Asked by Greg
The density of the dead sea is 1.24x10^3 kg m^-3. A wooden board with an area of 2.5m^2 is dropped in the Dead Sea. Calculate the proportion that would float above the surface.
Density of wood= 800kg m^-3
Density of wood= 800kg m^-3
Answers
Answered by
plumpycat
Proportion submersed = density of wood/density of Dead Sea
= 800/(1.24 *10^3)
~ 0.65 or 65 %
So the proportion above the surface would be 1 minus that.
= 800/(1.24 *10^3)
~ 0.65 or 65 %
So the proportion above the surface would be 1 minus that.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.