# Halliday resolvido

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• Publicado : 23 de março de 2013

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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 μm,

1km = 103 m = 103 m 106 μ m m = 109 μ m. The given measurement is 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 1centimeter. 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.

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2. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we obtain 1 inch 6picas ≈ 1.9 picas. 0.80 cm = ( 0.80 cm ) 2.54 cm 1 inch (b) With 12 points = 1 pica, we have
0.80 cm = ( 0.80 cm ) 1 inch 2.54 cm 6 picas 1 inch 12 points ≈ 23 points. 1 pica

3. 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 =

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

= 40 chains.

4. 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 .

5. Various geometric formulas are given in Appendix E. (a) Expressing the radius of the Earthas

R = ( 6.37 × 106 m )(10−3 km m ) = 6.37 × 103 km,
its circumference is s = 2π R = 2π (6.37 × 103 km) = 4.00 ×104 km. (b) The surface area of Earth is A = 4π R 2 = 4π ( 6.37 × 103 km ) = 5.10 × 108 km 2 .
2

(c) The volume of Earth is V =

4 π 3 4π R = 6.37 × 103 km 3 3

(

)

3

= 1.08 × 1012 km3 .

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
50.0 S = ( 50.0 S) 258 W = 60.8 W 212 S

(b) In units of Z, we have
50.0 S = ( 50.0 S ) 156 Z = 43.3 Z 180 S

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 π V = r2 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 )

102 cm = 3000 × 102 cm 1m
2

which yields V =

π 2000 × 105 cm 2

(

) ( 3000 × 10
2

cm = 1.9 × 1022 cm3 .

)

8. We make use of Table1-6. (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. 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 =
1 48

cahiz, or 2.08 × 10−2 cahiz.

Continuing in this way, theremaining 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

= 0.500 for the last entry.

(e) Since the conversiontable 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...