Answer
(a) $5.0\frac{m}{s}$
(b) $1.1atm$
Work Step by Step
(a) We can find the speed of water in the nozzle as
$A_1v_1=A_2v_2$
This can be rearranged as:
$v_2=\frac{A_1}{A_2}v_1=(\frac{d_1}{d_2})^2$
We plug in the known values to obtain:
$v_2=(\frac{0.63}{0.25})^2(0.78)$
$v_2=5.0\frac{m}{s}$
(b) The pressure in the nozzle can be determined as
$P_2=1.2(1.01\times 10^5)+\frac{1}{2}(1000)((0.78)^2-(5.0)^2)$
$P_2=(109\times 10^9)(\frac{1atm}{1.01\times 10^5}Pa)=1.1atm$
