At what depth of the lake is the pressure 500 kPa? Atmospheric pressure 100 kPa water density 1000 kg / m3, g = 10m / s2

Initial data: P (pressure at the desired depth of the lake) = 500 kPa (500 * 10 ^ 3 Pa).

Reference values: according to the Rathm condition. (atmospheric pressure) = 100 kPa (100 * 10 ^ 3 Pa); ρ (density of lake water) = 1000 kg / m3; g (acceleration due to gravity) ≈ 10 m / s2.

The desired depth can be expressed from the equality: P = Ratm. + Рв = Rathm. + ρ * g * h, whence h = (P – Ratm.) / (ρ * g).

Calculation: h = (500 * 10 ^ 3 – 100 * 10 ^ 3) / (1000 * 10) = 40 m.

Answer: The pressure will be 500 kPa at a depth of 40 m.