 # 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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1. The whole human body functions through signals carried by the nervous system from the brain to the heart, digestive system, lungs,etc and knows how the processes should work . If I cut off my arm, the arm will no longer be part of the body and can be treated as an ordinary object no signals on it, if I lift it it will press the knees just like the rock, and although it is human body on human body " upper part on knees "lifting another human will also behave as the rock because the other human is not part of the body.
2. The knees do not press, the mass does. The object presses normal but the body knows what it presses, if it his own knees it will press slightly so the human survives with such massive 40 kg. Think of a human jumps 5 years, walk 40 years, stand 30 years, run 20 years ..... How the alignment will help him to do all of this carrying a massive body of 40 kg for 70 years ? and think of the rock aligned perfectly as the body , the force on the knees still the weight of the rock which is massive for the knees to bear for hours
3. In other words replacing the body above the knees with any other object will press much harder than the body does both the body above the knees and the object have the same mass. That why the knees can bear mere body for years but if replaced by any other equivalent object the knees will fail after few hours.
4. The human joints bear an average human body of 60 kg for years without joints damage. Let's say the body above the knees for a human is 40 kg. An average human knees bear a 40 kg body above the knees for years without knees damage. If an object like a rock of 40 kg is fixed to the upper part, the knees will bear the rock of 40 kg for a short period of time, minutes, hours, days, before the knee's damage. The time the knees bear the upper part with no rock is years, the time the knees bear the body with the rock is several hours. First the knees bear 40 kg upper part for years, then the knees bear a double of 80 kg for hours. Even though the mass doubled, the time of bearing must double as well, but it actually multiplies by years or thousands of hours which is a very big number compared to only several hours. The force of human upper part on the knees is very small compared to the force of any object of the same mass on the knees. That why human walks on his knees carrying his upper part for years, but he walks on his knees carrying an object of the same mass for only several hours. 1) I have an upper part of 60 kg and I lift a rock of 60 kg : I put the rock on stomach and back equally, I have 60 kg upper part before putting the rock and 120 kg after putting the rock. The period of time my knees bear the rock plus my upper part or 120 kg can be approximately 5 hours. The time my knees bear when I remove the rock should not exceed approximately 10 hours because I removed half of the load. But when I remove the rock, the time my knees bear is years. I left with upper part body alone, and human knees bear a 60 kg human upper part for years. This difference in time is because a human body alone presses knees with tiny force and this tiny force make knees bear this upper part for years even though the bearing should not exceed 10 hours The difference between knees bearing 60 kg upper part for years and knees bearing 120 kg for 5 hours is very big. 2) I have an upper part of 40 kg and I lift a rock of 40 kg : I put the rock equally on back and stomach. The total weight I carry is 80 kg, it is the rock 40 kg plus my body above the knees 40 kg. Now we have a person of upper part 80 kg, this person does not carry any load. I will lift a load of 80 kg, which is my upper part 40 kg plus the rock 40 kg, the person will lift an 80 kg load which is his mere upper part. Carrying a rock of 40 kg" 40 kg rock plus my upper part 40 kg or 80 kg" for a day will damage the knees. However, the person's knees do not injure even if he carries his upper part of 80 kg for many years.
5. It is all here Thread.
6. 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.
7. 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-(v12/ c2))=5 kg so K.E=5c2-4c2= c2 joules The total mass 5 kg did not change when it moved with the v1 velocity it is internal Mass/Energy conversion then I can consider the 5 kg as rest mass . If I can consider the 5 kg as rest mass then the kinetic energy of the 5 kg is also zero , the above equation gives the 5 kg kinetic energy of c2 and that means a according to the S.R equations free energy of c2 joules is generated.
8. What I mean is only part of the 2 kg energy pushed the mass and this can be a physical situation
9. 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 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 joules Because the whole energy of the system with speed v1 does not change " energy from inside mass" then I can consider the 5 kg as rest mass 1 kg energy and 4 kg mass and both as rest mass because energy and mass are equivalent in which K.E=2mc^2 or 2*5c^2 = 10c^2 joules instead of 9c^2 joules S.R K.E
10. 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 joules The actual mass is 1+4 kg, and the energy of this system must be: E=mc^2 = 5c^2 joules instead of 9c^2 joules
11. 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.
12. 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 energy of the rest mass or m0c^2 plus energy due to motion: K.E= m0c²/√(1-v1²/c²) + m0c² K.E= 9c²/√(1-v1²/c²)+9c² According to * and because m0=9 kg : 9 / √(1-v1²/c²)=10 kg K.E= 10c²+9c² or 19c² joules : However the whole energy of the system must be the converted mass energy 1 kg plus the remained rest mass 9 kg which is 10 kg or E=10c² joules but K.E is 19c² joules which also must be the whole energy of the moving system This means the system generates 9c^2 joules free energy while it moves with some speed v1
13. I PM you with some suggestions.
14. 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 mass as a smaller mass "alternative mass " but when the rock pushes me I will treat my mass as the actual mass. Both forces are equal.
15. 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.
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