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What is E/mc2?


ryan2006

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[math]E=mc^2[/math]

 

[math]\frac{1}{mc^2} \ \cdot \ E=mc^2 \ \cdot \ \frac{1}{mc^2}[/math]

 

[math]\frac{E}{mc^2} = \cancel{mc^2} \ \cdot \ \frac{1}{\cancel{mc^2}}[/math]

 

[math]\frac{E}{mc^2} = \frac{1}{1} = 1[/math]

 

So, yah, 1.

 

Mathematically speaking,

Buffy

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what is the value you place on E=MC2

 

c squared is the speed of light in vacuum, or 300,000,000 m/second, times itself. So it's a huge figure: 8.98755179*10^16 m^2/s^2. Using this as a constant in the equation means that E and m are variables, and the implication is that a *lot* of energy is yielded by a tiny amount of matter. It also takes a *lot* of energy to create matter.

 

The formula says nothing about the speed or method of conversion, nor does it give any values for E or m.

 

So the examples of 1 and 0 above are just figurative. Since E=mc^2 is basically the same as saying x=y, then 1=1 is *equally valid* - but it is not the *same* equation.

 

I'm probably just confusing this more so I'll shut up now. :pirate:

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although there is a slightly philosophical twist to that one.
Slightly philosophical? :sherlock:

 

...the implication is that a *lot* of energy is yielded by a tiny amount of matter. It also takes a *lot* of energy to create matter.
No, it's that we use such tiny units of energy, except for those such as gramme, kilogramme etc.

 

Anyway the answer to the original question is: 1 according to the rest frame and more than 1 according to other frames.

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OK Guys - Let's not let this get out of hand. It's pretty simple:

E divided by MC2 = 1; E minus MC2 = 0

 

There, now both one and zero are involved in variants on Einstein....Ryan should be happy, and the rest of you might consider not thinking 'zebras' every time you hear hoof beats.

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