fisica1-youn-e-freedman-exercicios-resolvidos

1.43: a) The magnitude of A
r + B
r
is
o
o
( 2.80 cm) cos 60.0 + ( 1.90 cm)
cos 60.0 )
o
+ ( 2.80 cm) sin 60.0 − ( 1.90 cm)
2
⎛
⎞
⎜
⎟ = 2.48 cm
⎜
o 2
( sin 60.0 )
⎟
⎝
⎠
and the angle is
o
o
⎛ ( 2.80 cm) sin 60.0 − ( 1.90 cm)
sin 60.0 ⎞ o
arctan ⎜
= 18
o
o
( 2.80 cm) cos 60.0 ( 1.90 cm)
cos60.0
⎟
⎝
+
⎠
b) The magnitude of A
r − B
r
is
o
o
( 2.80 cm) cos 60.0 − ( 1.90 cm)
cos 60.0 )
o
+ ( 2.80 cm) sin 60.0 + ( 1.90 cm)
2
⎛
⎞
⎜
⎟ = 4.10 cm
⎜
o 2
( sin 60.0 )
⎟
⎝
⎠
and the angle is
o
o
⎛ ( 2.80 cm) sin 60.0 + ( 1.90 cm)
sin 60.0 ⎞ o
arctan ⎜
= 84
o
o
( 2.80 cm) cos 60.0 ( 1.90 cm)
cos60.0
⎟
⎝
−
⎠
r r r r
o
B − A = − A − B ; the magnitude is 4.10 cm and the angle is 84 + 180
o o
c) ( ) = 264 .
1.44:
→
A = (–12.0 m) î . More precisely,
r
B =
r
A =
r
o
o
( 12. 0 m)( cos180 ) i + ( 12.
0 m)( sin 180 ) j.
o
( )( ) iˆ
o
18 . 0 m cos 37 + ( 18.
0 m)( sin 37 ) ˆj
= ( 14.4 m) iˆ
+ ( 10.8 m)j
ˆ
r
r
ˆ
o
o
1.45: A = ( 12 .0 m) sin 37.0 i + ( 12.0 m) cos 37.0 j = ( 7.2 m) i + ( 9.6 m) j
r
o
B ( ) iˆ
o
= 15.0 m cos 40.0 − ( 15.0 m) sin 40.0 ˆj
= ( 11.5 m) iˆ
− ( 9.6 m)
ˆj
r
o
C ( 6.0 m) cos 60.0 iˆ
o
= −
− ( 6.0 m) sin 60.0 ˆj
= −( 3.0 m) iˆ
− ( 5.2 m)j
ˆ
ˆ
ˆ
ˆ