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A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied by the club is 115 N. What is the speed of the ball as it leaves the tee

Respuesta :

machmillerg

Answer:

v = 50.5 m/s

Explanation:

F = (m)(^v/^t)

115N = (0.04551kg)(v/(0.020s))

2,526.917161 m/s² = v/(0.020s)

v = 50.53834322 m/s

v = 50.5 m/s

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