞⎜⎟==⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎜⎟⎝⎠⎝⎠=×≈×25.(a)33328002.81011010=×≈×=(b)344586.30108.63010101010×=×≈×=(c)3320.00767.610101010−−−=×≈×=(d)899915 0101 51011010.×=.×≈×=26.The 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 then24141 book(3500 m )7.010books.(0.25 m)(0.05 m)⎛⎞=×⎜⎟⎝⎠With 8 shelves of books, the total number of booksstored is as follows:()45books7 0108 shelves610booksshelf level⎛⎞.×≈×⎜⎟⎝⎠27.The distance across the U.S. is about 3000 miles.(3000 mi)(1 km/0.621 mi)(1 h/10 km)500 h≈Of 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.4(70 yr)(365 d/1 yr)(2 L/1 d)510L≈×29.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, or25500 m .We assume the mower has a cutting width of

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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:2area5500 m11000 m11 kmwidth0.5 md====At a speed of 4.5 km/h, it will take about1 h11 km2.5 h4.5 km×≈to mow the field.30.There are about8310×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.8111 automobile12,000 mi/auto1 gallon(310people)110gal/yr2 people1 yr20 mi⎛⎞⎛⎞⎛⎞×≈×⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠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:52 visits/yr1 yr1 dentist(810people)700 dentists1 person225 workdays10 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 about30 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 distance200 mh=above 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 distancedaway fromthe balloon. Use the Pythagorean theorem.()22222222222644()2222(6.410m)(200 m)200 m5.110m510m (80 mi)rhrdrrhhrdrhhddrhhd+=+→++=++=→=+=×+=×≈×≈33.At $1,000 per day, you would earn $30,000 in the 30 days. With the other pay method, you wouldget1$0.01(2)t−on thetth day. On the first day, you get1 1$0.01(2)$0.01.−=On the second day,you get2 1$0.01(2)$0.02.−=On the third day, you get3 1$0.01(2)$0.04.−=On the 30th day, youget30 16$0.01(2)$5.410 ,−=×which is over 5 million dollars. Get paid by the second method.hrrd

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Introduction, Measurement, Estimating1-9To 1st sunsetTo 2nd sunsetABEarth centerRRhdθθ34.In the figure in the textbook, the distancedis perpendicular to the radius that is drawn approximatelyvertically. Thus there is a right triangle, with legs ofdandR, and a hypotenuse of.Rh+Since2,2.hR hRh22222222226()2222(4400 m)6.510m2(1.5 m)ddRRhRRhhdRhhdRhRh→→→+=+=++=+≈===×A better measurement gives66.3810m.R=×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 height130 cm.h=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 angleθis the angle through which the Sunappears to move relative to the Earth during the timeto be measured. The distancedis the distance fromyour eyes when standing to point B.Use the Pythagorean theorem for the following relationship:2222222()22dRRhRRhhdRhh+=+=++→=+The distancehis much smaller than the distanceR, so22hRhwhich leads to22.dRh≈We alsohave from the same triangle that/tan,d Rθ=sotan.dRθ=Combining these two relationships gives2222tan,dRhRθ≈=so22.tanhRθ=The angleθcan 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.112o26o2ooo222(1.3 m)tantan(3 6610)tan6.3810msec3600 s36024 h1 h3600 s(3 6610)3600 s24 h24 h8.8 s1 h1 h360360hhRRttθθθθ−−−−=→===.××=→×⎛⎞⎛⎞⎛⎞.×⎛⎞=×=×=⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠36.3mass unitsDensity unitsvolume unitsML⎡⎤==⎢⎥⎣⎦

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1-10Chapter 137.(a)For the equation3,AtBtυ =−the units of3Atmust be the same as the units of.υSo the unitsofAmust be the same as the units of3/,tυwhich would be4/.L TAlso, the units ofBtmust bethe same as the units of.υSo the units ofBmust be the same as the units of/ ,tυwhich wouldbe2/.L T(b)ForA, the SI units would be4m/s,and for B, the SI units would be2m/s.38.(a)The quantity2tυhas units of2(m/s)(s )m s,=iwhich do not match with the units of meters forx.The quantity 2athas units2(m/s )(s)m/s,=which also do not match with the units of meters forx.Thus this equationcannot be correct .(b)The quantity0tυhas units of(m/s)(s)m,=and212athas units of22(m/s )(s )m.=Thus, sinceeach term has units of meters, this equationcan be correct .(c)The quantity0tυhas units of(m/s)(s)m,=and22athas units of22(m/s )(s )m.=Thus, sinceeach term has units of meters, this equationcan be correct.39.Using the units on each of the fundamental constants (c,G, andh), we find the dimensions of thePlanck length. We use the values given for the fundamental constants to find the value of the Plancklength.[]3223235233333[/][/][/]PGhLMTMLTL L T MLLLcL TMT LL⎡⎤⎡⎤⎡⎤=→====⎢⎥⎢⎥⎣⎦⎢⎥⎢⎥⎣⎦⎣⎦A113234235383(6.6710m /kg s )(6.6310kg m /s)4 0510m(3.0010m/s)PGhc−−−××===.××iiAThus the order of magnitude is3510m .−40.The percentage accuracy is572 m100%110% .210m−×=××The distance of 20,000,000 m needs to bedistinguishable from 20,000,002 m, which means that 8 significant figures are needed in the distancemeasurements.41.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.5chips1 wafer250 mmchips4003.310wafer0.300 mm1 cylindercylinder⎛⎞⎛⎞⎛⎞=×⎜⎟⎜⎟⎜⎟⎝⎠ ⎝⎠⎝⎠42.Assume that the alveoli are spherical and that the volume of a typical human lung is about 2 liters,which is30.002 m .The diameter can be found from the volume of a sphere,343.rπ

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Solution Manual for Physics: Principles with Applications, 7th Edition - Page 15 preview image

Introduction, Measurement, Estimating1-1133344331 333833348(/2)66(210)(310 )210mm210m6310/drdddπππππ−−−==⎡⎤××=×→==×⎢⎥×⎢⎥⎣⎦43.We assume that there are 40 hours of work per week and that the typist works 50 weeks out of theyear.1241 char1 min1 hour1 week1 year(1.010bytes)4.62910years1 byte180 char60 min40 hour50 weeks46,000 years××××××=×≈44.The volume of water used by the people can be calculated as follows:3343351200 L/day365 days1000 cm1 km(410people)4.3810km /yr4 people1 yr1 L10cm−⎛⎞⎛⎞⎛⎞⎛⎞×=×⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠⎝⎠The depth of water is found by dividing the volume by the area.335524.3810km /yrkm10cm8.76108.76 cm/yr9 cm/yryr1 km50 kmVdA−−⎛⎞⎛⎞×===×=≈⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠45.We approximate the jar as a cylinder with a uniform cross-sectional area. In counting the jelly beansin the top layer, we find about 25 jelly beans. Thus we estimate that one layer contains about 25 jellybeans. In counting vertically, we see that there are about 15 rows. Thus we estimate that thereare2515375400 jelly beans×=≈in the jar.46.The volume of a sphere is given by343,Vrπ=so the radius is1/33.4Vrπ⎛⎞=⎜⎟⎝⎠For a 1-ton rock, thevolume is calculated from the density, and then the diameter from the volume.332000 lb1 ft(1 T)10.8 ft1 T186 lbV⎛⎞⎛⎞==⎜⎟⎜⎟ ⎜⎟⎝⎠ ⎝⎠1/31/3333(10.8 ft )2222.74 ft3 ft44Vdrππ⎡⎤⎛⎞====≈⎢⎥⎜⎟⎝⎠⎢⎥⎣⎦47.We do a “units conversion” from bytes to minutes, using the given CD reading rate.668 bits1 s1 min(783.21610bytes)74.592 min75 min1 byte60 s1.410bits××××=≈×48.A pencil has a diameter of about 0.7 cm. If held about 0.75 m from the eye, it can just block out theMoon. The ratio of pencil diameter to arm length is the same as the ratio of Moon diameter to Moondistance. From the diagram, we have the following ratios.

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Solution Manual for Physics: Principles with Applications, 7th Edition - Page 16 preview image

1-12Chapter 135Pencil diameterMoon diameterPencil distanceMoon distancePencil diameter710mMoon diameter(Moon distance)(3.810km)3500 kmPencil distance0.75 m−=→×==×≈The actual value is 3480 km.49.To calculate the mass of water, we need to find the volume of water and then convert the volume tomass. The volume of water is the area of the city2(48 km )times the depth of the water (1.0 cm).2532553310cm10kg1 metric ton(48 km )(1.0 cm)4.810metric tons510 metric tons1 km1 cm10kg−⎡⎤⎛⎞⎛⎞⎛⎞⎢⎥=×≈×⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎜⎟⎢⎥⎝⎠⎝⎠⎝⎠⎣⎦To find the number of gallons, convert the volume to gallons.252883310cm1 L1 gal(48 km )(1.0 cm)1.2710gal110gal1 km3.78 L110cm⎡⎤⎛⎞⎛⎞ ⎛⎞⎢⎥=×≈×⎜⎟⎜⎟ ⎜⎟⎜⎟⎜⎟⎢⎥×⎝⎠⎝⎠⎝⎠⎣⎦50.The person walks4 km/h,12 hours each day. The radius of the Earth is about 6380 km, and thedistance around the Earth at the equator is the circumference,Earth2.RπWe assume that the personcan “walk on water,” so ignore the existence of the oceans.1 h1 day2(6380 km)835 days800 days4 km12 hπ⎛⎞ ⎛⎞=≈⎜⎟ ⎜⎟⎝⎠ ⎝⎠51.The volume of the oil will be the area times the thickness. The area is22(/2) .rdππ=3323101 m1000 cm100 cm(/2)22310m(210m)VVdtdtπππ−⎛⎞⎜⎟⎝⎠=→===××This is approximately 2 miles.52.578 s1 yr100%310%1 yr3.15610s−⎛⎞⎛⎞×=×⎜⎟⎜⎟⎜⎟×⎝⎠⎝⎠53.(a)10ooo910m1 nm1.0 A1.0 A0.10 nm10m1A−−⎛⎞ ⎛⎞⎛⎞ ⎜⎟==⎜⎟⎜⎟⎜⎟⎜⎟⎝⎠⎝⎠⎝⎠MoonPencilPencildistanceMoondistance

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