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The absolute pressure below the surface of a freshwater lake is 3.51 x 10^5 Pa. At what depth does this pressure occur? Assume that atmospheric pressure is 1.01 x 10^5 Pa. and the density of the water is 1.00 x 10^3 kg/m^3

Respuesta :

AlpenGlow
The general formula for absolute pressure is:

[tex] P_{total} = P_{atm} + (rgh)[/tex]

Where: [tex] P_{total} [/tex] = Absolute pressure
             [tex] P_{atm} [/tex]  = atmospheric pressure
             r = density
             g = [tex] 9.8 \frac{m}{{s^2}} [/tex]
             h = depth

We can use this formula to derive our formula for h:

[tex] P_{total} = P_{atm} + (rgh)[/tex]      transpose atmospheric pressure
[tex] P_{total} - P_{atm} [/tex] =  [tex](rgh)[/tex]     transpose r and g

[tex] P_{total} [/tex] - [tex] P_{atm} [/tex]
-------------------------      = h
         rg

Now let us input our given into our new formula:

3.52 x  [tex] 10^{5}[/tex] - 1.01 x [tex] 10^{5}[/tex]
----------------------------------------              = h
1.00 x [tex] 10^{3} [/tex] x 9.8 [tex] \frac{m}{ s^{2} } [/tex]   

[tex] \frac{2.51 x {10^5}}{9.8 x {10^3} \frac{m}{s^2}} = h [/tex]

[tex] 25.61 m = h [/tex]



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