A liquid of density rho is coming out of a ho | vedclass

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(B) Let the mass flow rate be $\frac{dm}{dt} = \rho A v$. The force exerted on the mesh is equal to the rate of change of momentum of the liquid.For t

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$\frac{3}{4} \rho v^2$

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(B) Let the mass flow rate be $\frac{dm}{dt} = \rho A v$. The force exerted on the mesh is equal to the rate of change of momentum of the liquid.For the $50\%$ of liquid passing through: $\Delta p = 0$,so $F_1 = 0$.For the $25\%$ of liquid that loses all momentum: $\Delta p = (0.25 \frac{dm}{dt})v - 0 = 0.25 \rho A v^2$.For the $25\%$ of liquid that bounces back: $\Delta p = (0.25 \frac{dm}{dt})v - (0.25 \frac{dm}{dt})(-v) = 0.5 \rho A v^2$.The total force $F = F_1 + F_2 + F_3 = 0 + 0.25 \rho A v^2 + 0.5 \rho A v^2 = 0.75 \rho A v^2 = \frac{3}{4} \rho A v^2$.The pressure $P = \frac{F}{A} = \frac{3}{4} \rho v^2$.

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