Asked by Sumita
A 60 g golf ball leaves the face of a golf club with a velocity of 75 m/s. If the club exerted an average force of 3.0 x 10^4 N, what was the time of impact between the club and the ball?
Answers
Answered by
Jack
f=m*a
30000N = 60g * a
a = (change in velocity) / (change in time)
30000N = 60g * (Vf - Vi) / (change in time)
500 = (75 - 0) / (change in time)
change in time = .15 sec
the change in time is the time that the ball was in contact with the club
30000N = 60g * a
a = (change in velocity) / (change in time)
30000N = 60g * (Vf - Vi) / (change in time)
500 = (75 - 0) / (change in time)
change in time = .15 sec
the change in time is the time that the ball was in contact with the club
Answered by
Jack
the mass you use is not 60g, it is .06kg. Sorry. Answer is .00015 sec.
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