/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A lost shipping container is fou... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 46

A lost shipping container is found resting on the ocean floor and completely submerged. The container is \(6.1 \mathrm{m}\) long, \(2.4 \mathrm{m}\) wide, and \(2.6 \mathrm{m}\) high. Salvage experts attach a spherical balloon to the top of the container and inflate it with air pumped down from the surface. When the balloon's radius is \(1.5 \mathrm{m},\) the shipping container just begins to rise toward the surface. What is the mass of the container? Ignore the mass of the balloon and the air within it. Do not neglect the buoyant force exerted on the shipping container by the water. The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The mass of the container is approximately 53582.5 kg.

Step by step solution

01

Calculate Volume of the Container

The volume of the shipping container can be found using the formula for the volume of a rectangular prism: \(V = \text{length} \times \text{width} \times \text{height}\). Substitute the given dimensions to get \(V = 6.1 \times 2.4 \times 2.6 = 38.064 \, \text{m}^3\).
02

Calculate Volume of the Balloon

The volume of a sphere is calculated using the formula \(V = \frac{4}{3} \pi r^3\). Substitute the radius of the balloon to find the volume: \(V = \frac{4}{3} \pi (1.5)^3 \approx 14.137 \, \text{m}^3\).
03

Calculate Total Buoyant Force

The buoyant force, which results from the displacement of water by both the balloon and the container, is given by \(F_{buoyant} = \rho_{water} \times g \times V_{total}\), where \(\rho_{water}\) is the density of seawater \(1025 \, \text{kg/m}^3\), \(g = 9.81 \, \text{m/s}^2\), and \(V_{total} = V_{container} + V_{balloon}\). Thus, \(V_{total} = 38.064 + 14.137 = 52.201 \, \text{m}^3\) and \(F_{buoyant} = 1025 \times 9.81 \times 52.201 \approx 525610.8 \, \text{N}\).
04

Equate Buoyant Force to Weight of the Container

When the container begins to rise, the buoyant force equals the gravitational force (weight) of the container: \(F_{buoyant} = m_{container} \times g\). Rearrange to solve for the mass of the container: \(m_{container} = \frac{F_{buoyant}}{g} = \frac{525610.8}{9.81} \approx 53582.5 \, \text{kg}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Volume Calculation
Calculating the volume of objects is fundamental when dealing with problems involving buoyancy because the force of buoyancy depends on the volume of water displaced. For a shipping container shaped like a rectangular prism, the formula to calculate volume is straightforward: multiply the length, width, and height together.

In this particular case, the dimensions were given as 6.1 meters in length, 2.4 meters in width, and 2.6 meters in height. By substituting these values into the formula, we find the container's volume:
  • \( V = \text{Length} \times \text{Width} \times \text{Height} \)
  • \( V = 6.1 \times 2.4 \times 2.6 = 38.064 \, \text{m}^3 \)
Remember that each unit of volume, or cubic meter, will affect the amount of water displaced, thus impacting the buoyant force.
Buoyant Force
The buoyant force is the upward force exerted on objects submerged in a fluid, such as seawater. This force is a result of the pressure exerted by the fluid on the object. The concept of buoyancy is derived from Archimedes' principle, which states that the force on an object submerged in a fluid is equal to the weight of the fluid displaced by the object.

To calculate the buoyant force, the equation is:
  • \( F_{\text{buoyant}} = \rho_{\text{water}} \times g \times V_{\text{total}} \)
where:
  • \( \rho_{\text{water}} \) is the density of the fluid (in this case, seawater)
  • \( g \) is the acceleration due to gravity
  • \( V_{\text{total}} \) is the total volume of water displaced, which includes both the container and the additional volume due to the balloon
In this exercise, we considered both the container's volume and the volume of an attached spherical balloon to find the total volume of displaced water.
Density of Seawater
Density is a measure of mass per unit volume and plays a crucial role in buoyancy calculations. Seawater is denser than fresh water, largely due to its salt content, which affects how objects float or sink in the ocean. In this problem, seawater density is given as 1025 kg/m鲁, which is typical for seawater.

Knowing the density is important because it enables us to compute the buoyant force exerted by the seawater on the submerged object.
  • The higher the density of the fluid, the greater the buoyant force.
  • This force determines whether the object will rise to the surface or remain submerged.
By multiplying the seawater density by the volume of displaced water and the acceleration due to gravity, we arrive at the buoyant force acting on the submerged container.

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

A fuel pump sends gasoline from a car's fuel tank to the engine at a rate of \(5.88 \times 10^{-2} \mathrm{kg} / \mathrm{s} .\) The density of the gasoline is \(735 \mathrm{kg} / \mathrm{m}^{3},\) and the radius of the fuel line is \(3.18 \times 10^{-3} \mathrm{m} .\) What is the speed at which the gasoline moves through the fuel line?

Spring Gun. A hydraulic press is used to compress a spring that will then be used to project a \(25.0-\mathrm{kg}\) steel ball. The system is similar, but smaller in scale, to the car jack illustrated in Figure \(11.14 .\) In this case, the smaller cylinder has a diameter of \(d_{1}=0.75 \mathrm{cm}\) and has a manually operated plunger. The larger cylinder has a diameter of \(d_{2}=10.0 \mathrm{cm},\) and its piston compresses the spring. The idea is that the gun operator pulls a lever that pushes the plunger on the small cylinder, which transmits a pressure to the larger piston that, in turn, exerts a force on the spring and compresses it. Once the spring is compressed, the steel ball is loaded and the spring is released, ejecting the ball. (a) If a force of \(950 \mathrm{N}\) is exerted on the primary (smaller) piston to compress the spring \(1.25 \mathrm{m}\) from its equilibrium (uncompressed) position, what is the spring constant \(k\) of the spring? (b) What is the velocity of the steel ball just after it is ejected? (c) Neglecting air resistance, what is the maximum range of this "spring gun"?

Water is circulating through a closed system of pipes in a two-floor apartment. On the first floor, the water has a gauge pressure of \(3.4 \times 10^{5} \mathrm{Pa}\) and a speed of \(2.1 \mathrm{m} / \mathrm{s}\). However, on the second floor, which is \(4.0 \mathrm{m}\) higher, the speed of the water is \(3.7 \mathrm{m} / \mathrm{s}\). The speeds are different because the pipe diameters are different. What is the gauge pressure of the water on the second floor?

A full can of black cherry soda has a mass of \(0.416 \mathrm{kg}\). It contains \(3.54 \times 10^{-4} \mathrm{m}^{3}\) of liquid. Assuming that the soda has the same density as water, find the volume of aluminum used to make the can.

A small crack occurs at the base of a 15.0 -m-high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Nuclear Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Engineering Physics

Read Explanation

Modern Physics

Read Explanation

Astrophysics

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.