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

You are preparing some apparatus for a visit to a newly discovered planet Caasi having oceans of glycerine and a surface acceleration due to gravity of 5.40 m/s2. If your apparatus floats in the oceans on earth with 25.0% of its volume submerged, what percentage will be submerged in the glycerine oceans of Caasi?

Short Answer

Expert verified
19.8% of the apparatus's volume will be submerged in the glycerine oceans of Caasi.

Step by step solution

01

Understand Archimedes' Principle

To solve this problem, we must apply Archimedes' principle, which states that a floating object displaces a weight of fluid equal to its own weight. This principle is vital for calculating the percentage of the volume submerged in any fluid.
02

Calculate Submersion on Earth

On Earth, 25.0% of the apparatus's volume is submerged. This implies that the buoyant force equals the weight of the displaced fluid, which in this case is water. Let's denote the volume of the apparatus as V. The weight of water displaced is equal to the weight of the apparatus. Therefore, 0.25×V×ρwater×g=Weight of apparatus.
03

Expression for Weight of Apparatus

Since on Earth the weight of the apparatus is balanced by the weight of the displaced water, write: 0.25×V×ρwater×g=Vapp×g, where Vapp is the volume of the apparatus, ρwater=1000 kg/m3, and g=9.81 m/s2. Thus, Vapp=0.25×V×1000.
04

Calculate Submersion on Caasi

On Caasi, the same apparatus floats in glycerine with surface gravity g=5.40 m/s2. Let x×V be the submerged volume of apparatus. Then x×V×ρglycerine×g=Vapp×g. Using ρglycerine=1260 kg/m3, we deduce x=VappV×1260.
05

Final Calculation

Substitute Vapp from earlier into x=0.25×V×1000V×1260 to find x=0.25×10001260=25012600.198. Hence, about 19.8% of its volume will be submerged.

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.

Buoyancy and Its Role in Floating
Buoyancy is the force that allows objects to float in fluids, such as liquids or gases. This force acts in the opposite direction to gravity, making things appear lighter in water or any fluid. This phenomenon is central to understanding Archimedes' Principle, which is essential when dealing with floating objects.

According to Archimedes' Principle, the buoyant force on an object submerged in fluid is equal to the weight of the fluid that the object displaces. If an object is partially submerged, it means that it displaces a volume of fluid whose weight equals the object's weight.

This principle helps us calculate how much of an object remains underwater based on its density and the fluid's density. If the buoyant force is greater than the object's weight, it will float. If lesser, it will sink. It’s fascinating to see how this principle applies to different fluids and gravitational conditions – like in the scenario with the planet Caasi.
Understanding Fluid Displacement
Fluid displacement happens when an object is submerged in a fluid, whether partially or fully. It pushes away or "displaces" a certain amount of that fluid. The displaced fluid's weight determines the upward buoyant force exerted on the object.

To determine how much of an object is submerged, you must consider the object's weight and the fluid's density. On Earth, the apparatus displaced enough water such that 25% of its volume was submerged, revealing the exact point where the force of gravity and the buoyancy force balance out.
  • Denser objects displace more fluid to achieve buoyancy.
  • If the fluid is denser, less volume needs to be displaced to equal the object's weight.
  • This concept is crucial when calculating submersion percentages in different fluids or planets.
Experiments like these that involve different fluids and planetary conditions illustrate how versatile Archimedes' principle truly is.
Gravity Effects on Buoyancy
Gravity significantly impacts how buoyant an object is. While buoyant force mainly depends on fluid density and volume displaced, gravity dictates the weight of the displaced fluid, thereby affecting the floating balance.

The surface gravity on planets affects how objects float, as seen in the planet Caasi with a lower gravity than Earth (5.4 m/s2 compared to 9.81 m/s2). This lower gravitational pull means that objects need to displace a relatively smaller volume of more dense fluid, like glycerine, to balance their weight.
  • Stronger gravity would require more displaced volume to achieve buoyancy.
  • The relationship between gravity and buoyancy forces determines submersion levels.
  • Understanding these gravity effects helps in predicting how objects behave in different environments.
By accounting for gravity, you can better predict buoyancy and better prepare when working with floating devices on other planets.

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

There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. E12.17 ) because as the depth increases, so does the pressure difference, which tends to collapse the diver's lungs. Since the snorkel connects the air in the lungs to the atmosphere at the surface, the pressure inside the lungs is atmospheric pressure. What is the external internal pressure difference when the diver's lungs are at a depth of 6.1 m (about 20 ft)? Assume that the diver is in freshwater. (A scuba diver breathing from compressed air tanks can operate at greater depths than can a snorkeler, since the pressure of the air inside the scuba diver's lungs increases to match the external pressure of the water.)

A closed and elevated vertical cylindrical tank with diameter 2.00 m contains water to a depth of 0.800 m. A worker accidently pokes a circular hole with diameter 0.0200 m in the bottom of the tank. As the water drains from the tank, compressed air above the water in the tank maintains a gauge pressure of 5.00 × 103 Pa at the surface of the water. Ignore any effects of viscosity. (a) Just after the hole is made, what is the speed of the water as it emerges from the hole? What is the ratio of this speed to the efflux speed if the top of the tank is open to the air? (b) How much time does it take for all the water to drain from the tank? What is the ratio of this time to the time it takes for the tank to drain if the top of the tank is open to the air?

A single ice cube with mass 16.4 g floats in a glass completely full of 420 cm3 of water. Ignore the water's surface tension and its variation in density with temperature (as long as it remains a liquid). (a) What volume of water does the ice cube displace? (b) When the ice cube has completely melted, has any water overflowed? If so, how much? If not, explain why this is so. (c) Suppose the water in the glass had been very salty water of density 1050 kg/m3. What volume of salt water would the 9.70-g ice cube displace? (d) Redo part (b) for the freshwater ice cube in the salty water.

A rock with density 1200 kg/m3 is suspended from the lower end of a light string. When the rock is in air, the tension in the string is 28.0 N. What is the tension in the string when the rock is totally immersed in a liquid with density 750 kg/m3?

A 950-kg cylindrical can buoy floats vertically in seawater. The diameter of the buoy is 0.900 m. Calculate the additional distance the buoy will sink when an 80.0-kg man stands on top of it.

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