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Chapter 9: Problem 12

At the surface of a freshwater lake the air pressure is 1.0 atm. At what depth under water in the lake is the water pressure 4.0 atm?

Short Answer

Expert verified
Answer: Approximately 51.6 meters.

Step by step solution

01

Recall the hydrostatic pressure formula

The hydrostatic pressure formula is given by \(P = \rho gh\), where \(P\) is the pressure at a certain depth, \(\rho\) is the density of the fluid, \(g\) is the acceleration due to gravity, and \(h\) is the depth below the surface.
02

Define the variables and given values

We know the following values: - Atmospheric pressure, \(P_{atm} = 1 \, atm\) - Water pressure we want to find the depth for, \(P_{water} = 4 \, atm\) - Total pressure at the depth, \(P_{total} = P_{atm} + P_{water}\) - Density of freshwater, \(\rho = 1000 \, kg/m^3\) - Acceleration due to gravity, \(g = 9.81 \, m/s^2\) - Depth underwater, \(h\): This is what we want to find.
03

Calculate the total pressure at the depth

First, we need to find the total pressure at the depth we want to find. This is the sum of the atmospheric pressure and the water pressure at the depth. So, \(P_{total} = P_{atm} + P_{water} = 1 \, atm + 4 \, atm = 5 \, atm\). To convert this pressure to Pascals (Pa), we need to multiply by the conversion factor of \(1 \, atm = 101325 \, Pa\). Therefore, \(P_{total} = 5 \, atm \times 101325 \, Pa/atm = 506625 \, Pa\).
04

Rearrange the hydrostatic pressure formula and solve for depth

Now, we can use the formula \(P_{total} = \rho gh\) to find the depth \(h\). We want to solve for \(h\), so we can rearrange this equation as follows: \(h = \frac{P_{total}}{\rho g}\) Now, plug in the values we know: \(h = \frac{506625 \, Pa}{1000 \, kg/m^3 \times 9.81 \, m/s^2} \approx 51.6 \, m\)
05

Interpret the result

So, the depth under the surface of the freshwater lake at which the water pressure is 4.0 atm is approximately 51.6 meters.

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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)?
A hypodermic syringe is attached to a needle that has an internal radius of \(0.300 \mathrm{mm}\) and a length of \(3.00 \mathrm{cm} .\) The needle is filled with a solution of viscosity $2.00 \times 10^{-3} \mathrm{Pa} \cdot \mathrm{s} ;\( it is injected into a vein at a gauge pressure of \)16.0 \mathrm{mm}$ Hg. Ignore the extra pressure required to accelerate the fluid from the syringe into the entrance of the needle. (a) What must the pressure of the fluid in the syringe be in order to inject the solution at a rate of $0.250 \mathrm{mL} / \mathrm{s} ?$ (b) What force must be applied to the plunger, which has an area of \(1.00 \mathrm{cm}^{2} ?\)
If you watch water falling from a faucet, you will notice that the flow decreases in radius as the water falls. This can be explained by the equation of continuity, since the cross-sectional area of the water decreases as the speed increases. If the water flows with an initial velocity of $0.62 \mathrm{m} / \mathrm{s}\( and a diameter of \)2.2 \mathrm{cm}$ at the faucet opening, what is the diameter of the water flow after the water has fallen $30 \mathrm{cm} ?$
A nozzle of inner radius \(1.00 \mathrm{mm}\) is connected to a hose of inner radius \(8.00 \mathrm{mm} .\) The nozzle shoots out water moving at $25.0 \mathrm{m} / \mathrm{s} .$ (a) At what speed is the water in the hose moving? (b) What is the volume flow rate? (c) What is the mass flow rate?
The density of platinum is \(21500 \mathrm{kg} / \mathrm{m}^{3} .\) Find the ratio of the volume of \(1.00 \mathrm{kg}\) of platinum to the volume of $1.00 \mathrm{kg}$ of aluminum.
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