Why Does Multiplying Pressure by \(10^5\) Help in Calculating Depth?

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Multiplying pressure by \(10^5\) converts it from atmospheres to Pascals, which is necessary for consistency in the equation \(P = \rho gh\). The user initially attempted to solve for depth using incorrect unit conversions, leading to an incorrect answer. It is crucial to ensure that all units in the equation are compatible to achieve accurate results. Proper unit management helps clarify calculations and avoids confusion. Understanding these unit conversions is essential for solving problems related to fluid pressure and depth accurately.
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Homework Statement


a1qqh1.png

PART B

Homework Equations


P = \rho gh


The Attempt at a Solution


Well I tried doing 412 = (1025)(9.8)h.. Divide by 10045 (9.8 x 1025). And ended up getting .041..
I found out that I need to use 412 x 105 for the pressure which I have no idea why.

Any help?

Also, the answer is 4100m
 
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iRaid said:

Homework Statement


a1qqh1.png

PART B

Homework Equations


P = \rho gh


The Attempt at a Solution


Well I tried doing 412 = (1025)(9.8)h.. Divide by 10045 (9.8 x 1025). And ended up getting .041..
I found out that I need to use 412 x 105 for the pressure which I have no idea why.

Any help?

Also, the answer is 4100m

You are mixing your units. "Atmospheres" is not an mks unit. You should write units in your equations as you go, to make sure that they are consistent (LSH units = RHS units, and the units of any quantities you are adding are the same).
 
In your equation P=ρgh, P is measured in Pascals (Pa)
 
Oh right thanks
 
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