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(D) The rate of change of momentum of the water is equal to the force exerted by the pipe on the water. Let the initial velocity be $\vec{v}_i = V \ha

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$\sqrt{2} \rho VL$

Vedclass · Answer

(D) The rate of change of momentum of the water is equal to the force exerted by the pipe on the water. Let the initial velocity be $\vec{v}_i = V \hat{j}$ and the final velocity be $\vec{v}_f = V \hat{i}$. The mass flow rate is $\frac{dm}{dt} = \rho \frac{dV}{dt} = \rho L$. The force exerted by the pipe on the water is $\vec{F}_{pw} = \frac{d\vec{p}}{dt} = \frac{dm}{dt} (\vec{v}_f - \vec{v}_i) = \rho L (V \hat{i} - V \hat{j}) = \rho VL \hat{i} - \rho VL \hat{j}$. The magnitude of this force is $|\vec{F}_{pw}| = \sqrt{(\rho VL)^2 + (-\rho VL)^2} = \sqrt{2} \rho VL$. By Newton's third law,the force exerted by the water on the pipe corner is equal in magnitude and opposite in direction to the force exerted by the pipe on the water. Therefore,the magnitude of the force exerted by the water on the corner is $\sqrt{2} \rho VL$.

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