## Essential University Physics: Volume 1 (3rd Edition)

From the Bernoulli equation, we know: $P_B + 0$ and $P_A+\frac{1}{2}\rho v^2$ are both equal. Thus, we find: $P_B =P_A+\frac{1}{2}\rho v^2 \\ P_B -P_A=\frac{1}{2}\rho v^2 \\\Delta P =\frac{1}{2}\rho v^2 \\ v = \sqrt{\frac{2\Delta P}{\rho}}$