Fluid pressure physics homework

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The discussion revolves around calculating the mass of liquid in a vat with a diameter of 0.800 m and a depth of 2.70 m, where the pressure at the bottom is 1.20 atm. A user presents their calculations using a force equation to arrive at a mass of approximately 10,400 kg. Other participants confirm the approach and suggest that using the pressure equation could also yield the same result. The importance of verifying the atmospheric pressure value is mentioned, ensuring accurate calculations. Overall, the calculations and methods discussed are validated as correct.
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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Welcome to PF!

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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