What Depth Achieves Sixfold Atmospheric Pressure in a Lake?

[cavery4]

At what depth beneath the surface of a lake is the absolute pressure 6 times the atmospheric pressure of 1.01 x 105 Pa that acts on the lake's surface?

pressure = density * g * h

I thought to multiply 6 * 1.10 X 10^5 pa. Then divide that answer by 1000 kg/m3 (density of water, right?) * 9.8 m/s/s.

This was marked wrong my webassign.
Can someone help me figure out what I did wrongly?

Thank you.

 

[SpaceTiger]

At what depth beneath the surface of a lake is the absolute pressure 6 times the atmospheric pressure of 1.01 x 105 Pa that acts on the lake's surface?

You were almost right, but you calculated the depth at which the change in pressure was six times the surface pressure. This makes the total pressure at that depth slightly more than six times the surface value.

$$h\ =\ depth$$

$$P-P_s=\Delta P=\rho gh$$

$$\frac{P+\Delta P}{P}=6=1+\frac{\rho gh}{P}$$

$$h=\frac{5P}{\rho g}$$

Just a little smaller than your value.

 

[thepatientmental]

To find the depth at which the absolute pressure is 6 times the atmospheric pressure, we can use the formula for pressure: P = ρgh, where P is pressure, ρ is density, g is acceleration due to gravity, and h is the depth. We know that the atmospheric pressure acting on the surface of the lake is 1.01 x 10^5 Pa. So, to find the depth at which the absolute pressure is 6 times this atmospheric pressure, we can set up the equation:

6 * 1.01 x 10^5 Pa = ρ * 9.8 m/s^2 * h

Now, we need to find the density of water at the given depth. The density of water changes with depth due to the increasing pressure, so we cannot simply use the density of water at the surface. Instead, we can use the fact that the density of water at a depth of 1 meter is approximately 1000 kg/m^3. This means that at a depth of h meters, the density of water would be approximately 1000 kg/m^3 * h. Substituting this into our equation, we get:

6 * 1.01 x 10^5 Pa = 1000 kg/m^3 * h * 9.8 m/s^2 * h

Simplifying, we get:

6.06 x 10^5 Pa = 9800 kg/m^3 * h^2

Solving for h, we get:

h = √(6.06 x 10^5 Pa / 9800 kg/m^3) = 7.17 meters

Therefore, at a depth of 7.17 meters below the surface of the lake, the absolute pressure would be 6 times the atmospheric pressure. It seems like you may have made a calculation error in your answer, which is why it was marked wrong. I hope this explanation helps you understand the correct way to solve this problem.