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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
(a) Force on window: \(3.49 \times 10^6 \, \mathrm{N}\). (b) Jetliner's weight: \(1.18 \times 10^6 \, \mathrm{N}\).

Step by step solution

01

Calculate the Pressure at the Depth of the Trench

To find the pressure exerted by the water at a depth of 11000 m, we use the formula for pressure due to the liquid column: \( P = \rho \cdot g \cdot h \), where \( \rho = 1025 \, \mathrm{kg/m^3} \) (density of seawater), \( g = 9.81 \, \mathrm{m/s^2} \) (acceleration due to gravity), and \( h = 11000 \, \mathrm{m} \) (depth). Thus, the pressure is \( P = 1025 \times 9.81 \times 11000 = 1.11 \times 10^8 \, \mathrm{Pa} \) (Pascals).
02

Calculate the Area of the Window

The window of the underwater vehicle is circular, and its radius is given. The area \( A \) of the circular window can be calculated using the formula for the area of a circle: \( A = \pi r^2 \), where \( r = 0.10 \, \mathrm{m} \). So, the area is \( A = \pi \times (0.10)^2 = 0.0314 \, \mathrm{m^2} \).
03

Calculate the Force on the Window by the Water

The force exerted by the water on the window is calculated using the relationship between force, pressure, and area: \( F = P \cdot A \). Using the pressure calculated in Step 1 and the area from Step 2, we have \( F = 1.11 \times 10^8 \, \mathrm{Pa} \times 0.0314 \, \mathrm{m^2} = 3.49 \times 10^6 \, \mathrm{N} \).
04

Calculate the Weight of the Jetliner

The weight of the jetliner is calculated using the formula \( W = m \cdot g \), where the mass \( m = 1.2 \times 10^5 \, \mathrm{kg} \) and \( g = 9.81 \, \mathrm{m/s^2} \). Thus, the weight is \( W = 1.2 \times 10^5 \times 9.81 = 1.18 \times 10^6 \, \mathrm{N} \).

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Key Concepts

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

Seawater Density
Seawater density is a crucial factor to understand when studying fluid pressure at great ocean depths. The density of seawater is typically about 1025 kg/m³. This value can vary slightly depending on factors like temperature and salinity, but it remains roughly in this range.

Density is mass per unit volume, which means in a cubic meter of seawater, there are approximately 1025 kilograms of mass. This density contributes significantly to the pressure exerted by seawater at any depth, including the deepest parts of the ocean. The deeper you go, the more seawater there is above, hence greater weight and pressure.
  • Denser fluids apply more pressure at a given depth.
  • The consistent density allows calculations of pressure at various depths.
  • It helps predict the force exerted by seawater on submerged objects.
Understanding seawater density simplifies calculations for applications like marine engineering and exploration.
Force Calculation
Calculating the force that seawater exerts on objects is essential, especially when designing vehicles to explore deep-sea environments. The fundamental formula linking force, pressure, and area is vital here: \[ F = P \times A \]where \( F \) is the force, \( P \) is the pressure, and \( A \) is the area.

To determine the force on a specific part, like the window of an underwater vehicle, you calculate the pressure first using the density (\( \rho \)) and depth (\( h \)): \[ P = \rho \times g \times h \]where \( g \) is the gravitational acceleration, approximately 9.81 m/s².
  • Once pressure is known, calculate the area of the window or object surface.
  • Multiply this area by the calculated pressure to find the force.
  • The "force" helps evaluate the structural integrity needed for deep-sea equipment.
This method ensures that vehicles are safe and can withstand extreme pressures.
Weight Comparison
Weight comparison is used to understand the scale of forces involved when analyzing the effects of ocean pressure. For example, comparing the force exerted on an underwater vehicle's window with the weight of a common object like a jetliner can provide perspective.

The weight of an object is the force due to gravity acting on its mass, usually computed as: \[ W = m \times g \]where \( m \) is the mass of the object and \( g \) is gravitational acceleration.
  • In our example, a jetliner with a large mass has significant weight.
  • Comparing this weight with forces from ocean pressure helps grasp the magnitude.
  • It highlights the engineering challenges in both aerospace and marine design.
Such comparisons are essential for appreciating the massive scales and forces in physics.
Depth Pressure
Depth pressure is the pressure exerted by a fluid due to its weight at a certain depth. In the context of seawater, this pressure increases linearly with depth.

For a depth like the Mariana Trench at 11000 meters, pressure can be calculated using the formula:\[ P = \rho \cdot g \cdot h \]Here, \( \rho \) is the density of seawater, \( g \) is gravity, and \( h \) is the depth. The pressure at such a depth would be extremely high, creating intense forces.
  • Each meter of seawater adds approximately 1025 kg/m² of pressure.
  • Understanding this pressure is crucial for designing submersibles.
  • It ensures that equipment can perform at great depths safely.
By comprehensively analyzing depth pressure, we can better prepare for exploring and working in the world’s oceans.

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Most popular questions from this chapter

An antifreeze solution is made by mixing ethylene glycol \(\left(\rho=1116 \mathrm{kg} \mathrm{m}^{3}\right)\) with water. Suppose that the specific gravity of such a solution is \(1.0730 .\) Assuming that the total volume of the solution is the sum of its parts, determine the volume percentage of ethylene glycol in the solution.

Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds number Re is less than about \(2000: \mathrm{Re}=2 \bar{v} \rho R / \eta\) Here \(\bar{v}, \rho,\) and \(\eta\) are, respectively, the average speed, density, and viscosity of the fluid, and \(R\) is the radius of the pipe. Calculate the highest average speed that blood \(\left(\rho=1060 \mathrm{kg} / \mathrm{m}^{3}, \eta=4.0 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s}\right)\) could have and still remain in laminar flow when it flows through the aorta \(\left(R=8.0 \times 10^{-3} \mathrm{m}\right)\).

A pipe is horizontal and carries oil that has a viscosity of \(0.14\) \(\mathrm{Pa} \cdot \mathrm{s}\) The volume flow rate of the oil is \(5.3 \times 10^{-5} \mathrm{m}^{3} / \mathrm{s} .\) The length of the pipe is \(37\) \(\mathrm{m},\) and its radius is \(0.60 \mathrm{cm} .\) At the output end of the pipe the pressure is atmospheric pressure. What is the absolute pressure at the input end?

A Venturi meter is a device that is used for measuring the speed of a fluid within a pipe. The drawing shows a gas flowing at speed \(v_{2}\) through a horizontal section of pipe whose cross-sectional area is \(A_{2}=\) \(0.0700 \mathrm{m}^{2} .\) The gas has a density of \(\rho=1.30 \mathrm{kg} / \mathrm{m}^{3} .\) The Venturi meter has a cross-sectional area of \(A_{1}=0.0500 \mathrm{m}^{2}\) and has been substituted for a section of the larger pipe. The pressure difference between the two sections is \(P_{2}-P_{1}=120\) Pa. Find (a) the speed \(v_{2}\) of the gas in the larger, original pipe and (b) the volume flow rate \(Q\) of the gas.

The construction of a flat rectangular roof \((5.0 \mathrm{m} \times 6.3 \mathrm{m})\) allows it to withstand a maximum net outward force that is \(22000\) \(\mathrm{N}\). The density of the air is \(1.29\) \(\mathrm{kg} / \mathrm{m}^{3} .\) At what wind speed will this roof blow outward?

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