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Topic Physics, Density.
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Original Post
woolo Posted 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?

Replies
ElephantStone Posted 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.

woolo Posted 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. :(
Violently Happy Posted at 11:05 am on Aug. 24, 2008
Don't you have notes that could help you?
BodomChick Posted at 11:02 am on Aug. 24, 2008
that's rather difficult.
woolo Posted 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.
Total Destruction Posted at 10:56 am on Aug. 24, 2008
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Baron Samedi Posted at 10:54 am on Aug. 24, 2008
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LittleItaly Posted at 10:54 am on Aug. 24, 2008
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