/*! 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 127 When a water polo ball is submer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

When a water polo ball is submerged below the surface in a swimming pool and released from rest, it is observed to pop out of the water. How would you expect the height to which it rises above the water to vary with depth of submersion below the surface? Would you expect the same results for a beach ball? For a table-tennis ball?

Short Answer

Expert verified
The height to which water polo ball rises increases with the depth of submersion. This is due to the increased potential energy at deeper depths. The same phenomenon wouldn't apply to a beach ball because its large volume and light weight would cause it to displace a large amount of water resulting in a high buoyant force, making the height it can reach largely independent of the depth of submersion. For a table-tennis ball, the buoyant force is lower due to its small volume and mass, which means the depth of submersion will affect the height it can reach, but to a lesser degree than the water polo ball.

Step by step solution

01

Understand the forces involved

When an object is submerged in a fluid, it experiences a buoyant force. This force is equal to the weight of the fluid displaced by the object \(F_{buoyant} = \text{weight of the fluid displaced} = \text{density of fluid} \times \text{volume of object} \times g\). When the object is released, this buoyant force propels it upwards.
02

Energy considerations

The depth to which the object is submerged determines the potential energy it has. The deeper it is, the more potential energy it has. \(PE_{object} = m \times g \times h\), where \(m\) is the mass of the object, \(g\) is the acceleration due to gravity, and \(h\) is the depth to which it is submerged. When the object is released, this potential energy is converted to kinetic energy, which propels the object upwards.
03

Compare different objects

Different objects have different masses and volumes, which affects the amount of fluid they displace, and thus their buoyant force. A beach ball, for instance, has a large volume and a small mass, so it displaces a lot of water for a small weight, resulting in a high buoyant force. A table-tennis ball, however, has a small volume and a small mass, so it displaces less water and has a lower buoyant force.
04

Conclusion

For objects that are denser than the fluid, the height to which they rise after being released increases as the depth of submersion increases, due to the increased potential energy. However, for objects that are less dense than the fluid, such as a beach ball, the height does not change significantly with depth of submersion since the buoyant force remains largely constant.

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!

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

Assuming the bulk modulus is constant for seawater, derive an expression for the density variation with depth, \(h\) below the surface. Show that the result may be written \\[ \rho \approx \rho_{0}+b h \\] where \(\rho_{0}\) is the density at the surface. Evaluate the constant \(b .\) Then, using the approximation, obtain an equation for the variation of pressure with depth below the surface. Determine the depth in feet at which the error in pressure predicted by the approximate solution is 0.01 percent.

A sphere of radius 1 in., made from material of specific gravity of \(\mathrm{SG}=0.95,\) is submerged in a tank of water. The sphere is placed over a hole of radius 0.075 in., in the tank bottom. When the sphere is released, will it stay on the bottom of the tank or float to the surface?

A reservoir manometer has vertical tubes of diameter \(D=18 \mathrm{mm}\) and \(d=6 \mathrm{mm} .\) The manometer liquid is Meriam red oil. Develop an algebraic expression for liquid deflection \(L\) in the small tube when gage pressure \(\Delta p\) is applied to the reservoir. Evaluate the liquid deflection when the applied pressure is equivalent to \(25 \mathrm{mm}\) of water (gage).

A proposed ocean salvage scheme involves pumping air into "bags" placed within and around a wrecked vessel on the sea bottom. Comment on the practicality of this plan, supporting your conclusions with analyses.

Consider the cylindrical weir of diameter \(3 \mathrm{m}\) and length \(6 \mathrm{m} .\) If the fluid on the left has a specific gravity of \(1.6,\) and on the right has a specific gravity of \(0.8,\) find the magnitude and direction of the resultant force.

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.