College Physics (4th Edition)

(a) The speed of the water through the hose is $0.391~m/s$ (b) The volume flow rate is $7.86\times 10^{-5}~m^3/s$ (c) The mass flow rate is $0.0786~kg/s$
(a) We can find the speed $v_2$ that the water moves through the hose: $A_2~v_2 = A_1~v_1$ $v_2 = \frac{A_1~v_1}{A_2}$ $v_2 = \frac{\pi~r_1^2~v_1}{\pi~r_2^2}$ $v_2 = \frac{r_1^2~v_1}{r_2^2}$ $v_2 = \frac{(0.0010~m)^2~(25.0~m/s)}{(0.0080~m)^2}$ $v_2 = 0.391~m/s$ The speed of the water through the hose is $0.391~m/s$ (b) We can find the volume flow rate: $\frac{V}{t} = A~v$ $\frac{V}{t} = \pi~r_2^2~v_2$ $\frac{V}{t} = (\pi)~(0.0080~m)^2~(0.391~m/s)$ $\frac{V}{t} = 7.86\times 10^{-5}~m^3/s$ The volume flow rate is $7.86\times 10^{-5}~m^3/s$ (c) We can find the mass flow rate: $\frac{m}{t} = \rho~\frac{V}{t}$ $\frac{m}{t} = (10^3~kg/m^3)~(7.86\times 10^{-5}~m^3/s)$ $\frac{m}{t} = 0.0786~kg/s$ The mass flow rate is $0.0786~kg/s$