Ask a New Question

Asked by Anonymous

A golf ball of mass 50.0 grams is at rest on a tee. It is struck by a golf club with an average force of 850 Newtons. The ball accelerates to a speed of 68.0 m/sec. How long was the club in contact with the ball?
8 years ago

Answers

Henry
F = M*a.
a = F/M = 850/0.05kg = 17,000 m/s^2.

V = Vo + a*t = 68 m/s.
0 + 17,000t = 68,
t = 0.004 s.

8 years ago

Related Questions

The mass of a golf ball is 45.9g . If it leaves the tee with a speed of 80.0m/s , what is its corres... A golf ball of mass 4.6 g is struck by a golf club at a speed of 50 m/s. The ball has inital velocit... A golf ball of mass 0.045 kg is hit off the tee at a speed of 48 m/s. The golf club was in contact w... A golf ball (mass 0.045 kg) is hit with a club from a tee. The plot shows the force on the ball as a... A golf ball of mass 4.50 10-2 kg is struck by a club. Contact lasts 2.35 10-4 s, and the ball leav... A golf ball of mass 30.0 grams is struck by a 4.00 kg golf club for a time of 3.00 ms. If the golf b... A golf ball, mass 0.048 kg, rests on a tee. It is struck by a golf club with an effective mass of 0.... A golf ball with a mass of 0.080kg initially at rest is given a speed of 50 m.s-1 when it is struck...
Ask a New Question
Archives Contact Us Privacy Policy Terms of Use