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Chapter 13: Problem 49

A fountain sends water to a height of \(100 . \mathrm{m}\). What is the difference between the pressure of the water just before it is released upward and the atmospheric pressure?

Short Answer

Expert verified
Answer: The difference between the pressure of the water just before it is released and the atmospheric pressure is \(9.81 \times 10^5 \, \mathrm{Pa}\).

Step by step solution

01

Identify the known values and unknowns

We are given the height of the water at the top of the fountain (\(h = 100\) meters). We need to find the difference between the pressure at the base of the fountain and atmospheric pressure.
02

Find the potential energy at the top

Using the given height, we can find the potential energy at the top of the fountain. The potential energy (\(U\)) is given by the formula: \( U = mgh \) where \(m\) is the mass of the water, \(g\) is the acceleration due to gravity (which is approximately \(9.81 \, \frac{m}{s^2}\)), and \(h\) is the height of the water (\(100 \, m\)).
03

Find the kinetic energy at the top

Since the water reaches its maximum height, its vertical velocity at the top is \(0\). Therefore, its kinetic energy (\(K\)) at the top is also \(0\).
04

Apply Bernoulli's equation

Bernoulli's equation is given by: \(P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2\) At the top of the fountain (\(P_2\), \(v_2\) and \(h_2\)), the pressure is equal to atmospheric pressure, the velocity is zero, and the height is \(100 \,m\). At the base of the fountain, we need to find the pressure \(P_1\), the velocity \(v_1\) and the height \(h_1\) (which is \(0\)). Substitute the known values into Bernoulli's equation: \(P_1 + \frac{1}{2} \rho v_1^2 = P_{atm} + \rho g h\) Since the kinetic energy at the top is \(0\), the equation becomes: \(P_1 = P_{atm} + \rho g h\)
05

Calculate the difference in pressure

To find the difference between the pressure at the base of the fountain (just before the water is released) and the atmospheric pressure, we can subtract \(P_{atm}\) from both sides of the equation: \(\Delta P = P_1 - P_{atm} = \rho g h\) We don't have the density of water in the problem, but it is approximately \(\rho = 1000 \, \frac{kg}{m^3}\). Plug in the values for density, gravity, and height: \(\Delta P = 1000 \, \frac{kg}{m^3} \cdot 9.81 \, \frac{m}{s^2} \cdot 100 \, m\) So, \(\Delta P = 9.81 \times 10^5 \, \mathrm{Pa}\) The difference between the pressure of the water just before it is released and the atmospheric pressure is \(9.81 \times 10^5 \, \mathrm{Pa}\).

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Most popular questions from this chapter

A very large balloon with mass \(M=10.0 \mathrm{~kg}\) is inflated to a volume of \(20.0 \mathrm{~m}^{3}\) using a gas of density \(\rho_{\text {eas }}=\) \(0.20 \mathrm{~kg} / \mathrm{m}^{3}\). What is the maximum mass \(m\) that can be tied to the balloon using a \(2.00 \mathrm{~kg}\) piece of rope without the balloon falling to the ground? (Assume that the density of air is \(1.30 \mathrm{~kg} / \mathrm{m}^{3}\) and that the volume of the gas is equal to the volume of the inflated balloon).

A sealed vertical cylinder of radius \(R\) and height \(h=0.60 \mathrm{~m}\) is initially filled halfway with water, and the upper half is filled with air. The air is initially at standard atmospheric pressure, \(p_{0}=1.01 \cdot 10^{5} \mathrm{~Pa}\). A small valve at the bottom of the cylinder is opened, and water flows out of the cylinder until the reduced pressure of the air in the upper part of the cylinder prevents any further water from escaping. By what distance is the depth of the water lowered? (Assume that the temperature of water and air do not change and that no air leaks into the cylinder.)

Calculate the ratio of the lifting powers of helium (He) gas and hydrogen (H \(_{2}\) ) gas under identical circumstances. Assume that the molar mass of air is \(29.5 \mathrm{~g} / \mathrm{mol}\).

Given two springs of identical size and shape, one made of steel and the other made of aluminum, which has the higher spring constant? Why? Does the difference depend more on the shear modulus or the bulk modulus of the material?

Many altimeters determine altitude changes by measuring changes in the air pressure. An altimeter that is designed to be able to detect altitude changes of \(100 \mathrm{~m}\) near sea level should be able to detect pressure changes of a) approximately \(1 \mathrm{~Pa}\). d) approximately \(1 \mathrm{kPa}\). b) approximately 10 Pa. e) approximately \(10 \mathrm{kPa}\). c) approximately \(100 \mathrm{~Pa}\).
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