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Chapter 9: Problem 41

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

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

Step by step solution

01

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.
02

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. $$ A_{1}v_{1} = A_{2}v_{2} $$
03

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 \(v_{2}\): $$ v_{2} = \frac{r_{1}^2}{r_{2}^2}v_{1} $$ $$ v_{2} = \frac{(1.0 \text{ cm})^2}{(0.20 \text{ cm})^2}(2.0 \text{ m/s}) $$
04

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 \(25.0 \text{ m/s}\).

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Most popular questions from this chapter

A very large vat of water has a hole \(1.00 \mathrm{cm}\) in diameter located a distance \(1.80 \mathrm{m}\) below the water level. (a) How fast does water exit the hole? (b) How would your answer differ if the vat were filled with gasoline? (c) How would your answer differ if the vat contained water, but was on the Moon, where the gravitational field strength is $1.6 \mathrm{N} / \mathrm{kg} ?$
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