t = 0 t 3.5 s vo = (2.5i + 3.5j) m/s i = (-2.51) m/s %3D The velocity of a 3.0-kg object is shown at two times. What was the magnitude of the average force exerted on the object between t 0 and t 3.5s? な。

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Chapter1: Units, Trigonometry. And Vectors
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vo
t = 0
t = 3.5 s
vo = (2.5i + 3.5j) m/s
i = (-2.51) m/s
The velocity of a 3.0-kg object is shown at two times. What was the magnitude of the average force
exerted on the object between t= 0 and t = 3.5s?
O 1.74 N
O 6.26 N
O 14.7 N
O 3.23 N
O 0.845 N
Transcribed Image Text:vo t = 0 t = 3.5 s vo = (2.5i + 3.5j) m/s i = (-2.51) m/s The velocity of a 3.0-kg object is shown at two times. What was the magnitude of the average force exerted on the object between t= 0 and t = 3.5s? O 1.74 N O 6.26 N O 14.7 N O 3.23 N O 0.845 N
Expert Solution
Step 1

t = 0 sec t1 = 3.5 secchange in time = 3.5 sec mass = 3 kg vo = 2.5i^ + 3.5 j^vf = -2.5 i^change in velocity = vo -vf  =  -2.5 i^-( 2.5i^ + 3.5 j)^ = - 5 i^ - 3.5j^

 

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