A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver’s speed immediately after leaving the raft is 3.8 m/s. If the ti

Question

A 72 kg swimmer dives horizontally off a raft floating in a lake. The diver’s speed immediately after leaving the raft is 3.8 m/s. If the time interval of the interaction between the diver and the raft is 0.25 s, what is the magnitude of the average horizontal force by diver on the raft?

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Thiên Hương 5 years 2021-07-14T18:54:46+00:00 1 Answers 50 views 1

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  1. binhan
    1
    2021-07-14T18:56:22+00:00
    Reply

    Answer:

    F = 1094.4 N

    Explanation:

    From impulse – momentum theorem, we now that ;

    Impulse = momentum

    Where;

    Formula for impulse = force (F) × time(t)

    Momentum = mass(m) × velocity(v)

    Now, we are given;

    Mass of swimmer; m = 72 kg

    Speed; v = 3.8 m/s

    Time; t = 0.25 s

    Thus;

    F × t = mv

    F = mv/t

    F = (72 × 3.8)/0.25

    F = 1094.4 N

    This value of force is the magnitude of the average horizontal force by diver on the raft.

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