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vector is therefore
22
22
2
(3.00m/s )
( 5.00m/s )
5.83 m/s
a
=+
−
=
,
and the force is
obtained from this by multiplying with the mass (
m
= 2.00 kg).
The result is
F
=
ma
=11.7 N.
(b) The direction of the force is the same as that of the acceleration:
θ
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Unformatted text preview: = tan –1 [(–5.00 m/s 2 )/(3.00 m/s 2 )] = –59.0 ° . 21. (a) The slope of each graph gives the corresponding component of acceleration. Thus, we find a x = 3.00 m/s 2 and a y = –5.00 m/s 2 . The magnitude of the acceleration...
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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, Acceleration, Force

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