/*! 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 25 The human lungs can function sat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The human lungs can function satisfactorily up to a limit where the pressure differ- ence between the outside and inside of the lungs is one-twentieth of an atmosphere. If a diver uses a snorkel for breathing, how far below the water can she swim? Assume the diver is in salt water whose density is 1025 \(\mathrm{kg} / \mathrm{m}^{3}\)

Short Answer

Expert verified
The diver can swim approximately 0.505 meters below the water using a snorkel.

Step by step solution

01

Understanding the Pressure Difference

The pressure difference that the lungs can withstand is one-twentieth of an atmosphere. An atmospheric pressure is approximately 101325 Pa (Pascals). Therefore, the pressure difference, \( \Delta P \), is \( \frac{101325}{20} \).
02

Calculate the Pressure Difference

\[ \Delta P = \frac{101325}{20} = 5066.25 \text{ Pa} \]
03

Relate to Hydrostatic Pressure

The hydrostatic pressure experienced by the diver depends on the depth \( h \) of the water, given by the equation \( P = \rho g h \), where \( \rho \) is the density of the water, \( g \) is the acceleration due to gravity (approximately 9.8 m/s²), and \( h \) is the depth in meters.
04

Set Hydrostatic Pressure Equal to Lungs Tolerance

Since \( P = \Delta P \) at the maximum depth the diver can endure, we set \( \rho g h = 5066.25 \). Substitute \( \rho = 1025 \text{ kg/m}^3 \) and \( g = 9.8 \text{ m/s}^2 \) into the equation.
05

Solve for Depth \( h \)

\[ h = \frac{5066.25}{1025 \times 9.8} \]
06

Perform the Calculation

Calculate \( h \) using the equation: \[ h = \frac{5066.25}{1025 \times 9.8} = 0.505 \text{ meters} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Lung Pressure Tolerance
Our lungs, incredible as they are, have their limits when it comes to handling different pressures. Specifically, the human lung can tolerate a pressure difference of up to one-twentieth of an atmosphere. This might sound technical, but it boils down to the maximum pressure your lungs can safely handle while snorkeling. An atmosphere, when measured in Pascals — the unit for pressure — is roughly 101325 Pa. But what exactly does one-twentieth mean? Simply put, it's dividing that total pressure by twenty, resulting in 5066.25 Pa. This value signifies the maximum pressure difference your lungs can safely withstand while snorkeling. Understanding this limit is fundamental to ensuring safety when exploring underwater worlds.
Density of Salt Water
In order to calculate how deep one can snorkel safely, understanding the properties of saltwater is crucial. One such property is the density of the water, a measure of how much mass is contained within a given volume. Saltwater is a bit denser than freshwater due to the salt content. Specifically, the density of saltwater is about 1025 kg/m³. But why does this matter for snorkeling?
  • This density impacts the pressure experienced by a diver underwater.
  • Denser water means more mass is pressing down as you dive deeper, which results in increased pressure.
Knowing the water's density allows us to calculate how far one can dive without exceeding the lung pressure tolerance. This simple fact of denser water resulting in higher pressure is a crucial element in safely determining the maximum snorkeling depth.
Snorkeling Depth Calculation
Let's delve into the math needed to find out how deep you can snorkel while staying safe. The pressure you experience under water increases the deeper you go. This increase is known as hydrostatic pressure, calculated using the formula:\[ P = \rho g h \]where:
  • \( P \) is the pressure at a certain depth.
  • \( \rho \) is the density of the water (for saltwater, 1025 kg/m³).
  • \( g \) is the acceleration due to gravity (approximately 9.8 m/s²).
  • \( h \) is the depth of the water in meters.
When snorkeling, you need to ensure that this pressure \( P \) does not exceed the lung pressure tolerance of 5066.25 Pa. Setting \( \rho g h = 5066.25 \) Pa, and plugging the known values into this equation, you get:\[ h = \frac{5066.25}{1025 \times 9.8} \approx 0.505 \text{ meters} \]This means the maximum safe snorkeling depth in saltwater, given the lung pressure tolerance, is about 0.505 meters. Remember, this is a theoretical limit; in practice, always prioritize safety and personal comfort.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A spring is attached to the bottom of an empty swimming pool, with the axis of the spring oriented vertically. An 8.00-kg block of wood \(\left(\rho=840 \mathrm{kg} / \mathrm{m}^{3}\right)\) is fixed to the top of the spring and compresses it. Then the pool is filled with water, completely covering the block. The spring is now observed to be stretched twice as much as it had been compressed. Determine the percentage of the block's total volume that is hollow. Ignore any air in the hollow space.

A \(0.10-\mathrm{m} \times 0.20-\mathrm{m} \times 0.30-\mathrm{m}\) block is suspended from a wire and is completely under water. What buoyant force acts on the block?

The drawing shows a hydraulic system used with disc brakes. The force \(\overrightarrow{\mathbf{F}}\) is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius \(=9.50 \times 10^{-3} \mathrm{m} )\) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii \(=1.90 \times 10^{-2} \mathrm{m}\) ), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of \(\overrightarrow{\mathbf{F}}\) is 9.00 \(\mathrm{N}\) . Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc.

Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.020 m. Water enters the hydrant through an underground pipe of radius 0.080 m. In this pipe the water has a speed of 3.0 m/s. (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

A \(58-\mathrm{kg}\) skier is going down a slope oriented \(35^{\circ}\) above the horizontal. The area of each ski in contact with the snow is 0.13 \(\mathrm{m}^{2}\) Determine the pressure that each ski exerts on the snow.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.