1 Chapter 1 1. 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 36.37 10 m 10 km m 6.37 10 km,ER the corresponding circumference, surface area and volume are: 23 42 , 4 ,3E E E C R A R V R . The geometric formulas are given in Appendix E. ANALYZE (a) Using the formulas given above, we find the circumference to be 342 2 (6.37 10 km) 4.00 10 km.ECR (b) Similarly, the surface area of Earth is 22 3 8 24 4 6.37 10 km 5.10 10 kmEAR
, (c) and its volume is 33 3 12 3 446.37 10 km 1.08 10 km .33EVR LEARN From the formulas given, we see that ECR , 2
EAR
, and 3
EVR . The ratios of volume to surface area, and surface area to circumference are / /3EV A R and /2E A C R
. 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 220.50 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). CHAPTER 1 2 (a) Since 1 km = 1 103 m and 1 m = 1 106 m, 3 3 6 91km 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 = 102 m, 2 2 6 41cm =10 m = 10 m 10 m m 10 m. We conclude that the fraction of one centimeter equal to 1.0 m is 1.0 104. (c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, 651.0yd = 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 1inch 6 picas0.80 cm = 0.80 cm 1.9 picas.2.54 cm 1inch (b) With 12 points = 1 pica, we have 1inch 6 picas 12 points0.80 cm = 0.80 cm 23 points.2.54 cm 1inch 1pica 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 rod1.0 furlong 201.168 m (201.168 m) 40 rods,5.0292 m and 1 chain1.0 furlong 201.168 m (201.168 m) 10 chains20.117 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 rods4.0 furlongs 4.0 furlongs 160 rods,1 furlongd 3 (b) and in chains to be 10 chains4.0 furlongs 4.0 furlongs 40 chains.1 furlongd 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 8.33 102 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = 1
48 cahiz, or 2.08 102 cahiz. Continuing in this way, the remaining entries in the first column are 6.94 103 and 33.47 10
. (b) In the second (“fanega”) column, we find 0.250, 8.33 102, and 4.17 102 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) Usin
g the value (1 almude = 6.94 103 cahiz) found in part (a), we conclude that 7.00 almudes is equivalent to 4.86 102 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
12 fanega = 7.00
12 (55501 cm3) = 3.24 104 cm3. 7. We use the conversion factors found in Appendix D. 231 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(26 km )(1/6 ft) (26 km )(3281ft/km) (1/6 ft) 4.66 10 ft .V Thus, V
466 10
43560 1011107
43 .
..ft
ftacreftacre ft.3
3 CHAPTER 1 4 8. From Fig. 1-4, we see that 212 S is equivalent to 258 W and 212 – 32 = 180 S is e
quivalent to 216 – 60 = 156 Z. The information allows us to convert S to W or Z. (a
) In units of W, we have 258 W50.0S 50.0S 60.8 W212S (b) In units of Z, we have 156 Z50.0S 50.0S 43.3 Z180S 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 = r2/2, where r is the radius. Therefore, the volume is 2
2V r z where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have 32
5 10 m 10 cm2000km 2000 10 cm.1km 1mr In these units, the thickness becomes 2
2 10 cm3000m 3000m 3000 10 cm1mz which yields 25 2 22 32000 10 cm 3000 10 cm 1.9 10 cm .2V 10. 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. 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
