-- Posted by woolo at 10:53 am on Aug. 24, 2008 I'm stuck on a physics question, any help would be great! A cylindrical cone of semi-angle α, height H and mass M is submerged in a fluid of density ρ. Givecn that the volume of a cone is 1/3 area of base x height, find the depth to which it is submerged. Any ideas? -- Posted by woolo at 10:58 am on Aug. 24, 2008 I guess the answer should be in algebra form. I just realy don't know what it's asking me. -- Posted by BodomChick at 11:02 am on Aug. 24, 2008 that's rather difficult. -- Posted by Violently Happy at 11:05 am on Aug. 24, 2008 Don't you have notes that could help you? -- Posted by woolo at 11:06 am on Aug. 24, 2008 No sadly all the notes i have include mass as a variable. However this does not include any hint to the mass of the object. :( -- Posted by ElephantStone at 11:14 am on Aug. 24, 2008 Lets think...what is density...density is the amount of matter in an object, if Im correct so that is the volume of the cone which is (1/3)(pie.r^2)(H).... There are 2 things that can happen: the cone can sink, or float, depending on whether it is less dense or more dense than the fluid... So (1/3)(pie.r^2)(H)>p Now thats a test, lets see if it can be proved... archimedies principle applies here I think.. well change density to pressure (1/3)(pie. r^2) H = row.g.H cancelt he Hs (1/3)(pie.r^2)=row.9.8 (pie.r^2)/row=9.8/1.333 ditto            =7.351 I lost myself now, but it goes something liek that Im sure. Hopefully my notes might be of some use. Ill come back later and try.