Fisicaimpulsivaingles

Introductory RelationshipClarifying note.- to make the lecture easy for the young, thepresent clarifying note is put forward, due that the previouswriting was not clear enough, was lacking of didactics andthe deductive issues that have to be filled by the readerwere not done.These very short calculations based on the theory ofaccretion metric variable nonlinear, discontinuous functionsmultifaceted and multidimensional, is a concern that I had 42years ago and nobody took me in mind, also in June last oneyear and previously traveled 15600 km to be heard but notgave me the slightest attention, to this public I expose,what follows, in early years I was opposed from thetheoretical point of view and the formal to the constructionof the Hadron accelerator, considering a waste of money andtime.The total failure of the General Theory of Relativity and theBig Bang, as well as the uncertainties of quantum mechanicsand other theories, of: strings, Global, etc.. make me to dothis compelling parentheses, and the present work isdedicated to the young.Now see what Mr. Einstein said.Relativity Theory Restricted and General said:1º. Invarivility of the speed of light.This is untrue because the current speed is a reduced speed cthe 6º degree and c = 0 , which can be prove by the xnaccelerator of Hadron. See experiment of 310 times the speedof light.2º. The equation that relates mass speed: Ε = mc 2 , is not suchsince Ε = δc = nhγx n© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 1

E= total energy, if you take in account m= what? This m makethe equation void, because of m=?.3º. The perihelion of Mercurys orbit is given by:drack f′ = f′ s m forces alveolar a′ a′ a´sm s ∗ n = mm ∗ m s s = sol ; m = Mercury = drackaccelerations xs n xm α sn a ′ mn x snα1 = s = ∗ n a′ m ms xmwhere :α1 = variation of the perihelionα1 = sec onds ∗ 1.296.000′′ ∗ 100 yearsthen : 3.30 ∗ 10 23 x snα1 = ∗ 1296000 n ∗ 100 years 1.99 ∗ 10 30 xm 2 3∗ 2∗ 3α1 = 21.4914572863 ∗ 1 3∗ 3α1 = 21.4914572863 ∗ 2α1 = 42.9829145726′′4º. The red shift is due to the realignment of photons formingthe axons.The actual light is and helix that is formed by the treedynamics forces of the intrinsic thiedron, which evolutesforms a straight line. λfγ = shift γ1 = γ c λc λfγ1 = γ c = 1.10γ c λc© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 2

Namely:The frequency varies towards red, infrared due to therecomposing of the photons, say the light had variable photondensitiy at the beginning of time and at end of therecomposing it fixed at given value.γ 1 = 1.10γ c i = (1,2,3,4,5)According to the leak λ f speed, frequency of light γ5º. The diversion of light from the stars is no due to gravity,or formation of space-time metrics.If CE passes near another, attempt to engage with the newsource CS=CE n XECE = CS n XSCE =αCS n XE 2α= n = XS 2 3α = 3 = 1.732050807561 6º. The existence of time and its variation with velocity ismore popular folklore and utilitarian, as anyone who has acar can see, if the gears or the resistance of hisspeedometer is failing in any case it can be assumed that isa variation at entropy, with depends on absolute temperatureαt, like it is shawn in the background of microwaves, thereis no time without variation of entropy, same physics havethe direction of entropy inverted now a days.7º. Gravitational and inertial systems are equivalent since inthe onset of the matter it organization equals to a dracksvalue of 1 so the matters is neutral with respect to the twoforces giving value of 1.The Big Bang is an absurd theory as the development ofgravitational, magnetic, fields,etc, it was achieve at the1010 at 109 years into the drivers resonance axon, creatingdrack fixed structures, as centers of energy accumulation,nebulae, galaxies, etc and then on another occasion I willgive the structure and dynamics of dracks structures and© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 3

other as well . It is to show that having a fixed force theproblem with the inertial cosmic are is solve by itselfQuantum mechanics uncertainty principle which is not such asħ is an expression of the metrics premater and there is notuncertainty as one can develop non-invasive tests that givethe approach of ∆x and ∆h with the approximation one whithmeet another time we discuss these further experiments;interference experiment is invasive and involves arealignment of photons. See experiments of cooling atoms.The desire to build a house made of bricks falling up andstanding alienated is absurd.String theory, global and so on. Provide no longerperspective, there are no hidden dimensions.I note that this brief introduction is only the first of aseries of avance work that I have and will enter on theInternet. Your Truly J.A.G.Ch. H.N.D Mech Eng© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 4

Applying the mathematics of the non-linear and discontinuousvariable metric multiforme to a space within which there areisotropic energy at Planck radio.We define:δ = Formative density forces drack dinas s.c.n = isotropy indexn = 1 isotropy n ≠1 Anisotropicr = radius Planck = 10-31 mt, 10-29 cmtxn = conformatory drack metricsλ = wavelengthγ = Frequencyh = Planck constant dracks widespread to dracks spaces.δ = drack conformative force.The elemental impulsive force δ it is descompose in sin φ,cos φ and variability of xn generating the intrincictrihidron of force giving rise to impulsivenes δ = ƒ(xn)n Then: xn = M0 iD φM0 = drack moduleiD = metric absolute drackφ = sen, cos mass √3 1 φ Energy √2 associated field, ∆ impulsive force ∆ variation o force© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 5

M0=n 3 n 2 2φ = Cosφ = , Cos nφ = n 3 3 1 1φ = Senφ = , Sen φ = n n 3 3iD = 3n n = 0,1,2,3,4,5,6, etc.iD = 30 = 1 31= 3 32= 9 33= 27 34= 8135= 243 36= 729 37= 2187 38=6561 39=19683310=59049 311=177147 312=531441 313=1594323314=4782969 315= 14348907316=43046721 317=129140163But we have to impulsivity dracks δcc= drack speedδc c c = nhγ here = 0 =c x n = Μ 0 i Dφxn x n 3Key roleδc = nhγ 30 = 1 x n = 30 , 31 , 32 , Μ0 φn=1 isotropy indexIf the drive drack exerts its action in space Planck we have:δ= 2π impulsors πrxn2πrxnc=nhγ n=12πrxnλγ=hγ2πrxnλ=h 2πr=c c=1 r= λ© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 6

1But here since λ = has to have the dimension of the radio 2πof Planck. 1 iD =0.729 x1032πrx n =h 2π x n =0.729 x103 M 0φ D irx n = h 8 24 2 2h = rx n x n =0.729 x103 6 18 22 3 3 3 8 24 2 2h =10 −29 0,729.10 3 6 18 22 3 3 3Assuming drack space is made up by the drakianas Planckspace. c δ = nhγ xn = 1 xn 2πrλγ = nhγ n =1 2πrλ = h h λ= = 6.6263799116 x10 − 27 2πr λ = 1.05462111773 x10 −4 cmt c = λγWe have to have that r = γ resonance in a drack slightlyanisotropy 1.7014491868 x1034γ= cmt 1.05462111773x10 − 4γ = 1.61332743882 x1038 c / sHowever:2π γ n λ=nhγ0 r xγ =1 hγ 0r= 2πx n λr = 1.61332743882 x1015 cmtr = 1.61332743882 x107 Kmt© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 7

Diameter that is confirmativeBut in conformity drack we have:δ =nhγ cn =12π n λ = hγ rx γ2π n λ = h rx 6.6263799116 10 −27 λ= x −31 6.28318530716 10 x19683λ = 0.535803037009cmtThen :δc = nhγn =1δλγ = hγx n δλ = h hδ= n λx 6.6263799116 x10 − 27δ= 0.535803037009 x19683δ = 6.28318530727 x10 −31 ergHere M 0φCosφ =1 being subject to a drive of 90ºNow we have the formation of photons, electrons, protons,neutrons, helium, and so on. Is a process due to theimpulsive activity drack must: mass √3 1 φ Energy √2 associated field,∆ impulsive force ∆ variation of force© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 8

cδ = nhγ xnn =1 cδ n = hγ xδλγ = hγx nδλ = hx nδx n = field rotation of fasor hm= xn m = δx n λphotónE1 = 6.28318530727 2 M =1 2 elemental fasorE1 = 8.88572028853 gr.c / s.10 −31Ie impulsivity drack becomes photon energy that drives andcreates the wave guide is no wave particle duality. 2E1 = 6.28318530718 3.10 −3 3E2 = 15.3906203948 gr.c / s.10 −31E3 = 51.0975267325.10 −31 gr.c / sE4 = 165.400453161.10 −3 gr.c / sPhotons are arranged around the axon in packs lists ofparticles equidistantly and synchronous, wich evolute is astraight lineBut the rotation of modulus of acting fasor Sen φ© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 9

1 Senφ = creates 3 Electron δx n = 6.28318530718[36 ].2M 0φ The equivalent drack  1 3 6 constant of matter δx n = 6.28318530718 x0.729 x10 3  3 6 .2 equals 1  3 3 312 Electron mass = 6.28318530718 x0.729 x 2 masa = 9.16088417786 x10 −28 gr Proton Proton [ ][ ] mp = 6.28318530718 x 2 312 x 313 313 mp = 6.28318530718 x 2.657205 x10 gr 6 mp = 1.669557114141x10 −24 gr 3: quark masses as you call then 3: electron or neutrino mass due Neutron to drack activity [ ][ ] [ ] mn = 6.28318530718 x 2 312 x 313 x3 36 mn = 1.67094527404 x10 −24 grThe typical mass Helio or any element of the periodic tableis: Helio mH = 2( proton + neutron) N º δ N º = Avogadro number N º = 6.022045 x10 23 mH = 6.283180718 x10 −31 N 0 2(5.316597) x10 6 mH = 4.02334803592 gr© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 10

Furthermore we have the impulsivity drack decreased theeffect of group fasores up.Modules and the speeds are acting fasores: 4 2 230 = 2 , 31 = 3 , 32 = 9 4 , 3 3 6 8 2 233 = 27 6 , 34 = 81 8 3 3Where a reduce module and argument 10 10 2 21x3 x9 x 27 x81x 8 = 59.049 8 3 3Giving a reduction of the speed of light: c = 1.7014491868 *10 34 cmt / s c c0 = n x 1 1.7014491868 x10 34 x mt 100 1.7014491868 x10 32 mt 0.56337770478 x10 5 c1 = 3.020086120493 x10 27 c2 = 5.36067738369 x10 22 c3 = 9.51524587893 x1017 c4 = 16.8896386885 x1012 c5 = 29.9792457976 x10 7 c5 = 2.99792457976 x10 8 mt / sSo the light is a reduce small order of 6 or multiples of itthereof.Xn drack spaces form argument:© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 11

30 = 1.4142135627731 = 2.4494897427932 = 8.1655118316933 = 25.665655273634 = 78.9731234754Where is ∑xn=116.667993886 impIf:c 1.7014491868 x10 32 = fr =c0 2.99786427756 x108f r = 0.56754093487 x10 24 fotonIe reduced impulsivity drack creates photons and maintainsthe current wave. fotón x n .5.4 = n = 1x10 22 foton axón ∑xThe distribution of these photons will be: x n 3o f1 = 4∑ x n f 1 = 1.70728225 x10 20 fotonf 2 = 2.95708329 x10 20 fotonf 3 = 9.8570348 x10 20 fotonf 4 = 30.98420015 x10 20 fotonf 5 = 95.3382658 x10 20 fotonIe the distribution is done in packages according to theimpulsiveness of fasores.But the formation of photons electrons, protons, neutronsmust be developed at lower temperatures due to the highimpulsivity drack energy© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 12

Stefan-Boltzmann Lawδc = σεΤ4c=1σ = 0.567x10-4 erg.cm-2.grado-4.s-1If this is a constant value must be maintained at drackcm-2 ° drack-2Photons have the temperature of formation of δc Τ4 = ε =1 σε 6.28318530716 x10 −34 c Τ4 = 0.567 x10 −4 c 1.7014491868 *10 34 cmt c0 = = xn 2 c1 = 1.20310625783 x10 34 cmt 1.7014491868 c2 = x10 34 cmt 2 3 c 2 = 6.9446137219 x10 33 cmt c3 = 2.0837018203 x10 33 cmt 6.28318530716 x10 −31 x1.7014491868 x10 34 Τ4 = 0.567 x10 −4 Τ 4 = 13.3321685405 x10 7 Τ = 1.07454646258 ∗10 2 Τ = 107.454646258 0 K Τ24 = 76.973310957 ∗10 6 0 K Τ2 = 93.666661658 0 K Τ3 = 23.090449294 ∗10 6 0 K Τ3 = 69.319926869 K0© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 13

As the speeds are a function of the drack space: c 1.7014491868c0 = n = 1 1 x  2 ( ) 2  24 8 2 2 8 2 x0.729  6   8   24 3 3 ∗10 3    3  3    1.7014491868 ∗10 34c0 = 1.4218551598 ∗10 3Speed electronc3 = 1.19664030128 ∗10 31 cmt 6.28318530716 ∗1.19664030128Τ4 = 0.567 ∗10 −4Τ4 = 13.2605163297 ∗10 4Τ = 19.0827125616 0 K electrónIn addition the frequency varies with drakianas, so we have:δc = nhγ n =1 δc 6.28318530718 ∗10 −31 ∗1.19664030128 ∗10 31γ= = h 6.6263759116 ∗10 −27γ = 1.13456824491 ∗10 27 c / s para el electrónfotón1 = 1.1329389656 ∗10 30 c / sfotón2 = 1.99237810378 ∗10 29 c / sThese are values training impulsivity is not due to reducedvalues.The wavelengths are associated: γδλ = nhγ n =1 xn hδλ = h λ= δλ f = 1.05462111716 ∗1.414213566273λ f = 2.46286912954 ∗10 −4 cmtBut due to the reduction photonics activity© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 14

cc = λγ γ = λγ 1 = 6.92864817601 ∗ 1014 c / s violet lightγ 2 = 6.20691398797 ∗ 10 c / s 14 blue lightγ 3 = 5.62135606452 ∗ 10 c / s 14 green lightγ 4 = 5.13675640384 ∗ 10 c / s 14 yellow lightγ 5 = 4.72907732417 ∗ 1014 c / s orange lightγ 6 = 4.3813115033 ∗ 1014 c / s red lightγ 7 = 4.08125851264 ∗ 1014 c / s infrared lightThe ratio of frequenciesγ1 = 1.11627906976 = R1γ2R2 = 1.10416666667R3 = 1.09433962262all R ≅ 1.10The electrostatic field of an electron is cδ 1 = λhγ n =1 x λγ Qδ n = hγ para δ = f (xn ) δ= x xn 3 ( ) Q λ =h(x n )3 x n h( x n ) 4Q= λ 6.6263799116 ∗10 −27 ∗ ( x n ) 4Q= λFor the electron, photon, protons, neutrons© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 15

6.6263799116 ∗10 −27 ∗ 2(0.729) 4 φM 8 ∗1012Q= 1.05462111772 ∗10 4Q = 3.54911402739 ∗10 −11φM 04 1 1  8 2  2  24 2  2 ( )Q = 3.54911402739  6   18   24 3   3 8   3  3  Q = 4.87901528331 ∗10 −11 g / cm 3 s to electronBut the gravitational field or specific gravity G reverse is:.δc = λhγ n =1 λγ λδ n = hγ δ n =h x xIf we call:δ = G( x n ) 3 3 λG( x n ) n = h N º n º Avogadro x hN ºG= λ( x n ) 2 6.6263799116 ∗10 −27 ∗ 6.022045 ∗10 23G= ( 1.054621117734 ∗10 −4 ∗ 2 0.5314417 ∗10 6 φM 4 ) 71.198166233 ∗10 −11 8 2G= φM 4 = 32 2φM 0 4 8 3 71.198166233 ∗10 −11 cmt grG= ∗10 2 2 ∗10 3 8 2 M Kgr 2 ∗ 5.19615262697 8 3 71.198166233 ∗10 −11G= 10.3923052534 ∗1.02566539646 M3G = 6.67961159114 ∗10 −11 s Kg© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 16

But if we have to hN ºG= λ( x n ) 2Gλ( x n ) 2 =h NºThe forces are an a plane and are the δ coplanar coincidentwe must have:δ forcesG ( x n ) δδ 2 λ 2 N = hδδN 2 Nº hGδx nδx n N º = δNδN λGm1m2 = δ N 3 3Being coplanar forces the vectors: acts ona plane giving:: x1 x2 m1 m2d 2 = x12 + x2 and that the forces acting 2according to its gravitational axisformation, the centers will be line withthe center of mass and create d2 y1 y2Gm1m2 = fd 2 thatF = ∑ f1, f 2, f 3 Gravitational centerFurthermore, the two forces are coplanar and third force at90 degrees from the other two, is independent. The other twoforces are coplanar system represented a system according tothe value of f1, f2, f3This system of two forces say f2, f3 develops a conic curvewith regards to their values, the 3 forces is a alveolar.© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 17

f1 been the third force tangent, is the alveolar 90 of theother two.But the interaction between a mass drack and drakianas willbe: λγ fδ n = γ δ = n′ f = force x x fλ n′ n = x n′ = var iation of impulsive dracks spacex Nº x  f = x n N º x n′ Nº xn = m λ λ n′f = mx where a x n′ = d (i DφM 0 )That is the variability of the argument in drakianas, formand frequency.If x n = a called when the modulus and the argument, get afixed frequency:f = ma ( Newton force)That is having an argument, modulus and fixed frequency,acceleration drackiana becomes a variable accelerationNewtonian gravitational field, having a fixed o the previousfield is constant q.But the force between two particles of electrostatic chargeswill be:q= λ (x ) h n 4 x1 x2q1 q 2 = ( x n ) q 2 h 4 λq1 q 2 N º N º = N º ( x n ) ( xn )4 h 4 h f1 f2 λ λQ1Q2 = f1 f 2 ( x n ) ( x n ) d 2 4 4d 2 = x12 x 2 2f1, f2 can be attractive or repulsive depending on the value ofthese f due to the variable field© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 18

f1, f2 resulting from the forces acting according to its axleloads can be attractive or repulsive depending on the shapeand position of fasores of drakianas. 1 1(x ) n 4 4  8 2  2  24 2  2 = 2( 0.729 ) ∗1012  6   18  ( 3) 24 8  3  3(x ) (x ) n 4 n 44 3 = 79.356703007c / s ∗10 24ε 0 = 79.356703007 ∗10 24 c / sQ1Q2 = f1, 2ε 0d 2But the specific magnetic permeability will be:δc = nhγ n =1 λγ λδ n = hγ δ n =h x x fBut the electric field, E = then we have: qδc = nhx n n =1δc λγ = hγ δ n = hγxn xδ = f (xn ) f f h λ = hx n = xnx n x n λ© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 19

We derivate about xn( x )′ n f =  ∗1 (x ) n 2 λ fE= =  xn ( ) 2 λ = conformativ q λ q 6.6263799116 ∗10 −27 ∗1.16220586581 ∗10 6E= 1.05462111772 ∗10 4 ∗ 4.87901043989 ∗10 −11 E 1.49668768083 ∗10 −14 EV0 = = = Volts d 1.344602848737 ∗10 −16 dV0 = 1.1119286812 ∗10 2 Vλ = 1239852 ∗10 −6 eV 1.239852 ∗10 −6V1 = = 1.1756373726 ∗10 −2 ∗100 1.05462111772 ∗10 4V1 = 1.1756373726eVV0 = Volts ∗ eV = 1.1119286812 ∗10 2 ∗1.1756373726 131.587540692 131.587540692V0 = ∗1 = ′ ( ) 2 2 xn  1   1  20 4 −  8  ∗  24  ∗ 3 ∗ 3 ∗ 4 3     ( )  3  3DerivateCosφ = −SenφV0 = −13.650613879eV© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 20

However, we find the equivalent volume drack πγ = 4πγ 3 ( x n )  04 3 2  0 = radio eq.3γ 3 = 3γ 2 ( x n )  0 2 ( 3) 8 2 20 = 4∗3 6 24 3 0 = 11.1124669808cmt to proton + electronγ 3 = 3 * 10 −58 ( 2.658663) * 11.1124669808cm 3 * (10 6 ) 2 2γ 3 = 21.2054668426 * 10 −46 * 11.1124669808cm 3γ 3 = 235.6450501cm 3 * 10 − 46The volume to be an atom:V = πγ 3 ( x n ) 4 3 3V = π ( 235.6450501x10 − 48 ) ( x n ) 4 3 3 3V = 4.18879020476 * ( 235.6450501) ∗ 10 −48 * (2.658663) 3 *1018V = 18.5 * 10 − 26however mρ= mass = m Volume = V V 1.6695711414 * 10 − 24 grρ= 1.85 *10 −26 cm 3 grρ = 0.90005308252 3 x10 − 4 for hydrogen cm© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 21

Now lets look at the magnetic field, so we have: δ = f ( xn ) cδ n = nγ x fm  nδλ = x n = x (x ) n 5 λfm = δ ( xn ) 6 1 1 − 31 6  8 2  2  24 2  2 24 1f m = 6.283185307 *10 * 2(0.729)  6   8   3  3  32 ( )     − 31f m = 6.283185307 *10 (0.15009463529) *10 *1.65089130707 18f m = 1.55691003883 *10 −13 dinas field √3 1 φ Energy √2 associated field, ∆ impulsive forcef m = ϕ 0 effort magnetisedBut if we consider a charged particle in its speed with aspeed v we have: f qf m ⊗δ = ϕδ fm ⊗ = ϕ0δ δ= xn (x ) n 3 f qfm ⊗ = ϕ0 x n ( x n )3f m ⊗ f ( x n ) = ϕ0 q 2© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 22

but V=volumen=v  2fm ⊗ q f ( x) 2 = ϕ 0 V Vf m ⊗ f ( x)  2 q 2 = ϕ0 v s 0fm ⊗ f ( xn )  2 1 2 = ϕ0 I v 0q =Isf m ⊗ f ( x n ) = ϕ 0 i 2f m ⊗ f ( x n ) = ϕ 0i 0φ 2a 90 0 f γ = ϕ 0i 0φ v= sI =iSeS e = seg =  0φφ = senφ cos φ© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 23

Now if we consider that cδ n = nγ n =1 xV0 = voltage d = dis tan ceδc = γx n  n f m x n δx n qfm xn = x = f m = ϕ0 δ = λ qd qd (x ) n 3 q xnϕ = (x ) n 3 = 1 ϕ E = = V0qd qd (x ) n 2 d qd q ϕ( xn ) 2 ϕ 1V0 = voltage V0 = = = qd ( x n ) 2 d qV0 q ϕ 0 ( x n ) 2 q = s = área =I s s s φ0 ( x n ) 2 φ φ0 n 2 φ V0 I = Vo ICos = ( x ) Cos s Se s Se φ0 n 2 φ0 V0i cos φ = s (x ) γ s = magnetic excitation fmc γ n 2 xn = γ x n fm = c (x ) fm = δ ( xn ) 2© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 24

Before turning to the drakianas fixed structure and variableseeNº (Avogadro number)= Mγ x Moϕ 8 24 2 36 2Nº=0.56337770478x10 x 24 3 8 3 30 3Nº=6.0218977618x1023 atom x gram atomIe the expression is translated into the reduction of numberof atoms in a mole or gram atom, the difference is due to theexperimental value of the fasors and the initial velocity ofthe axon.It is noteworthy that∂c ∂c1 = if we multiplyx n x1n∂cx1n = ∂c1 x n ∂x1n = mass of proton ∂x = mass of electron n n n cx c x1 c cmp n = me 1 n as n = 1n x x1 x x1i.e 1.66957114141x10−24 x1.458 = 2.657.205 x106 gr 9.16088417786 x10−28 g24.3423472417 x10−22 gr dr = 24.3423472417 x10−22 gr drIe orbitals are distributed according to the masses anddrakians, it is expected that the electrical loads andmoments are compensated equally, and then we develop thistheme.Further clarifying note.- Also being ϑ fixed is constant andonly depend of the masses and distance that meant that the socall Mach principal if does not exist due that the drackconstant of mass is one, the so call inertia exist has not© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 25

relevance due that 1 is 1 in any part of the universe and thecosmological principle is demonstrated due to Ε = δc = nhγx nIt means the universe only depends on xnBut due that n=1 were matter is present will be equivalent.Newton first principle of inertia is for a world withoutmasses, but any way was rectified when gravitation fixed wasput forward and the action – reaction principle was named.Sorry not written more as I promised time is a moody thingdue to the entropy je je.© 2009 Copyright Antonio Gutiérrez Chávez fisica.impulsiva@gmail.com 26

Fisicaimpulsivaingles

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