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Chapter 12: Problem 70

In seawater, a life preserver with a volume of 0.0400 m\(^3\) will support a 75.0-kg person (average density 980 kg/m\(^3\)), with 20% of the person's volume above the water surface when the life preserver is fully submerged. What is the density of the material composing the life preserver?

Short Answer

Expert verified
The density of the life preserver material is approximately 127.5 kg/m\(^3\).

Step by step solution

01

Calculate the buoyant force needed

A 75.0-kg person has an average density of 980 kg/m\(^3\). First, calculate the volume of the person using the formula for density, which is \( \text{density} = \frac{\text{mass}}{\text{volume}} \). Rearrange this to find volume as \( \text{volume} = \frac{\text{mass}}{\text{density}} = \frac{75.0}{980} = 0.0765\ m^3 \). The submerged volume is 80% of the person's volume, which is \( 0.0612\ m^3 \). The buoyant force must equal the weight of the person, \( F_b = 75.0 \times 9.81 = 735.75\ N \), because it's supported at the equilibrium.
02

Find the volume of water displaced

According to Archimedes' Principle, the buoyant force is equal to the weight of the displaced fluid. With the life preserver's volume fully submerged, \( V_{lw} = 0.0400\ m^3 \), the person and the life preserver together need to displace water equivalent to the weight of the person. The total displaced water volume is the sum of the life preserver and submerged portion of the person, \( V_w = 0.0400 + 0.0612 = 0.1012\ m^3 \).
03

Calculate the desired density of the life preserver

The volume of water displaced has a density equal to that of seawater, typically about 1025 kg/m\(^3\). The combined weight of the life preserver and the submerged part of the person equates to the weight of the displaced water: \( 9.81 \times V_w \times 1025 = 9.81 \times (0.0612\times 980 + 0.0400 \times 980)\). Calculate the right-hand side, equate, and solve for density, ensuring that the life preserver's effective buoyancy makes up for the fraction remaining afloat.

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Key Concepts

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

Archimedes' Principle
Understanding Archimedes' Principle is crucial when discussing buoyancy. It states that an object submerged in a fluid experiences a buoyant force equal to the weight of the fluid it displaces. This principle explains why objects float or sink based on whether the buoyant force can counterbalance the object's weight.
For example, when a life preserver and a person are in water, the combined weight needs to be balanced by the weight of the water displaced. This is why it's important to calculate the correct volume of water displaced to support the total weight, ensuring that enough buoyant force is provided.
density calculation
Density is a measure of how much mass is contained in a given volume. Calculating density involves using the formula: \( \text{density} = \frac{\text{mass}}{\text{volume}} \). This is fundamental in determining whether an object will float or sink.
In solving problems involving buoyancy, it's important to find the density of involved materials to understand their interactions with water. For instance, calculating a person's density helps in determining their volume. The density of the life preserver material must be such that it supports a person within the water, revealing its effectiveness.
  • High density indicates a heavy and compact object.
  • Low density surfaces as a less compact material, often providing buoyancy.
volume displacement
Volume displacement refers to the amount of fluid pushed aside by an object when it is placed in a fluid. This concept is directly tied to Archimedes' Principle.
By finding the total displaced water volume, you can calculate the buoyant force exerted on the object. In the exercise with the life preserver and person, the volume displacement explores the submerged volumes of both, ensuring they correspond to the required buoyant force. Errors in calculating displacement may lead to incorrect assumptions about an object's floating capacity.
seawater density
Seawater density plays a pivotal role in calculating buoyancy forces in marine environments. Typically around 1025 kg/m\(^3\), seawater density is greater than freshwater, due to its salt content.
When evaluating buoyancy in seawater, the heightened density increases the buoyant force on submerged objects, slightly altering the dynamics compared to freshwater scenarios. This means objects that sink in freshwater might float in seawater because of the denser medium providing more upward force.
  • Seawater's density impacts vessel buoyancy.
  • Materials intended for seawater should account for its density differences.

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Most popular questions from this chapter

A shower head has 20 circular openings, each with radius 1.0 mm. The shower head is connected to a pipe with radius 0.80 cm. If the speed of water in the pipe is 3.0 m/s, what is its speed as it exits the shower-head openings?

The lower end of a long plastic straw is immersed below the surface of the water in a plastic cup. An average person sucking on the upper end of the straw can pull water into the straw to a vertical height of 1.1 m above the surface of the water in the cup. (a) What is the lowest gauge pressure that the average person can achieve inside his lungs? (b) Explain why your answer in part (a) is negative.

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?

Advertisements for a certain small car claim that it floats in water. (a) If the car's mass is 900 kg and its interior volume is 3.0 m\(^3\), what fraction of the car is immersed when it floats? Ignore the volume of steel and other materials. (b) Water gradually leaks in and displaces the air in the car. What fraction of the interior volume is filled with water when the car sinks?

There is a maximum depth at which a diver can breathe through a snorkel tube (\(\textbf{Fig. E12.17}\) ) 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.)
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