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A crane drops a 0.25kg steel ball onto a steel plate. The ball's speeds just before impact and after are 5.0m/s and 4.7m/s, respectively. If the ball is in contact with the plate for 0.035s, what is the magnitude of the average force that the ball exerts on the plate during impact?

Sagot :

abidemiokin

Answer:

2.14N

Explanation:

According to Newtons second law

F = ma

F = m(v-u)/t

m = 0.25kg

v= 5.0m/s

u = 4.7m/s

t = 0.035s

Substitute

F = 0.25(5-4.7)/0.035

F = 0.25(0.3)/0035

F = 0.075/0.035

F = 2.14N

Hence the required force is 2.14N

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