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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/s\(^2\). 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, \( 0.25 \times V \times \rho_{\text{water}} \times g = \text{Weight of apparatus} \).
03

Expression for Weight of Apparatus

Since on Earth the weight of the apparatus is balanced by the weight of the displaced water, write: \( 0.25 \times V \times \rho_{\text{water}} \times g = V_{\text{app}} \times g \), where \( V_{\text{app}} \) is the volume of the apparatus, \( \rho_{\text{water}} = 1000 \text{ kg/m}^3 \), and \( g = 9.81 \text{ m/s}^2 \). Thus, \( V_{\text{app}} = 0.25 \times V \times 1000 \).
04

Calculate Submersion on Caasi

On Caasi, the same apparatus floats in glycerine with surface gravity \( g' = 5.40 \text{ m/s}^2 \). Let \( x \times V \) be the submerged volume of apparatus. Then \( x \times V \times \rho_{\text{glycerine}} \times g' = V_{\text{app}} \times g' \). Using \( \rho_{\text{glycerine}} = 1260 \text{ kg/m}^3 \), we deduce \( x = \frac{V_{\text{app}}}{V \times 1260} \).
05

Final Calculation

Substitute \( V_{\text{app}} \) from earlier into \( x = \frac{0.25 \times V \times 1000}{V \times 1260} \) to find \( x = \frac{0.25 \times 1000}{1260} = \frac{250}{1260} \approx 0.198 \). Hence, about 19.8% of its volume will be submerged.

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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/s\(^2\) compared to 9.81 m/s\(^2\)). 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.

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Most popular questions from this chapter

A rock with density 1200 kg/m\(^3\) 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/m\(^3\)?

A hollow plastic sphere is held below the surface of a freshwater lake by a cord anchored to the bottom of the lake. The sphere has a volume of 0.650 m\(^3\) and the tension in the cord is 1120 N. (a) Calculate the buoyant force exerted by the water on the sphere. (b) What is the mass of the sphere? (c) The cord breaks and the sphere rises to the surface. When the sphere comes to rest, what fraction of its volume will be submerged?

A firehose must be able to shoot water to the top of a building 28.0 m tall when aimed straight up. Water enters this hose at a steady rate of 0.500 m\(^3\)/s and shoots out of a round nozzle. (a) What is the maximum diameter this nozzle can have? (b) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

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 \(\times\) 10\(^3\) 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?

The Environmental Protection Agency is investigating an abandoned chemical plant. A large, closed cylindrical tank contains an unknown liquid. You must determine the liquid's density and the height of the liquid in the tank (the vertical distance from the surface of the liquid to the bottom of the tank). To maintain various values of the gauge pressure in the air that is above the liquid in the tank, you can use compressed air. You make a small hole at the bottom of the side of the tank, which is on a concrete platform\(-\)so the hole is 50.0 cm above the ground. The table gives your measurements of the horizontal distance \(R\) that the initially horizontal stream of liquid pouring out of the tank travels before it strikes the ground and the gauge pressure \({p_g}\) of the air in the tank. (a) Graph \({R^2}\) as a function of \({p_g}\). Explain why the data points fall close to a straight line. Find the slope and intercept of that line. (b) Use the slope and intercept found in part (a) to calculate the height \(h\) (in meters) of the liquid in the tank and the density of the liquid (in kg/m\(^3\)). Use \(g\) \(=\) 9.80 m/s\(^2\). Assume that the liquid is nonviscous and that the hole is small enough compared to the tank's diameter so that the change in h during the measurements is very small.

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