the energy of a proton moving at 0.999999991c

[Random Variable]

It's 7 teraelelectronvolts (or 1.12152352 × 10-6 joules), but I can't seem to derive it correctly.

E = [(mc^2)^2 +(pc)^2]^(1/2) = [(mc^2)^2+(mv/((1-v^2/c^2)^(1/2))*c)^2]^(1/2)

where m is the mass of the proton, c is the speed of light, and v is the speed of the proton? Is that right?

This thread of forum continuum has lost its momentum and is hereby stopped. - Moderator Rubycon

 

[buck]

Jeebus, another thread?

 

[Vehemence]

wat

 

[buck]

I don't see how this post helps me fit into my diesel jeans....

 

[LemonHead]

Originally posted by: Random Variable
It's 7 teraelelectronvolts (or 1.12152352 × 10-6 joules), but I can't seem to derive it correctly.

E = [(mc^2)^2 +(pc)^2]^(1/2) = [(mc^2)^2+(mv/((1-v^2/c^2)^(1/2))*c)^2]^(1/2)

where m is the mass of the proton, c is the speed of light, and v is the speed of the proton? Is that right?

You forgot to factor in the power derived from the Flux Capacitor and the 1.21 Jigawatts!

 

[Random Variable]

Originally posted by: buck
Jeebus, another thread?

yeah, but a serious one

 

[darkxshade]

Originally posted by: Random Variable
It's 7 teraelelectronvolts (or 1.12152352 × 10-6 joules), but I can't seem to derive it correctly.

E = [(mc^2)^2 +(pc)^2]^(1/2) = [(mc^2)^2+(mv/((1-v^2/c^2)^(1/2))*c)^2]^(1/2)

where m is the mass of the hardon, c is the speed of collusion, and v is the speed of the proton? Is that right?

now it should make more sense

 

[buck]

Originally posted by: Random Variable

Originally posted by: buck
Jeebus, another thread?

yeah, but a serious one

Why so serious?

 

[dighn]

the stated figure is the kinetic energy. you have the total.

 

[IceBergSLiM]

is the energy on a treadmill or not?

 

[firewolfsm]

Wait...what's the speed of dark?