Sea water of density 1300kg m-3 exerts a pressure of 104-105Pa on the floor- Calculate the depth of the sea at that place g-10m s-2

Area= 1m ^ 2 S=1300kg/ m ^ 3 P = Po + Sgh 104× 10 ^ 5 Pa= 10 ^ 5 Pa+1300× 10 h 103× 10 ^ 5 Pa=13000h #160; #160; #160; #160; ...

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Pressure in Fluids

Sea water of ...

Question

Sea water of density $1300 k g$ $m^{- 3}$ exerts a pressure of $104 \times 10^{5} P a$ on the floor. Calculate the depth of the sea at that place ( $g = 10 m$ $s^{- 2}$ )

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Calculate the pressure at a depth of $50 m$ below the surface of the sea. (The density of sea water is $1024 k g m^{- 3}$ ).

Calculate the pressure at a depth of 50m below the surface of sea. The density of sea water is 1024 $\frac{k g}{m^{3}}$ .

Calculate the pressure at a depth of $50 m$ below sea level.

Given:
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And, acceleration due to gravity $g = 10 m / s^{2}$

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