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A bowler releases a 7.0kg bowling ball from rest to a final velocity of 8.0m/s. What is the magnitude of the change in momentum of the bowling ball?

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Respuesta :

MathPhys

Answer:

56 kg m/s

Explanation:

Δp = mΔv

Δp = (7.0 kg) (8.0 m/s − 0 m/s)

Δp = 56 kg m/s

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