A0.0450-kg golf ball initially at rest is given a speed of 25.2 m/s when a club strikes. part a part complete if the club and ball are in contact for 1.95 ms , what average force acts on the ball? f = 582 n submitprevious answers correct significant figures feedback: your answer 581 n was either rounded differently or used a different number of significant figures than required for this part. part b is the effect of the ball's weight during the time of contact significant?

We are given information:
m = 0.0450 kg
Δv = 25.2 m/s
Δt = 1.95 ms = 0.00195s

To find force we use formula:
F = m * a

a is acceleration. To find it we use formula:
a = Δv / Δt 
a = 25.2 / 0.00195
a = 12923.1 m/s^2

Now we can find force:
F = 0.0450 * 12923.1
F = 581.5 N 

To check the effect of the ball's weight on this movement we need to calculate it and then compare it to this force.
W = m * g
W = 0.0450 * 9.81
W = 0.44145 N 

We can see that weight is much smaller than the applied force so it's influence in negligible.

The answer is b hope this