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

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

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(net) displacement. That is, it has the same magnitude (5.69 
m) but points in the opposite direction (25° south of west). 
65. The two vectors a& and b
&
are given by 
\u2c6 \u2c6 \u2c6 \u2c63.20(cos 63 j sin 63 k) 1.45 j 2.85 k
\u2c6 \u2c6 \u2c6 \u2c61.40(cos 48 i sin 48 k) 0.937i 1.04 k
a
b
= ° + ° = +
= ° + ° = +
&
&
The components of &a are ax = 0, ay = 3.20 cos 63° = 1.45, and az = 3.20 sin 63° = 2.85. 
The components of 
&
b are bx = 1.40 cos 48° = 0.937, by = 0, and bz = 1.40 sin 48° = 1.04. 
(a) The scalar (dot) product is therefore 
& &
a b a b a b a bx x y y z z\u22c5 = + + = + + =0 0 937 145 0 2 85 104 2 97b gb g b gb g b gb g. . . . . . 
(b) The vector (cross) product is 
( ) ( ) ( )
( ) ( ) )( ) ( ) ( )( )) ( )( )( )
\u2c6 \u2c6 \u2c6i + a j + k
\u2c6 \u2c6 \u2c61.45 1.04 0 i + 2.85 0.937 0 j 0 1.45 0.937 k
\u2c6 \u2c6 \u2c61.51i + 2.67 j 1.36k .
y z z y z x x z x y y xa b a b a b b a b a b a b× = \u2212 \u2212 \u2212
= \u2212 \u2212 + \u2212
= \u2212
&&
(c) The angle \u3b8 between &a and 
&
b is given by 
( ) ( )
1 1 2.96cos cos 48 .
3.30 1.40
a b
ab
\u3b8 \u2212 \u2212
§ ·§ ·
\u22c5
= = = °¨ ¸¨ ¸ ¨ ¸© ¹ © ¹
&&
66. The three vectors given are 
\u2c6 \u2c6 \u2c6
 5.0 i 4.0 j 6.0 k
\u2c6 \u2c6 \u2c62.0 i 2.0 j 3.0 k
\u2c6 \u2c6 \u2c6
 4.0 i 3.0 j 2.0 k
a
b
c
= + \u2212
= \u2212 + +
= + +
&
&
&
(a) The vector equation r a b c= \u2212 +
&& & &
 is
\u2c6 \u2c6 \u2c6[5.0 ( 2.0) 4.0]i (4.0 2.0 3.0) j ( 6.0 3.0 2.0)k
\u2c6 \u2c6 \u2c6
=11i+5.0j 7.0k.
r = \u2212 \u2212 + + \u2212 + + \u2212 \u2212 +
\u2212
&
(b) We find the angle from +z by \u201cdotting\u201d (taking the scalar product) &r with \ufffdk. Noting 
that 2 2 2 = | | = (11.0) + (5.0) + ( 7.0) = 14,r r \u2212& Eq. 3-20 with Eq. 3-23 leads to 
( ) ( )k 7.0 14 1 cos 120 .r \u3c6 \u3c6\u22c5 = \u2212 = Ÿ = $&&
(c) To find the component of a vector in a certain direction, it is efficient to \u201cdot\u201d it (take 
the scalar product of it) with a unit-vector in that direction. In this case, we make the 
desired unit-vector by 
( )2 2 2
\u2c6 \u2c6 \u2c62.0i+2.0 j+3.0k
\u2c6
.| | 2.0 (2.0) (3.0)
bb
b
\u2212
= =
\u2212 + +
&
&
We therefore obtain 
( )( ) ( )( ) ( )( )
( )2 2 2
5.0 2.0 4.0 2.0 6.0 3.0
\u2c6 4.9 .
2.0 (2.0) (3.0)
ba a b
\u2212 + + \u2212
= \u22c5 = = \u2212
\u2212 + +
&
(d) One approach (if all we require is the magnitude) is to use the vector cross product, as 
the problem suggests; another (which supplies more information) is to subtract the result 
in part (c) (multiplied by \ufffdb ) from &a . We briefly illustrate both methods. We note that if 
a cos \u3b8 (where \u3b8 is the angle between &a and 
&
b ) gives ab (the component along \ufffdb ) then 
we expect a sin \u3b8 to yield the orthogonal component: 
a
a b
b
sin .\u3b8 =
×
=
& &
7 3 
(alternatively, one might compute \u3b8 form part (c) and proceed more directly). The second 
method proceeds as follows: 
&
a a bb\u2212 = \u2212 + \u2212 \u2212 \u2212 \u2212 \u2212
+ \u2212
\ufffd
. .
\ufffd
. .
\ufffd
. .
\ufffd
\ufffd
.
\ufffd
.
\ufffd
50 2 35 4 0 2 35 6 0 353
6 35 2 47
b g b gc h b g b gc hi j + k
= 2.65i j k
This describes the perpendicular part of &a completely. To find the magnitude of this part, 
we compute 
2 65 6 35 2 47 7 32 2 2. . ( . ) .+ + \u2212 =
which agrees with the first method. 
67. Let A denote the magnitude of 
&
A ; similarly for the other vectors. The vector equation 
is
& & &
A B C = + where B = 8.0 m and C = 2A. We are also told that the angle (measured 
in the \u2018standard\u2019 sense) for &A is 0° and the angle for &C is 90°, which makes this a right 
triangle (when drawn in a \u201chead-to-tail\u201d fashion) where B is the size of the hypotenuse. 
Using the Pythagorean theorem, 
2 2 2 28.0 4B A C A A= + Ÿ = +
which leads to A = 8 / 5 = 3.6 m. 
68. The vectors can be written as \u2c6 \u2c6i and ja a b b= =&& where , 0.a b >
(a) We are asked to consider 
&
b
d
b
d
=
F
HG
I
KJ
\ufffdj
in the case d > 0. Since the coefficient of \ufffdj is positive, then the vector points in the +y
direction. 
(b) If, however, d < 0, then the coefficient is negative and the vector points in the \u2013y
direction. 
(c) Since cos 90° = 0, then 0a b\u22c5 =
&&
, using Eq. 3-20. 
(d) Since 
&
b d/ is along the y axis, then (by the same reasoning as in the previous part) 
( / ) 0a b d\u22c5 =
&&
.
(e) By the right-hand rule, &
&
a b× points in the +z-direction. 
(f) By the same rule, 
& &b a× points in the \u2013z-direction. We note that b a a b× = \u2212 ×
& && &
 is true 
in this case and quite generally. 
(g) Since sin 90° = 1, Eq. 3-27 gives | |a b ab× =&& where a is the magnitude of &a .
(h) Also, | | | | a b b a ab× = × =& && & .
(i) With d > 0, we find that ( / )a b d×
&&
 has magnitude ab/d.
(j) The vector ( / )a b d× && points in the +z direction. 
69. The vector can be written as \u2c62.5 m jd =& , where we have taken \u2c6j to be the unit vector 
pointing north. 
(a) The magnitude of the vector &
&
a d= 4.0 is (4.0)(2.5) = 10 m. 
(b) The direction of the vector &
&
a d = 4.0 is the same as the direction of 
&
d (north). 
(c) The magnitude of the vector = 3.0c d\u2212
&&
 is (3.0)(2.5) = 7.5 m. 
(d) The direction of the vector = 3.0c d\u2212
&&
 is the opposite of the direction of 
&
d . Thus, the 
direction of &c is south. 
70. We orient \ufffdi eastward, \ufffdj northward, and \ufffdk upward. 
(a) The displacement in meters is consequently \u2c6 \u2c6 \u2c61000 i 2000 j 500k+ \u2212 .
(b) The net displacement is zero since his final position matches his initial position. 
71. The solution to problem 25 showed that each diagonal has a length given by a 3 , 
where a is the length of a cube edge. Vectors along two diagonals are \u2c6 \u2c6 \u2c6i j kb a a a= + +&
and \u2c6 \u2c6 \u2c6i j k.c a a a= \u2212 + +& Using Eq. 3-20 with Eq. 3-23, we find the angle between them: 
cos .\u3c6 = + + = \u2212 + + =b c b c b c
bc
a a a
a
x x y y z z
2 2 2
23
1
3
The angle is \u3c6 = cos\u20131(1/3) = 70.5°. 
72. The two vectors can be found be solving the simultaneous equations. 
(a) If we add the equations, we obtain 2 6& &a c= , which leads to \u2c6 \u2c63 9 i 12 ja c= = +& & .
(b) Plugging this result back in, we find 
& &b c= = +3 4\ufffd \ufffdi j .
73. We note that the set of choices for unit vector directions has correct orientation (for a 
right-handed coordinate system). Students sometimes confuse \u201cnorth\u201d with \u201cup\u201d, so it 
might be necessary to emphasize that these are being treated as the mutually 
perpendicular directions of our real world, not just some \u201con the paper\u201d or \u201con the 
blackboard\u201d representation of it. Once the terminology is clear, these questions are basic 
to the definitions of the scalar (dot) and vector (cross) products. 
(a) \u2c6 \u2c6i k=0\u22c5 since \u2c6 \u2c6i k\u22a5
(b) \u2c6 \u2c6( k) ( j)=0\u2212 \u22c5 \u2212 since \u2c6 \u2c6k j\u22a5 .
(c) \u2c6 \u2c6j ( j)= 1.\u22c5 \u2212 \u2212
(d) \u2c6 \u2c6 \u2c6k j= i (west).× \u2212
(e) \u2c6 \u2c6 \u2c6( i) ( j)= k (upward).\u2212 × \u2212 +
(f) \u2c6 \u2c6 \u2c6( k) ( j)= i (west).\u2212 × \u2212 \u2212
74. (a) The vectors should be parallel to achieve a resultant 7 m long (the unprimed case 
shown below),
(b) anti-parallel (in opposite directions) to achieve a resultant 1 m long (primed case 
shown),
(c) and perpendicular to achieve a resultant 3 4 52 2+ = m long (the double-primed case 
shown).
In each sketch, the vectors are shown in a \u201chead-to-tail\u201d sketch but the resultant is not 
shown. The resultant would be a straight line drawn from beginning to end; the beginning 
is indicated by A (with or without primes, as the case may be) and the end is indicated by 
B.
75. A sketch of the displacements is shown. The resultant (not shown) would be a straight 
line from start (Bank) to finish (Walpole). With a careful drawing, one should find that