~SuicideOfReason

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About Me Member Long-Time Deviant Meg19/Female/United States

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Mon Feb 9, 2009, 11:57 AM

Mood: Insecure
Listening to: Dreamer - Ozzy Osbourne

I guess sometimes the ground can shift between your feet
sometimes your footing slips, you stumble...
and sometimes you grab what's close to you
and hold on as tight as you can...

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Devious Info

Current Residence: Pennsylvania, USA
Favourite band or musician: Frederic Chopin
Favourite genre of music: classical :) I'm a piano geek.
Favourite artist: Salvador Dali
Operating System: Mac OS X
MP3 player of choice: iPod
Favourite game: Gears of War / Halo
Favourite gaming platform: XBox
Favourite cartoon character: Jessica Rabbit
Personal Quote: These are the times that try men's souls...
Tools of the Trade: Imagination, pent-up emotion

deviantID

I am an 19 year old with the soul of an ancient and the heart of a child.
I am mess of a writer, musician, fixer, lover, fighter, muse, artist, sinner, saint, deviant, learner, dreamer, creator, mastermind, & believer.
I am a lot of things and I am nothing.
Define me as you see fit.

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Comments

*rosscaughers

Feb 15, 2009, 1:36:37 PM

Thank you for the watch

--
[link] - profile

[link] - gallery

[link] - prints

~the-final-I

Jul 16, 2007, 7:36:47 AM

Oh dear, seems like you've left.

--
:
E^2= (p^2c^2)+([Δm^2]c^4)
which implies that
E= [mc^2]/[<sq. root>1-(v^2/c^2)]
when p=0 (in eq 1, or in 2 rearranged to 1])
E=mc^2
[as rest mass cannot be attained, p can never=0, so E should, not always be calculated as = mc^2].

~WinGzx18

Apr 2, 2007, 10:23:15 PM

I miss you megan. How are you?

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I'm a Photographer.
Click the link! : [link]

~the-final-I

Mar 22, 2007, 2:45:39 AM

What up, girl?

--
Did you know that:
E^2 = (p^2c^2)+([<delta>m^2]c^4).
which implies that:
E= [mc^2]/[<sq. root> 1-(v^2/c^2)].
when p=0 (in eq 1, or in 2 rearranged [which gives 1])
E=mc^2....
(In reality, p can never=0, so E should, realistically not be =mc^2, )