Question a/b pressure variation equation

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    Pressure Variation
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SUMMARY

The discussion centers on the application of the pressure variation equation in fluid mechanics, specifically regarding how atmospheric pressure affects fluid height in a straw and the maximum depth organisms can survive underwater. The professor's solutions demonstrate the calculations using the formula P = Patm + ρgd, leading to a maximum height of 1.0 m for water suction and a maximum depth of 10 km for organisms under 1000 atm pressure. The confusion arises from understanding the pressure difference calculations, particularly the interpretation of atmospheric pressure in these contexts.

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  • Understanding of fluid mechanics principles
  • Familiarity with the pressure variation equation
  • Knowledge of units for pressure (kPa, atm) and density (kg/m³)
  • Basic algebra for solving equations
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Hi everyone, i have a question abuot how my professor is using the pressure variation equation and I would really appreciate help with it!

Homework Statement



How high can you suck water up a strw? The pressure in the lungs can be reduced to about 10 kPa below atmospheric pressure

2. Professor's Solution

P=Patm+ρgd --> d=(P-Patm)/ρg --> d=(10 kPa)/(1000 kg/m^3 * 9.8 m/s^2) = 1.0 m

3. My attempt at the problem

P=Patm+ρgd --> d=(P-Patm)/ρg --> d=(101.3 kPa-(1atm-10 kPa)/(1000 kg/m^3] * 9.8 m/s^2) =

What is wrong with my attempt at the solution.

Another example of my professor's solution is in this problem:

Homework Statement



The maximum pressure most organisms can survive is abotu 1000 times atmospheric pressure. ONly small, simple organisms such as tadpoles and bacteria can survive such high pressure. What then is the maximum depth at whic these organisms can live under the sea (assuming that the density of seawater is 1025 kg/m^3?

2. Professors Solution

P=Patm+ρgd --> d=(P-Patm[/SUB)/ρg --> d=(1000 atm)/(1025 kg/m3 * 9.8 m/s2) * (1.013e5 Pa/atm) = 10 km

Why does 1000 atm= P-Patm[/SUB) ? wouldn't P-Patm[/SUB) be (1000*1 atm)-(1atm)?


I would really appreciate some help with this problem! Thank you!
 
Physics news on Phys.org
If the pressure in your lungs can be reduced 10 kPa below atmospheric, what is the difference in pressure between the atmosphere and the inside of your lungs?
 
Thank you very mcuh for your reply

1 atm - 10 kPa?
 

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