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Chapter 9: Problem 85

If you watch water falling from a faucet, you will notice that the flow decreases in radius as the water falls. This can be explained by the equation of continuity, since the cross-sectional area of the water decreases as the speed increases. If the water flows with an initial velocity of $0.62 \mathrm{m} / \mathrm{s}\( and a diameter of \)2.2 \mathrm{cm}$ at the faucet opening, what is the diameter of the water flow after the water has fallen $30 \mathrm{cm} ?$

Short Answer

Expert verified
Answer: After falling 30 cm, the diameter of the water flow is approximately 1.086 cm.

Step by step solution

01

Find the initial cross-sectional area

We are given the diameter of the water flow at the faucet opening. Let's find the initial cross-sectional area using the formula for the area of a circle: \(A = \pi r^2\). The radius of the water flow at the faucet opening is half the diameter, which is \((2.2 / 2) \mathrm{cm} = 1.1 \mathrm{cm}\) or \(0.011 \mathrm{m}\). The initial cross-sectional area is then: \(A_1 = \pi (0.011)^2 = 3.8 \times 10^{-4} \mathrm{m}^2\)
02

Find the final velocity of the water flow

We are given the initial velocity of the water flow and the distance it has fallen. We can use the conservation of energy principle to find the final velocity of the water flow: \(v_2^2 = v_1^2 + 2gh\) Where \(v_1\) is the initial velocity, \(v_2\) is the final velocity, \(h\) is the height the water has fallen, and \(g\) is the acceleration due to gravity (\(9.81 \mathrm{m} / \mathrm{s}^2\)). Plugging in the given values, we get: \(v_2^2 = (0.62)^2 + 2(9.81)(0.3)\) Solving for \(v_2\), we get: \(v_2 = \sqrt{(0.62)^2 + 2(9.81)(0.3)} = 2.56 \mathrm{m} / \mathrm{s}\)
03

Use the equation of continuity to find the final cross-sectional area

The equation of continuity states that the product of the cross-sectional area and the velocity remains constant: \(A_1v_1 = A_2v_2\) Where \(A_1\) and \(v_1\) are the initial cross-sectional area and velocity, and \(A_2\) and \(v_2\) are the final cross-sectional area and velocity. We can solve for \(A_2\): \(A_2 = \frac{A_1v_1}{v_2}\) Plugging in the known values, we get: \(A_2 = \frac{(3.8 \times 10^{-4})(0.62)}{2.56} = 9.3 \times 10^{-5} \mathrm{m}^2\)
04

Find the final diameter of the water flow

Now that we have the final cross-sectional area, we can find the final diameter. First, we need to find the final radius: \(A_2 = \pi r_2^2\) \(r_2 = \sqrt{\frac{A_2}{\pi}}\) \(r_2 = \sqrt{\frac{9.3 \times 10^{-5}}{\pi}} = 5.431 \times 10^{-3} \mathrm{m}\) Finally, we can find the final diameter, which is twice the radius: \(D_2 = 2r_2 = 2(5.431 \times 10^{-3}) = 1.086 \times 10^{-2} \mathrm{m}\) Converting the final diameter to centimeters, we get: \(D_2 = 1.086 \times 10^{-2} \times 100 = 1.086 \mathrm{cm}\) So, the diameter of the water flow after it has fallen 30 cm is approximately \(1.086 \mathrm{cm}\).

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Most popular questions from this chapter

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