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A fountain sends a stream of water straight up into the air to a maximum height of 5.00 \(\mathrm{m}\) . The effective cross-sectional area of the pipe feeding the fountain is \(5.00 \times 10^{-4} \mathrm{m}^{2}\) . Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (Note: 1 gal \(=3.79 \times 10^{-3} \mathrm{m}^{3} . )\)

Short Answer

Expert verified
The fountain uses approximately 78.6 gallons per minute.

Step by step solution

01

Understanding the Problem

We need to find the flow rate of water in gallons per minute given the maximum height reached by water in a fountain and the cross-sectional area of the feeding pipe. We'll use physics principles related to projectile motion and fluid flow to solve this problem.
02

Determining the Initial Velocity

Using the kinematic equation for motion under gravity, the maximum height (h) 5.00 meters is given by: \[ h = \frac{v^2}{2g} \]where \(v\) is the initial velocity and \(g = 9.81 \, \mathrm{m/s^2}\) is the acceleration due to gravity. Solving for \(v\), we have:\[ v = \sqrt{2gh} = \sqrt{2 \times 9.81 \, \mathrm{m/s^2} \times 5.00 \, \mathrm{m}} = \sqrt{98.1} \, \mathrm{m/s} \approx 9.90 \, \mathrm{m/s} \]
03

Calculating the Volume Flow Rate

The volume flow rate \(Q\) is given by the product of the cross-sectional area \(A\) and the initial velocity \(v\):\[ Q = A \times v = 5.00 \times 10^{-4} \, \mathrm{m^2} \times 9.90 \, \mathrm{m/s} \approx 4.95 \times 10^{-3} \, \mathrm{m^3/s} \]
04

Converting to Gallons per Minute

To convert the volume flow rate to gallons per minute, we need to use the conversion \(1 \, \mathrm{gal} = 3.79 \times 10^{-3} \, \mathrm{m^3}\):1. Convert cubic meters per second to gallons per second:\[ \frac{4.95 \times 10^{-3} \, \mathrm{m^3/s}}{3.79 \times 10^{-3} \, \mathrm{m^3/gal}} \approx 1.31 \, \mathrm{gal/s} \]2. Convert gallons per second to gallons per minute by multiplying by 60:\[ 1.31 \, \mathrm{gal/s} \times 60 = 78.6 \, \mathrm{gal/min} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fountain Physics
The magical display of water in a fountain is not just a whimsical addition to a garden; it is a complex interplay of physics. When we observe water being shot upwards and reaching a maximum height, we're witnessing kinetic energy transforming into potential energy. The force that's working here is gravity, which ultimately pulls the water back down.
In the context of physics, fountains serve as an excellent way to demonstrate concepts like conservation of energy and fluid dynamics. The maximum height the water reaches is directly influenced by the speed at which it exits the fountain and the energy supplied by the system.
By understanding fountain physics, we can predict and manipulate the water's behavior, making it a fascinating blend of art and science. Therefore, when designing or analyzing fountains, keeping these principles in mind can help optimize their function and aesthetic appeal.
Projectile Motion
Projectile motion is a cornerstone concept in physics that helps us understand how objects move under the influence of gravity. When water is ejected from a fountain, it essentially becomes a projectile. We can analyze its motion using equations of kinematics.
In our fountain example, the water reaches its highest point when its velocity reaches zero just for a moment before gravity takes it back down. The formula used here, \( h = \frac{v^2}{2g} \), is derived from these kinematic equations. Here, \(h\) represents height, \(v\) is the initial velocity, and \(g\) is the acceleration due to gravity.
This understanding of projectile motion helps us calculate vital aspects like initial velocity, which in turn can help us determine the volume flow rate of the water in the fountain.
Unit Conversion
When calculating fluid flow rates or any other physics-related measurements, unit conversion is crucial. Different systems of measurement are used across the world, and converting these units accurately is essential for correct calculations.
In our exercise, the fountain's flow rate was ultimately sought in gallons per minute, a common measurement in the United States, while other initial calculations were done in meters and seconds, typical in scientific work.
The conversion process involved knowing that 1 gallon equals \(3.79 \times 10^{-3} \) cubic meters. By converting cubic meters per second to gallons per second, and then to gallons per minute, we effectively translate scientific results into practical and familiar terms, ensuring they are meaningful and accessible.
Cross-Sectional Area Calculation
In examining fluid dynamics, the cross-sectional area of a pipe or opening is a critical factor that influences flow rates. Essentially, it is the size of the "window" through which the fluid is passing.
This area, combined with the velocity of the fluid, determines the volume flow rate, \(Q\), using the equation \(Q = A \times v\). Here, \(A\) is the cross-sectional area and \(v\) is the velocity of the fluid.
For the fountain problem, knowing the effective cross-sectional area allowed us to calculate the exact rate at which water is expelled. This calculation is foundational because it affects how much water is exiting and reaching the observed maximum height. Hence, accurately determining this area is vital for precise calculations and effective design in any hydraulic system.

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Most popular questions from this chapter

Poiseuille’s law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about \(2000 : \mathrm{Re}=2 \overline{v} \rho R / \eta .\) Here \(\overline{v},\) \(\rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{m}\right)\)

Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.020 m. Water enters the hydrant through an underground pipe of radius 0.080 m. In this pipe the water has a speed of 3.0 m/s. (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. The center of one of these holes is located twice as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole is located nearest the surface of the water. (b) Calculate the ratio of the radius of the larger hole to the radius of the smaller hole.

(a) The volume flow rate in an artery supplying the brain is \(3.6 \times 10^{-6} \mathrm{m}^{3} / \mathrm{s} .\) If the radius of the artery is \(5.2 \mathrm{mm},\) determine the average blood speed. (b) Find the average blood speed at a constriction in the artery if the constriction reduces the radius by a factor of 3. Assume that the volume flow rate is the same as that in part (a).

A full can of black cherry soda has a mass of 0.416 \(\mathrm{kg}\) . It contains \(3.54 \times 10^{-4} \mathrm{m}^{3}\) of liquid. Assuming that the soda has the same density as water, find the volume of aluminum used to make the can.

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