Pressure in a Lake: Find Depth for 4.5 atm

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SUMMARY

The discussion focuses on calculating the depth in a freshwater lake where the water pressure reaches 4.5 atm. The equation used is P = a + hgp, where h is the depth, P is the total pressure, a is the atmospheric pressure, g is the acceleration due to gravity, and p is the density of water. The correct calculation shows that the depth h can be derived as h = (4.5 x 10^5 - 1 x 10^5) / (9.8 x 1000), resulting in a depth of approximately 35.71 meters. The importance of converting all pressures to pascals and ensuring the density of freshwater is in kg/m³ is emphasized to avoid unit errors.

PREREQUISITES
  • Understanding of fluid pressure equations, specifically P = a + hgp
  • Knowledge of unit conversion, particularly between atmospheres and pascals
  • Familiarity with the density of freshwater, typically 1000 kg/m³
  • Basic grasp of gravitational acceleration, approximately 9.8 m/s²
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  • Learn about fluid statics and hydrostatic pressure calculations
  • Study unit conversion techniques between different pressure units
  • Explore the implications of water density variations in different conditions
  • Investigate the effects of atmospheric pressure on underwater measurements
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Students in physics or engineering courses, educators teaching fluid mechanics, and anyone interested in understanding pressure dynamics in fluids.

Manda_24
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Homework Statement


At the surface of a freshwater lake the air pressure is 1 atm. At what depth under water in the lake is the water pressure 4.5 atm?

Homework Equations


P=a+hgp
h=(P-a)/(g*p)

The Attempt at a Solution


h=(4.5x105 - 1x105)/(9.8*1000)=35.71
I also did the equation with 9,000 Pa converted to atm, which I think it what I'm supposed to do but I still get the incorrect answer.
 
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Convert everything to pascals before using the equations. Also, be sure that the density you have for freshwater is in kg/m3. I'm not seeing anything wrong in your method, so I'm assuming the problem is in the units.
 
"h=(4.5x105 - 1x105)/(9.8*1000)=35.71" Appears to be correct. If I am not mistaken, 10 m of water is roughly = to 1 atm. But, I am not 100% sure off the top of my head. If that's the case, it would make sense that 35 m of water would be about 3.5 atm plus the 1 atm from the atmosphere = 4.5 atm.

If you know, what is the answer supposed to be?
 
Gear300 said:
Convert everything to pascals before using the equations. Also, be sure that the density you have for freshwater is in kg/m3. I'm not seeing anything wrong in your method, so I'm assuming the problem is in the units.

Thanks, I just needed to convert the top to pascals, I thought I only had to do it to the bottom.
 

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