Instructor’s Solution Manual For Fundamentals Of Physics .

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Instructor’s Solution ManualforFundamentals of Physics, 6/Eby Halliday, Resnick, and WalkerJames B. WhitentonSouthern Polytechnic State University

ii

PrefaceThis booklet includes the solutions relevant to the EXERCISES & PROBLEMS sections of the 6th editionof Fundamentals of Physics, by Halliday, Resnick, and Walker. We also include solutions to problems inthe Problems Supplement. We have not included solutions or discussions which pertain to the QUESTIONSsections.I am very grateful for helpful input from J. Richard Christman, Meighan Dillon, Barbara Moore, and JearlWalker regarding the development of this document.iii

ivPREFACE

Chapter 11. The metric prefixes (micro, pico, nano, . . .) are given for ready reference on the inside front cover of thetextbook (see also Table 1-2).(a) Since 1 km 1 103 m and 1 m 1 106 µm,1 km 103 m (103 m)(106 µm/m) 109 µm .The given measurement is 1.0 km (two significant figures), which implies our result should bewritten as 1.0 109 µm.(b) We calculate the number of microns in 1 centimeter. Since 1 cm 10 2 m,1 cm 10 2 m (10 2 m)(106 µm/m) 104 µm .We conclude that the fraction of one centimeter equal to 1.0 µm is 1.0 10 4 .(c) Since 1 yd (3 ft)(0.3048 m/ft) 0.9144 m,1.0 yd (0.91 m)(106 µm/m) 9.1 105 µm .2. The customer expects 20 7056 in3 and receives 20 5826 in3 , the difference being 24600 cubic inches,or 2.54 cm 31L324600 in 403 L1 inch1000 cm3where Appendix D has been used (see also Sample Problem 1-2).3. Using the given conversion factors, we find(a) the distance d in rods to bed 4.0 furlongs (4.0 furlongs)(201.168 m/furlong) 160 rods ,5.0292 m/rod(b) and that distance in chains to bed (4.0 furlongs)(201.168 m/furlong) 40 chains .20.117 m/chain4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain 1 inch6 picas12 points(0.80 cm) 23 points ,2.54 cm1 inch1 pica(b) and(0.80 cm) 1 inch2.54 cm 6 picas1 inch1 1.9 picas .

2CHAPTER 1.5. Various geometric formulas are given in Appendix E.(a) Substituting R 6.37 106 m 10 3 km/m 6.37 103 kminto circumference 2πR, we obtain 4.00 104 km.(b) The surface area of Earth is(c) The volume of Earth is 24πR2 4π 6.37 103 km 5.10 108 km2 . 34π 3 4πR 6.37 103 km 1.08 1012 km3 .336. (a) Using the fact that the area A of a rectangle is width length, we findAtotal (3.00 acre) (25.0 perch)(4.00 perch) (40 perch)(4 perch)(3.00 acre) 100 perch21 acre580 perch2 .We multiply this by the perch2 rood conversion factor (1 rood/40 perch2 ) to obtain the answer:Atotal 14.5 roods.(b) We convert our intermediate result in part (a): 216.5 ftAtotal (580 perch2 ) 1.58 105 ft2 .1 perchNow, we use the feet meters conversion given in Appendix D to obtain 2 1m25Atotal 1.58 10 ft 1.47 104 m2 .3.281 ft7. The volume of ice is given by the product of the semicircular surface area and the thickness. Thesemicircle area is A πr2 /2, where r is the radius. Therefore, the volume isV π 2r z2where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have 3 210 cm10 mr (2000 km) 2000 105 cm .1 km1mIn these units, the thickness becomes102 cmz (3000 m)1m Therefore,V 3000 102 cm . 2 π2000 105 cm3000 102 cm 1.9 1022 cm3 .28. The total volume V of the real house is that of a triangular prism (of height h 3.0 m and base areaA 20 12 240 m2 ) in addition to a rectangular box (height h′ 6.0 m and same base). Therefore, 1h′′V hA h A h A 1800 m3 .22

3(a) Each dimension is reduced by a factor of 1/12, and we find3Vdoll 1800 m 112 3 1.0 m3 .(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore,3Vminiature 1800 m 1144 3 6.0 10 4 m3 .9. We use the conversion factors found in Appendix D.1 acre · ft (43, 560 ft2 ) · ft 43, 560 ft3 .Since 2 in. (1/6) ft, the volume of water that fell during the storm isV (26 km2 )(1/6 ft) (26 km2 )(3281 ft/km)2 (1/6 ft) 4.66 107 ft3 .Thus,V 4.66 107 ft3 1.1 103 acre · ft .4.3560 104 ft3 /acre · ft10. The metric prefixes (micro (µ), pico, nano, . . .) are given for ready reference on the inside front cover ofthe textbook (also, Table 1-2). 100 y365 day24 h60 min1 µcentury 10 6 century1 century1y1 day1h 52.6 min .The percent difference is therefore52.6 min 50 min 5.2% .50 min11. We use the conversion factors given in Appendix D and the definitions of the SI prefixes given in Table 12 (also listed on the inside front cover of the textbook). Here, “ns” represents the nanosecond unit, “ps”represents the picosecond unit, and so on.(a) 1 m 3.281 ft and 1 s 109 ns. Thus, 3.281 ft s 3.0 108 m3.0 108 m/s 0.98 ft/ns .sm109 ns(b) Using 1 m 103 mm and 1 s 1012 ps, we find 3 3.0 108 m10 mms3.0 108 m/s sm1012 ps 0.30 mm/ps .12. The number of seconds in a year is 3.156 107. This is listed in Appendix D and results from the product(365.25 day/y)(24 h/day)(60 min/h)(60 s/min) .(a) The number of shakes in a second is 108 ; therefore, there are indeed more shakes per second thanthere are seconds per year.

4CHAPTER 1.(b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humanshave existed is given by106 10 4 u-day ,1010which we may also express as 86400 u-sec10 4 u-day 8.6 u-sec .1 u-day13. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterionfor judging their quality for measuring time intervals. What is important is that the clock advance bythe same amount in each 24-h period. The clock reading can then easily be adjusted to give the correctinterval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected sinceit would impossible to tell what the correction should be. The following gives the corrections (in seconds)that must be applied to the reading on each clock for each 24-h period. The entries were determined bysubtracting the clock reading at the end of the interval from the clock reading at the beginning.CLOCKABCDESun.-Mon. 16 3 58 67 70Mon.-Tues. 16 5 58 67 55Tues.-Wed. 15 10 58 67 2Wed.-Thurs. 17 5 58 67 20Thurs.-Fri. 15 6 58 67 10Fri.-Sat 15 7 58 67 10Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relativeto WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. Thecorrection for clock C is less than the correction for clock D, so we judge clock C to be the best andclock D to be the next best. The correction that must be applied to clock A is in the range from 15 sto 17s. For clock B it is the range from 5 s to 10 s, for clock E it is in the range from 70 s to 2 s.After C and D, A has the smallest range of correction, B has the next smallest range, and E has thegreatest range. From best the worst, the ranking of the clocks is C, D, A, B, E.14. The time on any of these clocks is a straight-line function of that on another, with slopes 6 1 andy-intercepts 6 0. From the data in the figure we deducetC tB 2594tB 7733662tA .405These are used in obtaining the following results.(a) We findt′B tB 33 ′(t tA ) 495 s40 Awhen t′A tA 600 s.(b) We obtaint′C tC 2 ′2(tB tB ) (495) 141 s .77(c) Clock B reads tB (33/40)(400) (662/5) 198 s when clock A reads tA 400 s.(d) From tC 15 (2/7)tB (594/7), we get tB 245 s.

515. We convert meters to astronomical units, and seconds to minutes, using1000 m 1 AU 60 s 1 km1.50 108 km1 min .Thus, 3.0 108 m/s becomes 3.0 108 m1 kmAU60 s 0.12 AU/min .s1000 m1.50 108 kmmin16. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to changelongitude by 360 /24 15 before resetting one’s watch by 1.0 h.17. The last day of the 20 centuries is longer than the first day by(20 century)(0.001 s/century) 0.02 s .The average day during the 20 centuries is (0 0.02)/2 0.01 s longer than the first day. Since theincrease occurs uniformly, the cumulative effect T isT (average increase in length of a day)(number of days) 0.01 s365.25 day (2000 y)dayy 7305 sor roughly two hours.18. We denote the pulsar rotation rate f (for frequency).f 1 rotation1.55780644887275 10 3 s(a) Multiplying f by the time-interval t 7.00 days (which is equivalent to 604800 s, if we ignoresignificant figure considerations for a moment), we obtain the number of rotations: 1 rotationN (604800 s) 388238218.41.55780644887275 10 3 swhich should now be rounded to 3.88 108 rotations since the time-interval was specified in theproblem to three significant figures.(b) We note that the problem specifies the exact number of pulsar revolutions (one million). In thiscase, our unknown is t, and an equation similar to the one we set up in part (a) takes the formN1 106 ft 1 rotation1.55780644887275 10 3 s twhich yields the result t 1557.80644887275 s (though students who do this calculation on theircalculator might not obtain those last several digits).(c) Careful reading of the problem shows that the time-uncertainty per revolution is 3 10 17 s.We therefore expect that as a result of one million revolutions, the uncertainty should be ( 3 10 17 )(1 106 ) 3 10 11 s.

6CHAPTER 1.19. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, thenME N m or N ME /m. We convert mass m to kilograms using Appendix D (1 u 1.661 10 27 kg).Thus,ME5.98 1024 kgN 9.0 1049 .m(40 u)(1.661 10 27 kg/u)20. To organize the calculation, we introduce the notion of density (which the students have probably seenin other courses):m.ρ V(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With densityρ 19.32 g/cm3 and mass m 27.63 g, the volume of the leaf is found to beV m 1.430 cm3 .ρWe convert the volume to SI units:1.430 cm3 1m100 cm 3 1.430 10 6 m3 .And since V A z where z 1 10 6 m (metric prefixes can be found in Table 1-2), we obtainA 1.430 10 6 m3 1.430 m2 .1 10 6 m(b) The volume of a cylinder of length ℓ is V Aℓ where the cross-section area is that of a circle:A πr2 . Therefore, with r 2.500 10 6 m and V 1.430 10 6 m3 , we obtainℓ V 7.284 104 m .πr221. We introduce the notion of density (which the students have probably seen in other courses):ρ mVand convert to SI units: 1 g 1 10 3 kg.(a) For volume conversion, we find 1 cm3 (1 10 2 m)3 1 10 6 m3 . Thus, the density in kg/m3is 3 10 kgcm31g31 g/cm 1 103 kg/m3 .cm3g10 6 m3Thus, the mass of a cubic meter of water is 1000 kg.(b) We divide the mass of the water by the time taken to drain it. The mass is found from M ρV(the product of the volume of water and its density):3M (5700 m3 )(1 103 kg/m ) 5.70 106 kg .The time is t (10 h)(3600 s/h) 3.6 104 s, so the mass flow rate R isR M5.70 106 kg 158 kg/s .t3.6 104 s

722. The volume of the water that fell isV(26 km2 )(2.0 in.) 2 1000 m0.0254 m(26 km2 )(2.0 in.)1 km1 in. (26 106 m2 )(0.0508 m)1.3 106 m3 . We write the mass-per-unit-volume (density) of the water as:ρ m3 1 103 kg/m .VThe mass of the water that fell is therefore given by m ρV : 3m 1 103 kg/m1.3 106 m3 1.3 109 kg .23. We introduce the notion of density (which the students have probably seen in other courses):ρ mVand convert to SI units: 1000 g 1 kg, and 100 cm 1 m.(a) The density ρ of a sample of iron is therefore 1 kg 100 cm 33ρ 7.87 g/cm1000 g1mwhich yields ρ 7870 kg/m3 . If we ignore the empty spaces between the close-packed spheres, thenthe density of an individual iron atom will be the same as the density of any iron sample. That is,if M is the mass and V is the volume of an atom, thenV 9.27 10 26 kgM 29 m3 .3 1.18 10ρ7.87 103 kg/m(b) We set V 4πR3 /3, where R is the radius of an atom (Appendix E contains several geometryformulas). Solving for R, we findR 3V4π 1/3 3(1.18 10 29 m3 )4π 1/3 1.41 10 10 m .The center-to-center distance between atoms is twice the radius, or 2.82 10 10 m.24. The metric prefixes (micro (µ), pico, nano, . . .) are given for ready reference on the inside front cover of thetextbook (see also Table 1-2). The surface area A of each grain of sand of radius r 50 µm 50 10 6 mis given by A 4π(50 10 6)2 3.14 10 8 m2 (Appendix E contains a variety of geometry formulas).We introduce the notion of density (which the students have probably seen in other courses):ρ mVso that the mass can be found from m ρV , where ρ 2600 kg/m3 . Thus, using V 4πr3 /3, the massof each grain is 3 ! 4π 50 10 6 mkgm 2600 3 1.36 10 9 kg .3m

8CHAPTER 1.We observe that (because a cube has six equal faces) the indicated surface area is 6 m2 . The number ofspheres (the grains of sand) N which have a total surface area of 6 m2 is given byN 6 m2 1.91 108 .3.14 10 8 m2Therefore, the total mass M is given byM N m 1.91 108 1.36 10 9 kg 0.260 kg . 25. From the Figure we see that, regarding differences in positions x, 212 S is equivalent to 258 W and180 S is equivalent to 156 Z. Whether or not the origin of the Zelda path coincides with the origins ofthe other paths is immaterial to consideration of x.(a) x (50.0 S) 258 W212 S 60.8 W 43.3 Z(b) x (50.0 S) 156 Z180 S26. The first two conversions are easy enough that a formal conversion is not especially called for, but inthe interest of practice makes perfect we go ahead and proceed formally:(a)(11 tuffet) 2 peck1 tuffet 22 peck(b)(11 tuffet) 0.50 bushel1 tuffet 5.5 bushel(c)(5.5 bushel) 36.3687 L1 bushel 200 L27. We make the assumption that the clouds are directly overhead, so that Figure 1-3 (and the calculationsthat accompany it) apply. Following the steps in Sample Problem 1-4, we haveθt 36024 hwhich, for t 38 min 38/60 h yields θ 9.5 . We obtain the altitude h from the relationd2 r2 tan2 θ 2rhwhich is discussed in that Sample Problem, where r 6.37 106 m is the radius of the earth. Therefore,h 8.9 104 m.

928. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where x 0.05 mis the increase in horizontal depth per step) 4.57x Nsteps x (0.05) 1.2 m .0.19However, we can approach this more carefully by noting that if there are N 4.57/.19 24 rises thenunder normal circumstances we would expect N 1 23 runs (horizontal pieces) in that staircase. Thiswould yield (23)(0.05) 1.15 m, which – to two significant figures – agrees with our first result.29. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25 wp/11 barnalong with appropriate conversion factors: 4047 m2 hide110 acre(25 wp) 1001 wp1 hide1 acre 1 1036 .1 10 28 m2(11 barn)1 barn30. It is straightforward to compute how many seconds in a year (about 3 107 ). Now, if we estimateroughly one breath per second (or every two seconds, or three seconds – it won’t affect the result) thento within an order of magnitude, a person takes 107 breaths in a year.31. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so(3.7 m)(106 µm/m) 3.1 µm/s .(14 day)(86400 s/day)32. The mass in kilograms is(28.9 piculs) 100 gin1 picul 16 tahil1 gin 10 chee1 tahil 10 hoon1 chee 0.3779 g1 hoon which yields 1.747 106 g or roughly 1750 kg.33. (a) In atomic mass units, the mass of one molecule is 16 1 1 18 u. Using Eq. 1-9, we find 1.6605402 10 27 kg(18 u) 3.0 10 26 kg .1u(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number ofwater molecules:1.4 1021 5 1046 .3.0 10 2634. (a) We find the volume in cubic centimeters 3231 in32.54 cm(193 gal) 7.31 105 cm31 gal1 inand subtract this from 1 106 cm3 to obtain 2.69 105 cm3 . The conversion gal in3 is given inAppendix D (immediately below the table of Volume conversions).(b) The volume found in part (a) is converted (by dividing by (100 cm/m)3 ) to 0.731 m3 , which corresponds to a mass of 1000 kg/m3 0.731 m2 731 kgusing the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in731 kg 4.06 105 min0.0018 kg/minwhich we convert to 0.77 y, by dividing by the number of minutes in a year (365 days)(24 h/day)(60 min/h).

10CHAPTER 1.35. (a) When θ is measured in radians, it is equal to the arclength divided by the radius. For very largeradius circles and small values of θ, such as we deal with in this problem,the arcs may beapproximated as.R Sun .straight lines –. . . . . . . .which for our.purposes corre.dD. . . θspond to the di. . . . .ameters d and. . . . .D of the Moon. . .R Moon.and Sun, respectively. Thus, θ Dd R MoonR Sun R SunD R Moondwhich yields D/d 400.(b) Various geometric formulas are given in Appendix E. Using rs and rm for the radius of the Sun andMoon, respectively (noting that their ratio is the same as D/d), then the Sun’s volume divided bythat of the Moon is 343rs3 πrs 4003 6.4 107 .43rπrmm3(c) The angle should turn out to be roughly 0.009 rad (or about half a degree). Putting this into theequation above, we get d θRMoon (0.009) 3.8 105 3.4 103 km .36. (a) For the minimum (43 cm) case, 9 cubit converts as follows: 0.43 m(9 cubit) 3.9 m .1 cubitAnd for the maximum (43 cm) case we obtain 0.53 m(9 cubit) 4.8 m .1 cubit(b) Similarly, with 0.43 m 430 mm and 0.53 m 530 mm, we find 3.9 103 mm and 4.8 103 mm,respectively.(c) We can convert length and diameter first and then compute the volume, or first compute the volumeand then convert. We proceed using the latter approach (where d is diameter and ℓ is length).Vcylinder, min π 2ℓ d 28 cubit34 0.43 m 3328 cubit1 cubit 2.2 m3 .Similarly, with 0.43 m replaced by 0.53 m, we obtain Vcylinder, max 4.2 m3 .37. (a) Squaring the relation 1 ken 1.97 m, and setting up the ratio, we obtain1.972 m21 ken2 3.88 .1 m21 m2

11(b) Similarly, we find1 ken31.973 m3 7.65 .31m1 m3(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus,πr2 h π(3.00)2 (5.50) 155.5 ken3 .(d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) 1.19 103 m3 .38. Although we can look up the distance from Cleveland to Los Angeles, we can just as well (for an order ofmagnitude calculation) assume it’s some relatively small fraction of the circumference of Earth – whichsuggests that (again, for an order of magnitude calculation) we can estimate the distance to be roughlyr, where r 6 106 m is the radius of Earth. If we take each toilet paper sheet to be roughly 10 cm(0.1 m) then the number of sheets needed is roughly 6 106 /0.1 6 107 108 .39. Using the (exact) conversion 2.54 cm 1 in. we find that 1 ft (12)(2.54)/100 0.3048 m (which alsocan be found in Appendix D). The volume of a cord of wood is 8 4 4 128 ft3 , which we convert(multiplying by 0.30483 ) to 3.6 m3 . Therefore, one cubic meter of wood corresponds to 1/3.6 0.3 cord.40. (a) When θ is measured in radians, it is equal to the arclength s di

This booklet includes the solutions relevant to the EXERCISES & PROBLEMS sections of the 6th edition of Fundamentals of Physics, by Halliday, Resnick, and Walker. We also include solutions to problems in the Problems Supplement. We have not included solutions or discussions which pertain to the QUESTIONS sections.

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