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Chapter 12: Problem 20

The pressure inside a champagne bottle can be quite high and can launch a cork explosively. Suppose you open a bottle at sea level. The absolute pressure inside a champagne bottle is 6 times atmospheric pressure; the cork has a mass of 7.5 g and a diameter of \(18 \mathrm{mm}\). Assume that once the cork starts to move, the only force that matters is the pressure force. What is the acceleration of the cork?

Short Answer

Expert verified
The acceleration of the cork is \(16933.33 \, m/s^2\).

Step by step solution

01

Calculate the Area of the Cork

The cross-sectional area \(A\) of the cork is given by the formula of the area of a circle \(A = \pi r^2\), where \(r\) is the radius of the cork. The radius of the cork is half of the diameter, and converting to meters for SI units, \(r = 18 \, mm / 2 = 0.009 \, m\). Therefore, the area \(A = \pi (0.009)^2 = 0.000254 \, m^2\).
02

Calculate the Force Acting on the Cork

The force on the cork is the difference in pressure between the inside and the outside of the bottle times the cross-sectional area of the cork. The atmospheric pressure at sea level is approximated as \(1 \times 10^5 \, N/m^2\). The pressure inside the bottle is 6 times the atmospheric pressure, which is \(6 \times 10^5 \, N/m^2\). Therefore, the force \(F = (6 \times 10^5 - 1 \times 10^5) \times 0.000254 = 127 \, N\).
03

Calculate the Acceleration of the Cork

From Newton's second law, the acceleration \(a\) of the cork is the force divided by the mass of the cork. Converting the mass of the cork to kilograms for SI units, \(m = 7.5 \, g = 0.0075 \, kg\). Therefore, the acceleration \(a = 127 \, N / 0.0075 \, kg = 16933.33 \, m/s^2\).

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Key Concepts

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

Understanding Pressure and Force
Pressure is a measure of force exerted per unit area. It is an essential concept in physics because it explains how forces are distributed over surfaces. When it comes to pressurized systems, like a champagne bottle, pressure can build up and create a significant force over a surface. In our example of a champagne cork, the pressure inside the bottle is much higher than the outside atmospheric pressure. This pressure difference creates a net force pushing on the cork.

The net force exerted by the pressure on the cork is imperative to understand because it initiates the movement of the cork. Once we calculate this force, we can use it to determine how quickly the cork will accelerate when the bottle is opened. This is directly related to Newton's second law, which connects force, mass, and acceleration.
Applying Newton's Second Law
Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. The formula for this is \( F = ma \), where \( F \) is force, \( m \) is mass, and \( a \) is acceleration. This law is at the heart of classical mechanics and can be seen in action when a cork is shot from a champagne bottle.

The force causing the cork's motion comes from the pressure difference calculated previously. The only resistance comes from the cork's mass. By dividing the net force by the cork's mass, as done in our step by step solution, we determine the cork's acceleration. This reveals just how rapidly the cork will start moving, which, given the high pressure inside the bottle, will be quite fast.
Cross-Sectional Area Calculation
The cross-sectional area of an object is the area of a slice taken perpendicular to the object's extension. It's crucial in problems involving pressure because the force exerted by a fluid is distributed over this area. For the champagne cork, determining this area helps us understand the total force it experiences due to the pressure inside the bottle.

To calculate the area of the cork, we treat it as a circle because we're dealing with its end, which is shaped as such. We use the formula \( A = \pi r^2 \), converting diameter to radius, and making sure our units are in the SI standard meter for consistency in our calculations. It's vital to be accurate in this step since the area greatly affects the force calculation, and thus the final acceleration of the cork.

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Most popular questions from this chapter

Many fish maintain buoyancy with a gas-filled swim bladder. The pressure inside the swim bladder is the same as the outside water pressure, so when a fish descends to a greater depth, the gas compresses. Adding gas to restore the original volume requires energy. A fish at a depth where the absolute pressure is 3.0 atm has a swim bladder with the desired volume of \(5.0 \times 10^{-4} \mathrm{m}^{3} .\) The fish now descends to a depth where the absolute pressure is 5.0 atm. a. The gas in the swim bladder is always the same temperature as the fish's body. What is the volume of the swim bladder at the greater depth? b. The fish remains at the greater depth, slowly adding gas to the swim bladder to return it to its desired volume. How much work is required?

A 5.0-m-diameter garden pond holds \(5.9 \times 10^{3} \mathrm{kg}\) of water. \(400 \mathrm{W} / \mathrm{m}^{2}\) of solar energy strikes the pond's surface, meaning each square meter of the pond's surface is absorbing \(400 \mathrm{W}\). How many hours will it take for the pond to warm from \(15^{\circ} \mathrm{C}\) to \(25^{\circ} \mathrm{C} ?\)

The maximum amount of water an adult in temperate climates can perspire in one hour is typically \(1.8 \mathrm{L}\). However, after several weeks in a tropical climate the body can adapt, increasing the maximum perspiration rate to \(3.5 \mathrm{L} / \mathrm{h}\). At what rate, in watts, is energy being removed when perspiring that rapidly? Assume all of the perspired water evaporates. At body temperature, the heat of vaporization of water is \(L_{\mathrm{v}}=24 \times 10^{5} \mathrm{J} / \mathrm{kg} .\)

Some jellyfish can eject stingers with remarkable force and speed. One species does this by building up a 15 MPa gauge pressure that pushes against the base of a \(2.0-\mu \mathrm{m}\) -diameter stylet, forcing it outward. a. What force does the excess pressure exert on the stylet? b. The mass set into motion is a tiny \(1.0 \times 10^{-12} \mathrm{kg} .\) What is the resulting acceleration?

0.10 mol of argon gas is admitted to an evacuated \(50 \mathrm{cm}^{3}\) container at \(20^{\circ} \mathrm{C}\). The gas then undergoes an isothermal expansion to a volume of \(200 \mathrm{cm}^{3}\). a. What is the final pressure of the gas? b. Show the process on a \(p V\) diagram. Include a proper scale on both axes.
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