## College Physics (4th Edition)

The gauge pressure of the water in the hose is $3.125\times 10^5~N/m^2$
Let $P_1$ be the pressure inside the hose and let $P_2$ be the pressure outside the hose, which is the atmospheric pressure. We can use Bernoulli's equation to find the pressure difference between the inside and outside of the hose: $P_1 + \frac{1}{2}\rho~v_1^2 = P_2+\frac{1}{2}\rho~v_2^2$ $P_1 + 0 = P_2+\frac{1}{2}\rho~v_2^2$ $P_1 - P_2 = \frac{1}{2}\rho~v_2^2$ $P_1 - P_2 = (\frac{1}{2})(1000~kg/m^3)(25~m/s)^2$ $P_1 - P_2 = 3.125\times 10^5~N/m^2$ The pressure difference between the inside and outside of the hose is $3.125\times 10^5~N/m^2$. Since this pressure difference is the gauge pressure, the gauge pressure of the water in the hose is $3.125\times 10^5~N/m^2$.