Acube of wood having an edge dimension of 20.1 cm and a density of 645 kg/m3 floats on water. (a) what is the distance from the horizontal top surface of the cube to the water level?

7.1355 cm

Explanation:

Principle of flotation

If an object of density is floating on a liquid of density , the weight of the displaced liquid equals the weight of the object. It's also known as the Archimedes' principle.

The object will float if its density is less than the density of the liquid and will sink if it's greater. In the special case it floats, part of its volume (and therefore, mass) will keep floating over the liquid's surface (let's call it ), and the rest will be sunk (). The total volume of the object will be:  

Applying the principle of flotation

But equals the sunk volume of the object

This relation gives us

This is the proportion of the object sunk in liquid .

The wooden cube has an edge dimension of L=20.1 cm and a density . The density of water is  . So the underwater volume of the cube is

Assuming the cube floats with one side parallel to the face of the water

Where h is the length of the side of the cube submerged in the water. Replacing those formulas we get

Simplifying

Finally, the distance from the horizontal top surface of the cube to the water level is

d) the number of neutrons plus the number of protons

ill do it for 100, also this isnt 80 its 40