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awaterpon

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About awaterpon

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    Baryon
  • Birthday 02/14/1987

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  • Favorite Area of Science
    Physics
  • Biography
    My name is Yahya A.Sharif I'm interested in inventing new theories

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  1. I did what all these parrot professors could not do. I have my own discovery with observations and experiments . Even though my discovery is fundamental no previous scientist was able to discover it. What evidence do you need than a clear experiment you did it yourself ? I am a great discoverer , greater than Newton or Einstein " both were not be able to discover this fundamental discovery" So why disrespect? Yes. And I completed what Newton started.
  2. I will use the equations wrong and they will be invalid for sure and I will use the equations right and they turn out to be wrong as I say Two kg of a 6 kg mass exploded and the rest 4 kg will be moved with part of this exploded energy: The rest mass is 4 kg the energy pushing it is the energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v2/c2) I will have a velocity v1 m=4/sqr(1-v12/c2) But m is 1+4 or 5 kg Then 4/sqr(1-v12/c2)=5 kg relativistic kinetic energy: K.E=m0c2/sqr(1-v2/c2)-m0c2 =4c2/sqr(1-(v12/c2))-4c2 But: =4/sqr(1-(v
  3. What I mean is only part of the 2 kg energy pushed the mass and this can be a physical situation
  4. I wrote the kinetic energy of the system at rest 5c^2 instead of S.R 2*5C^2 but this still give similar results I will rewrite it giving K.E at rest 2*5c^2 instead of 5c^2: I also would like to add this question: How a system contains only m0c^2 joules at rest have K.E of 2m0c^2? A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the
  5. It is a physical situation with few details. consider this: A rest mass of 6 kg, 2 kg of it exploded,it pushed the 4 kg with 1 kg energy and the other 1 kg went away. Now I have a new rest mass of 4kg and energy of 1 kg: relativistic mass m is= m=m0/sqr (1-v^2/c^2) I will have a velocity v1 m=4/sqr(1-v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1-v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1-v^2/c^2)+m0c^2 =4c^2/sqr(1-(v1^2/c^2))+4c^2 But: =4/sqr(1-(v1^2/c^2))=5 kg so K.E=5c^2+4c^2= 9c^2 jou
  6. Nothing, it is a matter of equations and that why I refer to the energy as pushing the system.The energy of 1 kg is a kinetic energy that the system can have. And it is true considering the atomic bomb energy.
  7. Suppose I have a rest mass m0 of 10 kg, part of this mass 1 kg converts to energy, E=mc² , E=e joules. In this case I will have a new rest mass m0 which is 10-1 kg or 9 kg . pushed by the 1 kg energy This system of rest mass 9 kg will move by the 1 kg energy with speed v1 in which the relativistic mass for the system will be : m=m0/√(1-v²/c²) or m=9 / √(1-v1²/c²) Because no energy comes from out side, then m will be 10 kg and this for speed v1 and rest mass 9 kg * Now let's calculate the kinetic energy of the system : The relativistic kinetic energy of the system is energ
  8. The concept is not just F=ma but also include gravitational equation F=GMm/r^2 If I push a wall with force "F" and got acceleration "a" then "m" will be my alternative mass: F=ma , m=F/a There must be gravity force that opposes when I lift myself so I can use this mass in the gravitational equation to get the small force of gravity on me that the scale shows or my alternative weight. F=GMm/r^2 Where F is my alternative weight and m is my alternative mass And the minimum force I lift my body with equals my alternative weight If I push a rock in space I can treat my m
  9. The ratio between the alternative mass and the actual mass is constant. I can use Newton's laws separately one for a person pushing himself and the other is another person pushing him. Newtonian equation F=ma deals with any mass ,acceleration and force ,I have separate equation one has actual mass ,force and actual acceleration, the other is for alternative mass,force and alternative acceleration. So: F1=m1a1 This when a person pushes himself F=ma This is when another external force pushes him.
  10. You will need to upload it in a site like YouTube copy the video URL insert it by one of the upper icons ,you also can give it a title in the text blank.
  11. The alternative weight is the force by gravity on a human which opposes human lifting himself Alternative mass is the mass with inertia that opposes body moving himself Unfortunately I live in a poor country ,these scales are rear only in clinics and hospitals ,and in the market it is too expensive for me to buy
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