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Instructor’s Solution Manual for Fundamentals of Physics, 6/E by Halliday, Resnick, and Walker James B. Whitenton Southern Polytechnic State University ii Preface 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 Dillon, Barbara Moore, and Jearl Walker regarding the development of this document. iii iv PREFACE Chapter 1 1. The metric prefixes (micro, pico, nano,...) are given for ready reference on the inside front coverof the textbook (see also Table 1-2). (a) Since 1km =1 103m and 1m =1 106µm, × × 1km=103m=(103m)(106µm/m)=109µm. The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 109 µm. × (b) We calculate the number of microns in 1 centimeter. Since 1cm=10 2m, − 1cm=10 2m=(10 2m)(106µm/m)=104µm. − − We conclude that the fraction of one centimeter equal to 1.0µm is 1.0 10 4. − × (c) Since 1yd =(3ft)(0.3048m/ft)=0.9144m, 1.0yd=(0.91m)(106µm/m)=9.1 105µm. × 2. The customer expects 20 7056 in3 and receives 20 5826in3, the difference being 24600cubic inches, × × or 2.54 cm 3 1 L 24600 in3 =403 L 1 inch 1000 cm3 (cid:18) (cid:19) (cid:18) (cid:19) (cid:0) (cid:1) where Appendix D has been used (see also Sample Problem 1-2). 3. Using the given conversionfactors, we find (a) the distance d in rods to be (4.0furlongs)(201.168m/furlong) d=4.0furlongs= =160rods, 5.0292m/rod (b) and that distance in chains to be (4.0furlongs)(201.168m/furlong) d= =40chains . 20.117m/chain 4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain 1 inch 6 picas 12 points (0.80 cm) 23 points , 2.54 cm 1 inch 1 pica ≈ (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) (b) and 1 inch 6 picas (0.80 cm) 1.9 picas . 2.54 cm 1 inch ≈ (cid:18) (cid:19)(cid:18) (cid:19) 1 2 CHAPTER 1. 5. Various geometric formulas are given in Appendix E. (a) Substituting R= 6.37 106m 10−3km/m =6.37 103km × × into circumference=2πR, we(cid:0)obtain 4.00 (cid:1)(cid:0)104 km. (cid:1) × (b) The surface area of Earth is 4πR2 =4π 6.37 103km 2 =5.10 108 km2 . × × (c) The volume of Earth is (cid:0) (cid:1) 4πR3 = 4π 6.37 103km 3 =1.08 1012 km3 . 3 3 × × (cid:0) (cid:1) 6. (a) Using the fact that the area A of a rectangle is width length, we find × A = (3.00 acre)+(25.0perch)(4.00perch) total (40perch)(4perch) = (3.00acre) +100perch2 1 acre (cid:18) (cid:19) = 580 perch2 . We multiply this by the perch2 rood conversion factor (1rood/40perch2) to obtain the answer: → A =14.5 roods. total (b) We convert our intermediate result in part (a): 2 16.5 ft A =(580 perch2) =1.58 105 ft2 . total 1 perch × (cid:18) (cid:19) Now, we use the feet meters conversiongiven in Appendix D to obtain → 2 1 m A = 1.58 105 ft2 =1.47 104 m2 . total × 3.281 ft × (cid:18) (cid:19) (cid:0) (cid:1) 7. The volume of ice is given by the product of the semicircular surface area and the thickness. The semicircle area is A=πr2/2, where r is the radius. Therefore, the volume is π V = r2z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have 103m 102cm r=(2000km) =2000 105 cm . 1km 1m × (cid:18) (cid:19)(cid:18) (cid:19) In these units, the thickness becomes 102cm z =(3000m) =3000 102cm . 1m × (cid:18) (cid:19) Therefore, V = π 2000 105cm 2 3000 102cm =1.9 1022cm3 . 2 × × × (cid:0) (cid:1) (cid:0) (cid:1) 8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area A=20 12=240 m2) in addition to a rectangular box (height h =6.0 m and same base). Therefore, ′ × 1 h V = hA+hA= +h A=1800 m3 . ′ ′ 2 2 (cid:18) (cid:19) 3 (a) Each dimension is reduced by a factor of 1/12,and we find 3 1 V = 1800 m3 1.0 m3 . doll 12 ≈ (cid:18) (cid:19) (cid:0) (cid:1) (b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore, 3 1 V = 1800 m3 6.0 10 4 m3 . miniature − 144 ≈ × (cid:18) (cid:19) (cid:0) (cid:1) 9. We use the conversionfactors found in Appendix D. 1acre ft=(43,560ft2) ft=43,560ft3. · · Since 2in.=(1/6)ft, the volume of water that fell during the storm is V =(26km2)(1/6ft)=(26km2)(3281ft/km)2(1/6ft)=4.66 107ft3. × Thus, 4.66 107ft3 V = × =1.1 103acre ft. 4.3560 104ft3/acre ft × · × · 10. The metric prefixes (micro(µ), pico, nano, ...) are given for ready reference on the inside front cover of the textbook (also, Table 1-2). 100y 365day 24h 60min 1µcentury = 10 6century − 1century 1y 1day 1h (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) = (cid:0)52.6min . (cid:1) The percent difference is therefore 52.6min 50min − =5.2%. 50min 11. WeusetheconversionfactorsgiveninAppendixDandthedefinitionsoftheSIprefixesgiveninTable1- 2(alsolistedonthe insidefrontcoverofthe textbook). Here,“ns”representsthe nanosecondunit, “ps” represents the picosecond unit, and so on. (a) 1m=3.281ft and 1s=109ns. Thus, 3.0 108m 3.281ft s 3.0 108m/s= × =0.98ft/ns. × s m 109ns (cid:18) (cid:19)(cid:18) (cid:19)(cid:16) (cid:17) (b) Using 1m=103mm and 1s=1012ps, we find 3.0 108m 103mm s 3.0 108m/s = × × s m 1012ps (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) (cid:19) = 0.30 mm/ps . 12. Thenumberofsecondsinayearis3.156 107. ThisislistedinAppendixDandresultsfromtheproduct × (365.25day/y)(24h/day)(60min/h)(60s/min) . (a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than there are seconds per year. 4 CHAPTER 1. (b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans have existed is given by 106 =10 4 u-day , − 1010 which we may also express as 86400 u-sec 10−4 u-day =8.6 u-sec . 1 u-day (cid:18) (cid:19) (cid:0) (cid:1) 13. None of the clocks advance by exactly 24hin a 24-hperiod but this is not the most important criterion for judging their quality for measuring time intervals. What is important is that the clock advance by the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct interval. Iftheclockreadingjumpsaroundfromone24-hperiodtoanother,itcannotbecorrectedsince itwouldimpossibletotellwhatthecorrectionshouldbe. Thefollowinggivesthecorrections(inseconds) that must be applied to the reading oneachclock for each24-hperiod. The entries were determined by subtracting the clock reading at the end of the interval from the clock reading at the beginning. CLOCK Sun. Mon. Tues. Wed. Thurs. Fri. -Mon. -Tues. -Wed. -Thurs. -Fri. -Sat A 16 16 15 17 15 15 − − − − − − B 3 +5 10 +5 +6 7 − − − C 58 58 58 58 58 58 − − − − − − D +67 +67 +67 +67 +67 +67 E +70 +55 +2 +20 +10 +10 ClocksC andD areboth goodtimekeepers in the sense thateachis consistentin its daily drift (relative to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The correction for clock C is less than the correction for clock D, so we judge clock C to be the best and clock D to be the next best. The correction that must be applied to clock A is in the range from 15s to 17s. For clock B it is the range from 5s to +10s, for clock E it is in the range from 70s to 2s. − − − After C and D, A has the smallest range of correction, B has the next smallest range, and E has the greatest 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= 1 and 6 y-intercepts=0. From the data in the figure we deduce 6 2 594 t = t + C B 7 7 33 662 t = t . B A 40 − 5 These are used in obtaining the following results. (a) We find 33 t t = (t t )=495 s ′B− B 40 ′A− A when t t =600 s. ′A− A (b) We obtain 2 2 t t = (t t )= (495)=141 s . ′C − C 7 ′B − B 7 (c) Clock B reads t =(33/40)(400) (662/5) 198 s when clock A reads t =400 s. B A − ≈ (d) From t =15=(2/7)t +(594/7), we get t 245 s. C B B ≈− 5 15. We convert meters to astronomical units, and seconds to minutes, using 1000m = 1km 1AU = 1.50 108km × 60s = 1min . Thus, 3.0 108m/s becomes × 3.0 108m 1km AU 60s × =0.12AU/min. s 1000m 1.50 108km min (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) × (cid:19)(cid:18) (cid:19) 16. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to change ◦ longitude 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 (20century)(0.001s/century)=0.02s. The average day during the 20 centuries is (0+0.02)/2 = 0.01 s longer than the first day. Since the increase occurs uniformly, the cumulative effect T is T = (average increase in length of a day)(number of days) 0.01s 365.25day = (2000y) day y (cid:18) (cid:19)(cid:18) (cid:19) = 7305 s or roughly two hours. 18. We denote the pulsar rotation rate f (for frequency). 1 rotation f = 1.55780644887275 10 3s × − (a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore significant figure considerations for a moment), we obtain the number of rotations: 1rotation N = (604800s)=388238218.4 1.55780644887275 10 3s (cid:18) × − (cid:19) which should now be rounded to 3.88 108 rotations since the time-interval was specified in the × problem to three significant figures. (b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form N = ft 1 rotation 1 106 = t × 1.55780644887275 10 3s (cid:18) × − (cid:19) which yields the result t = 1557.80644887275s (though students who do this calculation on their calculator 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. − − × ± × 6 CHAPTER 1. 19. IfM isthemassofEarth,mistheaveragemassofanatominEarth,andN isthenumberofatoms,then E M =NmorN =M /m. WeconvertmassmtokilogramsusingAppendixD(1u=1.661 10 27kg). E E − × Thus, M 5.98 1024kg N = E = × =9.0 1049 . m (40u)(1.661 10 27kg/u) × × − 20. To organize the calculation, we introduce the notion of density (which the students have probably seen in other courses): m ρ= . V (a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density ρ=19.32 g/cm3 and mass m=27.63 g, the volume of the leaf is found to be m V = =1.430 cm3 . ρ We convert the volume to SI units: 1 m 3 1.430 cm3 =1.430 10 6 m3 . − 100 cm × (cid:18) (cid:19) (cid:0) (cid:1) And since V =Az where z =1 10 6 m (metric prefixes can be found in Table 1-2), we obtain − × 1.430 10 6m3 A= × − =1.430 m2 . 1 10 6m × − (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 6m3, we obtain − − × × V ℓ= =7.284 104 m . πr2 × 21. We introduce the notion of density (which the students have probably seen in other courses): m ρ= V and convert to SI units: 1g=1 10 3kg. − × (a) For volume conversion, we find 1cm3 =(1 10 2m)3 =1 10 6m3. Thus, the density in kg/m3 − − × × is 1g 10 3kg cm3 1g/cm3 = − =1 103 kg/m3 . cm3 g 10 6m3 × (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19) Thus, 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): M =(5700m3)(1 103kg/m3)=5.70 106 kg . × × The time is t=(10h)(3600s/h)=3.6 104s, so the mass flow rate R is × M 5.70 106kg R= = × =158 kg/s . t 3.6 104s ×