Question

A garden hose of inner radius \(1.0 \mathrm{cm}\) carries water at $2.0 \mathrm{m} / \mathrm{s} .\( The nozzle at the end has radius \)0.20 \mathrm{cm} .$ How fast does the water move through the nozzle?

Short answer

Answer: The speed of the water through the nozzle is 25.0 m/s.

Step-by-step solution

Calculate the cross-sectional area of the hose and the nozzle

We can use the formula for the area of a circle to calculate the cross-sectional areas of both the hose and the nozzle by plugging in their respective radii. The area of the hose (\(A_{1}\)) is given by: $$ A_{1} = \pi r_{1}^2 $$ where \(r_1 = 1.0 \text{ cm}\) is the inner radius of the hose. The area of the nozzle (\(A_{2}\)) is given by: $$ A_{2} = \pi r_{2}^2 $$ where \(r_2 = 0.20 \text{ cm}\) is the inner radius of the nozzle.

Apply the principle of continuity

According to the principle of continuity, the product of the cross-sectional area and the speed of the fluid should remain constant in the hose and the nozzle. Let \(v_{1}\) be the speed in the hose, and \(v_{2}\) be the speed in the nozzle. $$ A_{1}v_{1} = A_{2}v_{2} $$

Plug in the given values and solve for \(v_{2}\)

We are given \(v_{1} = 2.0 \text{ m/s}\) and the inner radii of the hose and the nozzle. Plug in these values and the expressions for \(A_{1}\) and \(A_{2}\) and solve for \(v_{2}\): $$ (\pi r_{1}^2)v_{1} = (\pi r_{2}^2)v_{2} $$ Divide both sides by \(\pi\) and solve for \(v_{2}\): $$ v_{2} = \frac{r_{1}^2}{r_{2}^2}v_{1} $$ $$ v_{2} = \frac{(1.0 \text{ cm})^2}{(0.20 \text{ cm})^2}(2.0 \text{ m/s}) $$

Convert units and calculate the speed of the water through the nozzle

Convert the radii to meters before calculating \(v_{2}\): $$ v_{2} = \frac{(0.01 \text{ m})^2}{(0.002 \text{ m})^2}(2.0 \text{ m/s}) $$ Calculate the speed of the water through the nozzle: $$ v_{2} = 25.0 \text{ m/s} $$ The water moves through the nozzle at a speed of \(25.0 \text{ m/s}\).