Prefixos métricos

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 μ m, ()() 3 3 6 9 1km 10 m 10 m 10 m m 10 m. == = μμ

The given measurement is 1.0 km (two significant figures), which implies our result should be written as 1.0 × 109 μ 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. −− = μμ

We conclude that the fraction of one centimeter equal to 1.0 μ m is 1.0 × 10−4.

(c) Since 1 yd = (3 ft)(0.3048 m/ft) = 0.9144 m, ()() 65 1.0yd = 0.91m 10 m m 9.1 10 m. =× μμ

2. (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 § ·§ · ≈ ¨ ¸¨ ¸ © ¹© ¹

(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 § ·§ ·§ · ≈ ¨ ¸¨ ¸¨ ¸ © ¹© ¹© ¹

(b) and that distance in chains to be ()() 4.0 furlongs 201.168 m furlong 40 chains. 20.117 m chain d ==

3. Using the given conversion factors, we find

(a) the distance d in rods to be

()() 4.0 furlongs 201.168 m furlong 4.0 furlongs = 160 rods, 5.0292 m rod d ==

4. The conversion factors 1 gry 1/10 line = , 1 line=1/12 inchand 1 point = 1/72 inch 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 22 0.50 gry = 0.18 point .

5. Various geometric formulas are given in Appendix E.

(a) Expressing the radius of the Earth as () () 63 3 6.37 10 m 10 km m 6.37 10 km,R −=× = ×

its circumference is 34 2 2 (6.37 10 km) 4.00 10 km. sR ππ = = × = × (b) The surface area of Earth is () 2 2 3 8 2 4 4 6.37 10 km 5.10 10 km . AR = π = π × = × (c) The volume of Earth is () 3 3 3 12 3 446.37 10 km 1.08 10 km . 33 VR ππ = = × = ×

6. From Figure 1.6, 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

() 258 W50.0 S 50.0 S 60.8 W 212 S §· == ¨¸ ©¹

(b) In units of Z, we have

() 156 Z50.0 S 50.0 S 43.3 Z 180 S §· == ¨¸ ©¹

2

2 V r z π= wherez is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have

() 32 510 m 10 cm2000km 2000 10 cm. 1km 1m

r

§ · § · ==× ¨ ¸ ¨ ¸ © ¹ © ¹

In these units, the thickness becomes () 2 210 cm3000m 3000m 3000 10 cm 1m z §· == = × ¨¸ ©¹ which yields ()() 2 5 2 22 3 2000 10 cm 3000 10 cm 1.9 10 cm . 2 V π = × × = ×

7. The volume of ice is given by the product of the semicircular surface area and the thickness. The area of the semicircle is A = πr2/2, where r is the radius. Therefore, the volume is

8. 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