Re: Wheres weirdo?

On 19 Oct, 16:17, Prof Weird <pol...@xxxxxxxxxxxxxxxxxxxxx> wrote:
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:

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<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 trouble
keeping 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.



It is ***YOU*** & **WILLIAM** who are **CONFUSING** this whole
process.
You **CAN'T** comprehend the thought experiment & so change the rules
to argue.

You **TRY** to incorperate compton scattering, which **WILL NOT** work
here!!


See this below?

--------------------------------------------------------------

^ That (up above) is called a **STRAIGHT LINE**!.


Forces like **GRAVITY** act in **STRAIGHT LINES**!

YOURS & WILLIAMS setups ***DO NOT** act in straight lines.



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.


ABSOLUTELY IN NO WAY SHAPE OR FORM IS THAT CORRECT!


Your 0.06Kg's of deflected PHOTONS ***ARE NOT** deflected back in a
straight line.


The rocket has

p = sqrt ((0.51KgC^2 - 0.14KgC^2)/C^2) = 0.490407993KgC

NOW 0.49KgC **IS NOT** 0.490407993KgC

Using the "law of Parallelogram's" what you propose is
**IMPOSSIBLE**!!

As the **MAXIMUM** allowed momentum is

0.43 + 0.06 = 0.49 and is **LESS** than 0.490407993

SO YOUR PROPOSAL IS IMPOSSIBLE!


Now, THIS ROCKET HAS TO TRAVEL IN A STRAIGHT LINE JUST AS THE LINES OF
THE FIELD DO.

THE ONLY WAY TO DO IT IS THE WAY **** I *** PURPOSED!


1KgC^2 Rest
|
<--0.51KgC^2 = E' |
|
-p = 0.49KgC | p = 0.49KgC


V = P/m' = 0.960784314C

Rest mass = 0.141421356Kg


Kapiche??

YOUR 0.06KgC DOES NOT GO RIGHT, it goes off at some angle,
Your rocket DOES NOT go left. It goes off at some angle.

































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