Asked by wanda
At the instant a race began, a 60-kg sprinter exerted a force of 669-N on the starting block at a 24 angle with respect to the ground. If this force was exerted for 0.44-s, with what speed (in m/s) did the sprinter leave the starting block?
Answers
Answered by
Damon
Fx = 669 cos 24
change in x momentum or impulse = Fx t
so
669 cos 24 * .44 = 60 v
change in x momentum or impulse = Fx t
so
669 cos 24 * .44 = 60 v
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