Fundamentals of Physics Extended 10th Edition Instructor's S

Fundamentals of Physics Extended 10th Edition Instructor's S1. THINK In this problem we’re given the radius of Earth, and asked to compute its circumference, surface area and volume. EXPRESS Assuming Earth to be a sphere of radius    6 3 3 6.37 10 m 10 km m 6.37 10 km, RE      the corresponding circumference, surface area and volume are: 2 3 4 2 , 4 , 3 C R A R V R E E E       . The geometric formulas are given in Appendix E. ANALYZE (a) Using the formulas given above, we find the circumference to be 3 4 2 2 (6.37 10 km) 4.00 10 km. C R        E (b) Similarly, the surface area of Earth is   2 2 3 8 2 A R      4 4 6.37 10 km 5.10 10 km   E , (c) and its volume is   3 4 4 3 3 12 3 6.37 10 km 1.08 10 km . 3 3 V RE        LEARN From the formulas given, we see that C RE , 2 A RE , and 3 V RE . The ratios of volume to surface area, and surface area to circumference are / / 3 V A R  E and / 2 A C R  E . 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 2 2 0.50 gry = 0.18 point . 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). 2 CHAPTER 1 (a) Since 1 km = 1  103 m and 1 m = 1  106 m,    3 3 6 9 1km 10 m 10 m 10 m m 10 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,    2 2 6 4 1cm = 10 m = 10 m 10 m m 10 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,    6 5 1.0 yd = 0.91m 10 m m 9.1 10 m.     4. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain   1 inch 6 picas 0.80 cm = 0.80 cm 1.9 picas. 2.54 cm 1 inch           (b) With 12 points = 1 pica, we have   1 inch 6 picas 12 points 0.80 cm = 0.80 cm 23 points. 2.54 cm 1 inch 1 pica              5. THINK This problem deals with conversion of furlongs to rods and chains, all of which are units for distance. EXPRESS Given that 1 furlong 201.168 m,  1rod 5.0292 m  and 1chain 20.117 m  , the relevant conversion factors are 1 rod 1.0 furlong 201.168 m (201.168 m) 40 rods, 5.0292 m    and 1 chain 1.0 furlong 201.168 m (201.168 m) 10 chains 20.117 m    . Note the cancellation of m (meters), the unwanted unit. ANALYZE Using the above conversion factors, we find (a) the distance d in rods to be   40 rods 4.0 furlongs 4.0 furlongs 160 rods, 1 furlong d    3 (b) and in chains to be  10 chains 4.0 furlongs 4.0 furlongs 40 chains. 1 furlong d    LEARN Since 4 furlongs is about 800 m, this distance is approximately equal to 160 rods ( 1 rod 5 m  ) and 40 chains ( 1 chain 20 m  ). So our results make sense. 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 12 cahiz, or 8.33  102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1 48 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 3.47 10  . (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 2 = 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. m 3 or 55501 cm3 . Thus, 7.00 almudes = 7.00 12 fanega = 7.00 12 (55501 cm3 ) = 3.24  104 cm3 . 7. We use the conversion factors found in Appendix D. 2 3 1 acre ft = (43,560 ft ) ft = 43,560 ft   Since 2 in. = (1/6) ft, the volume of water that fell during the storm is 2 2 2 7 3 V  (26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) 4.66 10 ft .    Thus, V        4 66 10 4 3560 10 11 10 7 4 3 . . .