Answer to Question #4178 in Mechanics | Relativity for moha
A half inch garden hose can discharge five hundred gallons of water per hour under normal water pressure conditions.
What is the velocity of water leaving the garden hose under these conditions, in ft/min?
We know that the garden hose can discharge five hundred gallons of water per hour.So, V / t = 5 gallons / h, where V is the volume. Suppose, the hose has a cylindrical form. Then V = S*L, where S is the cross-sectional area, S=pi*R*R. We know the diameter d = 0.5 inch, then R = 0.25 inch. We need to calculate L / t but before we do this, we must convert all the measurements to the standard form:
R = 0.25 inch = 0.00635 m, V / t = 5 gallons/h = 0.01893 m^3 / h.
So, L / t = V /(S t) = 0.01893 / (pi*0.00635^2) = 149.4 m / h = 8.2 ft/min.