All the rants and raves I can think of?
Published on April 5, 2009 By mickeko In WinCustomize Talk

Ask me anything, I answer all questions (*).

You may get a stupid answer, an incorrect answer, or maybe even the RIGHT answer... Who knows...

Any answer given that is true and correct (as far as I know or in my opinion where applicable) will be marked with a  .

 

(*) - Exceptions: Personal questions or any other questions I don't feel like answering may not be answered. Being polite might help in these cases...


Comments (Page 4)
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on Apr 07, 2009

voidcore
Swami bootcamp?

If that's how you choose to see it.

on Apr 07, 2009

mickeko
Quoting voidcore, reply 20Swami bootcamp?

If that's how you choose to see it.

on Apr 07, 2009

LOL Doc!

I don't see the future in that crystal ball though, it's my touch screen.

on Apr 07, 2009

vStyler
Is it Oops! or Whoops!! ?

Yes, it most definetely is.

on Apr 07, 2009

Doc...I'm thinking more along the lines of "Carnac the Magnificent"

 

on Apr 07, 2009

Was he great or what?!!!

on Apr 07, 2009

CarGuy1
Doc...I'm thinking more along the lines of "Carnac the Magnificent"

Charlatan! May a bloated yak change the temperature of his jacuzzi!!

on Apr 07, 2009

Johnny's gone, I'm afraid...but while he was here, things were merry!

Link 

on Apr 07, 2009

Youtube is littered with Johnny too. As with all american talkshows, alot of the political jokes are lost on me, but there's tons of greatness to see even for me.

on Apr 07, 2009

Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

on Apr 07, 2009

the answer is always 42

on Apr 07, 2009

HG_Eliminator
Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

Clearly, as R->0, g-> infinity. The maximum will be achieved as a "Black Hole" on it's event horizon, as a singularity. As R->0, M/V will also approach infinity, as V->0.

on Apr 07, 2009

HG_Eliminator
Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

Why ask a question where the answer is already included? Just do the math. I'm not a calculator.

on Apr 07, 2009

Nothing unreal exists

on Apr 07, 2009

HG_Eliminator
Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

Max g = (4/5)(15/4)1/3(π)2/3M/V2/3

Let R be the max diameter of body then its shape will be given by relating distance r from the point, where there is max g,

as a function of the angle θ from the axis of symmetry.

r2 = R2cos(θ)

 

 

 

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