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Pressure ?by cragar
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#1
Jun309, 09:06 AM

P: 2,466

if I want to calculate the pressure at a given depth in the ocean or a lake
1 atmosphere of pressure will support a column of water 32 feet high right? so if the lake is 20,000 feet deep can we just divide 20,000 by 32 and get 625 atmospheres at the depth? i see that the pressure equation is p=(rho)gh is this measured in Pascal’s . This is not a homework question. 


#2
Jun309, 09:18 AM

P: 2,157

Yes, but
P = rho g h is an equation that is valid in any arbitrary system of units. The whole point of units is that you can choose whatever units are most convenient for you. 


#3
Jun309, 09:21 AM

P: 477

The pressure under water [tex]P = \rho gh + P_{\text{atm}}[/tex] is measured in Pascal.



#4
Jun309, 10:41 AM

P: 2,157

Pressure ?
g = 9.81 m/s^2, then it is still unit independent, because the meters and seconds in this equation will transform covariantly when changing to other units (e.g. to miles and weeks from meters and seconds). Only if you were to substitute numerical values like g = 9.81 ommitting the meters/second^2, then the equaion becomes unit dependent, it will only be valid in SI units. You are then effectively putting meter = second = kilogram = 1, analogous to what we do in theoretical physics when we put hbar = c = G = 1. The answer you get for P will then be a number and then the unit Pascal has to be inserted by hand, it doesn't come out of the equation itself. You can also say that since we have put meter = second = kilogram = 1, you can multiply by any arbitrary combination of meters, seconds and kilograms, but only one combination is dimensionally correct. So, the correct unit, Pascal, was set to 1. We can thus multiply by Pascal as that is equal to 1 without changing anything. But then the equation has become dimensionally correct again, and therefore this is the correct way to put back the units. 


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