/*! 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] | 91Ó°ÊÓ

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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 \(1.03 \mathrm{~g} / \mathrm{cm}^{3}\) \(P=\) Atmospheric pressure \(+\) Pressure of wate $$ \begin{array}{l} =1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\rho g h=1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+\left(1030 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(120 \mathrm{~m}) \\ =1.01 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}+12.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=13.1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.31 \mathrm{MPa} \end{array} $$

Short Answer

Expert verified
The total pressure is 1.31 MPa.

Step by step solution

01

Understanding Required Pressure Components

The total pressure on the submarine is the sum of atmospheric pressure and the pressure due to the water column. We denote total pressure as \( P \): \( P = \text{Atmospheric Pressure} + \text{Pressure from Water} \).
02

Calculate Atmospheric Pressure

The atmospheric pressure, which is the pressure exerted by the weight of air, is given as \( 1.01 \times 10^5 \; \text{N/m}^2 \).
03

Calculate Pressure Exerted by Water

The pressure exerted by the water column is calculated using the formula \( \rho g h \), where \( \rho \) is the density of seawater \( (1030 \; \text{kg/m}^3) \), \( g \) is acceleration due to gravity \( (9.81 \; \text{m/s}^2) \), and \( h \) is the depth \( (120 \; \text{m}) \). Substitute the values into the formula: \( \rho g h = 1030 \times 9.81 \times 120 = 12.1 \times 10^5 \; \text{N/m}^2 \).
04

Add the Pressures Together

Add the atmospheric pressure and the water pressure to find the total pressure: \( 1.01 \times 10^5 + 12.1 \times 10^5 = 13.1 \times 10^5 \; \text{N/m}^2 \).
05

Convert to Megapascals

Since the answer might also be required in megapascals (MPa), convert \( 13.1 \times 10^5 \; \text{N/m}^2 \) to MPa by dividing by \( 10^6 \), resulting in \( 1.31 \; \text{MPa} \).

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Key Concepts

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

Atmospheric Pressure
When we think about pressure, atmospheric pressure is the force exerted by air molecules acting on every surface. Atmospheric pressure is essential for understanding many physical concepts, and it forms a part of any total pressure calculation.For most calculations, the standard atmospheric pressure is approximately \(1.01 \times 10^{5}\, \text{N/m}^2\) (or Pascals). This is roughly the average pressure at sea level, caused by Earth's atmosphere.It acts equally in all directions and decreases with height above sea level.Understanding this is crucial in scenarios where objects are submerged, like in the case of the submarine exercise we are examining.Atmospheric pressure is only one part of the total pressure a submerged object experiences, with the rest coming from the liquid surrounding it.
Pressure from Water
The second major component in our total pressure calculation is the pressure from water.Pressure increases with water depth due to the weight of water above pushing down with force.This pressure is calculated using Hydrostatic Pressure Formula: \( \rho g h \).
  • \( \rho \) is the density of the fluid (for seawater, \(1030\, \text{kg/m}^3\)).
  • \( g \) is the acceleration due to gravity (\(9.81\, \text{m/s}^2\)).
  • \( h \) is the depth of the object in the fluid (\(120\, \text{m}\) for our submarine).
Knowing how to calculate depth-related pressure helps in assessing stresses on structures like submarines and oceans' creatures that withstand high pressures deep underwater.
Density of Seawater
The density of seawater is a key factor when calculating the pressure from water. Density (\(\rho\)) is the mass per unit volume of a substance, measured as \( \text{kg/m}^3 \) in the SI unit system.In this exercise, we have a seawater density of \(1030\, \text{kg/m}^3\).It's vital to use the correct density value, as errors could significantly affect pressure calculations.Seawater density can vary slightly with location and depth due to temperature and salinity differences, but we generally use a standard value for these calculations.
Acceleration due to Gravity
Gravity plays a critical role in various physical phenomena, including how pressure increases with depth. Acceleration due to gravity, denoted as \( g \), is approximately \(9.81\, \text{m/s}^2\) at the Earth's surface.This constant value describes the gravitational pull that Earth exerts on objects, causing them to accelerate towards its center.In the context of pressure calculations, gravity provides the force that causes water pressure to increase with depth.By multiplying gravity with the fluid's density and depth, we find the pressure that water exerts due to its weight, a necessary step for calculating total pressure on submerged objects.

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Most popular questions from this chapter

A glass tube is bent into the form of a U. A \(50.0-\mathrm{cm}\) height of olive oil in one arm is found to balance \(46.0 \mathrm{~cm}\) of water in the other. What is the density of the olive oil?

A tank containing oil of sp 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 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.

A vertical test tube has \(2.0 \mathrm{~cm}\) of oil \(\left(\rho=0.80 \mathrm{~g} / \mathrm{cm}^{3}\right)\) floating on \(8.0 \mathrm{~cm}\) of water. What is the pressure at the bottom of the tube due to the liquid in it? $$ \begin{aligned} P &=\rho_{1} h h_{1}+\rho_{2} h_{2}=\left(800 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2} \times 0.020 \mathrm{~m}\right)+\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.080 \mathrm{~m}) \\ &=0.94 \mathrm{kPa} \end{aligned} $$

A wooden cylinder has a mass \(m\) and a base area \(A\). It floats in water with its axis vertical. Show that the cylinder undergoes SHM if given a small vertical displacement. Find the frequency of its motion. When the cylinder is pushed down a distance \(y\), it displaces an additional volume Ay of water. Because this additional displaced volume has mass \(A y_{\rho w}\), an additional buoyant force \(A y_{\rho w g}\) acts on the cylinder, where \(\rho_{w}\) is the density of water. This is an unbalanced force on the cylinder and is a restoring force. In addition, the force is proportional to the displacement and so is a Hooke's Law force. Therefore, the cylinder will undergo SHM, as described in Chapter 11 . Comparing \(F_{B}=A \rho_{w} g y\) with Hooke's Law in the form \(F=k y\), we see that the elastic constant for the motion is \(k=A \rho_{w} g .\) This, acting on the cylinder of mass \(m\), causes it to have a vibrational frequency of $$ f=\frac{1}{2 \pi} \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{A \rho_{w} g}{m}} $$

A 2.0-cm cube of metal is suspended by a fine thread attached to a scale. The cube appears to have a mass of \(47.3 \mathrm{~g}\) when measured submerged in water. What will its mass appear to be when submerged in glycerin, sp gr = 1.26? [Hint: Find \(\rho\) too.]

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