posted by Physics Problem on .
Hi, I need some help with this problem:
The density of ice is 920 kg/m3, and that of sea water is 1030 kg/m3. What fraction of the total volume of an iceberg is exposed?
I took the density of the iceberg divided by the density of the water and multiplied this by 100, but this did not give me the correct answer.
I don't know why you mulitiplied by 100, however...
Consider 1000 kg of ice. It has a volume of 1000/920 m^3.
To meed the bouyance of this, 1000 kg of sea water has to be displaced> a volume of 1000/1030 m^3.
How much is exposed: the difference..
exposed= 1000/920 - 1000/1030
or 1000(1030 - 920)/920*1030
fraction exposed = exposed/total ice volume
= 1000(1030 - 920)/920*1030*(1000/920)
= (1030 - 920) / *1030