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astenroo
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Homework Statement
Recently I stumbled on an exercise in a physics book. My problem is, that I am confident there is a fault somewhere in the textbook, as my numbers don't add up with the ones in the "Answers" segment. The assignment:
A ship that sank off the coast of Aland during the Crimean war lies on a 40 m depth. Researchers want to investigate the the ships cargo from a small submarine. In the sub there is a "pressure equalization chamber" (airlock?) from which entrance to the water is possible from 1.00 m² hatch. a) calculate the hydrostatic pressure at this depth b) What is the pressure in the pressure chamber if the hatch is influenced by a 50.0 kN force? [tex]\rho[/tex]=1.03*10³ kg/m³ and p0=101 kPa
Homework Equations
The hydrostatic pressure p=[tex]\rho[/tex]gh
Total pressure at depth h p=p0 + [tex]\rho[/tex]gh
Pressure p=F/A
The Attempt at a Solution
a) Hydrostatic pressure at 40 m : 101*10³ + 1.03*10³*9,81*40 = 505 kPa
The answer provided by the textbook is 392 kPa
b) I assume that the force acting on the hatch is directed outwards (towards the sea) creating a a slight overpressure in the chamber to keep it from flooding as soon as the hatch is opened.
Then: p=F/A, 50000 N/ 1.00 m² = 50 kPa. The total pressure in the chamber should then be 50 kPa + 505 kPa resulting in a total pressure of 555 kPa in the chamber. The pressure in the chamber according to the textbook is 455 kPa.
I know the hydrostatic pressure at 40 m can in no way be less than 5 atm (505 kPa). Am I right or am I thinking wrong somewhere?