Gauge pressure at the bottom?

  • Thread starter DrMcDreamy
  • Start date
  • #1
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Homework Statement



A test tube standing vertically in a test tube rack contains 3.8 cm of oil, whose density is 0.81 g/cm3 and 6.4 cm of water. What is the gauge pressure on the bottom of the tube? The acceleration of gravity is 9.8 m/s2. Answer in units of Pa

Homework Equations



P1-Po=[tex]\rho[/tex]oilgh1+[tex]\rho[/tex]H2Ogh2

The Attempt at a Solution



P1-Po= (810 kg/m3)(9.80 kg m /s2)(0.038 m) + (1000 kg/m3)(9.80 kg m/s2)(0.064 m)

P1-Po= 929 Pa

Is the work and answer right? TIA
 

Answers and Replies

  • #2
114
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They are indeed correct.
 
  • #3
68
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Thank you! :smile:
 
  • #4
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Actually, I wonder whether you should also consider atmospheric pressure.

Also, why is it P1-P0 and not +?
 
  • #5
68
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^It came out right with 929 Pa.
The way the prof had shown it in class is P1-P0:

P2=Po+[tex]\rho[/tex]oilgh1

P1=P2+[tex]\rho[/tex]H2Ogh2

P1=Po+[tex]\rho[/tex]gh1+[tex]\rho[/tex]gh2

P1-Po=[tex]\rho[/tex]gh1+[tex]\rho[/tex]gh2
 
  • #6
114
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Oh ok, it was down to notation, then.
When I thought you might need the atmospheric pressure is because I was calculating P1, but you needed P1-P0. :)
 

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