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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 water exerts a force of \(3.4776 \times 10^6 \text{ N}\) on the window, and the weight of the jetliner is \(1.176 \times 10^6 \text{ N}\).

Step by step solution

01

Understand Pressure Formula

To find the force exerted by water, begin by calculating the pressure on the window. Pressure at a depth is given by the formula: \( P = \rho \cdot g \cdot h \), where \( \rho \) is the density of water, \( g \) is the acceleration due to gravity \((9.8 \text{ m/s}^2)\), and \( h \) is the depth.
02

Calculate Pressure at Depth

Using the pressure formula: \( P = 1025 \cdot 9.8 \cdot 11000 = 1.1075 \times 10^8 \text{ Pa} \). This is the pressure at the depth the vehicle is exploring.
03

Calculate Force on Window

The force exerted by the water on the window is given by \( F = P \cdot A \), where \( A \) is the area of the window. The area is \( A = \pi r^2 = \pi (0.1)^2 = 0.0314 \text{ m}^2 \). Thus, the force is \( F = 1.1075 \times 10^8 \cdot 0.0314 = 3.4776 \times 10^6 \text{ N} \).
04

Determine Weight of Jetliner

The weight of the jetliner is given by \( W = m \cdot g = 1.2 \times 10^5 \cdot 9.8 = 1.176 \times 10^6 \text{ N} \). This is the weight of the jetliner for comparison.

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Key Concepts

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

Pressure Calculation
When dealing with fluids, it's important to understand how pressure changes with depth. In fluid mechanics, pressure is defined as the force exerted per unit area. It increases with depth because the deeper you go, the more fluid there is above you, and thus more weight is pressing down.

This pressure in fluids can be calculated using the formula:
  • \( P = \rho \cdot g \cdot h \)
Where:
  • \( \rho \) is the density of the fluid, here seawater, which is given as \(1025 \mathrm{kg/m}^3\).
  • \( g \) is the acceleration due to gravity, approximately \(9.8 \mathrm{m/s}^2\).
  • \( h \) is the depth of the fluid, in this case, \(11000 \) meters.
Insert these values into the formula to find the pressure at the bottom of the Mariana Trench, resulting in a high value due to the extreme depth.
Force on Submerged Objects
Calculating the force exerted by fluids on submerged objects involves understanding how pressure translates into force. Force is calculated using the equation:
  • \( F = P \cdot A \)
Where:
  • \( F \) is the force exerted by the fluid.
  • \( P \) is the pressure calculated as shown previously.
  • \( A \) is the area of the object, which, for a circular window, is \( A = \pi r^2 \).
In our example, where the window's radius is \(0.10\) m, the area is calculated to be \(0.0314 \mathrm{m}^2\). Multiply this area by the previously calculated pressure to obtain the force the seawater applies to the window. This process shows just how much force results from such deep-sea water pressure.
Weight Comparison
To put the force experienced by items submerged in deep ocean waters into perspective, comparing it with familiar forces like the weight of an object on land can be very helpful. Weight is defined by the equation:
  • \( W = m \cdot g \)
Where:
  • \( W \) is the weight.
  • \( m \) is the mass of the object.
  • \( g \) is the gravitational acceleration.
For instance, a large jetliner with a mass of \(1.2 \times 10^5 \mathrm{kg}\) has a weight of \(1.176 \times 10^6 \mathrm{N}\) when you multiply the mass by \(9.8 \mathrm{m/s}^2\). Comparing this with the force on the observation window of the underwater vehicle helps students appreciate the magnitude of forces in submarine environments.
Density of Water
Density is a critical concept in fluid mechanics. It refers to the mass of the fluid in a specific volume and is usually expressed in kilograms per cubic meter (\(\mathrm{kg/m}^3\)). The density of a fluid directly affects the pressure exerted by the fluid because it determines how much mass is present in a column of fluid above a point.

For water, this density is typically around \(1000\ \mathrm{kg/m}^3\), but seawater is denser due to dissolved salts, with a density of \(1025\ \mathrm{kg/m}^3\).

Understanding the density of the medium helps in accurately predicting pressures in fluid systems such as the ocean. The higher density of seawater means at the same depth, it will exert more pressure compared to fresh water. Recognizing this helps in designing equipment that is capable of withstanding the environmental stresses in these conditions.

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Most popular questions from this chapter

Water flows straight down from an open faucet. The crosssectional area of the faucet is \(1.8 \times 10^{-4} \mathrm{m}^{2},\) and the speed of the water is \(0.85 \mathrm{m} / \mathrm{s}\) as it leaves the faucet. Ignoring air resistance, find the crosssectional area of the water stream at a point \(0.10 \mathrm{m}\) below the faucet.

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