Thanks!
I have been having that problem all semester. What you said makes sense as well. Is it safe to assume then for most problems like this it should be the density of the liquid*volume of liquid(submerged part)*g=density of object*density total of object*g
Ok how about this...Archimedes' principle!
We have ρocean=1025kg/m^3 ρwood=? Vol(below)=65% Vol(wood)=100% our g's cancel out though.
This would give us ρoceanVol(below)=ρwoodVol(wood) 1025(65)=ρwood(100) ρwood=666.25kg/m^3 I really hope this is it. I know this is similar to what I did...
No and that is what has me worried. I know for buoyancy, an object whose density is greater than that of the fluid in which it is submerged tends to sink and if the object is either less dense than the liquid or is shaped appropriately (as in a boat), the force can keep the object afloat. I am...
Homework Statement
A piece of driftwood floating in the ocean has 35 percent of its volume above water. What is the density of the driftwood if the ocean has a density of 1,025 kg/m^3.
Homework Equations
ρvol(driftwood)=ρvol(ocean)
The Attempt at a Solution
Well based on a...