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Unformatted text preview: 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 cover of the textbook (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 be written 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 3 1L 3 24600 in = 403 L 1 inch 1000 cm3 where 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 be d = 4.0 furlongs = (4.0 furlongs)(201.168 m/furlong) = 160 rods , 5.0292 m/rod (b) and that distance in chains to be d= (4.0 furlongs)(201.168 m/furlong) = 40 chains . 20.117 m/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 (b) and (0.80 cm)  1 inch 2.54 cm  6 picas 1 inch 1  ≈ 1.9 picas . 2 CHAPTER 1. 5. Various geometric formulas are given in Appendix E. (a) Substituting   R = 6.37 × 106 m 10−3 km/m = 6.37 × 103 km into circumference = 2πR, we obtain 4.00 × 104 km. (b) The surface area of Earth is (c) The volume of Earth is 2 4πR2 = 4π 6.37 × 103 km = 5.10 × 108 km2 . 3 4π 3 4π R = 6.37 × 103 km = 1.08 × 1012 km3 . 3 3 6. (a) Using the fact that the area A of a rectangle is width×length, we find Atotal = = = (3.00 acre) + (25.0 perch)(4.00 perch)   (40 perch)(4 perch) (3.00 acre) + 100 perch2 1 acre 580 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):  2 16.5 ft Atotal = (580 perch2 ) = 1.58 × 105 ft2 . 1 perch Now, we use the feet → meters conversion given in Appendix D to obtain 2   1m 2 5 Atotal = 1.58 × 10 ft = 1.47 × 104 m2 . 3.281 ft 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 V = π 2 r z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have   3  2 10 cm 10 m r = (2000 km) = 2000 × 105 cm . 1 km 1m In these units, the thickness becomes 102 cm z = (3000 m) 1m  Therefore, V =  = 3000 × 102 cm . 2  π 2000 × 105 cm 3000 × 102 cm = 1.9 × 1022 cm3 . 2 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 + h A = + h A = 1800 m3 . 2 2 3 (a) Each dimension is reduced by a factor of 1/12, and we find 3 Vdoll = 1800 m   1 12 3 ≈ 1.0 m3 . (b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore, 3 Vminiature = 1800 m   1 144 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 is V = (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 · 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).       100 y 365 day 24 h 60 min 1 µcentury = 10−6 century 1 century 1y 1 day 1h = 52.6 min . The percent difference is therefore 52.6 min − 50 min = 5.2% . 50 min 11. 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 m 3.0 × 108 m/s = = 0.98 ft/ns . s m 109 ns (b) Using 1 m = 103 mm and 1 s = 1012 ps, we find   3   3.0 × 108 m 10 mm s 3.0 × 108 m/s = s m 1012 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 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 13. None of the clocks advance by exactly 24 h in a 24-h period 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. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since it 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 by subtracting the clock reading at the end of the interval from the clock reading at the beginning. CLOCK A B C D E Sun. -Mon. −16 −3 −58 +67 +70 Mon. -Tues. −16 +5 −58 +67 +55 Tues. -Wed. −15 −10 −58 +67 +2 Wed. -Thurs. −17 +5 −58 +67 +20 Thurs. -Fri. −15 +6 −58 +67 +10 Fri. -Sat −15 −7 −58 +67 +10 Clocks C and D are both good timekeepers in the sense that each is consistent in 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 15 s to 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 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 6= 1 and y-intercepts 6= 0. From the data in the figure we deduce tC = tB = 2 594 tB + 7 7 33 662 tA − . 40 5 These are used in obtaining the following results. (a) We find t′B − tB = 33 ′ (t − tA ) = 495 s 40 A when t′A − tA = 600 s. (b) We obtain t′C − tC = 2 ′ 2 (tB − tB ) = (495) = 141 s . 7 7 (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. 5 15. We convert meters to astronomical units, and seconds to minutes, using 1000 m = 1 AU = 60 s = 1 km 1.50 × 108 km 1 min . Thus, 3.0 × 108 m/s becomes      3.0 × 108 m 1 km AU 60 s = 0.12 AU/min . s 1000 m 1.50 × 108 km min 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 (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 the increase occurs uniformly, the cumulative effect T is T = (average increase in length of a day)(number of days)    0.01 s 365.25 day = (2000 y) day y = 7305 s or roughly two hours. 18. We denote the pulsar rotation rate f (for frequency). f= 1 rotation 1.55780644887275 × 10−3 s (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:   1 rotation N= (604800 s) = 388238218.4 1.55780644887275 × 10−3 s 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 1 × 10 6 = = ft  1 rotation 1.55780644887275 × 10−3 s  t which yields the result t = 1557.80644887275 s (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. If ME is the mass of Earth, m is the average mass of an atom in Earth, and N is the number of atoms, then ME = N m or N = ME /m. We convert mass m to kilograms using Appendix D (1 u = 1.661 × 10−27 kg). Thus, ME 5.98 × 1024 kg N= = = 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 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 V = m = 1.430 cm3 . ρ We convert the volume to SI units: 1.430 cm3   1m 100 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 obtain A= 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 obtain ℓ= V = 7.284 × 104 m . πr2 21. We introduce the notion of density (which the students have probably seen in other courses): ρ= m V and 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/m3 is    −3    10 kg cm3 1g 3 1 g/cm = = 1 × 103 kg/m3 . cm3 g 10−6 m3 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): 3 M = (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 is R= M 5.70 × 106 kg = = 158 kg/s . t 3.6 × 104 s 7 22. The volume of the water that fell is V (26 km2 )(2.0 in.)  2   1000 m 0.0254 m (26 km2 ) (2.0 in.) 1 km 1 in. = = (26 × 106 m2 )(0.0508 m) 1.3 × 106 m3 . = = We write the mass-per-unit-volume (density) of the water as: ρ= m 3 = 1 × 103 kg/m . V The mass of the water that fell is therefore given by m = ρV :    3 m = 1 × 103 kg/m 1.3 × 106 m3 = 1.3 × 109 kg . 23. We introduce the notion of density (which the students have probably seen in other courses): ρ= m V and 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 3 3 ρ = 7.87 g/cm 1000 g 1m which yields ρ = 7870 kg/m3 . If we ignore the empty spaces between the close-packed spheres, then the 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, then V = 9.27 × 10−26 kg M −29 = m3 . 3 = 1.18 × 10 ρ 7.87 × 103 kg/m (b) We set V = 4πR3 /3, where R is the radius of an atom (Appendix E contains several geometry formulas). Solving for R, we find R=  3V 4π 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 the textbook (see also Table 1-2). The surface area A of each grain of sand of radius r = 50 µm = 50×10−6 m is 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= m V so that the mass can be found from m = ρV , where ρ = 2600 kg/m3 . Thus, using V = 4πr3 /3, the mass of each grain is 3 !   4π 50 × 10−6 m kg m= 2600 3 = 1.36 × 10−9 kg . 3 m 8 CHAPTER 1. We observe that (because a cube has six equal faces) the indicated surface area is 6 m2 . The number of spheres (the grains of sand) N which have a total surface area of 6 m2 is given by N= 6 m2 = 1.91 × 108 . 3.14 × 10−8 m2 Therefore, the total mass M is given by M = 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 and 180 S is equivalent to 156 Z. Whether or not the origin of the Zelda path coincides with the origins of the other paths is immaterial to consideration of ∆x. (a) ∆x = (50.0 S)  258 W 212 S  = 60.8 W  = 43.3 Z (b) ∆x = (50.0 S)  156 Z 180 S 26. The first two conversions are easy enough that a formal conversion is not especially called for, but in the interest of practice makes perfect we go ahead and proceed formally: (a) (11 tuffet)  2 peck 1 tuffet  = 22 peck (b) (11 tuffet)  0.50 bushel 1 tuffet  = 5.5 bushel (c) (5.5 bushel)  36.3687 L 1 bushel  ≈ 200 L 27. We make the assumption that the clouds are directly overhead, so that Figure 1-3 (and the calculations that accompany it) apply. Following the steps in Sample Problem 1-4, we have θ t = ◦ 360 24 h which, for t = 38 min = 38/60 h yields θ = 9.5◦ . We obtain the altitude h from the relation d2 = r2 tan2 θ = 2rh which is discussed in that Sample Problem, where r = 6.37 × 106 m is the radius of the earth. Therefore, h = 8.9 × 104 m. 9 28. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where ∆x = 0.05 m is the increase in horizontal depth per step)   4.57 x = Nsteps ∆x = (0.05) = 1.2 m . 0.19 However, we can approach this more carefully by noting that if there are N = 4.57/.19 ≈ 24 rises then under normal circumstances we would expect N − 1 = 23 runs (horizontal pieces) in that staircase. This would 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 barn along with appropriate conversion factors:     4047 m2  hide 110 acre (25 wp) 100 1 wp 1 hide 1 acre   ≈ 1 × 1036 . 1×10−28 m2 (11 barn) 1 barn 30. It is straightforward to compute how many seconds in a year (about 3 × 107 ). Now, if we estimate roughly one breath per second (or every two seconds, or three seconds – it won’t affect the result) then to 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 gin 1 picul  16 tahil 1 gin  10 chee 1 tahil  10 hoon 1 chee  0.3779 g 1 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 of water molecules: 1.4 × 1021 ≈ 5 × 1046 . 3.0 × 10−26 34. (a) We find the volume in cubic centimeters   3 231 in3 2.54 cm (193 gal) = 7.31 × 105 cm3 1 gal 1 in and subtract this from 1 × 106 cm3 to obtain 2.69 × 105 cm3 . The conversion gal → in3 is given in Appendix 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 kg using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in 731 kg = 4.06 × 105 min 0.0018 kg/min which we convert to 0.77 y, by dividing by the number of minutes in a year (365 days)(24 h/day)(60 min/h). 10 CHAPTER 1. 35. (a) When θ is measured in radians, it is equal to the arclength divided by the radius. For very large radius circles and small values of θ, such as we deal with in this problem, the arcs may be approximated as .... R Sun ..................................................................................... . . . . straight lines – . . ... .............. . . . . . . . . . . . ............... ...... .......... ..... . ....... ...... ................. ..... .... ...... ........ ................. .... ....... ... . ................. . . . . which for our . . . . . . . . . . . . . . . . . . . . . . ... . ....... .......... . . . . . . . . . . . . . . . . . . . . ... . . . . . . ...... ....... . . . ... . . . . . . ................. . . . . ....... . ... purposes corre. . . . . . . . . ....... . ... . . . . . . . . . .... .. .... d D .......... ...... . ... .... ...... θ spond to the di.. . ....... ... .. . ....... ... . ... . .... ...... ................ . . ... . ....... ................ . ameters d and ... ................ ....... ... ... ................ .... ...... ... ... . ....... .......................... ..... .... ..... ....... ... D of the Moon ..... .... ................ .............. . . . . . . .......... R Moon ...................................... and Sun, respectively. Thus,   θ= D d = R Moon R Sun =⇒ R Sun D = R Moon d which yields D/d = 400. (b) Various geometric formulas are given in Appendix E. Using rs and rm for the radius of the Sun and Moon, respectively (noting that their ratio is the same as D/d), then the Sun’s volume divided by that of the Moon is  3 4 3 rs 3 πrs = = 4003 = 6.4 × 107 . 4 3 r πr m m 3 (c) The angle should turn out to be roughly 0.009 rad (or about half a degree). Putting this into the equation 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 cubit And 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 volume and then convert. We proceed using the latter approach (where d is diameter and ℓ is length). Vcylinder, min = = π 2 ℓ d = 28 cubit3 4    0.43 m 3 3 28 cubit 1 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 obtain 1.972 m2 1 ken2 = = 3.88 . 1 m2 1 m2 11 (b) Similarly, we find 1 ken3 1.973 m3 = = 7.65 . 3 1m 1 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 det...
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