At what depth beneath the surface of a lake is the absolute pressure 2 times the atmospheric pressure of 1.01 105 Pa that acts on the lake's surface?

Set

Po + (depth)(water density)(g) = 2 Po
(depth)(water density)(g) = Po
and solve for depth

To find the depth beneath the surface of the lake where the absolute pressure is 2 times the atmospheric pressure, we can use Pascal's law which states that the pressure at any depth in a fluid is equal to the pressure at the surface plus the pressure due to the weight of the fluid above it.

Let's denote the atmospheric pressure as P0 = 1.01 * 10^5 Pa. We want to find the depth (h) at which the pressure (P) is 2 times the atmospheric pressure.

Using Pascal's law, we have:

P = P0 + ρgh

Where:
P is the absolute pressure at depth h.
P0 is the atmospheric pressure.
ρ is the density of the fluid (water in this case).
g is the acceleration due to gravity (approximately 9.8 m/s^2).
h is the depth beneath the surface of the lake.

We need to solve for h. Rearranging the equation, we get:

h = (P - P0) / (ρg)

Let's substitute the values given:
P = 2P0
P0 = 1.01 * 10^5 Pa
ρ = density of water = 1000 kg/m^3
g = 9.8 m/s^2

Now we can calculate the depth h:

h = (2P0 - P0) / (1000 * 9.8)

h = P0 / (1000 * 9.8)

h = (1.01 * 10^5) / (1000 * 9.8)

h ≈ 10.3 meters

So, at a depth of approximately 10.3 meters beneath the surface of the lake, the absolute pressure is 2 times the atmospheric pressure of 1.01 * 10^5 Pa.