Instructor’s Solution Manual For Fundamentals Of Physics .

Transcription

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

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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 alt

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. I am very grateful for helpful input from J. Richard Christman, Meighan