Answer
a) $v_2=6.4m/s$
b) $P_2=91KPa$
Work Step by Step
(a) We can find the required speed as
$v_2=\frac{A_1}{A_2}v_1$
$\implies v_2=\frac{\frac{\pi}{4}d_1^2}{\frac{\pi}{4}d_2^2}v_1$
$\implies v_2=(\frac{d_1}{d_2})^2v_1$
$\implies v_2=(\frac{d_1}{\frac{1}{2}d_1})^2v_1$
$v_2=4v_1$
We plug in the known values to obtain:
$v_2=4(1.6m/s)=6.4m/s$
(b) We know that
$P_2=P_1+\frac{1}{2}\rho(v_1^2-v_2^2)$
We plug in the known values to obtain:
$P_2=110\times 10^3Pa+\frac{1}{2}(1000Kg/m^3)((1.6m/s)^2-(6.4m/s)^2)$
$P_2=91KPa$
