Question

a garden hose is attached to a water faucet on one end and a spray nozzle on the other end. the water faucet is turned on, but the nozzle is turned off so that no water flows through the hose. the hose lies horizontally on the ground, and a stream of water sprays vertically out of a small leak to a height of 0.68 m. what is the pressure inside the hose?

a. 1.08×10^15 pa
b. 2.11×10^14 pa
c. 6.43×10^11 pa
d. 9.32×10^10 pa
e. 4.25×10^13 pa

Explanation:

Step1: Use Torricelli's law

The speed of the water coming out of the leak $v = \sqrt{2gh}$, where $g = 9.8\ m/s^{2}$ and $h = 0.68\ m$.
$v=\sqrt{2\times9.8\times0.68}\approx 3.65\ m/s$.

Step2: Apply Bernoulli's equation

Since the hose is horizontal, the height terms cancel out. The pressure outside the hose is atmospheric pressure $P_{0}= 1.01\times 10^{5}\ Pa$, and the speed of water inside the hose $v_{1}=0$ (no - flow situation initially). Let the pressure inside the hose be $P$.
Bernoulli's equation $P + \frac{1}{2}
ho v_{1}^{2}+
ho gh_{1}=P_{0}+\frac{1}{2}
ho v^{2}+
ho gh_{2}$. With $v_{1} = 0$, $h_{1}=h_{2}$ (horizontal hose), we get $P=P_{0}+\frac{1}{2}
ho v^{2}$. The density of water $
ho = 1000\ kg/m^{3}$.
$P = 1.01\times 10^{5}+\frac{1}{2}\times1000\times(3.65)^{2}$
$P=1.01\times 10^{5}+ 6661.25\approx1.08\times 10^{5}\ Pa$.

Answer:

A. $1.08\times 10^{5}\ Pa$