# Halliday Resnick Walker 7ª edição, resolução dos exercícios em ingles

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```acres, we set up the
ratio 25 wp/11 barn along with appropriate conversion factors:
( ) ( ) ( ) ( )
( ) ( )
2
28 2
36
100 hide 110 acre 4047 m
1 wp 1 acre1 hide
1 10 m
1 barn
25 wp
1 10 .
11 barn
\u2212×
\u2248 ×
28. Table 7 can be completed as follows:
(a) It should be clear that the first column (under \u201cwey\u201d) is the reciprocal of the first
row \u2013 so that 910 = 0.900,
3
40 = 7.50 × 10
\u22122
, and so forth. Thus, 1 pottle = 1.56 × 10\u22123 wey
and 1 gill = 8.32 × 10\u22126 wey are the last two entries in the first column.
(b) In the second column (under \u201cchaldron\u201d), clearly we have 1 chaldron = 1 caldron (that
is, the entries along the \u201cdiagonal\u201d in the table must be 1\u2019s). To find out how many
chaldron are equal to one bag, we note that 1 wey = 10/9 chaldron = 40/3 bag so that 112
chaldron = 1 bag. Thus, the next entry in that second column is 112 = 8.33 × 10
\u22122
.
Similarly, 1 pottle = 1.74 × 10\u22123 chaldron and 1 gill = 9.24 × 10\u22126 chaldron.
(c) In the third column (under \u201cbag\u201d), we have 1 chaldron = 12.0 bag, 1 bag = 1 bag, 1
pottle = 2.08 × 10\u22122 bag, and 1 gill = 1.11 × 10\u22124 bag.
(d) In the fourth column (under \u201cpottle\u201d), we find 1 chaldron = 576 pottle, 1 bag = 48
pottle, 1 pottle = 1 pottle, and 1 gill = 5.32 × 10\u22123 pottle.
(e) In the last column (under \u201cgill\u201d), we obtain 1 chaldron = 1.08 × 105 gill, 1 bag = 9.02
× 103 gill, 1 pottle = 188 gill, and, of course, 1 gill = 1 gill.
(f) Using the information from part (c), 1.5 chaldron = (1.5)(12.0) = 18.0 bag. And since
each bag is 0.1091 m3 we conclude 1.5 chaldron = (18.0)(0.1091) = 1.96 m3.
29. (a) Dividing 750 miles by the expected \u201c40 miles per gallon\u201d leads the tourist to
believe that the car should need 18.8 gallons (in the U.S.) for the trip.
(b) Dividing the two numbers given (to high precision) in the problem (and rounding off)
gives the conversion between U.K. and U.S. gallons. The U.K. gallon is larger than the
U.S gallon by a factor of 1.2. Applying this to the result of part (a), we find the answer
for part (b) is 22.5 gallons.
30. (a) We reduce the stock amount to British teaspoons:
1
6 2 2 24
breakfastcup = 2 8 2 2 = 64 teaspoons
1 teacup = 8 2 2 = 32 teaspoons
6 tablespoons = teaspoons
1 dessertspoon = 2 teaspoons
× × ×
× ×
× × =
which totals to 122 British teaspoons, or 122 U.S. teaspoons since liquid measure is being
used. Now with one U.S cup equal to 48 teaspoons, upon dividing 122/48 \u2248 2.54, we find
this amount corresponds to 2.5 U.S. cups plus a remainder of precisely 2 teaspoons. In
other words,
122 U.S. teaspoons = 2.5 U.S. cups + 2 U.S. teaspoons.
(b) For the nettle tops, one-half quart is still one-half quart.
(c) For the rice, one British tablespoon is 4 British teaspoons which (since dry-goods
measure is being used) corresponds to 2 U.S. teaspoons.
(d) A British saltspoon is 12 British teaspoon which corresponds (since dry-goods
measure is again being used) to 1 U.S. teaspoon.
31. (a) Using the fact that the area A of a rectangle is (width) × (length), we find
( ) ( )( )
( ) ( )( )
total
2
2
3.00acre 25.0 perch 4.00 perch
40 perch 4 perch
3.00 acre 100perch
1acre
580 perch .
A = +
§ ·
= +¨ ¸
=
We multiply this by the perch2 \u2192 rood conversion factor (1 rood/40 perch2) to obtain the
(b) We convert our intermediate result in part (a):
( )
2
2 5 2
total
16.5ft580 perch 1.58 10 ft .
1perch
A
§ ·
= = ×¨ ¸
Now, we use the feet \u2192 meters conversion given in Appendix D to obtain
( )
2
5 2 4 2
total
1m1.58 10 ft 1.47 10 m .
3.281ft
A
§ ·
= × = ×¨ ¸
32. The customer expects a volume V1 = 20 × 7056 in3 and receives V2 = 20 × 5826 in3,
the difference being 31 2 =24600 inV V V\u2206 = \u2212 , or
( )
3
3
3
2.54cm 1L24600 in 403L
1 inch 1000 cm
V
where Appendix D has been used.
33. The metric prefixes (micro (µ), pico, nano, \u2026) are given for ready reference on the
inside front cover of the textbook (see also Table 1\u20132). The surface area A of each grain
of sand of radius r = 50 µm = 50 × 10\u22126 m is given by A = 4\u3c0(50 × 10\u22126)2 = 3.14 × 10\u22128
m2 (Appendix E contains a variety of geometry formulas). We introduce the notion of
density, /m V\u3c1 = , so that the mass can be found from m = \u3c1V, where \u3c1 = 2600 kg/m3.
Thus, using V = 4\u3c0r3/3, the mass of each grain is
( )36 9
3
4 50 10 m kg2600 1.36 10 kg.
3 m
m
\u3c0 \u2212
\u2212
§ ·× § ·¨ ¸
= = ×¨ ¸¨ ¸ © ¹
We observe that (because a cube has six equal faces) the indicated surface area is 6 m2.
The number of spheres (the grains of sand) N which have a total surface area of 6 m2 is
given by
2
8
8 2
6 m 1.91 10 .
3.14 10 m
N
\u2212
= = ×
×
Therefore, the total mass M is given by
( ) ( )8 91.91 10 1.36 10 kg 0.260 kg.M Nm \u2212= = × × =
34. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m
and base area A = 20 × 12 = 240 m2) in addition to a rectangular box (height h´ = 6.0 m
and same base). Therefore,
31 1800 m .
2 2
hV hA h A h A§ ·\u2032 \u2032= + = + =¨ ¸© ¹
(a) Each dimension is reduced by a factor of 1/12, and we find
Vdoll
3 3m m=
F
HG
I
KJ \u22481800
1
12
10
3
c h . .
(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144.
Therefore,
Vminiature
3m 6.0 10 m= FHG
I
KJ \u2248 ×
\u22121800 1
144
3
4 3c h .
35. (a) Using Appendix D, we have 1 ft = 0.3048 m, 1 gal = 231 in.3, and 1 in.3 = 1.639 ×
10\u22122 L. From the latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460
ft2/gal becomes
22
2 2460 ft 1 m 1 gal460 ft /gal 11.3 m L.
gal 3.28 ft 3.79 L
§ ·§ · § ·
= =¨ ¸¨ ¸ ¨ ¸
(b) Also, since 1 m3 is equivalent to 1000 L, our result from part (a) becomes
2
2 4 1
3
11.3 m 1000 L11.3 m /L 1.13 10 m .
L 1 m
\u2212
§ ·§ ·
= = ×¨ ¸¨ ¸
(c) The inverse of the original quantity is (460 ft2/gal)\u22121 = 2.17 × 10\u22123 gal/ft2.
(d) The answer in (c) represents the volume of the paint (in gallons) needed to cover a
square foot of area. From this, we could also figure the paint thickness [it turns out to be
about a tenth of a millimeter, as one sees by taking the reciprocal of the answer in part
(b)].
36. When the Sun first disappears while lying down, your line of sight to the top of the
Sun is tangent to the Earth\u2019s surface at point A shown in the figure. As you stand,
elevating your eyes by a height h, the line of sight to the Sun is tangent to the Earth\u2019s
surface at point B.
Let d be the distance from point B to your eyes. From Pythagorean theorem, we have
2 2 2 2 2( ) 2d r r h r rh h+ = + = + +
or 2 22 ,d rh h= + where r is the radius of the Earth. Since r h\ufffd , the second term can be
dropped, leading to 2 2d rh\u2248 . Now the angle between the two radii to the two tangent
points A and B is \u3b8, which is also the angle through which the Sun moves about Earth
during the time interval t = 11.1 s. The value of \u3b8 can be obtained by using
360 24 h
t\u3b8
=
°
.
This yields
(360 )(11.1 s) 0.04625 .(24 h)(60 min/h)(60 s/min)\u3b8
°
= = °
Using tand r \u3b8= , we have 2 2 2tan 2d r rh\u3b8= = , or
2
2
tan
h
r
\u3b8
=
Using the above value for \u3b8 and h = 1.7 m, we have 65.2 10 m.r = ×
37. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 =
0.3048 m (which also can be found in Appendix D). The volume of a cord of wood is 8 ×
4 × 4 = 128 ft3, which```