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2. (a) The position vector, according to Eq. 41, is
ˆˆ
= ( 5.0 m) i + (8.0 m)j
r
−
G
.
(b) The magnitude is
222
2
2
2
 
+
+
( 5.0 m)
(8.0 m)
(0 m)
9.4 m.
rx
y
z
==
−
+
+
=
G
(c) Many calculators have polar
↔
rectangular conversion capabilities which make this
computation more efficient than what is shown below. Noting that the vector lies in the
xy
plane and using Eq. 36, we obtain:
1
8.0 m
tan
58
or
122
5.0 m
θ
−
§·
−
°
°
¨¸
−
©¹
where the latter possibility (122° measured counterclockwise from the +
x
direction) is chosen since the signs of the components imply the vector is
in the second quadrant.
(d) The sketch is shown on the right. The vector is 122° counterclockwise
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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