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Calculate the depth in the ocean at which the
pressure is three times atmospheric pressure.
Atmospheric pressure is 1.013 × 10^5 Pa. The acceleration of gravity is 9.81 m/s^2 and the density of sea water is 1025 kg/m^3
Answer in units of m

Respuesta :

MathPhys

Explanation:

Assuming constant density, the pressure at a depth h is:

P = Patm + ρgh

3 (1.013×10⁵ Pa) = 1.013×10⁵ Pa + (1025 kg/m³) (9.81 m/s²) h

h = 20.1 m

The pressure is three times atmospheric pressure at a depth of 20.1 m.

Given information:

Atmospheric pressure is 1.013 × 10⁵ Pa. The acceleration due to gravity is 9.81 m/s² and the density ρ of sea water is 1025 kg/m³

Let the depth at which the pressure is three times the atmospheric pressure be h:

Then the pressure is given by:

[tex]P = P_{atm} + \rho gh\\\\3P_{atm}=P_{atm}+\rho gh\\\\3 (1.013\times10^5) = 1.013\times10^2 + (1025) (9.81) h\\\\h = 20.1 \;m[/tex]

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