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.)

Question

Neil Hull

Physics

26 November, 06:55

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.)

+1

Answers (1)

Know the Answer?

Allyson Warner · Answer 1

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

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