Question #8895

Water leaves a hose at a rate of 1.5 kg/s with a speed of 20 m/s and is aimed at the side of a car which stops it. Assuming no splashing back, what is the force exerted by the water?

Expert's answer

The rate of change of momentum is equal to the net force:

F = dp/dt.

This can be shown using Newton’s second law. So,

F = ΔP/Δt = (P_final - P_initial)/Δt = (0 - 30kg·m/s) / 1.0s = -30N.

F = dp/dt.

This can be shown using Newton’s second law. So,

F = ΔP/Δt = (P_final - P_initial)/Δt = (0 - 30kg·m/s) / 1.0s = -30N.

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