Answer
$P_2=14KPa$
Work Step by Step
According to Bernoulli's principle
$P_1+\frac{1}{2}\rho V_1^2=P_2+\frac{1}{2}\rho V_2^2$
This can be rearranged as:
$P_2=P_1+\frac{1}{2}\rho(V_1^2-V_2^2)$
We plug in the known values to obtain:
$P_2=16\times 10^3+\frac{1}{2}\times 1.06((35)^2-(175)^2)$
$P_2=16\times 10^3+(-15582\frac{g}{cm s^2})$
$P_2=16000-\frac{15582\times 10^{-3}Kg}{10^{-2}ms^2}$
This simplifies to:
$P_2=14KPa$
