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A spherical weather balloon is filled with hydrogen until its radius is \(3.00 \mathrm{~m}\). Its total mass including the instruments it carries is \(15.0 \mathrm{~kg}\). (a) Find the buoyant force acting on the balloon, assuming the density of air is \(1.29 \mathrm{~kg} / \mathrm{m}^{3}\). (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?

Short Answer

Expert verified
a) The buoyant force acting on the balloon is calculated by multiplying the volume of the air displaced by the balloon with the density of the air and acceleration due to gravity. b) The net force acting on the balloon is the difference between the buoyant force and the weight of the balloon and its instruments. c) As the balloon rises, the decreasing ambient air pressure causes the hydrogen gas inside the balloon to expand, hence increasing the volume and radius of the balloon.

Step by step solution

01

Calculate the Balloon's Volume

The volume V of a sphere can be calculated using the formula \(V = \frac{4}{3}\pi r^3\), where r is the radius. Substituting \(r = 3.00 m\) into the formula gives \(V = \frac{4}{3}\pi (3.00 m)^3\).
02

Calculate the Buoyant Force

The buoyant force F can be calculated using the formula \(F= \rho V g\), where \(\rho\) is the density of the fluid, V is the volume of the fluid displaced, and g is the acceleration due to gravity. Substituting \(\rho = 1.29 kg/m^3\), V from the previous step, and \(g = 9.81 m/s^2\) into the formula gives \(F = 1.29 kg/m^3 \cdot V \cdot 9.81 m/s^2\).
03

Calculate the Weight of the Balloon

The weight W of the balloon can be calculated using the formula \(W = m g\), where m is the mass and g is the acceleration due to gravity. Substituting \(m = 15.0 kg\) and \(g = 9.81 m/s^2\) into the formula gives \(W = 15.0 kg \cdot 9.81 m/s^2\).
04

Calculate the Net Force

The net force acting on the balloon F_net is the difference between the buoyant force and the weight of the balloon. Thus, \(F_net = F - W\).
05

Discuss the Change in Balloon's Volume with Altitude

As the balloon rises, the ambient air pressure drops. This causes the hydrogen gas inside the balloon to expand, hence increasing the volume and in turn radius of the balloon.

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Key Concepts

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

Spherical Weather Balloon
A spherical weather balloon is a perfect example of a 3D object that interacts with forces when in the air. These balloons are typically used for scientific purposes, such as measuring atmospheric conditions like pressure, temperature, and humidity. The round shape of the balloon is important because it minimizes drag, allowing it to rise smoothly through the atmosphere. The volume of a spherical balloon is crucial to its buoyancy. For example, the volume, which affects how much air it displaces, can be calculated using the formula: \[ V = \frac{4}{3} \pi r^3 \] where \(r\) is the radius of the balloon. When a balloon like this is filled with a lighter gas such as hydrogen, it can lift its own weight and that of any additional instruments due to the buoyant force exerted by the surrounding air.
Density of Air
The concept of air density plays a fundamental role in calculating the buoyant force. Air density is defined as the mass per unit volume of Earth's atmosphere and is usually denoted by \( \rho \). In the case of this exercise, the air density is given as \(1.29 \, \text{kg/m}^3\). Buoyant force, which causes the balloon to rise, can be calculated by the equation: \[ F = \rho V g \] with \(\rho\), the air density; \(V\), the volume of the balloon; and \(g\), the acceleration due to gravity (\(9.81 \, \text{m/s}^2\)). The higher the density of the air, the greater the buoyant force, meaning the balloon will rise more swiftly. A change in air density, due to weather conditions or altitude changes, can affect the overall performance of the balloon, influencing how high or fast it will ascend.
Net Force
Understanding the net force on the balloon helps us determine if and how fast it will rise. Net force is what you get when you subtract the weight of the balloon from the buoyant force. The weight of the balloon is the gravitational force pulling it down, which can be calculated using: \[ W = m g \] where \(m\) is the total mass of the balloon including instruments, and \(g\) is the gravitational acceleration. The net force equation appears as: \[ F_{\text{net}} = F - W \] Here, \(F\) is the buoyant force and \(W\) is the weight of the balloon. If \(F_{\text{net}}\) is positive, the balloon will rise. If it is negative, the balloon will not rise and may even descend. Calculating this net force is crucial in understanding whether the balloon will take off and maintain its trajectory.
Altitude Effects on Balloons
As a weather balloon rises, the effects of altitude cannot be ignored. Initially, the balloon will encounter higher atmospheric pressures, which exerts a stabilizing effect on the size of the balloon. However, as it ascends, the atmospheric pressure decreases. This decrease in pressure causes the gas inside the balloon to expand. Consequently, the volume of the balloon increases—as described by the principle of expanding gases when pressure decreases (Boyle's Law). With a larger volume, the buoyant force also increases, as more air is displaced. These effects can significantly impact the behavior of the balloon, requiring careful monitoring and making them a key consideration in designing and planning weather balloon releases. The impact can be summarized as:
  • Lower pressure at high altitude leads to gas expansion.
  • Increased volume results in a rise in buoyant force.
  • Monitoring such changes is critical for data accuracy and mission success.

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Most popular questions from this chapter

On October 21, 2001, Ian Ashpole of the United Kingdom achieved a record altitude of \(3.35 \mathrm{~km}\) (11 \(000 \mathrm{ft}\) ) powered by 600 toy balloons filled with helium. Each filled balloon had a radius of about \(0.50 \mathrm{~m}\) and an estimated mass of \(0.30 \mathrm{~kg}\). (a) Estimate the total buoyant force on the 600 balloons. (b) Estimate the net upward force on all 600 balloons. (c) Ashpole parachuted to Earth after the balloons began to burst at the high altitude and the system lost buoyancy. Why did the balloons burst?

A wooden block of volume \(5.24 \times 10^{-4} \mathrm{~m}^{3}\) floats in water, and a small steel object of mass \(m\) is placed on top of the block. When \(m=0.310 \mathrm{~kg}\), the system is in equilibrium, and the top of the wooden block is at the level of the water. (a) What is the density of the wood? (b) What happens to the block when the steel object is replaced by a second steel object with a mass less than \(0.310 \mathrm{~kg}\) ? What happens to the block when the steel object is replaced by yet another steel object with a mass greater than \(0.310 \mathrm{~kg}\) ?

Suppose a distant world with surface gravity of \(7.44 \mathrm{~m} / \mathrm{s}^{2}\) has an atmospheric pressure of \(8.04 \times 10^{4} \mathrm{~Pa}\) at the surface. (a) What force is exerted by the atmosphere on a disk-shaped region \(2.00 \mathrm{~m}\) in radius at the surface of a methane ocean? (b) What is the weight of a \(10.0-\mathrm{m}\) deep cylindrical column of methane with radius \(2.00 \mathrm{~m}\) ? (c) Calculate the pressure at a depth of \(10.0 \mathrm{~m}\) in the methane ocean. Note: The density of liquid methane is \(415 \mathrm{~kg} / \mathrm{m}^{3}\).

The approximate diameter of the aorta is \(0.50 \mathrm{~cm}\); that of a capillary is \(10 \mu \mathrm{m}\). The approximate average blood flow speed is \(1.0 \mathrm{~m} / \mathrm{s}\) in the aorta and \(1.0 \mathrm{~cm} / \mathrm{s}\) in the capillaries. If all the blood in the aorta eventually flows through the capillaries, estimate the number of capillaries in the circulatory system.

(a) Calculate the mass flow rate (in grams per second) of blood \(\left(\rho=1.0 \mathrm{~g} / \mathrm{cm}^{3}\right)\) in an aorta with a crosssectional area of \(2.0 \mathrm{~cm}^{2}\) if the flow speed is \(40 \mathrm{~cm} / \mathrm{s}\). (b) Assume that the aorta branches to form a large number of capillaries with a combined cross-sectional area of \(3.0 \times 10^{3} \mathrm{~cm}^{2}\). What is the flow speed in the capillaries?

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