/*! 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 35 A plastic sphere floats in water... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A plastic sphere floats in water with 50.0 percent of its vol. ume submerged. This same sphere floats in glycerin with 40.0 percent of its volume submerged. Determine the densities of the glycerin and the sphere.

Short Answer

Expert verified
The density of the plastic sphere is 0.50 g/cm^3, and the density of the glycerin is 1.25 g/cm^3.

Step by step solution

01

Establish Equations

The sphere displaces its own weight of water and glycerin when it floats. This establishes the following relationships:\( \rho_{water} \cdot V_{water} = \rho_{sphere} \cdot V_{sphere} \)For water, and\( \rho_{glycerin} \cdot V_{glycerin} = \rho_{sphere} \cdot V_{sphere} \)For glycerin, where \( \rho_{water} \) is the density of water (which we know to be 1 g/cm^3 or 1000 kg/m^3), \( V_{water} \) and \( V_{glycerin} \) are the volumes of water and glycerin the sphere displaces when it floats respectively, and \( \rho_{sphere} \) and \( V_{sphere} \) are the density and volume of the sphere.
02

Solve for the Density of the Sphere

Given that 50% of the sphere's volume is submerged in water when it floats, this volume displace an equal weight of water. We can re-arrange the buoyancy equation for the sphere in water and solve for the density of the sphere.\( \rho_{sphere} = \frac{\rho_{water} \cdot V_{water}}{V_{sphere}}Given that \( V_{water} = 0.50 V_{sphere} \), the equation reduces to\( \rho_{sphere} = \frac{0.50 \cdot \rho_{water}}{V_{sphere}} = 0.50 \cdot \rho_{water} = 0.50 \cdot 1 g/cm^3 = 0.50 g/cm^3 \)
03

Solve for the Density of the Glycerin

Knowing the density of the sphere, we can now solve for the density of the glycerin. Since 40% of the sphere's volume is submerged in glycerin when it floats, and given the buoyancy equation for the sphere in glycerin established in Step 1, we can re-arrange and solve for the density of the glycerin.\( \rho_{glycerin} = \frac{\rho_{sphere} \cdot V_{sphere}}{V_{glycerin}}Given that \( V_{glycerin} = 0.40 V_{sphere} \), the equation reduces to\( \rho_{glycerin} = \frac{\rho_{sphere}}{0.40} = \frac{0.50 g/cm^3}{0.40} = 1.25 g/cm^3 \)

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 91Ó°ÊÓ!

Key Concepts

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

Density
Density is a measure of how much mass is contained in a given volume. It is usually expressed in units such as grams per cubic centimeter (g/cm³) or kilograms per cubic meter (kg/m³). Density is an intrinsic property of a substance and influences many physical phenomena, including buoyancy.
Understanding the concept of density helps explain why certain objects float while others sink. A crucial point to remember is that if an object's density is less than that of the fluid it is placed in, it will float. If its density is higher, it will sink.
In the given exercise, the density of the plastic sphere is calculated using the fact that it sinks halfway in water. This is described by the equation:
\[ \rho_{sphere} = 0.50 \times \rho_{water} \]
Since water has a known density of 1 g/cm³, we conclude that the sphere's density is 0.50 g/cm³. Understanding density helps us describe and predict the behavior of materials in different environments.
Archimedes' Principle
Archimedes' Principle is a fundamental principle of fluid mechanics. It states that any object submerged in a fluid is buoyed up by a force equal to the weight of the fluid displaced by the object. This is the principle that allows objects to float.
To visualize this, imagine placing an object in water. The object pushes some of the water out of the way. According to Archimedes' Principle, the upward buoyant force on an object is the same as the weight of the water displaced.
In the context of our exercise, this principle helps calculate the densities of both the sphere and glycerin. When the sphere floats in water with half of its volume submerged, the buoyant force equals the weight of the sphere. Here, Archimedes' Principle is used to establish that the weight of the displaced water is equal to the weight of the sphere.
By applying this principle, the buoyant force is also described when the sphere floats in glycerin, where only 40% of its volume submerges. It allows us to calculate the glycerin's density as well. Understanding Archimedes' Principle is crucial in solving problems involving floating or submerged objects.
Floating Objects
Floating objects provide practical examples of the interplay between density and buoyancy. An object floats when its density is less than the density of the fluid it is in. The difference in densities results in an upward buoyant force that supports the object against gravity.
In our exercise, the plastic sphere floats in both water and glycerin, albeit with different proportions submerged. This occurs because their densities vary, thereby changing the degree of buoyancy experienced by the sphere.
When the sphere is in water, 50% of it is submerged because its density is 0.50 g/cm³. As the sphere's density is exactly half that of water, it displaces enough water to balance its own weight. On the other hand, when the sphere is in glycerin, only 40% is submerged. This indicates that glycerin has a higher density than the water but is lower than the solid sphere's average density.
Observing floating objects and understanding their dynamics with different liquids offers insight into larger natural and industrial processes, ranging from ships to hot air balloons. The patterns we observe when different objects float serve as practical demonstrations of fundamental principles in physics.

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

Water falls over a dam of height \(h\) with a mass flow rate of \(R,\) in units of \(\mathrm{kg} / \mathrm{s} .\) (a) Show that the power available from the water is $$\mathscr{P}=R g h$$ where \(g\) is the frec-fall acceleration. (b) Each hydroclectric unit at the Grand Coulce Dam takes in watcr at a rate of \(8.50 \times 10^{5} \mathrm{kg} / \mathrm{s}\) from a height of \(87.0 \mathrm{m} .\) The power developed by the falling water is converted to clectric power with an efficiency of \(85.0 \% .\) How much electric power is produced by each hydroelectric unit?

The small piston of a hydraulic lift has a cross-scrional area of \(3.00 \mathrm{cm}^{2},\) and its large piston has a cross-sectional area of \(\left.200 \mathrm{cm}^{2} \text { (Figure } 14.4\right) .\) What force must be applied to the small piston for the lift to raise a load of \(15.0 \mathrm{kN} ?\) (In service stations, this force is usually exerted by compressed air.)

The water supply of a building is fed through a main pipe \(6.00 \mathrm{cm}\) in diameter. A \(2.00-\) cm-diameter faucet tap, located \(2.00 \mathrm{m}\) above the main pipe, is observed to fill a \(25.0-\mathrm{I}\) container in \(30.0 \mathrm{s}\). (a) What is the speed at which the water leaves the faucet? (b) What is the gauge pressure in the 6-cm main pipe? (Assume the faucet is the only "leak" in the building.)

The true weight of an object can be measured in a vacuum, where buoyant forces are absent. An object of volume \(V\) is weighed in air on a balance with the use of weights of density \(\rho .\) If the density of air is \(\rho_{\text {air }}\) and the balance reads \(F_{g}^{\prime},\) show that the true weight \(F_{g}\) is $$F_{R}=F_{R}^{\prime}+\left(V-\frac{F_{R}^{\prime}}{\rho g}\right) \rho_{\operatorname{ain}} g$$

A cube of wood having an edge dimension of \(20.0 \mathrm{cm}\) and a density of \(650 \mathrm{kg} / \mathrm{m}^{3}\) floats on water. (a) What is the distance from the horizontal top surface of the cube to the water level? (b) How much lead weight has to be placed on top of the cube so that its top is just level with the water?

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.