How Much Air Must a Fish's Bladder Contain to Achieve Neutral Buoyancy?

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Homework Help Overview

The discussion revolves around a problem in buoyancy related to fish physiology, specifically how much air a fish's bladder must contain to achieve neutral buoyancy. The problem involves calculating the percentage of a fish's body that must be filled with air, given the densities of the fish and air.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations related to average density and mass, with attempts to isolate variables for volume of air and fish. Questions arise regarding the assumptions made about the volumes and densities involved.

Discussion Status

There is an ongoing exploration of different interpretations of the problem, particularly concerning the relationship between the volumes of air and fish. Some participants provide guidance on potential errors in assumptions, while others share their own calculations and results.

Contextual Notes

Participants note that the problem is part of an online quiz with limited attempts, which adds pressure to find the correct answer. There is also mention of a specific answer that was provided, which may not align with the calculations discussed.

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Homework Statement



The average density of the body of a fish is 1147.0 kg/m3. To keep from sinking, fish have an air bladder filled with air. If the density of air is 1.28 kg/m3, what percentage of a fish's body must be filled with air to be neutrally buoyant? (Note: Enter your percentage without a unit attached.)

Homework Equations



/rho avg = Total mass of object / Total volume of object
/rho avg = 1000 kg/m^3

The Attempt at a Solution



/rho avg = Mo / Vo (Total mass/ divided by total volume)

Mo = Ma + Mf (Ma= mass of air and Mf= mass of fish)

Ma= 1.28Va (density of air x volume of air)

Mf = 1147Vf (density of fish x volume of fish)

Now subsituting all the equations together I get...

/rho avg = 1000 = (1.28Va + 1147 (Vo-Va)) / Vo

and isolating for Va/ Vo = .128 x100% = 12.8 %

I also did this to get get Ma / Mo = 0.016% ... but neither answer is right :S. Can anyone tell me what I did wrong, or whether their answer is wrong? Its much appreciated thanks!
 
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I get pretty close to you, about 13%.

I think your (Vo-Va) is wrong and seems to be assuming the tissues of the fish are compressed. Not so, the tissues are incompressible and the fish actually has to get bigger!

What is the right answer supposed to be?
 
Well with the (Vo - Va) thing... i forgot to show my steps with that. Basically orginially its 1147 times the volume of the fish... and I assumed that volume of fish plus volume of air is equal to volume of object. Thus Vo = Va + Vf and I isolated for Vf and subsituted it into the equation.

As for the answer I am unsure, it is an online quiz thing... have 10 attempts to get the answer. So far I tried like those few ways.
 
The answer was 14.7%. Apparently it wanted the volume of the air to the volume of the fish, not relative to the total volume. Therefore instead of substituting to get rid of Vf I'd need to substitute to get rid of Vo and Isolate for Va/ Vf.

Thanks for the help though!
 

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