If the plug comes loose when the pressure hits 1.5 MPa, what was the force on the plug from the water in Newtons? $\huge \frac{v^2}{2} + gz + \frac{p}{\rho} = \text{const.} $ $ \frac{v^2}{2} + g $ $\huge P_{1} + \frac{\rho}{2}v^2 = P_{2} \\ $ $\huge 101 + 500v^2 = 1500\\ $ $ v = \sqrt{ \frac{1500-101}{500} * 1000 } $ $ v = \pi 0.15^2 * 1000 * 52.89 $ $\huge \frac{v^2}{2} + gz + \frac{p}{\rho} = \text{const.}$ $\huge \begin{align} \frac{v^2}{2} + \frac{294000}{1000} \end{align}$