Is (or not) all the E.M. energy of a mass "m", that "m" multiplied by the E.M. Unitary Energy?
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From: Romano Amodeo on 29 Jul 2010 07:04 Please, open: http://www.youtube.com/watch?v=Fkfy1LSVXXI It is clear that all the ENERGY of a E.M. mass "m", is obtained by: "m" times the UNITARY E.M. Energy. It is possible to get it by valuations. 10 dm = 1 m, and the unitary SPACE is formed by 10 dm^3 put in sequence, whose lenght is 1 m and whose section is 1 dm^2. Now THE ABSOLUTE, of dm^3 10 is given by dm^3 10^10 = m^3 10^7. It is the UNITARY PRESENCE of 1 m^3. When this m^3 10^7 of PRESENCE exists in 10 times of presence, we have that 10 times 10^7 m^3 are 10^8 m^3... in the unit of the time of 1 s. So 10^8 m^3 s^-1 are also THE SAME 10^8 m s^-1 in line, since the section of 1 m^2 having that speed. So the speed of 1 m^2 is of 10^8 m s^-1 in each of the 3 dimensions xyz of the space, referring the VOLUME of this m^3. Therefore, in all these 3 directions, the speed of 1 m^2 is 3 times 10^8 m s^-1. This C (THEORETICAL) speed (of 3 times 10^8 m s^-1) is as electric, as magnetic. The ELECTROMAGNETIC ENERGY is the product between the C and the equal C... so it is C^2. It is the simultaneous advancement Electric and Magnetic of the UNITARY MASS contained in 1 m^3 of Electromagnetism. Now, the C (REAL) isn't the speed of 1 m^2, but of 1,0006922855 m^2 containing 1 m^2 as SPACE-UNIT and +0,0006922855 as TIME-UNIT... the precise unitary time equal to m^2 1444,4905513100^-1, getting m^2 0,0006922855 (since m 10^-10 dimension, that is the unit Angstrom, of the atomic space). Romano Amodeo EXPLAINS IT IN ENGLISH LANGUAGE only in this shape on youtube: http://www.youtube.com/watch?v=Fkfy1LSVXXI It is of the MOST IMPORTANCE, because it corrects even a mistaken of EINSTEIN... and excuse me if it is so insignificant !!! Amoram |