Question

a diwali rocket is ejecting 0.05kg of gases per second at a velocity of 400m/s the accelerating force on the rocket is:100 N,22dynes,20dynes,20N

Answer 1

The answer might be 20 N
F = change in momentum
F = M × v(final) - v(initial)/time and hence it has to be
F = 0.05 × 400 = 20N

Answer 2

Answer:

0.05* 400 = 20N......

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