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Chapter 9: Problem 80

The deepest place in the ocean is the Marianas Trench in the western Pacific Ocean, which is over \(11.0 \mathrm{km}\) deep. On January \(23,1960,\) the research sub Trieste went to a depth of \(10.915 \mathrm{km},\) nearly to the bottom of the trench. This still is the deepest dive on record. The density of seawater is \(1025 \mathrm{kg} / \mathrm{m}^{3} .\) (a) What is the water pressure at that depth? (b) What was the force due to water pressure on a flat section of area \(1.0 \mathrm{m}^{2}\) on the top of the sub's hull?

Short Answer

Expert verified
Answer: (a) The water pressure at the depth of 10.915 km is approximately 109,565,816.75 Pa. (b) The force due to water pressure on a flat section of area 1.0 m² on the top of the sub's hull is approximately 109,565,816.75 N.

Step by step solution

01

Calculate the water pressure at the depth

Using the formula \(P = hρg\), we can find the water pressure at the depth of \(10.915 \mathrm{km}\). The density of seawater is given as \(1025 \mathrm{kg} / \mathrm{m}^{3}\) and the acceleration due to gravity is approximately \(9.81 \mathrm{m} / \mathrm{s}^{2}\). First, we need to convert the depth from kilometers to meters: \(10.915 \mathrm{km} * 1000 = 10915 \mathrm{m}\). Now we can plug these values into the formula: \(P = (10915 \mathrm{m})(1025 \mathrm{kg} / \mathrm{m}^{3})(9.81 \mathrm{m} / \mathrm{s}^{2})\).
02

Compute the pressure

Perform the calculation to find the pressure: \(P = (10915)(1025)(9.81) \approx 109565816.75 \mathrm{Pa}\). Therefore, the water pressure at the depth of \(10.915 \mathrm{km}\) is approximately \(109,565,816.75 \mathrm{Pa}\).
03

Calculate the force due to water pressure on a flat section of area \(1.0 \mathrm{m}^{2}\)

Using the formula \(F = PA\), we can now find the force due to water pressure on a flat section of area \(1.0 \mathrm{m}^{2}\) on the top of the sub's hull. We have already calculated the pressure, so we just need to multiply it by the area: \(F = (109565816.75 \mathrm{Pa})(1.0 \mathrm{m}^{2})\).
04

Compute the force

Perform the calculation to find the force: \(F = 109565816.75 \mathrm{N}\). Therefore, the force due to water pressure on a flat section of area \(1.0 \mathrm{m}^{2}\) on the top of the sub's hull is approximately \(109,565,816.75 \mathrm{N}\).

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Most popular questions from this chapter

A lid is put on a box that is \(15 \mathrm{cm}\) long, \(13 \mathrm{cm}\) wide, and \(8.0 \mathrm{cm}\) tall and the box is then evacuated until its inner pressure is \(0.80 \times 10^{5} \mathrm{Pa} .\) How much force is required to lift the lid (a) at sea level; (b) in Denver, on a day when the atmospheric pressure is \(67.5 \mathrm{kPa}\) ( \(\frac{2}{3}\) the value at sea level)?
An aluminum cylinder weighs \(1.03 \mathrm{N}\). When this same cylinder is completely submerged in alcohol, the volume of the displaced alcohol is $3.90 \times 10^{-5} \mathrm{m}^{3} .$ If the cylinder is suspended from a scale while submerged in the alcohol, the scale reading is \(0.730 \mathrm{N}\). What is the specific gravity of the alcohol? (tutorial: ball in beaker).
The pressure in a water pipe in the basement of an apartment house is $4.10 \times 10^{5} \mathrm{Pa},\( but on the seventh floor it is only \)1.85 \times 10^{5} \mathrm{Pa} .$ What is the height between the basement and the seventh floor? Assume the water is not flowing; no faucets are opened.
An air bubble of 1.0 -mm radius is rising in a container with vegetable oil of specific gravity 0.85 and viscosity \(0.12 \mathrm{Pa} \cdot \mathrm{s} .\) The container of oil and the air bubble are at \(20^{\circ} \mathrm{C} .\) What is its terminal velocity?
A manometer using oil (density \(0.90 \mathrm{g} / \mathrm{cm}^{3}\) ) as a fluid is connected to an air tank. Suddenly the pressure in the tank increases by \(0.74 \mathrm{cm}\) Hg. (a) By how much does the fluid level rise in the side of the manometer that is open to the atmosphere? (b) What would your answer be if the manometer used mercury instead?
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