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Ch 12: Fluid Mechanics
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 12: Fluid MechanicsProblem 37
Chapter 12, Problem 37

A shower head has 20 circular openings, each with radius 1.0 mm. The shower head is connected to a pipe with radius 0.80 cm. If the speed of water in the pipe is 3.0 m/s, what is its speed as it exits the shower-head openings?

Verified step by step guidance
1
First, understand that this problem involves the principle of conservation of mass, specifically the continuity equation for incompressible fluids, which states that the product of the cross-sectional area and the velocity of the fluid is constant along a streamline.
Calculate the cross-sectional area of the pipe using the formula for the area of a circle: \( A = \pi r^2 \). Convert the radius of the pipe from centimeters to meters before calculating.
Calculate the total cross-sectional area of all the shower head openings. First, find the area of one opening using the same formula \( A = \pi r^2 \), converting the radius from millimeters to meters. Then, multiply this area by the number of openings (20) to get the total area.
Apply the continuity equation: \( A_1 v_1 = A_2 v_2 \), where \( A_1 \) and \( v_1 \) are the area and velocity in the pipe, and \( A_2 \) and \( v_2 \) are the total area and velocity at the shower head openings. Solve for \( v_2 \), the velocity of water exiting the shower head.
Rearrange the equation to solve for \( v_2 \): \( v_2 = \frac{A_1 v_1}{A_2} \). Substitute the known values for \( A_1 \), \( v_1 \), and \( A_2 \) to find the speed of water as it exits the shower-head openings.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Continuity Equation

The continuity equation in fluid dynamics states that the mass flow rate must remain constant from one cross-section of a pipe to another. This is expressed as A1V1 = A2V2, where A is the cross-sectional area and V is the fluid velocity. It implies that if the cross-sectional area decreases, the velocity must increase to maintain the same flow rate.
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Cross-Sectional Area Calculation

The cross-sectional area of a circular opening is calculated using the formula A = πr², where r is the radius of the circle. This calculation is crucial for determining the area through which the fluid flows, which is needed to apply the continuity equation effectively.
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Unit Conversion

Unit conversion is essential in physics to ensure consistency in calculations. In this problem, converting the radius from millimeters to centimeters and meters is necessary to match the units of velocity, which is given in meters per second. Proper conversion ensures accurate application of formulas and principles.
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Related Practice
Textbook Question

A hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 m3 and the tension in the cord is 1120 N. The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

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Textbook Question

Home Repair. You need to extend a 2.50-inch-diameter pipe, but you have only a 1.00-inch-diameter pipe on hand. You make a fitting to connect these pipes end to end. If the water is flowing at 6.00 cm/s in the wide pipe, how fast will it be flowing through the narrow one?

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Textbook Question

Water is flowing in a pipe with a varying cross-sectional area, and at all points the water completely fills the pipe. At point 1 the cross-sectional area of the pipe is 0.070 m2, and the magnitude of the fluid velocity is 3.50 m/s. (c) Calculate the volume of water discharged from the open end of the pipe in 1.00 hour.

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Textbook Question

A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below the interface (Fig. E12.33). The density of the oil is 790 kg/m3. (a) What is the gauge pressure at the upper face of the block? (b) What is the gauge pressure at the lower face of the block?

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Textbook Question

A cubical block of wood, 10.0 cm on a side, floats at the interface between oil and water with its lower surface 1.50 cm below the interface (Fig. E12.33). The density of the oil is 790 kg/m3. (a) What is the gauge pressure at the upper face of the block? (b) What is the gauge pressure at the lower face of the block?

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Textbook Question

BIO. Artery Blockage. A medical technician is trying to determine what percentage of a patient's artery is blocked by plaque. To do this, she measures the blood pressure just before the region of blockage and finds that it is 1.20×104 Pa, while in the region of blockage it is 1.15×104 Pa. Furthermore, she knows that blood flowing through the normal artery just before the point of blockage is traveling at 30.0 cm/s, and the specific gravity of this patient's blood is 1.06. What percentage of the cross-sectional area of the patient's artery is blocked by the plaque?

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