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5 PoorAbout awaterpon

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 Birthday 02/14/1987
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My name is Yahya A.Sharif I'm interested in inventing new theories
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The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
It is all here Thread. 
The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
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. 31 replies

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The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
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 (1v2/c2) I will have a velocity v1 m=4/sqr(1v12/c2) But m is 1+4 or 5 kg Then 4/sqr(1v12/c2)=5 kg relativistic kinetic energy: K.E=m0c2/sqr(1v2/c2)m0c2 =4c2/sqr(1(v12/c2))4c2 But: =4/sqr(1(v 
The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
What I mean is only part of the 2 kg energy pushed the mass and this can be a physical situation 
The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
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 (1v^2/c^2) I will have a velocity v1 m=4/sqr(1v1^2/c^2) But at the 
The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
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 (1v^2/c^2) I will have a velocity v1 m=4/sqr(1v1^2/c^2) But at the same time m is 1+4 or5 kg Then 4/sqr(1v1^2/c^2)=5 kg relativistic kinetic energy: K.E=m0c^2/sqr(1v^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 
The problem of free energy in the special theory of relativity
awaterpon replied to awaterpon's topic in Speculations
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. 
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 101 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/√(1v²/c²) or m=9 / √(1v1²/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

A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
I PM you with some suggestions. 
A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
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 
A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
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. 
A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
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. 
A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
Flesh.Perhaps bones as well 
A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
Why not you use your scale and try it yourself? 
A mass can be be lifted with force less than its weight
awaterpon replied to awaterpon's topic in Speculations
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