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A horizontal jet of water with its cross-sectional area of 0.0028 m² has a fixed vertical plate with a velocity of 5 m/s. After impact the jet splits symmetrically in a plane parallel to the plane of the plate. The force of impact (in N) of the jet on the plate is

1. 90
2. 80
3. 70
4. 60

Correct answer: 70

Solution

Force on a stationary flat plate from a normal jet = rho*A*V^2 = 1000 * 0.0028 * 5^2 = 1000*0.0028*25 = 70 N. So the correct value is 70 N (index 2), not 90 N.

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