# Exercicios resolvidos

Disponível somente no TrabalhosFeitos
• Páginas : 719 (179614 palavras )
• Publicado : 22 de agosto de 2012

Amostra do texto
1. The metric prefixes (micro, pico, nano, …) are given for ready reference on the inside
(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 thenumber 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.

2. (a) Using the conversion factors 1 inch = 2.54 cm exactly and 6 picas = 1 inch, we
obtain1 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
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 thatdistance in chains to be
d=

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

= 40 chains.

4. The conversion factors 1 g ry = 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 Earth as

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 that212 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 ofthe 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 )

which yields V =

π
2000 × 105 cm
2

(

102 cm
= 3000 × 10 2 cm
1m

) (3000 × 10
2

2

)

cm = 1.9 × 1022 cm3 .

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 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 1cuartilla =

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
3.47 ×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 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...