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About Me Member Deviously Deviant ajayusmMale/India Recent Activity Deviant for 2 Years
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:iconthe-final-i:
Hi, and welcome to DA.

Where in India are you based? [If you don't mind me asking, that is].

--
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, )

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