Floating of wood - Archimedes law

What will be the volume of the floating part of a wooden (balsa) block with a density of 200 kg/m3 and a volume of 0.02 m3 that floats in alcohol? (alcohol density is 789 kg/m3)

Correct result:

V =  14930.2915 cm3


ρ1=200 kg/m3 V1=0.02 m3 ρ2=789 kg/m3  m1=m2 V ρ1=(V1V) (ρ2ρ1) V ρ1=V1 (ρ2ρ1)V (ρ2ρ1) V ρ2=V1 (ρ2ρ1)  v=V1 (ρ2ρ1)ρ2=0.02 (789200)7890.0149 m3  V=v cm3=v 1000000  cm3=0.0149 1000000  cm3=14930.292 cm3=14930.2915 cm3

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