A submarine is located 2.000 km below the surface of the water. What is the absolute pressure of the seawater on the outside of the submarine? (density of seawater is 1,025 kg/m3)

To solve this problem, we need to use the formula for pressure due to a fluid column, which is P = ρgh, where:

• P is the pressure,
• ρ is the fluid density,
• g is the acceleration due to gravity, and
• h is the height of the fluid column.

In this case, the fluid is seawater, the density (ρ) of which is given as 1,025 kg/m^3. The acceleration due to gravity (g) is approximately 9.81 m/s^2. The height (h) is the depth of the submarine below the surface of the water, which is 2,000 km or 2,000,000 m.

Substituting these values into the formula gives:

P = (1,025 kg/m^3) * (9.81 m/s^2) * (2,000,000 m)

Calculating this gives a pressure of about 20,097,000,000 Pascals, or 20.097 GigaPascals (GPa).

However, this is the gauge pressure, which is the pressure relative to the atmospheric pressure at the surface of the water. The absolute pressure is the gauge pressure plus the atmospheric pressure.

The atmospheric pressure at sea level is about 101,325 Pascals, or 0.101325 GPa.

So, the absolute pressure on the submarine is 20.097 GPa + 0.101325 GPa = 20.198325 GPa.