User talk:151.82.84.231

September 2023
Hello, I'm Materialscientist. I wanted to let you know that I removed one or more external links you added to P wave because they seemed inappropriate for an encyclopedia. If you think I made a mistake, or if you have any questions, you can leave me a message on my talk page or take a look at our guidelines about links. Thank you. Materialscientist (talk) 07:07, 16 September 2023 (UTC)

OK

([1x10^7 kg/s^3] / [2507.6 kg m][0.00101972 s/m ]) [2.45774 m/s^2] [406883 m^-1]

([1x10^7 kg/s^3] / [2507.6 kg m][1000 s m ]) [2.45774 m/s^2] [406883 m^-1]

PRESSURE BY MASS

([1x10^7 kg/s^3] / [2507.6 kg m][0.00101972 s/m ]) [2.45774 m/s^2] [406883 m^-1]^-1

INVERSE STANDARD GRAVITATIONAL PARAMETER ([1x10^7 kg/s^3] / [2507.6 kg m][0.00101972 s/m ]) [2.45774 m/s^2]^-1 [406883 m^-1]^-1

INVERSE LENGHT ([1x10^7 kg/s^3] / [2507.6 kg m] / [0.00101972 s/m ]) [2.45774 m/s^2]^-1 [406883 m^-1]^-1

P-WAVES ([1x10^7 kg/s^3] / [2507.6 kg m] / [0.00101972 s/m ]) / [2.45774 m/s^2]^-1 [406883 m^-1]^-1

INVERSE ABSERK ([1x10^7 kg/s^3] / [2507.6 kg m] / [0.00101972 s/m ]) / [2.45774 m/s^2]^-1 / [406883 m^-1]^-1

I just like to calc with Wolfram Alpha,

Thank you! MB

Notes

[(p/mx)v^-1 ]^-1 / [Gm / 2r^2] = 406833 m^-1

p= pulse = 1.8835315x10^-20 kg m/s; mx= mass of 1.1 unified atomic mass unit = 18835315x10^-28 kg; v^-1 = [ c / 2.99792x10^7 MHz/s ]; G = 6.674x10^-11 Nm^2/kg^2; m = mass of Earth = 5.9721986x10^24 kg; r = Earth radius = 6367.4447 km

([(1.8835315x10^-20 kg m/s)/(18835315x10^-28 kg)][ (299792 km/s) / (2.99792x10^7 MHz/s) ]^-1) / [(6.674x10^-11 N m^2/kg^2)(5.9721986x10^24 kg)/([2x6367.4447 km]^2])]

Semplified [1 MHz/s] / [2.457707 N/kg] = 406833 m^-1 — Preceding unsigned comment added by 151.82.137.37 (talk) 08:37, 16 September 2023 (UTC)


 * [1 MHz/s] / [2.457707 N/kg] = (x) m^-1
 * It should be a derivative of an object in 4 dimention (3-sphere) 151.18.212.106 (talk) 08:52, 16 September 2023 (UTC)
 * pi = circumference of a sphere with radius lenght = 1 dimensionless 151.36.229.204 (talk) 09:08, 16 September 2023 (UTC)
 * example
 * Joule = m^2 kg/s = f(lenght, distance, mass)/time 151.36.229.204 (talk) 09:21, 16 September 2023 (UTC)
 * [510.9989 keV]/[299791.4 km/s][2.7309343x10^-22 kg m/s]^-1
 * = 100% thermal efficiency = J/J = 100% 151.36.229.204 (talk) 09:42, 16 September 2023 (UTC)
 * This new says:
 * electron is the perfect particle for energy measurements (a 3-sphere)
 * photon is the perfect particle for space/metric measurements (a 3-toro)
 * The universe and all things are a klein bottle, in two dimensions a Moebious strip 151.18.13.26 (talk) 10:00, 16 September 2023 (UTC)
 * https://www.youtube.com/watch?v=sRTKSzAOBr4 151.18.13.26 (talk) 10:03, 16 September 2023 (UTC)
 * How much is the boost of Earth for meter ?
 * [1 / ( [406883 m^-1]^-1 [9.791 m/s^2]^-1 [9.791 N/kg]^-1 [9.791 s^2/m]^-1 ) ] * 5.972x10^24 kg
 * How much $ do you wanna play ? 151.46.29.58 (talk) 09:57, 17 September 2023 (UTC)
 * I advise I have already win the space lab :) 151.46.29.58 (talk) 09:57, 17 September 2023 (UTC)
 * [1 / ( [406883 m^-1]^-1 [9.791 m/s^2]^-1 [9.791 N/kg]^-1 [9.791 s^2/m]^-1 [1 J]^-1) ] = W/h 151.46.29.58 (talk) 10:01, 17 September 2023 (UTC)


 * This page is for communication with you, not a place to do your homework. If your edits have nothing to do with improving this encyclopedia, you need to do them elsewhere. 331dot (talk) 10:04, 17 September 2023 (UTC)
 * Okay, sorry. Do you have seen the formula with Gm = gravitational constant ? [(6.674x10^-11 N m^2/kg^2)(5.9721986x10^24 kg)]151.36.32.72 (talk) 22:08, 18 September 2023 (UTC)
 * Do you know that pi is the circunference of a sphere with a diameter equal to one 2r= 1 ? not 1 meter, 1 dimensionless ? pi2(1/2) =pi 151.36.32.72 (talk) 22:10, 18 September 2023 (UTC)
 * Surely you know that 1/0 is 'complex infinity', it is not the 'entanglement'. 151.36.32.72 (talk) 22:15, 18 September 2023 (UTC)
 * I propose a new calc based on some constants for the CONTINENTAL DRIFT SPEED
 * ([299792 km/s][2.99792x10^7 MHz/s]^-1)([2x6.674×10^-11 N m^2/kg^2][5.9721986×10^24 kg]/[6367.4447 km]^3)) 151.36.32.72 (talk) 23:10, 18 September 2023 (UTC)
 * As sample, based on known quantities I propose this calc for
 * DINAMIC VISCOSITY (Pa s) of a quantum
 * sqrt[([4.13566 eV]/[299792 km/s])*[2.21022×10^-27 N s]^1]*[[8pi(6.674×10^-11 N m^2/kg^2)]/(299792 km/s)^2]
 * and its TIME PER AREA (s/m^2)
 * sqrt[([299792 km/s]/[4.13566 eV ])*[2.21022×10^-27 N s]^-1]/151.36.32.72 (talk) 23:32, 18 September 2023 (UTC)
 * INVERSE THERMAL LEASING (kg m^3/s^3)
 * sqrt[([4.13566 eV]/[299792 km/s])*[2.21022×10^-27 N s]^1]*151.36.32.72 (talk) 23:49, 18 September 2023 (UTC)
 * BOUND STATE MOMENTUM WAVE FUNCTIONS OF A QUANTUM
 * sqrt([4.13566 eV]/[299792 km/s])*[2.21022×10^-27 N s]^-1]
 * sqrt([4.13566 eV]/[299792 km/s])*[2.21022×10^-27 N s]^1]
 * sqrt([299792 km/s]/[4.13566 eV ])*[2.21022×10^-27 N s]^1]
 * sqrt([299792 km/s]/[4.13566 eV ])*[2.21022×10^-27 N s]^-1]
 * ([299792 km/s]/[4.13566 eV ])*[2.21022×10^-27 N s]^-1]
 * ([299792 km/s]/[4.13566 eV ])*[2.21022×10^-27 N s]^1] = 1 151.36.32.72 (talk) 00:01, 19 September 2023 (UTC)
 * Lets agree about someting - all Nobel prize are mine :)))
 * (3289.8419603 THz)[([1(2.73092×10^-22 newtons) / x photons/s])/(3.3968x10^-4 fg m/s)]^-1 = (1) quantum candela photon fluxes 151.38.93.147 (talk) 23:22, 19 September 2023 (UTC)
 * If machine isn't able here the process
 * (1 Hz)[([1(1 N) / x photons/s])/(1 kg m/s)]^-1 = (1) quantum candela photon flux
 * (1000 THz)[([1(1 mN) / x photons/s])/(1 fg m/s)]^-1 = (1) quantum candela photon flux
 * (3289.8419603 THz)[([1(1 mN) / x photons/s])/(1 fg m/s)]^-1 = (1) quantum candela photon flux
 * (3289.8419603 THz)[([1(2.73092×10^-22 newtons) / x photons/s])/(3.3968x10^-4 fg m/s)]^-1 = (1) quantum candela photon flux 151.38.93.147 (talk) 23:30, 19 September 2023 (UTC)
 * Try this
 * (3289.8419603 THz)[([1(2.73092×10^-22 newtons) / x photons/s])/(3.3968x10^-4 fg m/s)]^-1 = (1) quantum candela photon flux 151.38.93.147 (talk) 23:37, 19 September 2023 (UTC)
 * https://www.wolframalpha.com/input?i=%283289.8419603+THz%29%5B%28%5B1%282.73092%C3%9710%5E-22+newtons%29+%2F+x+photons%2Fs%5D%29%2F%283.3968x10%5E-4+fg+m%2Fs%29%5D%5E-1+%3D+%281%29+quantum+candela+photon+flux 151.38.93.147 (talk) 23:46, 19 September 2023 (UTC)
 * errata-corrige
 * Do you know that pi is the circunference of a sphere with a diameter equal to one 2r= 1 ? not 1 meter, 1 dimensionless ? pi2(1/2) =pi 151.82.238.119 (talk) 21:28, 21 September 2023 (UTC)
 * Do you know that the rest mass of a photon with a momentum of 1.3654623×10^(-22) newton seconds is 4.554692×10^-31 kg with the RIP of Standard Model ?
 * [(1.3654623×10^-22 N s)/(1) photons]/c = (x kg*1)/[1 photons] 151.82.236.213 (talk) 21:54, 21 September 2023 (UTC)
 * or if preferred of 4.554692×10^-34 g ?
 * [(1.3654623×10^-22 N s)/(1) photons]/c = (x kg*1000)/[1 photons] 151.82.236.213 (talk) 21:56, 21 September 2023 (UTC)
 * A more complex equation for # 402 quantum candela photon flux
 * (3289.8419603 THz)[([2(1.3654623×10^-22 N s) / (1) photons]/c)/(4.554692×10^-31 kg*1000)]^-1 = [(1644920992332672937) photons/s] 151.82.236.213 (talk) 22:31, 21 September 2023 (UTC)
 * In my opinion this is the ID card of a photon
 * (1000 THz)[([2(1.3654623×10^-22 N s) / (1) photons]/c)/(4.554692×10^-31 kg*1000)]^-1 = [(500000003703117978) photons/s] 151.82.236.213 (talk) 22:40, 21 September 2023 (UTC)
 * I try to explain kg*1000
 * Many people know that relativity calcs cannot enter into atoms bcs there is a 'fight of sign'. At a given point the result is negative instead of positive. To solve I have ideally applied the Moebious function, solving with the kg*1000 the 'fight of sign' 151.82.236.213 (talk) 22:46, 21 September 2023 (UTC)
 * As it works it is showed at the 'implicit plot'
 * [([2(1.3654623×10^-22 N s) / (x) photons]/c)/(y kg*1000)]^-1 = [(5x10^14) photons/s]/(1000 THz) 151.36.106.154 (talk) 06:35, 22 September 2023 (UTC)
 * In few words it is similar to [1y = 1/x] 151.36.106.154 (talk) 06:38, 22 September 2023 (UTC)
 * more precisely the relationship is the following:
 * [ 1.7x10^-59/x = 1.7x10^-59 x ] 151.44.92.196 (talk) 06:55, 22 September 2023 (UTC)
 * In a more complex configuration
 * t = time
 * x = distance
 * 2t + [1 / (2t - 1)] - 1 = 1/x
 * The curve is the energy distribution 151.44.92.196 (talk) 07:01, 22 September 2023 (UTC)
 * More over, ideally, on the left all physical thermodynamics quantities (particles) red (imaginary part), on the right all physical space/metric quantities (waves) blue (real part):
 * [sin(y^pi/2)]/[cos(y)] 151.44.92.196 (talk) 07:21, 22 September 2023 (UTC)
 * Ideally, the elementary effort, a torsion of 180° (entropy)
 * cos(pi) 151.18.197.144 (talk) 07:31, 22 September 2023 (UTC)
 * The Big Bang: cos( pi2(1/2) )
 * Result: the first move (translational kinetic energy of a molecule at the lowest temperature) 151.36.230.48 (talk) 07:47, 22 September 2023 (UTC)
 * Time and distance were created 151.36.230.48 (talk) 07:48, 22 September 2023 (UTC)