Re: Wheres weirdo?
- From: Prof Weird <poland@xxxxxxxxxxxxxxxxxxxxx>
- Date: Mon, 19 Oct 2009 08:17:11 -0700 (PDT)
On Oct 17, 9:04 am, spintronic <spintro...@xxxxxxxxxxx> wrote:
On 15 Oct, 16:17, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:> On Oct 14, 8:30 am, spintronic <spintro...@xxxxxxxxxxx> wrote:
.
.
<snip>
GIGO
You UTTER RETARDED SPASDICK!
screamed spintwitty at his reflection
From:http://en.wikipedia.org/wiki/Compton_scattering
hf + m_0C^2 = hf' + sqrt((p_m_0C)^2 + (m_0C^2)^2)
*****************************
hf = 0.43KgC^2
m_0C^2 = 0.1400000KgC^2
From what orifice did you pull 'hf = 0.43 kgC^2' ? Having troublekeeping your masses straight ? Again ?
You sure you got that formula correct ? AND that it applies to
something other than hydrogen atoms ?
****************************
hf' = 0.06KgC^2
ERGO
p = ( hf - hf' )/C = 0.43KgC - 0.06KgC
Get THAT???????
You GIBBERTARD???
p = *******0.37KgC********
If the 0.14 kg mass WAS COMPOSED OF PHOTONS, you may have come close
to having a point (given that is what the Compton effect deals with -
the change in wavelength of PHOTONS); is a 0.14 kg mass a photon ?
In other words, you are changing the question to conform to your
delusions, rather than accept that you are so confused that you can't
tell the difference between momentum, mass, and kinetic energy !
Again, oaf : the 0.14 kg mass would have MOMENTUM of 0.49 kg c, and a
KINETIC ENERGY of 0.37 kg c^2 J.
Once again buffoon : 0.14 kg rest mass, 0.86 mass to be converted into
photons.
The 0.86 kg converted into 0.86 kg c^2 J of energy.
Ultimately, 0.49 kg c^2 J to the right, meaning a MOMENTUM of 0.49 kg
c goes right.
As momentum and energy MUST BE CONSERVED, there is a MOMENTUM of 0.49
kg c going left - the MOMENTUM of the 0.14 kg mass.
Which corresponds to a velocity of about 0.96078 c, meaning there is
0.51 kg c^2 J of ENERGY going left (the 0.14 kg mass, plus 0.37 kg c^2
J of KINETIC ENERGY). Energy and momentum are conserved.
.
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