## Essential University Physics: Volume 1 (3rd Edition)

$P_2=14KPa$
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: $P_2=14KPa$