Question

above the ground and fired. The water stream from the gun hits the ground a horizontal distance of 7.3 m from the muzzle. Find the gauge pressure of the water gun’s reservoir at the instant when the gun is fired. Assume that the speed of the water in the reservoir is zero and that the water flow is steady. Ignore both air resistance and the height difference between the reservoir and the muzzle.

Answer 1

Answer:

$P_g=1.1373Mpa$

Explanation:

A hand-pumped water gun is held level at a height of 0.75 m, this part miss of the question so now:

h=0.75 m

x=7.3 m

$h=\frac{1}{2}*a*t^2$

$a=g$

$t=\sqrt{\frac{2*h}{g}}=\sqrt{\frac{2*0.75m}{9.8m/s^2}}$

$t=0.153s$

$v=\frac{x}{t}=\frac{7.3m}{0.153s}=47.6m/s$

Gauge Pressure can be find

$P_g=\frac{1}{2}*1000*v^2$

$P_g=\frac{1}{2}*1000*(47.69m)^2$

$P_g=1.1373Mpa$