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Anika [276]
3 years ago
13

A 2.0-kg pistol fires a 1.0-g bullet with a muzzle speed of 1000 m/s. The bullet then strikes a 10-kg wooden block resting on a

horizontal frictionless surface. The block and the embedded bullet then slide across the surface. The explosive charge in the pistol acts for 0.001 s. What is the average force exerted on the bullet while it is being fired
Physics
1 answer:
Andreyy893 years ago
4 0

Answer:

1000 N

Explanation:

An impulse results in a change of momentum

FΔt = mΔv

F = 0.001 kg(1000 - 0) m/s / 0.001 s = 1000 N

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