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a submarine dives to a depth of 500 m. The density of seawater is 1025 kg/m^3.
how much absolute total pressure must it’s hull be able to withstand-how many times larger is this pressure than the pressure at the surface-
- billrussell42Lv 71 month agoFavourite answer
pressure is the weight of a 1 m² column of water.
in this case, volume is 500 x 1 = 500 m³
mass is 1025 kg/m³ x 500 m³ = 512500 kg
weight = 512500 kg x 9.8 N/kg = 5023000 N
pressure = 5023000 N/m² = 5023000 Pa = 5023 kPa
add air pressure of 101 kPa to that to get absolute pressure of 5124 kPa
but hull has internal air pressure, so pressure differential the hull sees is actually 5023 kPa
ratio is 5124/101 = 50.7
or 5023/101 = 49.7