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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?

Short Answer

Expert verified
The density of the life jacket is approximately 871 kg/m³.

Step by step solution

01

Understand the Given Data

We are given the mass of the person as \(81\, \mathrm{kg}\). The life jacket's volume is \(3.1 \times 10^{-2} \, \mathrm{m}^{3}\), and the submerged volume of the person's body is \(6.2 \times 10^{-2} \, \mathrm{m}^{3}\). We need to find the density of the life jacket.
02

Calculate Total Volume Underwater

The total volume underwater is the sum of the life jacket's volume and the person's submerged volume. Hence, it is \(3.1 \times 10^{-2} \, \mathrm{m}^{3} + 6.2 \times 10^{-2} \, \mathrm{m}^{3} = 9.3 \times 10^{-2} \, \mathrm{m}^{3}\).
03

Apply Archimedes' Principle

According to Archimedes' principle, the buoyant force equals the weight of the water displaced. This means the mass of the water displaced is equal to the mass of the person and the jacket together, which is \(81\, \mathrm{kg}\).
04

Calculate the Density of Water

We know that the density of water \(\rho_{\text{water}}\) is approximately \(1000\, \mathrm{kg/m}^{3}\). This information helps us find the volume of water displaced based on mass.
05

Establish Equation for Density of Life Jacket

Let \(\rho_{\text{jacket}}\) be the density of the life jacket. The total mass of water displaced is \(81\, \mathrm{kg}\), and the total volume submerged \((9.3 \times 10^{-2} \, \mathrm{m}^{3})\) equals the combined volume of the life jacket and the body. Using the volume, we can set up the equation for buoyancy.
06

Solve for Density of Life Jacket

The volume submerged by the jacket is \(3.1 \times 10^{-2} \, \mathrm{m}^{3}\). The weight of water displaced by just the life jacket is equal to the jacket's apparent weight, or gravity acts on \(\rho_{\text{jacket}} \times 3.1 \times 10^{-2} \, \mathrm{m}^{3}\). From step 5, solve for \(\rho_{\text{jacket}} = \frac{81\, \mathrm{kg}}{ 9.3 \times 10^{-2} \, \mathrm{m}^{3}}\) to find the density.

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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 cornerstone concept in the study of buoyancy. It states that an object, when submerged in a fluid such as water, experiences an upward force known as the buoyant force. This force is equal to the weight of the fluid that the object displaces.
For the person wearing the life jacket, Archimedes' principle explains why they float. The total volume submerged, which includes both the volume of the person’s submerged body and the jacket, displaces a certain amount of water.
According to the principle, this displacement results in a buoyant force that
  • counters the force of gravity pulling the person downward,
  • allows the person to float even though their weight exceeds the weight of the jacket itself.
This understanding is crucial in safety equipment design, like life jackets, ensuring they provide sufficient buoyant force to keep a person afloat.
density calculation
Calculating density is essential to understand the properties of objects like life jackets. Density is defined as mass per unit of volume, typically expressed in \[\rho = \frac{m}{V}\]where \(m\) is the mass and \(V\) is the volume.
This relationship helps us determine how 'heavy' an object is relative to its size, crucial in predicting whether it will float or sink.
In our exercise situation, knowing the density of the life jacket helps us see how it assists in flotation. Given the life jacket’s volume and the total mass of the person and jacket displacing water
  • we understand how much lighter the jacket is compared to the water,
  • which explains how it contributes to keeping the person afloat.
Calculating this density involves using the total displaced water's volume and the combined mass of the person and the life jacket.
submerged volume
The concept of submerged volume entails the part of an object or body that is under the surface of the fluid it is in. In our problem, both the life jacket and part of the person's body are submerged.
  • The submerged volume of the life jacket is critical for calculating buoyancy,
  • as this is the volume actively displacing water, contributing to the buoyant force according to Archimedes' principle.
In practical terms, knowing the submerged volume helps determine if the life jacket provides enough lift to keep a person afloat.
It’s important to understand that different objects have different submerged volumes based on their densities. Less dense objects like life jackets tend to displace enough water to create an adequate buoyant force, which is why they are essential safety gear in water activities.

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Most popular questions from this chapter

A pipe is horizontal and carries oil that has a viscosity of 0.14 \(\mathrm{Pa} \cdot \mathrm{s}\) The volume flow rate of the oil is \(5.3 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s}\) . The length of the pipe is \(37 \mathrm{m},\) and its radius is 0.60 \(\mathrm{cm}\) . At the output end of the pipe the pressure is atmospheric pressure. What is the absolute pressure at the input end?

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At a given instant, the blood pressure in the heart is \(1.6 \times 10^{4} \mathrm{Pa}\) If an artery in the brain is 0.45 m above the heart, what is the pressure in the artery? Ignore any pressure changes due to blood flow.

Two hoses are connected to the same outlet using a Y-connector, as the drawing shows. The hoses A and B have the same length, but hose B has the larger radius. Each is open to the atmosphere at the end where the water exits. Water flows through both hoses as a viscous fluid, and Poiseuille’s law \(\left[Q=\pi R^{4}\left(P_{2}-P_{1}\right) /(8 \eta L)\right]\) applies to each. In this law, \(P_{2}\) is the pressure upstream, \(P_{1}\) is the pressure down- stream, and \(Q\) is the volume flow rate. The ratio of the radius of hose \(B\) to the radius of hose A is \(R_{B} / R_{A}=1.50 .\) Find the ratio of the speed of the water in hose \(\mathrm{B}\) to the speed in hose A.

Multiple-Concept Example 8 presents an approach to problems of this kind. The hydraulic oil in a car lift has a density of \(8.30 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}\) . The weight of the input piston is negligible. The radii of the input piston and output plunger are \(7.70 \times 10^{-3} \mathrm{m}\) and \(0.125 \mathrm{m},\) respectively. What input force \(F\) is needed to support the 24500 \(\mathrm{-N}\) combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.30 m above that of the input piston?

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