/*! 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 28 There is a maximum depth at whic... [FREE SOLUTION] | 91Ó°ÊÓ

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There is a maximum depth at which a diver can breathe through a snorkel tube (Fig. 13.42\()\) , because as the depth increases, so does the pressure difference, which the difference, if any.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 \(\mathrm{m}\) (about 20 \(\mathrm{ft}\) ? Assume that the diver is in fresh- water. (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.)

Short Answer

Expert verified
The pressure difference is 5978 Pa at a depth of 6.1 meters.

Step by step solution

01

Understanding the Problem

We have a diver at the depth of 6.1m in freshwater. We need to find the difference between the external pressure at that depth and the atmospheric pressure inside the diver's lungs.
02

Finding External Pressure at Depth

The external pressure at a depth in fluid is given by the formula: \( P = P_0 + \rho gh \), where \( P_0 \) is atmospheric pressure (101,325 Pa), \( \rho \) is the density of the fluid (for freshwater, \( \rho = 1000 \, kg/m^3 \)), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the depth (6.1 m).
03

Calculating External Pressure

Using the formula, we calculate the external pressure at 6.1m depth: \[ P = 101325 \, \text{Pa} + (1000 \, \text{kg/m}^3)(9.8 \, \text{m/s}^2)(6.1 \, \text{m}) \] \[ P = 101325 \, \text{Pa} + 5978 \, \text{Pa} \] \[ P = 107303 \, \text{Pa} \]
04

Determining Pressure Difference

Since the pressure in the lungs is still atmospheric (101,325 Pa), the pressure difference is the external pressure minus the internal lung pressure: \[ \Delta P = 107303 \, \text{Pa} - 101325 \, \text{Pa} \] \[ \Delta P = 5978 \, \text{Pa} \]
05

Conclusion

The external-internal pressure difference when the diver is at a depth of 6.1 meters is 5978 Pa.

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Key Concepts

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

Hydrostatic Pressure
In fluid mechanics, hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases with depth because the weight of the fluid above increases as you go deeper. The formula used to calculate the hydrostatic pressure at a given depth in a fluid is \( P = P_0 + \rho gh \). Here, \( P_0 \) represents the atmospheric pressure at the surface, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth of the fluid. This formula shows that pressure increases in a linear relation with depth as long as the density remains constant. By understanding this principle, we can calculate how much pressure a diver experiences at a certain depth, including what happens when snorkeling at a depth of 6.1 meters.
Snorkeling Physics
Snorkeling involves breathing through a tube while swimming at the surface of the water or just below it. The physics of snorkeling limit how deep one can snorkel safely. This limit is due to the increasing hydrostatic pressure as you go deeper. When snorkeling, the pressure inside the lungs is equivalent to atmospheric pressure because the snorkel tube connects the lungs directly to the surface air. However, as a snorkeler goes deeper, the external water pressure around the body increases while the pressure inside the lungs remains constant. This difference in pressure can become dangerous. At certain depths, this pressure difference may collapse the lungs, making it difficult or impossible to breathe. This is why it's recommended to snorkel relatively close to the surface.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the air above us in the Earth's atmosphere. At sea level, this pressure is approximately 101,325 pascals (Pa). This is the baseline pressure our bodies are accustomed to. When underwater, this pressure continues to press down on the water surface. If a diver is only as deep as a snorkel tube allows, the lungs still experience this atmospheric pressure because the air comes directly from the surface. However, as depth increases, the water adds additional pressure, which is why the external-internal pressure difference rises with depth, posing potential risks for snorkelers.
Fluid Mechanics
Fluid mechanics is the branch of physics concerned with the behavior of fluids (liquids and gases) and the forces on them. Within this field, understanding the concepts of fluid density, buoyancy, and pressure is crucial. In the context of a diver or snorkeler, fluid mechanics helps explain how forces act on a body immersed in water. Water density influences buoyancy and the pressure exerted on a body. In freshwater, which has a density of approximately 1000 kg/m³, the pressure exerted by the water increases linearly with depth. These principles are useful for calculating the pressures involved when diving and understanding the potential impacts on the human body when in water at varying depths.

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Most popular questions from this chapter

An open cylindrical tank of acid rests at the edge of a table 1.4 \(\mathrm{m}\) above the floor of the chemistry lab. If this tank springs a small hole in the side at its base, how far from the foot of the table will the acid hit the floor if the acid in the tank is 75 \(\mathrm{cm}\) deep?

How big is a million dollars? At the time this problem was written, the price of gold was about \(\$ 1239\) per ounce, while that of platinum was about \(\$ 1508\) an ounce. The "ounce" in this case is the troy ounce, which is equal to 31.1035 g. The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 \(\mathrm{g} / \mathrm{cm}^{3}\) and that of platinum is 21.4 \(\mathrm{g} / \mathrm{cm}^{3} .\) (a) If you find a spherical gold nugget worth 1.00 million dollars, what would be its diameter? (b) How much would a platinum nugget of this size be worth?

Compressible fluids. Throughout this chapter, we have dealt only with incompressible fluids. But under very high pressure, fluids do, in fact, compress. (a) Show that the continuity condition for compressible fluids is \(\rho_{1} A_{1} v_{1}=\rho_{2} A_{2} v_{2}\) where \(\rho\) is the density of the fluid. (b) Show that your result reduces to the familiar result for incompressible fluids.

A barrel contains a 0.120 \(\mathrm{m}\) layer of oil of density 600 \(\mathrm{kg} / \mathrm{m}^{3}\) floating on water that is 0.250 \(\mathrm{m}\) deep. (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?

The piston of a hydraulic automobile lift is 0.30 \(\mathrm{m}\) in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 \(\mathrm{kg}\) ? Now express this pressure in atmospheres.

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