Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A waterproof ball made of rubber with a bulk modulus of \(6.309 \cdot 10^{7} \mathrm{~N} / \mathrm{m}^{2}\) is submerged under water to a depth of \(55.93 \mathrm{~m}\). What is the fractional change in the volume of the ball?

Short Answer

Expert verified
Answer: The fractional change in volume of the rubber ball when it is submerged to a depth of 55.93 meters underwater is approximately 0.8679%.

Step by step solution

01

Identify the variables given in the problem

We are given the bulk modulus of the rubber ball (\(B = 6.309 \cdot 10^{7} \mathrm{~N} / \mathrm{m}^{2}\)) and the depth at which it is submerged (\(h = 55.93 \mathrm{~m}\)).
02

Calculate the pressure at the depth

The pressure at the depth where the ball is submerged underwater can be calculated using the formula: \(P = \rho g h\), where \(P\) is the pressure at the depth, \(\rho\) is the density of water (approximated to be \(1000 \mathrm{~kg/m^3}\) for this problem), \(g\) is the acceleration due to gravity (\(9.81 \mathrm{~m/s^2}\)), and \(h\) is the depth. Now, calculate the pressure at the specified depth: \(P = (1000 \mathrm{~kg/m^3}) (9.81 \mathrm{~m/s^2})(55.93 \mathrm{~m})\) \(P = 5.473 \cdot 10^{5} \mathrm{~N/m^2}\)
03

Calculate the volumetric strain

The volumetric strain can be calculated using the formula: \(\frac{\Delta V}{V} = \frac{P}{B}\), where \(\Delta V\) is the change in volume, \(V\) is the initial volume, and \(B\) is the bulk modulus. Now, calculate the volumetric strain: \(\frac{\Delta V}{V} = \frac{5.473 \cdot 10^{5} \mathrm{~N/m^2}}{6.309 \cdot 10^{7} \mathrm{~N/m^2}}\) \(\frac{\Delta V}{V} = 8.679 \cdot 10^{-3}\)
04

Calculate the fractional change in volume

Now that we have the volumetric strain, we can determine the fractional change in volume by expressing it as a percentage: \(\boxed{\text{Fractional Change in Volume} = 8.679 \cdot 10^{-3} \times 100 \% = 0.8679 \%}\) Thus, the fractional change in volume of the rubber ball when it is submerged to a depth of \(55.93 \mathrm{~m}\) underwater is approximately \(0.8679 \%\).

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 Vaia!

Key Concepts

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

Volumetric Strain
Volumetric strain measures the deformation of an object in response to pressure or stress. It is defined as the fractional change in volume of a material and is a unitless quantity. In the context of a submerged object, like the rubber ball in our exercise, volumetric strain clarifies how much the ball's volume has changed due to the water pressure.

In physics, volumetric strain is mathematically represented as \(\frac{\Delta V}{V} = \frac{P}{B}\), where \(\Delta V\) is the change in volume, \(V\) is the original volume, \(P\) is the pressure exerted on the object, and \(B\) is the bulk modulus, indicative of the material's resistance to compression. In essence, a high bulk modulus means a material is less compressible, and thus will exhibit a smaller volumetric strain under a given pressure.

By understanding volumetric strain, we not only gauge how an object might behave when subjected to external forces, but it also helps in designing materials for structures and applications where volume consistency under pressure is crucial.
Submerged Object Physics
Objects behave differently when submerged in a fluid due to factors like buoyancy and pressure. The physics of submerged objects is crucial in a variety of applications, ranging from marine engineering to the manufacturing of submersibles.

When an object is submerged, it experiences pressure from the fluid around it, which increases with depth. This pressure exerts a force over the surface of the object, potentially deforming it, as per its material properties. In the context of our rubber ball exercise, understanding how the ball interacts with the water at a certain depth involves comprehending how pressure-induced forces lead to volumetric changes, considering the bulk modulus.

Moreover, the application of these principles is immense, found in the functioning of underwater equipment, designing of submarines, and even in the field of deep-sea exploration, where pressure can drastically change the shape and functionality of objects.
Pressure Calculation
Calculating pressure is fundamental in understanding how fluids interact with objects. For an object at rest within a fluid, the pressure at a particular depth is determined by the equation \( P = \rho g h \), where \( P \) denotes the pressure, \( \rho \) is the fluid's density, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the object beneath the fluid's surface.

In our exercise, by calculating the pressure exerted on the rubber ball at a specific depth, we can understand the forces acting on it. The pressure helps us determine how much the ball is compressed, leading to volumetric strain, given the ball's material represented by its bulk modulus.

This calculation is not only useful in academic exercises but also in practical scenarios, such as determining water pressure on dams, calculating the force on submarine hulls, or figuring the depth at which an underwater oil pipeline must withstand pressure.

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

Brass weights are used to weigh an aluminum object on an analytical balance. The weighing is done one time in dry air and another time in humid air. What should the mass of the object be to produce a noticeable difference in the balance readings, provided the balance's sensitivity is \(m_{0}=0.100 \mathrm{mg} ?\) (The density of aluminum is \(\rho_{\mathrm{A}}=2.70 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\); the density of brass is \(\rho_{11}=8.50 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\), The density of the dry air is \(1.2285 \mathrm{~kg} / \mathrm{m}^{3}\), and the density of the humid air is \(1.2273 \mathrm{~kg} / \mathrm{m}^{3}\).).

A tourist of mass 60.0 kg notices a chest with a short chain attached to it at the bottom of the ocean. Imagining the riches it could contain, he decides to dive for the chest. He inhales fully, thus setting his average body density to \(945 \mathrm{~kg} / \mathrm{m}^{3}\), jumps into the ocean (with saltwater density \(=1024 \mathrm{~kg} / \mathrm{m}^{3}\) ), grabs the chain, and tries to pull the chest to the surface. Unfortunately, the chest is too heavy and will not move. Assume that the man does not touch the bottom. a) Draw the man's free-body diagram, and determine the tension on the chain. b) What mass (in kg) has a weight that is equivalent to the tension force in part (a)? c) After realizing he cannot free the chest, the tourist releases the chain. What is his upward acceleration (assuming that he simply allows the buoyant force to lift him up to the surface)?

A 20 -kg chandelier is suspended from a ceiling by four vertical steel wires. Each wire has an unloaded length of \(1 \mathrm{~m}\) and a diameter of \(2 \mathrm{~mm},\) and each bears an equal load. When the chandelier is hung, how far do the wires stretch?

The average density of the human body is \(985 \mathrm{~kg} / \mathrm{m}\), and typical density of seawater is about \(1024 \mathrm{~kg} / \mathrm{m}^{3}\), a) Draw a free-body diagram of a human body floating in seawater and determine what percentage of the body's volume is submerged. b) The average density of the human body, after maximum inhalation of air, changes to \(9.45 \mathrm{~kg} / \mathrm{m}^{3}\). As a person floating in seawater inhales and exhales slowly, what percentage of his volume moves up out of and down into the water? c) The Dead Sea (a saltwater lake between Israel and Jordan) is the world's saltiest large body of water. Its average salt content is more than six times that of typical seawater, which explains why there is no plant and animal life in it. Two-thirds of the volume of the body of a person floating in the Dead Sea is observed to be submerged. Determine the density (in \(\mathrm{kg} / \mathrm{m}^{3}\) ) of the seawater in the Dead Sea.

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 rim. d) It depends on the difference in density between water and ice.

See all solutions

Recommended explanations on Physics Textbooks

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