## College Physics (4th Edition)

The speed of the air at point B is $1.82~m/s$
The area of the pipe at point A is $\frac{1}{9}$ the area of the pipe at point B. We can use the continuity equation to find an expression for the speed of the water at point A: $A_A~v_A = A_B~v_B$ $\frac{A_B}{9}~v_A = A_B~v_B$ $v_A = 9~v_B$ The pressure difference between point A and point B is equal to the gauge pressure of the water at point A. We can use Bernoulli's equation to find the speed of the air at point B: $P_A + \frac{1}{2}\rho_a~v_A^2 = P_B+\frac{1}{2}\rho_a~v_B^2$ $\frac{1}{2}\rho_a~(v_A^2-v_B^2) = P_B-P_A$ $\frac{1}{2}\rho_a~(v_A^2-v_B^2) = \rho_w~g~h$ $v_A^2-v_B^2 = \frac{2\rho_w~g~h}{\rho_a}$ $(9~v_B)^2-v_B^2 = \frac{2\rho_w~g~h}{\rho_a}$ $80~v_B^2 = \frac{2\rho_w~g~h}{\rho_a}$ $v_B^2 = \frac{2\rho_w~g~h}{80~\rho_a}$ $v_B = \sqrt{\frac{2\rho_w~g~h}{80~\rho_a}}$ $v_B = \sqrt{\frac{(2)(1000~kg/m^3)(9.80~m/s^2)(0.0175~m)}{(80)(1.29~kg/m^3)}}$ $v_B = 1.82~m/s$ The speed of the air at point B is $1.82~m/s$.