Answer to Question #4178 in Mechanics | Relativity for moha
What is the velocity of water leaving the garden hose under these conditions, in ft/min?
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.
Need a fast expert's response?Submit order
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!