1/5 of water?????
Just a guess though.
A cube of 10cm and mass 0.5kg floats in a liquid with only one-fifth of it's height about the liquid surface.what is the Relative density?
4 answers
It certainly is a strange guess.
volume of cube = s^3
volume of water displaced = 4/5 s^3 = 0.8 s^3
Archimedes says weight of water displaced = buoyancy
if floating then buoyancy = weight
so
density water *.8 s^3 = density block * s^3
density block / density water = 0.8
ratio of
mass of cube = x s^3
mass of water displaced = 0.8
volume of cube = s^3
volume of water displaced = 4/5 s^3 = 0.8 s^3
Archimedes says weight of water displaced = buoyancy
if floating then buoyancy = weight
so
density water *.8 s^3 = density block * s^3
density block / density water = 0.8
ratio of
mass of cube = x s^3
mass of water displaced = 0.8
Yeah it was.
Typo. Meant to say 4/5 of the density of water, since only 1/5 isn't submerged.
Typo. Meant to say 4/5 of the density of water, since only 1/5 isn't submerged.
volume of the cube= 10x10x10=1000cm³
mass of the cube= 0.5kg
density of d cube= 0.5/1000=0.0005kg/cm³
volume that floats in liquid=1/5*10*10*10=200cm³
mass of d cube in liquid= density*200cm³=0.0005kg/cm³*200cm³=0.1kg.
lost of mass=mass of the liquid=0.5kg-0.1kg=0.4kg.
R.D= mass of cube/mass of liquid=0.5/0.4=1.25
mass of the cube= 0.5kg
density of d cube= 0.5/1000=0.0005kg/cm³
volume that floats in liquid=1/5*10*10*10=200cm³
mass of d cube in liquid= density*200cm³=0.0005kg/cm³*200cm³=0.1kg.
lost of mass=mass of the liquid=0.5kg-0.1kg=0.4kg.
R.D= mass of cube/mass of liquid=0.5/0.4=1.25
1 answer