## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (4th Edition)

(a) $v_2 = 1.0~m/s$ $v_3 = 16~m/s$ (b) The volume flow rate is 0.314 L/s
(a) The volume flow rate is constant throughout the pipe. The volume flow rate is the speed of the liquid times the cross-sectional area of the pipe. We can find the speed $v_2$ in the second segment. $v_2~A_2 = v_1~A_1$ $v_2~\pi~r_2^2 = v_1~\pi~r_1^2$ $v_2 = \frac{v_1~r_1^2}{r_2^2}$ $v_2 = \frac{(4.0~m/s)(0.0050~m)^2}{(0.010~m)^2}$ $v_2 = 1.0~m/s$ We can find the speed $v_3$ in the third segment. $v_3~A_3 = v_1~A_1$ $v_3~\pi~r_3^2 = v_1~\pi~r_1^2$ $v_3 = \frac{v_1~r_1^2}{r_3^2}$ $v_3 = \frac{(4.0~m/s)(0.0050~m)^2}{(0.0025~m)^2}$ $v_3 = 16~m/s$ (b) The volume flow rate is the speed of the liquid times the cross-sectional area of the pipe. $flow~rate = v_1~A_1$ $flow~rate = v_1~\pi~r_1^2$ $flow~rate = (4.0~m/s)(\pi)(0.0050~m)^2$ $flow~rate = 3.14\times 10^{-4}~m^3/s$ We can convert the volume flow rate to units of L/s. $flow~rate = (3.14\times 10^{-4}~m^3/s)(\frac{1000~L}{1~m^3})$ $flow~rate = 0.314~L/s$ The volume flow rate is 0.314 L/s.