/*! 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 17 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. E12.17) because as the depth increases, so does the pressure difference, which 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 freshwater. (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 at a depth of 6.1 m is 59780 Pa.

Step by step solution

01

Understand the Problem

The problem involves calculating the pressure difference between the pressure outside of a diver's lungs at a certain depth in water and atmospheric pressure inside the lungs. We need to use the depth of the diver to find this pressure difference.
02

Recall the Pressure Formula

The pressure in a fluid increases with depth according to the formula: \( P = P_0 + \rho gh \), where \( P_0 \) is the atmospheric pressure, \( \rho \) is the density of the fluid (freshwater in this case), \( g \) is the acceleration due to gravity, and \( h \) is the depth.
03

Identify the Values

For freshwater, the density \( \rho \approx 1000 \, \mathrm{kg/m^3} \), \( g = 9.8 \, \mathrm{m/s^2} \), and the depth \( h = 6.1 \, \mathrm{m} \). Atmospheric pressure \( P_0 \approx 101325 \, \mathrm{Pa} \).
04

Calculate Pressure at Depth

The pressure at depth due to the water column is \( P_w = \rho gh = 1000 \, \mathrm{kg/m^3} \times 9.8 \, \mathrm{m/s^2} \times 6.1 \, \mathrm{m} \). Simplifying gives us \( P_w = 59780 \, \mathrm{Pa} \).
05

Determine the Pressure Difference

The external-internal pressure difference is the pressure at depth minus atmospheric pressure (internal pressure remains atmospheric). Thus, the pressure difference \( \Delta P = P_w - P_0 = 59780 \, \mathrm{Pa} \).

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Key Concepts

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

Snorkeling
Snorkeling is an enjoyable water activity where a person swims on or through the water wearing a diving mask and breathing through a snorkel tube. This allows for underwater observation for extended periods while remaining near the water's surface.
The snorkel itself is a simple piece of equipment—a tube that connects the diver’s mouth to the air above water.
  • The primary function is to enable breathing while the diver’s face is submerged.
  • It allows for easy exploration of underwater life.
However, there are limitations to how deep one can snorkel, largely due to pressure differences as one goes deeper into the water. This is a crucial factor as the human body can only withstand so much pressure exerted by water, which increases with depth.
Fluid Pressure Calculation
Calculating fluid pressure is key in understanding how pressures change in a submerged environment. The fundamental formula used is \( P = P_0 + \rho gh \). This formula helps us calculate the pressure exerted by a fluid at any given depth.

Let's break it down:
  • \( P_0 \): Atmospheric pressure at the surface.
  • \( \rho \): Density of the fluid—in this case, freshwater, approximately 1000 \( \mathrm{kg/m^3} \).
  • \( g \): Acceleration due to gravity, which is about \( 9.8 \, \mathrm{m/s^2} \).
  • \( h \): Depth of the fluid, or how far below the surface the object is—in this example, \( 6.1 \, \mathrm{m} \).
Using this formula provides the pressure due to the water column at a specific depth. Understanding how to apply this equation is essential for various applications in diving and marine activities.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the air in the atmosphere. It acts uniformly in all directions and decreases as we move to higher altitudes. At sea level, it is approximately \( 101325 \, \mathrm{Pa} \) (Pascal).

In the context of diving, atmospheric pressure is the baseline.
  • When a diver is at the surface, their lungs experience this pressure only.
  • Once submerged, the diver's lungs still have air at atmospheric pressure due to the air in the lungs or snorkel tube being from above the surface.
Understanding atmospheric pressure is vital since it is a constant factor that does not change much with minor height variations above the sea level. It is a reference point when considering pressure differences underwater.
Diving Physics
Diving physics revolves around the principles of pressure, buoyancy, and fluid dynamics as they apply to underwater activities.

When diving, the pressure a diver feels underwater is the combination of atmospheric pressure and the pressure from the water itself. The deeper you dive, the more pressure is exerted on your body due to the increased weight of the water column above.
  • This increased pressure affects your lungs, which naturally want to equalize pressure with the water outside.
  • For snorkelers, the pressure difference is more pronounced because they rely solely on atmospheric pressure air.
  • Scuba divers use compressed air tanks to breathe under higher pressure conditions, allowing them to dive deeper without pressure-related issues.
Understanding diving physics ensures safety and comfort in water activities by respecting the limits of natural pressure resistance the human body can safely handle.

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Most popular questions from this chapter

When an open-faced boat has a mass of 5750 kg, including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that 20\(\%\) of the boat's volume will be above water. How much mass should he throw out?

Dropping Anchor. An iron anchor with mass 35.0 \(\mathrm{kg}\) and density 7860 \(\mathrm{kg} / \mathrm{m}^{3}\) lies on the deck of a small barge that has vertical sides and floats in a freshwater river. The area of the bottom of the barge is 8.00 \(\mathrm{m}^{2} .\) The anchor is thrown overboard but is suspended above the bottom of the river by a rope; the mass and volume of the rope are small enough to ignore. After the anchor is overboard and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water? By what vertical distance?

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 \(\mathrm{m}\) . (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth. \((\mathrm{b})\) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 \(\mathrm{cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You can ignore the small variation of pressure over the surface of the window.)

Water stands at a depth \(H\) in a large, open tank whose side walls are vertical (Fig. \(\mathrm{Pl} 2.89 ) .\) A hole is made in one of the walls at a depth \(h\) below the water surface. (a) At what distance \(R\) from the foot of the wall does the emerging stream strike the floor? (b) How far above the bottom of the tank could a second hole be cut so that the stream emerging from it could have the same range as for the first hole?

In a lecture demonstration, a professor pulls apart two hemispherical steel shells (diameter \(D )\) with ease using their attached handles. She then places them together, pumps out the air to an absolute pressure of \(p,\) and hands them to a bodybuilder in the back row to pull apart. (a) If atmospheric pressure is \(p_{0},\) how much force must the bodybuilder exert on each shell? (b) Evaluate your answer for the case \(p=0.025\) atm, \(D=10.0 \mathrm{cm} .\)

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