Asked by Sam
A 0.7 kg object is at rest. A 6.5 N force to the right acts on the object during a time interval of 3.7 seconds. What is the velocity of the object at the end of this interval?
Answers
Answered by
Steve
F = ma
6.5 = .7a
a = 6.5/.7
v = at = 6.5/.7 * 3.7 = 34.357
6.5 = .7a
a = 6.5/.7
v = at = 6.5/.7 * 3.7 = 34.357
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