What is the density of the body in Archimedes' Principle problem?

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SUMMARY

The density of the body in the Archimedes' Principle problem is calculated using the relationship between the weight of the body in air and in water. Given that the weight of the body in water is half of its weight in air, and knowing the density of water is 1000 kg/m³, the density of the body is determined to be 2000 kg/m³. The formula used is derived from the equilibrium of forces acting on the body submerged in water.

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chmate
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Hello all!

Inside the water, body weights two times less than in air. Density of water is 1000kg/m^3. Find the density of body.


1. Well I tried this solution and I'm not sure if it's correct. I started from this principle Fa=W/2 (inside water force equals weight of the body divided by two). Then:


\rho 1 -> density of water
\rho 2 -> density of body
V -> volume

\rho 1 gV=\dfrac{\rho 2gV}{2}

\rho 1=\dfrac{\rho 2}{2}
\rho 2=2\rho 1

Anything wrong?

Thank you.
 
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Looks good to me.
 

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