Question 1020919
Note that L/min = {{{cm^3/min}}}

==> speed of water = (volume speed)/(cross-sectional area of pipe) = {{{(192.5 (cm^3/min))/((22/7)(3.5^2)(cm^2)) = 5 cm/min}}}, where we used {{{pi = 22/7}}}

==> speed of water = {{{(5cm/min)(1km/((10^5) cm))(60min/hr)}}} = 0.003 kph