Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 56

A basketball of circumference \(75.5 \mathrm{~cm}\) and mass \(598 \mathrm{~g}\) is forced to the bottom of a swimming pool and then released. After initially accelerating upward, it rises at a constant velocity, a) Calculate the buoyant force on the basketball. b) Calculate the drag force the basketball experiences while it is moving upward at constant velocity.

Short Answer

Expert verified
Question: Calculate the buoyant force and drag force acting on a basketball submerged in a swimming pool. The basketball has a circumference of 0.75 meters and a mass of 600 grams. The density of water is 1000 kg/m³. Answer: To calculate the buoyant and drag forces on the basketball, follow the steps outlined in the solution. In Step 1, find the volume of the basketball using its circumference. In Step 2, calculate the buoyant force using the Archimedes' principle. In Step 3, calculate the force of gravity on the basketball. Finally, in Step 4, calculate the drag force by balancing the buoyant force and the force of gravity when the basketball is moving at a constant velocity.

Step by step solution

01

Find the volume of the basketball

Using the given circumference of the basketball, we can find its radius and then calculate its volume. The formula for the circumference of a sphere is \(C = 2\pi r\). Therefore, the radius can be found by rearranging this equation as \(r = \frac{C}{2\pi}\). Hence, the volume of the basketball can be calculated using the formula for the volume of a sphere, which is \(V = \frac{4}{3}\pi r^3\).
02

Calculate the buoyant force

The buoyant force can be calculated using the Archimedes' principle, which states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid that the body displaces. The buoyant force formula is \(F_b = \rho_{fluid}Vg\), where \(\rho_{fluid}\) is the density of the fluid, \(V\) is the volume of the submerged object, and \(g\) is the acceleration due to gravity. Using the density of water as \(1000\, kg/m^3\), the volume of the basketball obtained in step 1, and the standard value for the acceleration due to gravity (\(9.81\, m/s^2\)), we can calculate the buoyant force on the basketball.
03

Calculate the force of gravity on the basketball

To calculate the force of gravity on the basketball, we can use the formula \(F_g = mg\), where \(m\) is the mass of the basketball and \(g\) is the acceleration due to gravity. We are given the mass of the basketball in grams, so we should convert it to kilograms before using it in the formula.
04

Calculate the drag force

When the basketball is moving upward at a constant velocity, the net force acting on it is zero. This means that the buoyant force and the force of gravity must be balanced by the drag force. We can calculate the drag force using the following equation: \(F_d = F_b - F_g\). With all of the information we have gathered in the previous steps, we can now perform the calculations needed to find the buoyant and drag forces on the basketball.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

You know from experience that if a car you are riding in suddenly stops, heavy objects in the rear of the car move toward the front. Why does a helium-filled balloon in such a situation move, instead, toward the rear of the car?

An approximately round tendon that has an average diameter of \(8.5 \mathrm{~mm}\) and is \(15 \mathrm{~cm}\) long is found to stretch \(3.7 \mathrm{~mm}\) when acted on by a force of \(13.4 \mathrm{~N}\). Calculate Young's modulus for the tendon.

In a horizontal water pipe that narrows to a smaller radius, the velocity of the water in the section with the smaller radius will be larger. What happens to the pressure? a) The pressure will be the same in both the wider and narrower sections of the pipe. b) The pressure will be higher in the narrower section of the pipe. c) The pressure will be higher in the wider section of the pipe d) It is impossible to tell.

You fill a tall glass with ice and then add water to the level of the glass's rim, so some fraction of the ice floats above the rim. When the ice melts, what happens to the water level? (Neglect evaporation, and assume that the ice and water remain at \(0^{\circ} \mathrm{C}\) during the melting process.) a) The water overflows the rim. b) The water level drops below the rim. c) The water level stays at the top of the rim. d) It depends on the difference in density between water and ice.

The calculation of atmospheric pressure at the summit of Mount Everest carried out in Example 13.3 used the model known as the isothermal atmosphere, in which gas pressure is proportional to density: \(p=\gamma \rho\), with \(\gamma\) constant. Consider a spherical cloud of gas supporting itself under its own gravitation and following this model. a) Write the equation of hydrostatic equilibrium for the cloud, in terms of the gas density as a function of radius, \(\rho(r) .\) b) Show that \(\rho(r)=A / r^{2}\) is a solution of this equation, for an appropriate choice of constant \(A\). Explain why this solution is not suitable as a model of a star.
See all solutions

Recommended explanations on Physics Textbooks

Space Physics

Read Explanation

Fields in Physics

Read Explanation

Particle Model of Matter

Read Explanation

Work Energy and Power

Read Explanation

Thermodynamics

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free