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Chapter 12: Problem 52

(a) As you can tell by watching them in an aquarium, fish are able to remain at any depth in water with no effort. What does this ability tell you about their density? (b) Fish are able to inflate themselves using a sac (called the swim bladder) located under their spinal column. These sacs can be filled with an oxygen-nitrogen mixture that comes from the blood. If a 2.75 -kg fish in freshwater inflates itself and increases its volume by \(10 \%,\) find the net force that the water exerts on it. (c) What is the net external force on it? Does the fish go up or down when it inflates itself?

Short Answer

Expert verified
(a) The density of the fish equals water's density. (b) The net force is 2.695 N upward. (c) The fish will go up.

Step by step solution

01

Density and Buoyancy

Fish can maintain a certain depth without effort, which implies that their density is equal to the density of the water around them. This balance of densities allows them to remain neutrally buoyant so that no additional energy is needed to stay at a given depth.
02

Volume Increase Calculation

The fish increases its volume by 10%. If we let \( V_f \) be its original volume, the new volume \( V' \) is \( V' = V_f + 0.10V_f = 1.10V_f \).
03

Weight of the Fish

The weight of the fish is calculated as \( W = mg \), where \( m = 2.75 \, \text{kg} \) is the mass of the fish and \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity. Thus, \( W = 2.75 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 26.95 \, \text{N} \).
04

Buoyant Force Calculation

The buoyant force, \( F_b \), is calculated using Archimedes' principle: \( F_b = \rho_{\text{water}} \cdot g \cdot V' \) where \( \rho_{\text{water}} = 1000 \, \text{kg/m}^3 \). Substitute \( V' = 1.10V_f \) to find the increase in buoyant force. Since the fish was initially neutrally buoyant, it's originally displacing its own weight of water: \( V_f = \frac{m}{\rho_{\text{water}}} \). Calculate \( F_b \) using \( V' = 1.10 \times V_f = 1.10 \times \frac{m}{\rho_{\text{water}}} = 1.10 \times \frac{2.75}{1000} \approx 0.003025 \, \text{m}^3 \). Now, \( F_b = 1000 \times 9.8 \times 0.003025 = 29.645 \, \text{N} \).
05

Net Force Calculation

To find the net force that the water exerts on the fish, we subtract the weight of the fish from the buoyant force. Therefore, the net force \( F_{\text{net}} = F_b - W = 29.645 \, \text{N} - 26.95 \, \text{N} = 2.695 \, \text{N} \).
06

Direction of Movement

Since the net force is positive, the net external force on the fish is upward, meaning the fish will rise when it inflates its swim bladder.

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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 helps us understand how objects float. It states that any object submerged in a fluid experiences a buoyant force equivalent to the weight of the fluid displaced by the object. This principle allows us to determine whether an object will float, sink, or remain neutrally buoyant in a fluid. For fish, this principle explains how they can hover at different depths in water. By adjusting their volume using their swim bladder, they change the amount of water displaced, which in turn affects the buoyant force acting on them.
density
Density plays a crucial role in determining whether an object will float or sink. It is defined as mass per unit volume and can be expressed as \( \rho = \frac{m}{V} \). When the density of an object is equal to the density of the fluid it is submerged in, the object achieves neutral buoyancy, meaning it neither sinks nor floats. Fish have adapted to adjust their density to match the water's surrounding them, allowing them to remain at a consistent depth effortlessly.
neutral buoyancy
Neutral buoyancy is achieved when the gravitational force and the buoyant force on an object are equal, resulting in no net force. For fish, being neutrally buoyant means they can stay at their desired depth without using their fins or expending energy. This efficient mechanism is vital for their survival, allowing them to adapt to various water environments. By leveraging their swim bladder, fish can fine-tune their buoyancy to remain at equilibrium within the water column.
swim bladder
The swim bladder is a specialized organ found in most fish, enabling them to control their buoyancy. Located under the spinal column, it functions like an expandable balloon that can be filled with gas, usually oxygen or nitrogen sourced from the fish's blood. By altering the amount of gas inside the swim bladder, the fish changes its volume and density, increasing or decreasing its buoyant force. This allows fish to ascend or descend in the water without active swimming, giving them a tactical advantage in conserving energy.
fish buoyancy mechanism
Fish have evolved a sophisticated buoyancy mechanism that allows them to adapt to different water depths efficiently. Their ability to modulate buoyancy hinges on the interaction between their body's density and the surrounding water's density, typically through the use of their swim bladder. When a fish needs to move to shallower waters, it increases the gas in its swim bladder, thereby increasing its volume and reducing its overall density. This change results in a higher buoyant force lifting the fish upward. Conversely, to descend, the fish releases gas from the swim bladder, decreasing its volume and increasing its density, making it sink. Efficient buoyancy regulation is crucial for fish as it impacts their survival, feeding, and social interactions.

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Most popular questions from this chapter

A plastic ball has radius 12.0 \(\mathrm{cm}\) and floats in water with 24.0\(\%\) of its volume submerged. (a) What force must you apply to the ball to hold it at rest totally below the surface of the water? (b) If you let go of the ball, what is its acceleration the instant you release it?

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 \(\mathrm{kg} / \mathrm{m}^{3} )\) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 \(\mathrm{Pat}\) is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 12.6\()\) of the fluid.

A hot-air balloon has a volume of 2200 \(\mathrm{m}^{3} .\) The balloon fabric (the envelope) weighs 900 \(\mathrm{N} .\) The basket with gear and full propane tanks weighs 1700 \(\mathrm{N} .\) If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A .\) The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \omega. Gravitational forces are negligible. Consider a volume element of the fluid of area \(A\) and thickness \(d r^{\prime}\) a distance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime}\) . (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0},\) show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .\) (c) An object of volume \(V\) and density \(\rho_{\mathrm{ob}}\) has its center of mass at a distance \(R_{\mathrm{cmob}}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{ob}} R_{\mathrm{cmob}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}} .\) (e) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) . What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

At one point in a pipeline the water's speed is 3.00 \(\mathrm{m} / \mathrm{s}\) and the gauge pressure is \(5.00 \times 10^{4}\) Pa. Find the gauge pressure at a second point in the line, 11.0 \(\mathrm{m}\) lower than the first, if the pipe diameter at the second point is twice that the first.
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