Bulletin of the American Physical Society
22nd Annual Meeting of the APS Northwest Section
Volume 67, Number 5
Friday–Saturday, June 3–4, 2022; Thompson Rivers University, Kamloops, British Columbia, Canada
Session H01: Poster Session (3:30-5:00pm, PT)
3:30 PM,
Friday, June 3, 2022
Abstract: H01.00010 : A great revolution in producing affordable, abundant, convenient, clean and… energy so that one kilogram of it, could supply the power of moving a machine for years!
Presenter:
Gh. Saleh
(Saleh Research Centre)
Author:
Gh. Saleh
(Saleh Research Centre)
ρ = me / Ve ⇒ ρ = (9×10-31) / (2.19×10-44) = 4.1×1013 (kg/m3)
It can be said that the product of velocity (s) in density (ρ) is an effective parameter in the high ability of electron to do work.
Eff = ρs
Now we calculate the energy of “n” electrons to do the work:
En = n (½ ms2)
En = n[½ ρs(sV)]
n=∆m / me
Where ∆m is the mass changes of the source before and after doing the work and me is the mass of one electron. So:
E_n=\frac{\mathrm{\Delta }\mathrm{m}}{{\mathrm{m}}_{\mathrm{e}}}\left[\frac{1}{2}E_{ff}\ sV_e\right]
s ≅ 3×108 m/s
Ve = 2.19×10-44 m3
Now by placing the density and speed of electron to calculate the energy of 1 gram of electrons:
En ≅ 1013 j
For 1 kg of electrons, we have:
En ≅ 1016 j
Now, consider a 2-ton car. It requires about 109 j energy to travel 100 km. Therefore 1 kg of electrons can easily provide the energy to travel more than 500,000 km. In this article we will explain the different applicable ways to produce such Electron Tank.
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