**Determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. Assume that the barometric pressure is 14.7 psia and the specific gravity of seawater is 1.03.**

#### Solution:

### We can find the density of water by multiplying its specific gravity by the density of sea water.

### \rho =SG\times \rho_{H_{2}O}=(1.03)(62.4lbm/ft^{3})=64.27lbm/ft^{3} (SG is specific gravity, density of water can be found through research)

### Now we will calculate the absolute pressure at 175ft below sea level as this is the location of the submarine.

### P=P_{atm}+pgh

### =(14.7psia)+(64.27lbm/ft^{3})(32.2ft/s^{2})(175ft)(\frac{1lbf}{32.2lbm\cdot ft/s^{2}})(\frac{1ft^{2}}{144in^{2}})

### =92.8psia

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Thank you so much. Jesus Christ bless u .he gives this mind for you.you

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