## College Physics (4th Edition)

The speed of the water through the point where the pipe tapers is $3.2~m/s$
We can find the speed $v_2$ that the water moves through the point where the pipe tapers: $A_2~v_2 = A_1~v_1$ $v_2 = \frac{A_1~v_1}{A_2}$ $v_2 = \frac{\pi~r_1^2~v_1}{\pi~r_2^2}$ $v_2 = \frac{r_1^2~v_1}{r_2^2}$ $v_2 = \frac{(0.010~m)^2~(0.20~m/s)}{(0.0025~m)^2}$ $v_2 = 3.2~m/s$ The speed of the water through the point where the pipe tapers is $3.2~m/s$.