#### Answer

(a) $v_2 = 1.0~m/s$
$v_3 = 16~m/s$
(b) The volume flow rate is 0.314 L/s

#### Work Step by Step

(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.