The density of sea water is 1025 kg m^- 3. How many times is the pressure at the depth of 3 km under the sea surface greater than the atmospheric pressure? (g = 9.8 ms^-2 )
Question
The density of sea water is 1025 kg m^- 3. How many times is the pressure at the depth of 3 km under the sea surface greater than the atmospheric pressure? (g = 9.8 ms^-2 )
Solution
To solve this problem, we need to use the formula for pressure in a fluid at a certain depth: P = ρgh, where:
- P is the pressure,
- ρ is the fluid density,
- g is the acceleration due to gravity, and
- h is the height (or depth in this case).
Given in the problem, we have:
- ρ = 1025 kg/m^3 (the density of sea water),
- g = 9.8 m/s^2 (the acceleration due to gravity), and
- h = 3000 m (the depth under the sea, converted from km to m).
Substituting these values into the formula, we get:
P = 1025 kg/m^3 * 9.8 m/s^2 * 3000 m = 30195000 Pa.
The atmospheric pressure at sea level is about 101325 Pa.
So, the pressure at the depth of 3 km under the sea surface is about 30195000 Pa / 101325 Pa = 29.8 times greater than the atmospheric pressure.
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