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A paperweight, when weighed in air, has a weight of \(W=6.9 \mathrm{N}\) . When completely immersed in water, however, it has a weight of \(W_{\text { in water }}=4.3 \mathrm{N}\) . Find the volume of the paperweight.

Short Answer

Expert verified
The volume of the paperweight is approximately 0.000265 cubic meters or 0.265 liters.

Step by step solution

01

Understand Buoyancy

When the paperweight is immersed in water, it experiences a buoyant force that reduces its weight. The buoyant force is equal to the weight of the water displaced by the paperweight.
02

Use Buoyant Force Formula

The buoyant force can be calculated using the formula: \( F_b = W - W_{\text{in water}} \), where \( F_b \) is the buoyant force, \( W \) is the weight in air, and \( W_{\text{in water}} \) is the weight in water. Substitute the given weights: \( F_b = 6.9 \text{ N} - 4.3 \text{ N} = 2.6 \text{ N} \).
03

Apply Archimedes' Principle

According to Archimedes' Principle, the buoyant force \( F_b \) is equal to the weight of the volume of water displaced. Therefore, \( F_b = \rho \cdot V \cdot g \), where \( \rho \) is the density of water (\( 1000 \text{ kg/m}^3 \)), \( V \) is the volume of the paperweight, and \( g \) is the acceleration due to gravity (\( 9.81 \text{ m/s}^2 \)).
04

Solve for Volume

Rearranging the formula for volume gives \( V = \frac{F_b}{\rho \cdot g} \). Substituting the values we have: \( V = \frac{2.6 \text{ N}}{1000 \text{ kg/m}^3 \cdot 9.81 \text{ m/s}^2} \approx 0.000265 \text{ m}^3 \).
05

Convert Volume to Relevant Units

While the volume is accurate in cubic meters, it might be more intuitive to express it in liters. Since 1 m³ = 1000 L, the volume in liters is \( 0.000265 \text{ m}^3 \times 1000 \text{ L/m}^3 = 0.265 \text{ L} \).

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Key Concepts

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

Archimedes' Principle
Archimedes' Principle is a fundamental law of physics that explains why objects float or sink in a fluid. It states that any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. This principle helps us understand the concept of buoyancy and how to calculate the buoyant force acting on an object. Its practical applications can be seen in everything from shipbuilding to understanding why icebergs float.
This principle was discovered by the ancient Greek scientist Archimedes of Syracuse, who famously ran through the streets shouting "Eureka!" upon discovering this revolutionary idea. When we apply Archimedes' Principle, we essentially weigh not the object itself but the fluid that it displaces. This weight of the displaced fluid is what gives us the buoyant force.
Buoyant Force
The buoyant force is the upward force exerted by a fluid that opposes the weight of an object immersed in it.
It explains why objects feel lighter in water than they do in the air. The equation for buoyant force is:
  • \( F_b = W - W_{\text{in water}} \)
Here, \( F_b \) is the buoyant force, \( W \) is the weight of the object in air, and \( W_{\text{in water}} \) is the weight of the object when submerged in water.
For an object in equilibrium in water, this buoyant force exactly balances the object's weight that acts downward.
This force is what you feel when you try to push a beach ball underwater, as it keeps trying to float back to the surface.
Density of Water
The density of water is a constant crucial in applying Archimedes' Principle. This density is typically \( 1000 \text{ kg/m}^3 \) for pure water at 4 degrees Celsius, which is used as a standard reference point because water is densest at this temperature.
This value is essential when calculating the buoyant force, because the buoyant force is directly proportional to both the volume of the fluid displaced and its density.
  • Formula for buoyant force: \( F_b = \rho \cdot V \cdot g \)
  • \( \rho \) is the density of the fluid (water in this case)
Understanding water's density helps solve problems involving buoyancy, ensuring accuracy when determining the volume of displaced water and thus the buoyant force.
Volume Calculation
Calculating the volume of an object submerged in water is a direct application of Archimedes' Principle and involves understanding the relationship between buoyant force and displaced water volume.
The formula to find volume using buoyant force is:
  • \( V = \frac{F_b}{\rho \cdot g} \)
Where \( V \) is the volume of the object, \( F_b \) is the buoyant force, \( \rho \) is the density of water, and \( g \) is the acceleration due to gravity.
By rearranging the formula for buoyant force, we solve for volume. In practice, once we find the volume in cubic meters, we might convert it to more intuitive units like liters (since 1 m³ = 1000 L), which often offer a clearer sense of scale for the calculated volumes. This ability to calculate volume is especially useful in crafting items that need to meet specific buoyancy criteria, such as life jackets or boats.

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Most popular questions from this chapter

An underground pump initially forces water through a horizontal pipe at a flow rate of 740 gallons per minute. After several years of operation, corrosion and mineral deposits have reduced the inner radius of the pipe to 0.19 m from 0.24 m, but the pressure difference between the ends of the pipe is the same as it was initially. Find the final flow rate in the pipe in gallons per minute. Treat water as a viscous fluid.

An \(81-\mathrm{kg}\) person puts on a life jacket, jumps into the water, and floats. The jacket has a volume of \(3.1 \times 10^{-2} \mathrm{m}^{3}\) and is completely submerged under the water. The volume of the person's body that is under water is \(6.2 \times 10^{-2} \mathrm{m}^{3} .\) What is the density of the life jacket?

An airplane wing is designed so that the speed of the air across the top of the wing is 251 \(\mathrm{m} / \mathrm{s}\) when the speed of the air below the wing is 225 \(\mathrm{m} / \mathrm{s}\) . The density of the air is 1.29 \(\mathrm{kg} / \mathrm{m}^{3} .\) What is the lifting force on a wing of area 24.0 \(\mathrm{m}^{2} ?\)

A small crack occurs at the base of a \(15.0-\mathrm{m}\) -high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

A full can of black cherry soda has a mass of 0.416 \(\mathrm{kg}\) . It contains \(3.54 \times 10^{-4} \mathrm{m}^{3}\) of liquid. Assuming that the soda has the same density as water, find the volume of aluminum used to make the can.

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