If an object of dimension 40cm×30cm×10cm is floating on a liquid of density 1g/cc.What height of the object will be exposed outside of the water given that the density of object is 0.92g/cc?

Answer 1

Volume of object = 40*30*10=12000 cm³
Principle of floatation says that the mass of liquid displaced by a floating object equals the mass of the object. (C.F. Archimedes)
For a density of 0.92 g/cc floating on a liquid of density 1.0 g/cc, the mass of liquid displaced is 0.92*12000=11040 g ≡ 11040 cc.
Therefore the volume of object exposed is 12000-11040=950.

The height of object exposed depends on the face completely immersed.
If the 40*30 cm^2 face is immersed, then the height of object exposed is 950/(40*30)=0.792 cm.

For other immersed faces, the height can be calculated similarly.

Answer 2

How is the density of an object determined?(1 point)
Responses

by dividing the mass of an object by its volume
by dividing the mass of an object by its volume

by multiplying the volume of an object by its mass
by multiplying the volume of an object by its mass

by subtracting the mass of an object from its volume
by subtracting the mass of an object from its volume

by adding the volume of an object to its mass