Volume of a object by being submerged

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    Submerged Volume
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Homework Help Overview

The discussion revolves around a physics problem involving the determination of the volume of a submerged rock based on its weight in air and apparent weight in water. The context includes concepts from fluid mechanics and buoyancy, particularly Archimedes' principle.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the weight of the rock in air and its apparent weight in water, exploring the application of Archimedes' law. There are attempts to derive the volume using different values for gravitational acceleration and questioning the assumptions behind these values.

Discussion Status

The discussion is active, with participants exploring different interpretations of the problem and questioning the assumptions made regarding gravitational acceleration. Some guidance has been offered regarding the relationship between the forces involved, but no consensus has been reached on the correct approach or final answer.

Contextual Notes

Participants note discrepancies in the value of gravitational acceleration used in calculations, as well as the implications of the apparent weight of the rock in water. There is also mention of multiple choice answers that may influence the direction of the discussion.

mikezietz
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I have a multiple choice question where a rock weighs 1400 N in air and 900 N apparent weight in fresh water with 998 kg/m^3 density (i think), The question is what is the volume of the rock. 2 of the choices are .14 m^3 and .051 m^3. I have tried
p = m / v and 1400 - 500 and division by g's and densities, i have come too both of these conclusions many different ways, coming to either answer about the same amount of times. Does anyone know the true formula for this problem.
 
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Well : 900 = 1400 - 500 that is OK. So let's take 10 for g, then the mass of the water that is moved by the submerged rock (you know, Archimedes' law) is 50kg. Divide this by the density of water and you get the volume that was moved, so this needs to be equal to the volume of the rock : answer 0.051...
marlon
 
why 10 for g, isn't g 9.8 and where does the 50 kg's come from
 
Well then take 9.8 for g...

500 N is the upward force and is equal to the weight of the replaced water. So you have that 500 = mg and if we take g to be 10, then m = 50

regards
marlon
 

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