Answer
(a) $53m/s$
(b) $19cm$
Work Step by Step
(a) We know that
$\frac{\Delta V}{\Delta t}=Av$
This can be rearranged as:
$v=\frac{\frac{\Delta V}{\Delta t}}{\frac{1}{4}\pi d^2}$
We plug in the known values to obtain:
$v=\frac{\frac{1.5L/s}{1000m^3}}{\frac{1}{4}\pi(0.0060m)^2}$
$v=53m/s$
(b) The height can be determined as follows:
$P_2-P_1=\frac{1}{2}\rho v_1^2$
$\implies P_2-P_1=\frac{1}{2}(1.29)(53)^2=1.815KPa$
Now $h=\frac{P_2-P_1}{\rho_wg}$
We plug in the known values to obtain:
$h=\frac{1.815KPa}{(1000Kg/m^3)(9.81m/s^2)}$
$h=0.19m=19cm$
