/*! 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 4 When a submarine dives to a dept... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 4

When a submarine dives to a depth of \(120 \mathrm{~m}\), to how large a total pressure is its exterior surface subjected? The density of seawater is about \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\). $$ \begin{aligned} P &=\text { Atmospheric pressure }+\text { Pressure of water } \\ &=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\rho g h=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\left(1030 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(120 \mathrm{~m}) \\ &=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+12.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=13.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.31 \mathrm{MPa} \end{aligned} $$

Short Answer

Expert verified
The total pressure is 1.31 MPa.

Step by step solution

01

Understanding the Problem

We need to calculate the total pressure exerted on the exterior surface of a submarine at a depth of 120 meters in seawater. This total pressure is the sum of atmospheric pressure and the pressure due to the column of water above the submarine.
02

Identify Known Values

The known values are: atmospheric pressure, which is \(1.01 \times 10^5 \, \text{N/m}^2\), seawater density \(\rho = 1.03 \, \text{g/cm}^3 = 1030 \, \text{kg/m}^3\), gravitational acceleration \(g = 9.81 \, \text{m/s}^2\), and depth \(h = 120 \, \text{m}\).
03

Calculate the Pressure Due to Water

Use the formula for the pressure due to water, \(P_{\text{water}} = \rho \cdot g \cdot h\). Substitute the values: \(1030 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 120 \, \text{m}\).
04

Perform the Calculation

Calculate the water pressure: \(1030 \times 9.81 \times 120 = 12.1 \times 10^5 \, \text{N/m}^2\).
05

Calculate Total Pressure

Add the atmospheric pressure to the water pressure: \(1.01 \times 10^5 \, \text{N/m}^2 + 12.1 \times 10^5 \, \text{N/m}^2\).
06

Determine Final Total Pressure

The sum of the two pressures is \(13.1 \times 10^5 \, \text{N/m}^2\), which can be converted to megaPascals as \(1.31 \, \text{MPa}\).

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the air in the earth's atmosphere. At sea level, this pressure is approximately equal to 101,325 pascals or 1.01 x 10鈦� N/m虏. This pressure decreases with altitude as there is less air above. Atmospheric pressure is crucial in pressure calculations, especially in underwater scenarios like in submarine physics. It serves as the baseline pressure that exists even at the surface level before an object, such as a submarine, dives into the ocean.
To understand why atmospheric pressure is added to water pressure, imagine the air above the surface of the ocean which pushes down on it. When you go underwater, this atmospheric pressure still affects your situation. That's why the total pressure exerted on an object submerged underwater includes both the atmospheric pressure and the pressure from the water column above it.
Hydrostatic Pressure
Hydrostatic pressure is the pressure exerted by a fluid at rest in response to the force of gravity. This pressure increases with depth due to the weight of the fluid above. Underwater, it's the pressure due to the water column that extends from the surface to the depth where the object is located.
The formula for calculating hydrostatic pressure is given by \(P = \rho g h\), where \( \rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the height of the fluid column.
For a submarine diving to a depth of 120 meters, this pressure calculation considers the density of seawater, gravitational force, and depth to determine how much water pressure is applied to the submarine.
Submarine Physics
In submarine physics, understanding pressure is critical for safety and functionality. Submarines are engineered to withstand immense pressures as they dive deep under the ocean. The pressure increases as submarines dive deeper, posing risks if not properly managed.
The outer surface of a submarine is subject to total pressure, containing both atmospheric and hydrostatic pressures. Engineers must design the submarine hull to be strong and resilient enough to resist these forces.
  • Structural integrity: Ensures the hull can resist pressure differentials.
  • Buoyancy: Managed by ballast tanks to maintain desired depth and stability.
  • Pressure-resistant materials: Used to construct the outer shell, safeguarding against pressure.
Density of Seawater
The density of seawater differs slightly from freshwater due to the dissolved salts. This density is typically around 1.03 g/cm鲁 or 1030 kg/m鲁. Density plays an integral role in pressure calculations, particularly in those involving hydrostatic pressure beneath the ocean's surface.
Density affects buoyancy, which is pivotal in submarine design. More dense fluids exert more pressure, which influences how weight and buoyancy are calculated for objects like submarines. Therefore, understanding seawater density enables accurate predictions of buoyant forces and pressures at various ocean depths.
Additionally, the density can vary with temperature, salinity, and depth due to thermal and solute dynamics, requiring precise measurements for applications in submarine physics.

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Most popular questions from this chapter

A tank containing oil of sp \(\mathrm{gr}=0.80\) rests on a scale and weighs \(78.6 \mathrm{~N}\). By means of a very fine wire, a \(6.0 \mathrm{~cm}\) cube of aluminum, sp \(\mathrm{gr}=2.70\), is submerged in the oil. Find \((a)\) the tension in the wire and \((b)\) the scale reading if none of the oil overflows.

In a hydraulic press such as the one shown in Fig. \(13-3\), the large piston has cross-sectional area \(A_{1}=200 \mathrm{~cm}^{2}\) and the small piston has cross-sectional area \(A_{2}=5.0 \mathrm{~cm}^{2} .\) If a force of \(250 \mathrm{~N}\) is applied to the small piston, find the force \(F_{1}\) on the large piston. By Pascal's principle, Pressure under large piston \(=\) Pressure under small piston \(\quad\) or \(\quad \frac{F_{1}}{A_{1}}=\frac{F_{2}}{A_{2}}\) so that $$F_{1}=\frac{A_{1}}{A_{2}} F_{2}=\frac{200}{5.0} 250 \mathrm{~N}=10 \mathrm{kN}$$ Note that atmospheric pressure acting on both pistons cancels out of the calculation.

A piece of pure gold \(\left(\rho=19.3 \mathrm{~g} / \mathrm{cm}^{3}\right)\) is suspected to have a hollow center. It has a mass of \(38.25 \mathrm{~g}\) when measured in air and \(36.22 \mathrm{~g}\) in water. What is the volume of the central hole in the gold? Remember that you go from a density in \(\mathrm{g} / \mathrm{cm}^{3}\) to \(\mathrm{kg} / \mathrm{m}^{3}\) by multiplying by 1000 . From \(\rho=m / V\), Actual volume of \(38.25 \mathrm{~g}\) of gold \(=\frac{0.03825 \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}}=1.982 \times 10^{-6} \mathrm{~m}^{3}\) Volume of displaced water \(=\frac{(38.25-36.22) \times 10^{-3} \mathrm{~kg}}{1000 \mathrm{~kg} / \mathrm{m}^{3}}=2.030 \times 10^{-6} \mathrm{~m}^{3}\) $$ \text { Volume of hole }=(2.030-1.982) \mathrm{cm}^{3}=0.048 \mathrm{~cm}^{3} $$

A spring whose composition is not completely known might be either bronze (sp gr 8.8) or brass (sp gr 8.4). It has a mass of \(1.26 \mathrm{~g}\) when measured in air and \(1.11 \mathrm{~g}\) in water. Which is it made of?

At a height of \(10 \mathrm{~km}\) ( \(33000 \mathrm{ft}\) ) above sea level, atmospheric pressure is about \(210 \mathrm{~mm}\) of mercury. What is the net resultant normal force on a \(600 \mathrm{~cm}^{2}\) window of an airplane flying at this height? Assume the pressure inside the plane is \(760 \mathrm{~mm}\) of mercury. The density of mercury is \(13600 \mathrm{~kg} / \mathrm{m}^{3}\).
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