A garden hose has an inside diameter of 10 mm, and water flows through it at 4.0 m/s. At what rate does water leave the nozzle?. Answer in m3/s

Area = pi r^2

Look at your units and figure out what needs to be multiplied by what.

To find the rate at which water leaves the nozzle, we can make use of the principle of conservation of mass. According to this principle, the mass of fluid entering a pipe should be equal to the mass of fluid leaving the pipe, assuming there are no leaks or changes in density.

Given:
Inside diameter of the garden hose (d) = 10 mm = 0.01 m
Water flow speed through the hose (v) = 4.0 m/s

To find the rate at which water leaves the nozzle (Q), we can use the equation of continuity:

A1v1 = A2v2

Where:
A1 and A2 are the cross-sectional areas of the hose and nozzle, respectively.
v1 and v2 are the velocities of water flow through the hose and nozzle, respectively.

The cross-sectional area of a circular pipe is given by the formula:

A = πr^2

Since we know the diameter (d) of the hose, we can calculate the radius (r) by dividing it by 2:

r = d/2 = 0.01/2 = 0.005 m

Now, let's calculate the cross-sectional areas:

A1 = π(0.005)^2 = 0.00007854 m^2
A2 is unknown (which is what we want to find).

Using the equation of continuity:

A1v1 = A2v2

0.00007854 m^2 × 4.0 m/s = A2 × v2

Simplifying the equation:

0.00031416 = A2 × v2

Now, we need to find the value of A2.

As the water leaves the nozzle, it converges to a smaller cross-sectional area. Assuming it forms a circular jet, we can use the same formula for the cross-sectional area, where the radius will be half the diameter of the nozzle.

Given that we do not have the diameter of the nozzle, we cannot provide an exact answer. However, if you have the diameter/dimensions of the nozzle, you can substitute it back into the formula above to calculate A2.

Once you have A2, you can substitute it back into the equation:

0.00031416 = A2 × v2

And solve for v2 to find the rate at which water leaves the nozzle in m^3/s.