/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 A fish maintains its depth in fr... [FREE SOLUTION] | 91Ó°ÊÓ

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A fish maintains its depth in fresh water by adjusting the air content of porous bone or air sacs to make its average density the same as that of the water. Suppose that with its air sacs collapsed, a fish has a density of \(1.08 \mathrm{~g} / \mathrm{cm}^{3}\). To what fraction of its expanded body volume must the fish inflate the air sacs to reduce its density to that of water?

Short Answer

Expert verified
The fish must inflate 7.41% of its volume with air.

Step by step solution

01

Understand the Problem

The fish needs to adjust its density to be equal to that of water, which is \(1.00 \text{ g/cm}^3\). Currently, its density is \(1.08 \text{ g/cm}^3\). We need to find the fraction of its volume that the fish must inflate the air sacs to match the water's density.
02

Set the Formula for Density

Density \(\rho\) is defined as mass \(m\) over volume \(V\), or \(\rho = \frac{m}{V}\). The fish needs to adjust its volume to change its density from \(1.08 \text{ g/cm}^3\) to \(1.00 \text{ g/cm}^3\).
03

Calculate the Desired Final Volume

To make the density \(1.00 \text{ g/cm}^3\), using the formula \(1.00 = \frac{m}{V_{\text{final}}}\), we rearrange to find \(V_{\text{final}} = m\) since \(1.00\) represents the density of water.
04

Express the Initial Volume

The initial volume \(V_{\text{initial}}\) is given by \(V_{\text{initial}} = \frac{m}{1.08}\). This is the volume of the fish when its density is \(1.08 \text{ g/cm}^3\).
05

Find the Inflation Volume

The required inflation volume \(V_{\text{inflate}}\) to achieve the density of water is \(V_{\text{inflate}} = V_{\text{final}} - V_{\text{initial}}\). Substituting the known expressions gives \(V_{\text{inflate}} = m - \frac{m}{1.08}\).
06

Calculate the Fraction

The fraction of the volume to be filled with air sacs is \(\frac{V_{\text{inflate}}}{V_{\text{final}}} = \frac{m - \frac{m}{1.08}}{m} = 1 - \frac{1}{1.08}\). Simplify this to find the fraction.
07

Simplify and Solve

Calculate \(1 - \frac{1}{1.08}\) to get the fraction: \[ 1 - \frac{1}{1.08} = 1 - 0.9259 \approx 0.0741 \].

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Key Concepts

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

Density Adjustment
Fish have the unique ability to control their buoyancy, which is their ability to float or sink in water. One way they do this is by adjusting their density. Density is a measure of how much mass is contained in a given volume. In the case of the fish, it manages its depth in water by changing the volume of air in its air sacs. By doing so, it adjusts its average density to match that of the surrounding water. This matching of densities allows the fish to maintain its desired depth without expending too much energy. When the air sacs are deflated, the fish’s density is higher than that of water. In our example, the fish has a density of 1.08 g/cm³ when its air sacs are collapsed. Since the density of water is 1.00 g/cm³, the fish must expand its volume until its density matches that of water. By increasing the volume (and therefore decreasing its density), the fish achieves neutral buoyancy. This is crucial for a fish's ability to float effortlessly at any depth. It can precisely control its position and movement, which is essential for efficient swimming and avoiding predators.
Boyle's Law
Boyle's Law is a fundamental principle in physics that describes how the pressure and volume of a gas are related, provided the temperature is constant. It states that pressure is inversely proportional to volume: if the volume increases, the pressure decreases, and vice versa. Mathematically, this principle is expressed as \( P_1 V_1 = P_2 V_2 \), where \( P \) is the pressure and \( V \) is the volume.When a fish inflates its air sacs, Boyle's Law plays a role. As the fish increases the volume of the air sacs, the pressure inside decreases. This expanded volume results in a decrease in the overall density of the fish, reaching a point where its density matches that of the water. By carefully modulating the amount of air in its sacs, and taking advantage of Boyle's Law, the fish can rise or sink in water as needed without expending significant energy.This gas law shows the fascinating adaptation of fish to their environment, allowing them to engage in complex movements. It's a beautiful demonstration of physics at work in biology!
Hydrostatics
Hydrostatics is the study of fluids at rest. In the context of fish buoyancy, it involves understanding the forces at play when a fish is submerged in water. The principle of buoyancy, which is a central concept in hydrostatics, dictates that an object will float in a fluid if the upward buoyant force is equal to the downward gravitational force. This buoyant force is determined by the fluid’s density and the volume of fluid displaced by the object, as described by Archimedes' principle. For a fish, achieving neutral buoyancy means making the downward gravitational force equal to this buoyant force. This is where the concept of density adjustment, as discussed earlier, is crucial. By expanding the air sacs and decreasing its density, the fish reduces the gravitational force effect until it matches the buoyant force, allowing it to maintain its depth without moving. Fish rely on hydrostatics not only to stay at a preferred depth but also to move efficiently through the water. Understanding these fundamental principles empowers students to appreciate the delicate balance of forces that nature achieves effortlessly.

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Most popular questions from this chapter

About one-third of the body of a person floating in the Dead Sea will be above the waterline. Assuming that the human body density is \(0.98 \mathrm{~g} / \mathrm{cm}^{3}\), find the density of the water in the Dead Sea. (Why is it so much greater than \(1.0 \mathrm{~g} / \mathrm{cm}^{3} ?\) )

Suppose that your body has a uniform density of \(0.95\) times that of water. (a) If you float in a swimming pool, what fraction of your body's volume is above the water surface? Quicksand is a fluid produced when water is forced up into sand, moving the sand grains away from one another so they are no longer locked together by friction. Pools of quicksand can form when water drains underground from hills into valleys where there are sand pockets (b) If you float in a deep pool of quicksand that has a density \(1.6\) times that of water, what fraction of your body's volume is above the quicksand surface? (c) In particular, are you submerged enough to be unable to breathe?

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