A child sprays her sister with water from a garden hose. The water is supplied to the hose at a rate of $0.131 \mathrm{~L} / \mathrm{s}$, and the diameter of the nozzle is $5.43 \mathrm{~mm}$. At what speed $v$ does the water exit the nozzle? $v=$
Solution
Step 1 ：Given the rate of water supply is 0.131 L/s. Convert this to m^3/s, which gives \(0.131 \times 10^{-3} \, m^3/s\).
Step 2 ：The diameter of the nozzle is given as 5.43 mm. Convert this to meters, which gives \(5.43 \times 10^{-3} \, m\).
Step 3 ：Calculate the cross-sectional area \(A2\) of the nozzle using the formula for the area of a circle, which is \(\pi d^2/4\). This gives \(A2 = \pi (5.43 \times 10^{-3} \, m)^2 / 4 = 2.32 \times 10^{-5} \, m^2\).
Step 4 ：Substitute \(A1v1\) and \(A2\) into the equation of continuity to find \(v2\): \(0.131 \times 10^{-3} \, m^3/s = 2.32 \times 10^{-5} \, m^2 \times v2\).
Step 5 ：Solve for \(v2\) to get \(v2 = (0.131 \times 10^{-3} \, m^3/s) / (2.32 \times 10^{-5} \, m^2) = 5.65 \, m/s\).
Step 6 ：\(\boxed{5.65 \, m/s}\) is the speed at which the water exits the nozzle.
