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17 December, 07:21

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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  1. 17 December, 07:46
    0
    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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