Question #42244

A rigid pipe of length $L = 5 \text{ m}$ is fixed such that it slopes upward and to the right at an angle $\theta = 37^\circ$ to the horizontal. The entire setup accelerates horizontally to the right with a constant acceleration $a = 5 \text{ m/s}^2$. An ideal incompressible fluid of density $\rho = 1000 \text{ kg/m}^3$ flows steadily up the pipe. At the lower end (point 1), the cross-sectional area is $A_1 = 20 \text{ cm}^2$ and the fluid velocity relative to the pipe is $v_1 = 2 \text{ m/s}$. At the upper end (point 2), the cross-sectional area is $A_2 = 10 \text{ cm}^2$. Assuming the flow is steady in the accelerating frame, what is the pressure difference $P_1 - P_2$ between the lower and upper ends? (Take $g = 10 \text{ m/s}^2$, $\sin 37^\circ = 3/5$)