ch05-p021

# Ch05-p021 - = 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

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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.

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