The density of ice is

ρ(ice) = 920 kg/m3,
and the average density of water is
ρ(water) = 1.00 g/cm3.
What fraction of the total volume of a block of ice is exposed when floating in water?

what does this even mean?

last one of the night, I promise. Physics really isn't my thing.

the ice is 92% as dense as the water

so 92% of the block is submerged when it floats

that leaves 8% exposed

Thank you SO much for all your help!!

Di/Dw = (920kg/m^3)/(1000kg/m^3) = 0.92 = 92% below the surface.

1.0-0.92 = 0.08 = 8% exposed.

To understand the question, we first need to know that when an object floats in a liquid, it displaces an amount of liquid equal to its own weight. This principle is known as Archimedes' principle.

In this case, we have a block of ice floating in water. The density of ice is given as 920 kg/m^3, while the average density of water is 1.00 g/cm^3 (which is equivalent to 1000 kg/m^3).

To find the fraction of the total volume of the ice block that is exposed when floating, we need to compare the densities.

The density of the ice is less than the density of water, which means that the ice is less dense and will float. When an object floats, it displaces an amount of liquid equal to its own weight.

To calculate the fraction of the ice block that is exposed when floating, we can use the formula:

Fraction exposed = (Density of ice) / (Density of water)

Substituting the given values:

Fraction exposed = (920 kg/m^3) / (1000 kg/m^3)

Simplifying:

Fraction exposed ≈ 0.92

Therefore, approximately 92% of the volume of the ice block will be exposed above the surface of the water when it is floating.