/*! 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 21 The Mariana trench is located in... [FREE SOLUTION] | 91Ó°ÊÓ

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The Mariana trench is located in the floor of the Pacific Ocean at a depth of about 11000 \(\mathrm{m}\) below the surface of the water. The density of seawater is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius \(=0.10 \mathrm{~m}\) )? (b) For comparison, determine the weight of a jetliner whose mass is \(1.2 \times 10^{5} \mathrm{~kg}\).

Short Answer

Expert verified
The force on the window is approximately \(3.476 \times 10^6 \, \text{N}\). The weight of the jetliner is approximately \(1.1772 \times 10^6 \, \text{N}\).

Step by step solution

01

Calculate the Pressure at Depth

The pressure exerted by the water at a depth is given by the hydrostatic pressure formula, which is \( P = \rho g h \). Here, \( \rho = 1025 \, \text{kg/m}^3 \) is the density of seawater, \( g = 9.81 \, \text{m/s}^2 \) is the acceleration due to gravity, and \( h = 11000 \, \text{m} \) is the depth. Calculating the pressure:\[ P = 1025 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 11000 \, \text{m} = 1.106925 \times 10^8 \, \text{Pa} \]
02

Determine the Force Exerted on the Observation Window

The force exerted by the water is given by \( F = P \times A \), where \( A \) is the area of the window. The area of a circle is \( A = \pi r^2 \), where \( r = 0.10 \, \text{m} \) is the radius of the window. Thus:\[ A = \pi \times (0.10 \, \text{m})^2 = 0.0314 \, \text{m}^2 \]The force is then:\[ F = 1.106925 \times 10^8 \, \text{Pa} \times 0.0314 \, \text{m}^2 = 3.476 \times 10^6 \, \text{N} \]
03

Calculate the Weight of the Jetliner

The weight of an object is calculated by \( W = m \times g \), where \( m = 1.2 \times 10^5 \, \text{kg} \) is the mass of the jetliner. Calculating the weight:\[ W = 1.2 \times 10^5 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 1.1772 \times 10^6 \, \text{N} \]

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Key Concepts

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

Underwater Vehicle
Underwater vehicles are specially designed vessels that allow us to explore the depths of our oceans safely. These vehicles, often referred to as submersibles or submarines, are built to withstand the intense pressures found in the deep sea. In areas like the Mariana Trench, pressure increases with depth, posing significant challenges for the construction and operation of underwater vehicles.
These vehicles are equipped with robust hulls, observation windows, and instrumentation that must resist the force exerted by the surrounding water. The ability to operate at such depths opens up areas of scientific study, resource exploration, and even bioprospecting for unique marine organisms that thrive in extreme environments.
It's essential for these underwater vehicles to have features like ballast systems for buoyancy control and thrusters for maneuverability. Designers must ensure that all materials and structures are capable of handling the severe conditions without compromising the safety of the mission and equipment. The exploration at such depth provides invaluable data on oceanic conditions, marine life, and geological formations.
Force Calculation
Calculating the force exerted by water on an underwater vehicle is crucial for its design and safety. At extreme depths, like the Mariana Trench, the force due to water pressure can be immense, particularly against flat surfaces such as observation windows.
To find the force, we use the formula:
  • The force, \( F \), equals the pressure \( P \) multiplied by the area \( A \)
  • .
  • In mathematical terms, \( F = P imes A \).
For curved surfaces, the calculation becomes more complex due to differences in force applications, but for flat surfaces, the formula simplifies the understanding of the forces at work.
By ensuring these force calculations are precise, engineers can design observation windows and other critical components that survive the harsh underwater environment, securing both the vehicle's integrity and the safety of any researchers or equipment inside.
Pressure Formula
The pressure experienced by an object submerged underwater increases with depth. This increase is captured by the hydrostatic pressure formula, which states:
  • Pressure \( P \) is the product of the density \( \rho \) of the fluid, gravitational acceleration \( g \), and the depth \( h \).
In formulaic terms, this is expressed as \( P = \rho g h \).
For the Mariana Trench, given seawater density \( \rho = 1025 \, \text{kg/m}^3 \), gravitational acceleration \( g = 9.81 \, \text{m/s}^2 \), and depth \( h = 11000 \, \text{m} \), we can calculate the pressure exerted at this depth as a staggering \( 1.106925 \times 10^8 \, \text{Pa} \) (Pascals).
Understanding how pressure builds with depth and mastery over using the pressure formula is foundational in solving physics problems related to fluid dynamics, especially in underwater exploration, where these massive pressures are the norm.
Physics Problem Solving
Physics problem solving often involves applying basic principles and formulas to real-world challenges, such as calculating forces on underwater vehicles. These problems may appear complex at first glance, but breaking them down into smaller, more manageable steps makes them more approachable.
  • First, identify the known variables and appropriate formulas needed for the problem.
  • Perform calculations step by step, maintaining attention to units and conversion factors.
  • Use logical reasoning to understand the underlying physics principles involved, such as force, pressure, and weight.
Tools like diagrams or sketches can also assist in visualizing the problem, aiding in the solution process.
Science and engineering students often practice these types of problems to develop critical thinking and analytical skills. This methodical approach not only applies to underwater vehicles but is also broadly used across different disciplines of physics and engineering, fostering a deeper understanding of the world.

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Most popular questions from this chapter

When an object moves through a fluid, the fluid exerts a viscous force \(\overrightarrow{\mathrm{F}}\) on the object that tends to slow it down. For a small sphere of radius \(R,\) moving slowly with a speed \(v\), the magnitude of the viscous force is given by Stokes' law, \(F=6 \pi \eta R v,\) where \(\eta\) is the viscosity of the fluid. (a) What is the viscous force on a sphere of radius \(R=5.0 \times 10^{-4} \mathrm{~m}\) falling through water \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) when the sphere has a speed of \(3.0 \mathrm{~m} / \mathrm{s} ?\) (b) The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of \(1.0 \times 10^{-5} \mathrm{~kg},\) what is its terminal speed?

One way to administer an inoculation is with a "gun" that shoots the vaccine through a narrow opening. No needle is necessary, for the vaccine emerges with sufficient speed to pass directly into the tissue beneath the skin. The speed is high, because the vaccine \(\left(\rho=1100 \mathrm{~kg} / \mathrm{m}^{3}\right)\) is held in a reservoir where a high pressure pushes it out. The pressure on the surface of the vaccine in one gun is \(4.1 \times 10^{6} \mathrm{~Pa}\) above the atmospheric pressure outside the narrow opening. The dosage is small enough that the vaccine's surface in the reservoir is nearly stationary during an inoculation. The vertical height between the vaccine's surface in the reservoir and the opening can be ignored. Find the speed at which the vaccine emerges.

A water bed has dimensions of \(1.83 \mathrm{~m} \times 2.13 \mathrm{~m} \times 0.229 \mathrm{~m}\). The floor of the bedroom will tolerate an additional weight of no more than \(6660 \mathrm{~N}\). Find the weight of the water in the bed and determine whether it should be purchased.

A suitcase (mass \(\mathrm{m}=16 \mathrm{~kg}\) ) is resting on the floor of an elevator. The part of the suitcase in contact with the floor measures \(0.50 \mathrm{~m}\) by \(0.15 \mathrm{~m}\). The elevator is moving upward, the magnitude of its acceleration being \(1.5 \mathrm{~m} / \mathrm{s}^{2}\). What pressure (in excess of atmospheric pressure) is applied to the floor beneath the suitcase?

Prairie dogs are burrowing rodents. They do not suffocate in their burrows, because the effect of air speed on pressure creates sufficient air circulation. The animals maintain a difference in the shapes of two entrances to the burrow, and because of this difference, the air \(\left(\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}\right)\) blows past the openings at different speeds, as the drawing indicates. Assuming that the openings are at the same vertical level, find the difference in air pressure between the openings and indicate which way the air circulates.

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