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 f

Question

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

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Thành Đạt 5 years 2021-07-18T21:23:00+00:00 1 Answers 96 views 0

Answers ( )

  1. thongdat2
    0
    2021-07-18T21:24:15+00:00
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    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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