TEST BANK For Fundamentals of Physics 10th Edition By Resnick, Walker and Halliday Chapters 1 - 44

1. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earth as R  6.37  106 m103 km m  6.37  103 km, its circumference is s  2 R 2 (6.37  103 km)  4.00104 km. (b) The surface area of Earth is A  4R 2  4 6.37  10 3 km 2  5.10  108 km2 . (c) The volume of Earth is V  4 R 3  4 6.37  10 3 km 3  1.08  10 12 km3 . 3 3 2. The conversion factors are: 1 gry 1/10 line , 1 line 1/12 inch and 1 point = 1/72 inch. The factors imply that 1 gry = (1/10)(1/12)(72 points) = 0.60 point. Thus, 1 gry2 = (0.60 point)2 = 0.36 point2 , which means that 0.50 gry2 = 0.18 point2 . 3. 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 m, 1km  103 m  103 m106  m m  109 m. The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0  109 m. (b) We calculate the number of microns in 1 centimeter. Since 1 cm = 102 m, 1cm = 102 m = 102m106  m m  104 m. We conclude that the fraction of one centimeter equal to 1.0 m is 1.0  104 . (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, 1 2 CHAPTER 1 12 48 1.0 yd = 0.91m106  m m  9.1  105 m. 4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain 0.80 cm = 0.80 cm  1 inch  6 picas  1.9 picas.  2.54 cm  1 inch  (b) With 12 points = 1 pica, we have    0.80 cm = 0.80 cm  1 inch  6 picas  12 points   23 points.  2.54 cm  1 inch  1 pica      5. Given that 1 furlong  201.168 m, 1 rod  5.0292 m and 1chain  20.117 m, we find the relevant conversion factors to be 1.0 furlong  201.168 m  (201.168 m) and 1 rod 5.0292  40 rods, 1.0 furlong  201.168 m  (201.168 m ) 1 chain 20.117 m 10 chains. Note the cancellation of m (meters), the unwanted unit. Using the given conversion factors, we find (a) the distance d in rods to be d  4.0 furlongs 4.0 furlongs 40 rods 1 furlong  160 rods, (b) and that distance in chains to be d  4.0 furlongs 4.0 furlongs 10 chains  40 chains. 1 furlong 6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1 fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. Thus, 1 fanega = 1 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1 cahiz, or 2.08  102 cahiz. Continuing in this way, the remaining entries in the first column are 6.94  103 and 3.47103 . m 3 2  180 S  (b) In the second (“fanega”) column, we find 0.250, 8.33  102 , and 4.17  102 for the last three entries. (c) In the third (“cuartilla”) column, we obtain 0.333 and 0.167 for the last two entries. (d) Finally, in the fourth (“almude”) column, we get 1 = 0.500 for the last entry. (e) Since the conversion table indicates that 1 almude is equivalent to 2 medios, our amount of 7.00 almudes must be equal to 14.0 medios. (f) Using the value (1 almude = 6.94  103 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86  102 cahiz. (g) Since each decimeter is 0.1 meter, then 55.501 cubic decimeters is equal to 0.055501 m3 or 55501 cm3 . Thus, 7.00 almudes = 7.00 fanega = 7.00 (55501 cm3 ) = 3.24  104 cm3 . 12 12 7. We use the conversion factors found in Appendix D. 1 acre ft = (43,560 ft 2 )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 )(3281ft/km)2 (1/6 ft)  4.66107 ft3 . Thus, V   1.1  103 acre ft. 8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is equivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a) In units of W, we have 50.0 S  50.0 S  258 W   212 S   60.8 W   (b) In units of Z, we have 50.0 S  50.0 S  156 Z   43.3 Z   9. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = r 2 /2, where r is the radius. Therefore, the volume is 4.66  107 ft3 4.3560  104 ft3 acre ft 4 CHAPTER 1 b14daygb86400s dayg 2 V   r 2 z 2 where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have 103 m   102 cm  5 r  2000 km      2000  10 cm.  1km   1m  In these units, the thickness becomes 102 cm  2 z  3000 m  3000 m    3000  10 cm  1m  which yields V   2000  10 5 cm 2 3000  10 2 cm  1.9  10 22 cm3 . 10. Since a change of longitude equal to 360 corresponds to a 24 hour change, then one expects to change longitude by360 / 24 15 before resetting one's watch by 1.0 h. 11. (a) Presuming that a French decimal day is equivalent to a regular day, then the ratio of weeks is simply 10/7 or (to 3 significant figures) 1.43. (b) In a regular day, there are 86400 seconds, but in the French system described in the problem, there would be 105 seconds. The ratio is therefore 0.864. 12. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so b3.7 mg c10 6  m mh  3.1  m s. 13. The time on any of these clocks is a straight-line function of that on another, with slopes  1 and y-intercepts  0. From the data in the figure we deduce t  2 t C 7 B  594 , 7 t  33 t B 40 A  662 . 5 These are used in obtaining the following results. (a) We find t  t  33 t  t   495 s B B 40 A A when t'A  tA = 600 s. 5      (b) We obtain t  t  2 bt  t g  2 b495g  141 s. C C 7 B B 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. 14. The metric prefixes (micro (), pico, nano, …) are given for ready reference on the inside front cover of the textbook (also Table 1–2). (a) 1 century  106 century  100 y   365 day   24 h   60 min   52.6 min.  1 century   1 y   1 day   1 h          (b) The percent difference is therefore 52.6 min  50 min  4.9%. 52.6 min 15. A week is 7 days, each of which has 24 hours, and an hour is equivalent to 3600 seconds. Thus, two weeks (a fortnight) is 1209600 s. By definition of the micro prefix, this is roughly 1.21  1012 s. 16. We denote the pulsar rotation rate f (for frequency). 1 rotation f 1.  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: N   1 rotation  604800 s  .4  1.  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 = ft, or 1  106   1 rotation  t  1.  103 s  6 CHAPTER 1 which yields the result t = 1557. 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17s . We therefore expect that as a result of one million revolutions, the uncertainty should be (  31017 )(1106 )=  31011 s . 17. 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 Sun. -Mon. Mon. -Tues. Tues. -Wed. Wed. -Thurs. Thurs. -Fri. 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 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 to worst, the ranking of the clocks is C, D, A, B, E. 18. 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