Fluid pressure physics homework

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Homework Help Overview

The discussion revolves around a fluid pressure problem involving a vat of liquid, where the original poster seeks to determine the mass of the liquid based on given dimensions and pressure at the bottom of the vat.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of pressure equations and force balance to find the mass of the liquid. There are attempts to confirm calculations and clarify the use of different equations related to fluid pressure.

Discussion Status

Some participants have provided hints and guidance on finding the density and using appropriate equations. There is an ongoing exploration of different methods to approach the problem, with no explicit consensus reached on the correctness of the calculations presented.

Contextual Notes

Participants are working under the constraints of not having a solution key and are questioning the accuracy of their calculations and assumptions regarding pressure values.

jason_r
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A 0.800 m-diameter vat of liquid is 2.70 m deep. The pressure at the bottom of the vat is 1.20 atm.

What is the mass of the liquid in the vat?
 
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jason_r said:
A 0.800 m-diameter vat of liquid is 2.70 m deep. The pressure at the bottom of the vat is 1.20 atm.

What is the mass of the liquid in the vat?

Hi jason_r! Welcome to PF! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:

Hint: find the density first.
 


tiny-tim said:
Hi jason_r! Welcome to PF! :smile:

Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:

Hint: find the density first.

so far i have :

PA=P_oA +mg
(1215600)(pi*(0.4)^2)= (1.013x10^5)(pi(0.4)^2) + 9.8m
m=1.04 x 10^4kg

I need confirmation on the answer and if I am doing this right
cause i don't have a key or anything
thanks
 
jason_r said:
PA=P_oA +mg
(1215600)(pi*(0.4)^2)= (1.013x10^5)(pi(0.4)^2) + 9.8m
m=1.04 x 10^4kg

Yes, that's fine:

(I haven't checked your value for 1 atm … and I'm assuming you meant 10^6, so that it matches the 1215600)

You've used a force equation … force on the bottom minus force on the top = weight … P-A - P+A = mg.

You could also have use the pressure equation (that's the one I'd have used): P- - P+ = rho gh.

But they both give the same result, and they're both fine. :smile:
 

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