# Answer to Question #4178 in Mechanics | Relativity for moha

Question #4178

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?

What is the velocity of water leaving the garden hose under these conditions, in ft/min?

Expert's answer

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.

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.

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