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A submarine has a "crush depth" (that is, the depth at which water pressure
will crush the submarine) of 250 m. What is the approximate pressure (water
plus atmospheric) at this depth? (Recall that the density of seawater is 1025
kg/m3 g = 9.81 m/s2, and 1 kg/(m.s) = 1 Pa = 9.8692 x 10-6 atm.)

Respuesta :

MathPhys

Answer:

25.8 atm

Explanation:

Total pressure is:

P = Patm + ρgh

where Patm is the atmosphere pressure,

ρ is the fluid density,

g is acceleration due to gravity,

and h is the depth of the fluid.

P = (1 atm) + (1025 kg/m³) (9.81 m/s²) (250 m) (9.8692×10⁻⁶ atm/Pa)

P = 1 atm + 24.8 atm

P = 25.8 atm

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