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How many cubic meters of helium are required to lift a balloon with a \(400-\mathrm{kg}\) payload to a height of \(8000 \mathrm{m} ?\) (Take \(\rho_{\mathrm{He}}=0.180 \mathrm{kg} / \mathrm{m}^{3} .\) ) Assume that the balloon maintains a constant volume and that the density of air decreases with the altitude \(z\) according to the expression \(\rho_{\text {air }}=\rho_{0} e^{-z / 8,000}\) where \(z\) is in meters and \(\rho_{0}=1.25 \mathrm{kg} / \mathrm{m}^{3}\) is the density of air at sca level.

Short Answer

Expert verified
After making the necessary calculations in Step 3, it is found that approximately 1019 cubic meters of helium are required to lift the payload to a height of 8000 meters.

Step by step solution

01

Calculate the Weight of the Payload.

To begin with, we need to calculate the weight of the payload. The weight \(W\) of an object is given by the product of its mass \(m\) and the gravitational acceleration \(g\), which on the surface of the Earth is approximately \(9.81 m/s^2\). Therefore, the weight of the payload is \(W = m \cdot g = 400 kg \cdot 9.81 m/s^2 = 3924 N\).
02

Derive the Relation for the Buoyant Force.

The buoyant force \(F\) on the balloon is given by Archimedes' principle, which states that it equals the weight of the air displaced. We can express this as \(F = \rho_{air} \cdot V \cdot g\), where \(\rho_{air}\) is the density of air, \(V\) is the volume of the balloon, and \(g\) is the gravitational acceleration. The density of air is given by \(\rho_{air} = \rho_{0} e^{-z/8000}\). Thus, the buoyant force becomes \(F = \rho_{0} e^{-z/8000} \cdot V \cdot g\). At equilibrium, this force equals the weight of the payload, thus, we have the equation \(\rho_{0} e^{-z/8000} \cdot V \cdot g = W\).
03

Solve for the Volume of the Balloon.

Next, we can rearrange the equation from Step 2 to solve for the volume \(V\) of the balloon. We obtain \(V = \frac{W}{\rho_{0} e^{-z/8000} \cdot g}\) and substituting the known values, we get \(V = \frac{3924 N}{1.25 kg/m^3 \cdot e^{-8000 m/8000} \cdot 9.81 m/s^2}\). Make the calculations to find the volume.

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Key Concepts

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

Archimedes' Principle
Archimedes' principle is a fundamental concept that describes the physical law of buoyancy. It states that any object, wholly or partially immersed in a fluid, is buoyed up by a force that is equal to the weight of the fluid displaced by the object. This principle is critical for understanding why objects float or sink and for calculating the buoyant force exerted by fluids.

For instance, when you place a toy boat in water, the boat pushes aside a certain volume of water. The force pushing the boat upwards (the buoyant force) is equivalent to the weight of that displaced water. This is why a heavy steel ship can float; it displaces a large volume of water, creating a buoyant force that supports the ship's weight.

In the context of our exercise, the buoyant force keeps the helium balloon afloat. The volume of helium required to lift the payload is directly related to the amount of air displaced, and therefore, the weight of that air. The larger the volume, the more air is displaced, and the greater the buoyant force.
Density of Air
The density of air is a crucial factor when calculating the buoyant force. It is the mass per unit volume of Earth's atmosphere and changes with altitude, temperature, and composition. Air is denser at lower altitudes due to the weight of the air above compressing it.

In our exercise, the density of air decreases exponentially with height, as described by the equation \( \rho_{\text{air }} = \rho_{0} e^{-z / 8,000} \), where \( z \) is the altitude, and \( \rho_{0} = 1.25 \text{kg/m}^3 \) is the air density at sea level. The exponential decrease means that as the balloon climbs to higher altitudes, the surrounding air becomes less dense, affecting the lift generated by the buoyant force.

Understanding how air density varies with altitude is important for applications ranging from aviation to meteorology. For the helium balloon in our exercise, it's essential to consider this variable air density to determine the precise volume of helium necessary to lift the payload.
Gravitational Acceleration
Gravitational acceleration is the acceleration of an object due to the pull of gravity towards Earth. On Earth's surface, this acceleration has an average value of \(9.81 \, m/s^2\), although it can vary slightly depending on altitude and geographical location. Gravitational acceleration is what gives weight to objects and is a vital component in calculating the force of gravity acting on a mass.

In the balloon problem, the gravitational force acting on the payload is a product of its mass and Earth's gravitational acceleration. This force must be balanced by the buoyant force for the balloon to float. Hence, the equation for the buoyant force includes gravitational acceleration, showing how it influences not only the payload's weight but also the force needed to lift it.

It is important to remember that even though gravitational acceleration is roughly constant for most terrestrial applications, any changes can affect calculations involving weight and buoyancy. Given the precision needed in applications like aviation and engineering, accurately accounting for gravitational acceleration is essential.

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Most popular questions from this chapter

Tor the cellar of a new house, a hole is dug in the ground, -with vertical sides going down \(2.40 \mathrm{m}\). A concrete foundamall is built all the way across the \(9.60-\mathrm{m}\) width of the excavation. This foundation wall is \(0.183 \mathrm{m}\) away from the front of the cellar hole. During a rainstorm, drainage from the strect fills up the spacc in front of the concrete wall, but not the cellar behind the wall. The water does not soak into the clay soil. Find the force the water causes on the foundation wall. For comparison, the weight of the water is given by \(2.40 \mathrm{m} \times 9.60 \mathrm{m} \times 0.183 \mathrm{m} \times 1000 \mathrm{kg} / \mathrm{m}^{3} \times\) \(9.80 \mathrm{m} / \mathrm{s}^{2}=41.3 \mathrm{kN}\)

A \(50.0-\mathrm{kg}\) woman balances on one hecl of a pair of highheeled shoes. If the heel is circular and has a radius of \(0.500 \mathrm{cm},\) what pressure does she exert on the floor?

Normal atmospheric pressure is \(1.013 \times 10^{5} \mathrm{Pa}\). The approach of a storm causes the height of a mercury barometer to drop by \(20.0 \mathrm{mm}\) from the normal height. What is the atmospheric pressure? (The density of mercury is \(\left.13.59 \mathrm{g} / \mathrm{cm}^{3} .\right)\)

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\- A \(10.0-\mathrm{kg}\) block of metal measuring \(12.0 \mathrm{cm} \times 10.0 \mathrm{cm} \times\) \(10.0 \mathrm{cm}\) is suspended from a scale and immersed in water as in Figure P14.25b. The 12.0-cm dimension is vertical, and the top of the block is \(5.00 \mathrm{cm}\) below the surface of the water. (a) What are the forces acting on the top and on the bottom of the block? (Take \(P_{0}=1.0130 \times 10^{5} \mathrm{N} / \mathrm{m}^{2}\) ) (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

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