# Estudante

Páginas: 26 (6368 palavras) Publicado: 9 de novembro de 2012
1. 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 )
5.0292 m rod

= 160 rods,

(b) and that distance in chains to be d =

( 4.0 furlongs )( 201.168 m furlong )
20.117 m chain

= 40 chains.

2. The conversion factors 1 gry = 1/10 line , 1 line=1/12 inch and 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 0.50 gry 2 = 0.18 point 2 .

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 µm,
1km = 103 m = 103 m 106 µ m m = 109 µ m.

(

)(

)

The given measurementis 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,
1cm = 10−2 m = 10 −2 m 106 µ m m = 104 µ m.

(

)(

)

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 = ( 0.91m ) 106 µ m m =9.1 × 105 µ 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 ) (b) With 12 points = 1 pica, we have 0.80 cm = ( 0.80 cm )

1 inch 2.54 cm

6 picas 1 inch

≈ 1.9 picas.

1 inch 2.54 cm

6 picas 1 inch

12 points ≈ 23 points. 1 pica

5. Various geometric formulas are given in Appendix E. (a)Substituting
R = 6.37 × 10 6 m 10 −3 km m = 6.37 × 10 3 km

c

hc

h

into circumference = 2πR, we obtain 4.00 × 104 km. (b) The surface area of Earth is
A = 4π R 2 = 4π 6.37 × 103 km

(

)

2

= 5.10 × 108 km 2 .

(c) The volume of Earth is
V= 4π 3 4π R = 6.37 × 103 km 3 3

(

)

3

= 1.08 × 1012 km3 .

6. We make use of Table 1-6. (a) We look at the first (“cahiz”) column: 1fanega is equivalent to what amount of cahiz? We note from the already completed part of the table that 1 cahiz equals a dozen fanega. 1 Thus, 1 fanega = 12 cahiz, or 8.33 × 10−2 cahiz. Similarly, “1 cahiz = 48 cuartilla” (in the already completed part) implies that 1 cuartilla = Continuing in this way, the remaining entries in the first column are 6.94 × 10−3 and 3.47 ×10−3 . (b) In the second(“fanega”) column, we similarly 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 1 48

cahiz, or 2.08 × 10−2 cahiz.

= 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 7.00 7.00 m3 or 55501 cm3. Thus, 7.00 almudes = 12 fanega = 12 (55501 cm3) = 3.24 × 104 cm3.

7. The volume of iceis given by the product of the semicircular surface area and the thickness. The are of the semicircle 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 r = ( 2000 km ) 103 m 1km 102 cm = 2000 × 105 cm. 1m

In these units, the thickness becomes z = 3000 m = ( 3000 m ) which yields,V = π 2000 × 105 cm 2

10 2 cm = 3000 × 102 cm 1m

(

) ( 3000 × 10
2

2

cm = 1.9 × 10 22 cm3 .

)

8. 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, 50.0 S = ( 50.0 S) (b) In units of Z, 50.0 S = ( 50.0 S) 156 Z = 43.3 Z 180 S 258 W = 60.8...

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