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To verify her suspicion that a rock specimen is hollow, a geologist weighs the specimen in air and in water. She finds that the specimen weighs twice as much in air as it does in water. The density of the solid part of the specimen is \(5.0 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3}\) . What fraction of the specimen's apparent volume is solid?

Short Answer

Expert verified
40% of the specimen's apparent volume is solid.

Step by step solution

01

Understanding the Problem

The geologist measured the weight of the rock specimen in air and in water. In air, the weight is twice as much as in water. This information allows us to calculate the buoyant force acting on the specimen when it's submerged in water.
02

Applying Archimedes' Principle

According to Archimedes' Principle, the buoyant force on an object submerged in a fluid (like water) is equal to the weight of the fluid displaced by the object. Let the specimen's weight in air be \( W \), so its weight in water is \( \frac{W}{2} \). The buoyant force is the difference, \( B = W - \frac{W}{2} = \frac{W}{2} \).
03

Expressing the Buoyant Force

The buoyant force is also given by \( B = \rho_{water} \cdot V_{specimen} \cdot g \), where \( \rho_{water} \) is the density of water (\( 1000 \, \text{kg/m}^3 \)), \( V_{specimen} \) is the apparent volume of the specimen, and \( g \) is the acceleration due to gravity. So, \( \frac{W}{2} = 1000 \cdot V_{specimen} \cdot g \).
04

Relating Weights and Volume

The weight of the specimen in air is \( W = \rho_{solid} \cdot V_{solid} \cdot g \), where \( \rho_{solid} = 5.0 \times 10^{3} \mathrm{kg/m}^3 \) and \( V_{solid} \) is the volume of the solid part. Therefore, \( W = 5000 \cdot V_{solid} \cdot g \).
05

Formulating Equation for Fraction of Solid Volume

Knowing \( \frac{W}{2} = 1000 \cdot V_{specimen} \cdot g \), substitute \( W = 5000 \cdot V_{solid} \cdot g \) into the buoyant force equation: \( \frac{5000 \cdot V_{solid} \cdot g}{2} = 1000 \cdot V_{specimen} \cdot g \).
06

Simplifying the Equation

Cancel \( g \) from the equation and solve for the fraction of the solid volume: \( \frac{5000 \cdot V_{solid}}{2} = 1000 \cdot V_{specimen} \). Thus, \( V_{solid} = \frac{2}{5} V_{specimen} \).
07

Calculating the Fraction of Solid Volume

The fraction of the specimen's apparent volume that is solid is \( \frac{V_{solid}}{V_{specimen}} = \frac{2}{5} \). This means 40% of the specimen's volume is solid, and 60% is hollow (filled with air).

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Key Concepts

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

Buoyant Force
When an object is submerged in a fluid, it experiences an upward force called the buoyant force. This force is a result of the pressure difference at the bottom and top of the object due to the fluid. Archimedes' Principle helps us understand that the buoyant force is equal to the weight of the fluid displaced by the object. This means that the fluid pushes up on the object with a force equal to the weight of the fluid that is displaced.
In our exercise, the geologist noticed a difference between the rock's weight in air and in water. When the rock is submerged, it experiences a buoyant force, thus reducing its apparent weight. This difference in weight reveals key information about the rock's inner composition. Recognizing this buoyant force allows us to understand that the rock possibly has hollow sections, allowing it to displace more water for its overall size.
Density of Solids
Density is defined as mass per unit volume, and is a fundamental property that affects an object's buoyancy. For solids, density can vary significantly, affecting how they interact with fluids. In the rock specimen example, we are given the density of its solid parts as \(5.0 \times 10^{3} \text{kg/m}^3\). This high density implies that the solid parts of the rock are much denser than the water, which is why it only partially affects the overall buoyancy.Understanding density helps us determine why certain objects float while others sink. Here, comparing the density of the rock to the fluid (water), we know heavier, denser solid portions sink unless balanced by hollow sections, impacting overall buoyancy.
Volume Calculation
Calculating volume is crucial in this problem as it provides insight into the object's overall structure. Volume defines the amount of space an object occupies, which, when combined with density, gives us the object's mass.In Archimedes鈥 principle, the apparent volume of the specimen plays a key role. To solve for the rock's solid fraction, we must relate its apparent volume to its weight changes in air and water. By formulating the equations involving buoyant force and density, we determined that the solid volume of the rock is \( \frac{2}{5} \) of the entire apparent volume.Efficiently calculating volume helps highlight the relationship between the hollow and solid parts, allowing for a deeper exploration of the material's physical state.
Displacement in Fluids
Displacement is the amount of fluid moved out of the way by an object immersed in it. This is well illustrated by Archimedes' Principle, which states any submerged object displaces a volume of fluid equal to its own volume. In the geologist's problem, displacement was measured by the difference in weight when the rock was submerged. This method tells us that the specimen displaces a volume of water corresponding to its apparent volume. Displacement helps derive the buoyant force, since the weight of the displaced fluid equals the buoyant force acting upon the object. Thus, understanding displacement is crucial for calculating and predicting how objects interact with fluids, which is a key part of many experimental and practical applications.

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Most popular questions from this chapter

The construction of a flat rectangular roof \((5.0 \mathrm{m} \times 6.3 \mathrm{m})\) allows it to withstand a maximum net outward force that is 22000 \(\mathrm{N}\) . The density of the air is 1.29 \(\mathrm{kg} / \mathrm{m}^{3}\) . At what wind speed will this roof blow outward?

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A log splitter uses a pump with hydraulic oil to push a piston, which is attached to a chisel. The pump can generate a pressure of \(2.0 \times 10^{7}\) Pa in the hydraulic oil, and the piston has a radius of 0.050 \(\mathrm{m}\) . In a stroke lasting 25 \(\mathrm{s}\) s, the piston moves 0.60 \(\mathrm{m} .\) What is the power needed to operate the log splitter's pump?

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