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

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In summary, the conversation discusses a problem involving finding the depth of water using the equation P=ρgh. The person attempting the problem initially made a mistake by mixing units, but ultimately arrived at the correct answer of 4100m.
  • #1
iRaid
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Homework Statement


a1qqh1.png

PART B

Homework Equations


[tex]P = \rho gh[/tex]


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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  • #2
iRaid said:

Homework Statement


a1qqh1.png

PART B

Homework Equations


[tex]P = \rho gh[/tex]


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).
 
  • #3
In your equation P=ρgh, P is measured in Pascals (Pa)
 
  • #4
Oh right thanks
 
  • #5
.

Sure, I can help with this problem! First, let's take a look at the formula P = ρgh. This is known as the hydrostatic pressure formula, where P represents pressure, ρ represents density, g represents acceleration due to gravity, and h represents height. In order to solve for h, we need to isolate it on one side of the equation.

To do this, we can divide both sides by ρg, giving us the equation h = P/ρg. Now, let's substitute in the given values. We know that P = 412 x 105 Pa and ρ = 1025 kg/m^3. We also know that g = 9.8 m/s^2. Plugging these values into the equation, we get:

h = (412 x 105 Pa) / (1025 kg/m^3 x 9.8 m/s^2)

Simplifying, we get:

h = 4100 m

So, the answer of 4100m is correct. It seems like you were on the right track with your attempt, but you may have made a calculation error. Remember to always double check your work and use proper units in your calculations.

I hope this helps! Good luck on your finals tomorrow. Remember to stay calm and focused, and you'll do great!
 

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

What is the formula for P = ρgh?

The formula for P = ρgh is used to calculate pressure (P) in a fluid based on its density (ρ), acceleration due to gravity (g), and the height (h) of the fluid column.

How do I solve for h in the formula P = ρgh?

To solve for h, you can rearrange the formula to h = P/(ρg). This means you divide the pressure (P) by the product of the density (ρ) and acceleration due to gravity (g).

What units should I use for the variables in P = ρgh?

The units used for the variables in P = ρgh will depend on the units used for pressure (P), density (ρ), and acceleration due to gravity (g). Generally, pressure is measured in Pascals (Pa), density in kilograms per cubic meter (kg/m3), and acceleration due to gravity in meters per second squared (m/s2).

Can I use this formula for any type of fluid?

Yes, the formula P = ρgh can be used for any type of fluid, whether it is a gas or a liquid. However, it is important to make sure that the units used for the variables are consistent with the specific type of fluid being analyzed.

Are there any assumptions made when using the formula P = ρgh?

One assumption made when using the formula P = ρgh is that the fluid is incompressible, meaning that the density does not change with pressure. Additionally, this formula assumes that the fluid is in a static state, meaning that there is no movement or flow.

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