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18 August, 22:00

3.6-kg chihuahua charges at a speed of 3.3 m/s what is the magnitude of the average force needed to bring the chihuahua to a stop in 0.50s?

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  1. 18 August, 22:29
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    Physics is my weakest subject, but it's worth a shot here. Hopefully the answer is right.

    velocity (final) = velocity (initial) + acceleration * time

    vf = vi + a * t

    We want to bring the chihuahua to a stop in half a second. So, velocity (final) = 0 m/s. Its velocity (initial) = 3.3 m/s.

    Solving for a, the acceleration.

    0 m/s = 3.3 m/s + (a) * 0.5 s

    a = - 6.6 m/s^2

    Force = mass * acceleration

    F = m * a

    We know the mass of the chihuahua and we just found the necessary acceleration.

    Plug those in.

    F = (3.6 kg) * (-6.6 m/s^2)

    F = - 23.76 N
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