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Chapter 9: Problem 38

The average density of a fish can be found by first weighing it in air and then finding the scale reading for the fish completely immersed in water and suspended from a scale. If a fish has weight \(200.0 \mathrm{N}\) in air and scale reading \(15.0 \mathrm{N}\) in water, what is the average density of the fish?

Short Answer

Expert verified
The average density of the fish is approximately 1081.34 kg/m鲁.

Step by step solution

01

Define the buoyancy force equation

Buoyancy force, which is the force exerted by the fluid (water in this case) that opposes the weight of an object immersed in it, is responsible for the difference in weight measurements. The buoyant force can be calculated using the equation: \[ F_b = W_{ ext{air}} - W_{ ext{water}} \]where \( F_b \) is the buoyant force, \( W_{ ext{air}} \)= 200.0 N is the weight of the fish in air, and \( W_{ ext{water}} \)= 15.0 N is the weight of the fish in water.
02

Calculate the buoyant force

Substitute the given values into the buoyancy equation:\[ F_b = 200.0 \, ext{N} - 15.0 \, ext{N} = 185.0 \, ext{N} \]This 185.0 N indicates the buoyant force acting on the fish.
03

Understand Archimedes' Principle

According to Archimedes' Principle, the buoyant force \( F_b \) is equal to the weight of the water displaced by the fish, which can be written as the formula:\[ F_b = \rho_{ ext{water}} \cdot V \cdot g \]where \( \rho_{ ext{water}} = 1000 \, \text{kg/m}^3 \) is the density of water, \( V \) is the volume of the fish, and \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity.
04

Solve for the volume of the fish

Rearrange the equation from Archimedes' Principle to solve for the volume \( V \):\[ V = \frac{F_b}{\rho_{ ext{water}} \cdot g} = \frac{185.0 \, \text{N}}{1000 \, \text{kg/m}^3 \cdot 9.81 \, \text{m/s}^2} \]Calculating this gives:\[ V \approx 0.01886 \, \text{m}^3 \]
05

Calculate the density of the fish

The density of the fish \( \rho_{ ext{fish}} \) can be calculated with the mass \( m \) of the fish and the volume \( V \) found above:- First, calculate the mass from the weight in air: \[ m = \frac{W_{ ext{air}}}{g} = \frac{200.0 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 20.39 \, \text{kg} \]- Now find the density:\[ \rho_{ ext{fish}} = \frac{m}{V} = \frac{20.39 \, \text{kg}}{0.01886 \, \text{m}^3} \approx 1081.34 \, \text{kg/m}^3 \]
06

Conclusion: Average density of the fish

The average density of the fish is approximately \( 1081.34 \, \text{kg/m}^3 \). This final answer takes into account the effects of buoyancy and the displaced water.

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Key Concepts

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

Archimedes' Principle
Imagine lifting something in the water, like a beach ball. It rises easier in water than in air, right? That's because when an object is submerged in a fluid, like water, it experiences a force called buoyant force. Archimedes' Principle explains this phenomenon.

The principle states that the buoyant force on an object is equal to the weight of the fluid displaced by the object. For example, if you submerge a fish into water, the amount of water it pushes aside appears lighter. This displacement notes how the fish seems lighter in water than when out of it.
  • Buoyant force = Weight of displaced fluid
  • Acts in the upward direction
  • Explains why things float or sink
Understanding Archimedes' Principle helps explain why the fish in our problem appears lighter in water. The water displaces a portion of the fish鈥檚 weight, resulting in a lower measured weight when submerged.
density calculation
Density shows us how tightly packed matter is in an object. For any material, it is defined as the mass per unit volume. In simpler terms, it tells us how much stuff there is in a particular space.

To calculate the density of the fish in the problem, we first found its volume using the buoyant force and Archimedes' Principle. Then, with the fish's mass already determined from its weight in air, we divided it by the volume. This gives us an idea of how dense the fish is compared to the fluid it displaces.
  • Density (\( \rho \)) = Mass (\( m \)) / Volume (\( V \))
  • Measured in kg/m鲁
  • Can determine if an object will float or sink
The fish we calculated had a higher density than water, helping it float at a lower level in water. But if it were less dense, it might float easily or need effort to stay submerged.
buoyant force
When you place an object in water, it typically feels lighter than in the air. This difference is due to buoyant force, which is a result of the fluid surrounding and exerting upward pressure on the object. The buoyant force depends on the weight of the liquid displaced.

In the example problem, the fish weighs 200 N in the air, but its weight appears to reduce to 15 N when submerged, indicating a buoyant force of 185 N. This force counters some of the fish's weight and helps us measure the volume of water displaced.
  • Buoyant Force = Subtracted weight in fluid from weight in air
  • Measured in Newtons (N)
  • Affected by fluid's density
Knowing the influence of this force is vital in designing ships or understanding why certain bodies float or sink, and how to manipulate the conditions, like density, to achieve buoyancy.

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Most popular questions from this chapter

A sphere of radius \(1.0 \mathrm{cm}\) is dropped into a glass cylinder filled with a viscous liquid. The mass of the sphere is \(12.0 \mathrm{g}\) and the density of the liquid is \(1200 \mathrm{kg} / \mathrm{m}^{3} .\) The sphere reaches a terminal speed of \(0.15 \mathrm{m} / \mathrm{s} .\) What is the viscosity of the liquid?
A manometer using oil (density \(0.90 \mathrm{g} / \mathrm{cm}^{3}\) ) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by \(0.74 \mathrm{cm}\) Hg. (a) By how much does the fluid level rise in the side of the manometer that is open to the atmosphere? (b) What would your answer be if the manometer used mercury instead?
The diameter of a certain artery has decreased by \(25 \%\) due to arteriosclerosis. (a) If the same amount of blood flows through it per unit time as when it was unobstructed, by what percentage has the blood pressure difference between its ends increased? (b) If, instead, the pressure drop across the artery stays the same, by what factor does the blood flow rate through it decrease? (In reality we are likely to see a combination of some pressure increase with some reduction in flow.)
A cube that is \(4.00 \mathrm{cm}\) on a side and of density $8.00 \times 10^{2} \mathrm{kg} / \mathrm{m}^{3}$ is attached to one end of a spring. The other end of the spring is attached to the base of a beaker. When the beaker is filled with water until the entire cube is submerged, the spring is stretched by \(1.00 \mathrm{cm} .\) What is the spring constant?

Scuba divers are admonished not to rise faster than their air bubbles when rising to the surface. This rule helps them avoid the rapid pressure changes that cause the bends. Air bubbles of 1.0 mm radius are rising from a scuba diver to the surface of the sea. Assume a water temperature of \(20^{\circ} \mathrm{C}\). (a) If the viscosity of the water is \(1.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s},\) what is the terminal velocity of the bubbles? (b) What is the largest rate of pressure change tolerable for the diver according to this rule?
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