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Chapter 14: Problem 8

Black Smokers. Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 \(\mathrm{m}\) to 3200 \(\mathrm{m}\) below the surface. What is the gauge pressure at a \(3200-\mathrm{m}\) deep vent, assuming the density of water does not vary? Express your answer in pascals and atmospheres.

Short Answer

Expert verified
The gauge pressure is approximately 32,144,800 Pascals or 317.31 atmospheres.

Step by step solution

01

Understanding Gauge Pressure

The gauge pressure at a certain depth in a fluid is given by the formula \( P = \rho g h \), where \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth.
02

Identify Known Values

For seawater, the density \( \rho \) is approximately \( 1025 \ \text{kg/m}^3 \). The acceleration due to gravity \( g \) is \( 9.81 \ \text{m/s}^2 \). The depth \( h \) is given as \( 3200 \ \text{m} \).
03

Calculate Gauge Pressure in Pascals

Substitute the known values into the equation: \[ P = \rho g h = 1025 \times 9.81 \times 3200 \]. Calculating this gives \( P \approx 32,144,800 \ \text{Pa} \).
04

Convert Pascals to Atmospheres

1 atmosphere is equivalent to \( 101325 \ \text{Pa} \). Therefore, the pressure in atmospheres is given by \( \frac{32,144,800}{101325} \approx 317.31 \ \text{atm} \).

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Key Concepts

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

Fluid Mechanics
Fluid mechanics is a branch of physics dealing with the behavior of fluids, both liquids and gases, and the forces acting on them. When we discuss fluids under pressure, concepts like pressure, buoyancy, and flow become crucial. For instance, gauge pressure at a depth in a fluid is essential in determining how much pressure is exerted by a fluid column. To calculate this, the formula used is
  • \( P = \rho g h \)
where \( P \) is the pressure, \( \rho \) is the density, \( g \) is the acceleration due to gravity, and \( h \) is the height or depth. In fluid mechanics, understanding how these variables interact helps predict fluid behavior under various conditions.
Fluid mechanics plays an essential role in understanding how the ocean environment works, from predicting ocean currents to exploring deep-sea vents like black smokers. Mastering the concepts in fluid mechanics paves the way for better predicting and reacting to phenomena related to fluids.
Seawater Density
Seawater density is a critical concept when calculating pressure at ocean depths. It's slightly higher than that of freshwater due to dissolved salts, which increase mass per unit volume. This higher density plays a crucial role in calculating gauge pressure. For seawater, average density is approximately \( 1025 \ \text{kg/m}^3 \). This value varies slightly depending on temperature and salinity, but is often used for simplifying calculations.
Understanding these nuances can help comprehend how small changes in seawater composition might affect the pressure calculations and thereby influence engineering tasks, such as designing submarines or structures on the seabed. Knowing how density impacts fluid pressure is essential not just in physics but in real-world applications like oceanography and marine biology.
Physics Education
Physics education aims to convey complex concepts like gauge pressure, density, and forces in fluids in an accessible manner. Effective teaching involves breaking down these concepts into manageable parts and providing students with tools to apply them in practical scenarios, such as understanding ocean depth pressures. By mastering these physics topics, students are empowered to analyze real-world problems, making calculations like determining pressure at ocean depths more intuitive.
Hands-on activities, thought-provoking problems, and real-world applications can significantly boost comprehension. For example, using exercises like calculating gauge pressure at different ocean depths helps students appreciate the application of theoretical knowledge to practical situations, enhancing both understanding and interest in physics.
Ocean Depth Pressure
Ocean depth pressure is a fascinating aspect of fluid mechanics, as it varies greatly with depth. With each meter you descend into the ocean, the pressure increases, largely due to the weight of the water above. This concept is crucial for any study related to marine environments. ***Black smokers,*** volcanic vents on the ocean floor, are located at very high pressures due to their depth, about 3200 meters below the surface.
The gauge pressure can be calculated using the formula previously mentioned, which reveals just how intense the conditions are at these depths. As we descend, higher pressures can influence everything from the type of organisms that thrive to the technology used to explore these regions. Understanding ocean depth pressure is a key aspect of oceanography and helps scientists ensure safety and precision in underwater exploration.

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Most popular questions from this chapter

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 \(\mathrm{Pa}\) , what 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 14.6 ) of the fluid.

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 \(d r^{\prime}\) adistance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .(\text { a) Apply }\) 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 \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .(\mathrm{c})\) An object of volume \(V\) and density \(\rho_{o b}\) has its center of mass at a distance \(R_{\text { cmob }}\) 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 \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{cb}} R_{\mathrm{cmod}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}}(\mathrm{e})\) 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?

A swimming pool is 5.0 \(\mathrm{m}\) long, 4.0 \(\mathrm{m}\) wide, and 3.0 \(\mathrm{m}\) deep. Compute the force exerted by the water against (a) the bottom; and (b) either end. (Hint: Calculate the force on a thin, horizontal strip at a depth \(h\) , and integrate this over the end of the pool.) Do not include the force due to air pressure.

A firehose must be able to shoot water to the top of a building 35.0 \(\mathrm{m}\) tall when aimed straight up. Water enters this hose at a steady rate of 0.500 \(\mathrm{m}^{3} / \mathrm{s}\) and shoots out of a round nozzle. (a) What is the maximum diameter this nozzle can have? (b) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

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 overbound and the barge has finally stopped bobbing up and down, has the barge risen or sunk down in the water? By what vertical distance?
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