Question #125370
submarine is 20m below the surface of the sea.The pressure due to the water at the depth is P.On anotherday, the submarine is 26m below the surface of the fresh water.The density of sea is 1.3times the density of fresh water.What is the pressure due to the fresh water at a depth of 26m?
1
Expert's answer
2020-07-06T15:13:00-0400

In general p=ρghp=\rho\cdot g\cdot h


The pressure due to the fresh water p0=ρgh0=ρg26ρg=p026p_0=\rho \cdot g\cdot h_0=\rho \cdot g \cdot 26 \to \rho \cdot g=\frac{p_0}{26}


The pressure due to the sea water p=1.3ρgh=1.3ρg20ρg=p1.326p=1.3\rho \cdot g\cdot h=1.3\rho \cdot g \cdot 20 \to \rho \cdot g=\frac{p}{1.3\cdot26}


We have p026=p1.326p0=p\frac{p_0}{26}=\frac{p}{1.3\cdot26} \to p_0=p


So, the pressure will be the same.








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