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94. (a) With
G
G
pm
v
==
−
⋅
16
#
,
j kg m s we take the vector cross product (using either Eq.
330 or, more simply, Eq. 1120 and the righthand rule):
2
ˆ
(3
2k
gm/
s
)
k
.
rp
=× =−
⋅
G
GG
A
(b) Now the axis passes through the point
G
R
=
40
.
#
j m, parallel with the
z
axis. With
G
G
G
′
=− =
rr
R
20
.
#
i m, we again take the cross product and arrive at the same result as
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 Spring '08
 Any
 Physics

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