A flat plate moves normally with a speed v_1 | vedclass

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(D) The force acting on the plate is given by the rate of change of momentum,$F = \frac{dp}{dt} = v_{rel} \left( \frac{dm}{dt} \right)$.The relative v

Vedclass · Accepted Answer

$\rho \left[ \frac{V}{v_2} \right] (v_1 + v_2)^2$

Vedclass · Answer

(D) The force acting on the plate is given by the rate of change of momentum,$F = \frac{dp}{dt} = v_{rel} \left( \frac{dm}{dt} ight)$.The relative velocity of the water jet with respect to the plate is $v_{rel} = v_1 + v_2$.The mass of water reaching the plate per second is $\frac{dm}{dt} = ho A v_{rel}$,where $A$ is the cross-sectional area of the jet.Since the jet discharges volume $V$ per second at speed $v_2$,the area $A = \frac{V}{v_2}$.Thus,$\frac{dm}{dt} = ho \left( \frac{V}{v_2} ight) (v_1 + v_2)$.Substituting these into the force equation: $F = (v_1 + v_2) 	imes \left[ ho \left( \frac{V}{v_2} ight) (v_1 + v_2) ight]$.Therefore,$F = ho \left[ \frac{V}{v_2} ight] (v_1 + v_2)^2$.

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