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Finding the TNT Equivalence with psi or pascals?

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Ok, so I would like to know what some of the values are in trying to solve psi for TNT equivalent.

I am using this calc with Ivan Drago https://vsbattles.fandom.com/wiki/User_blog:Kkapoios/Ivan_Drago's_punching_strength.

So I would like to know how the T= time was found before velocity was found, what do the triangles mean just after Force was found.

I would like to know more but I would like to find all the values that would equate to the Kinetic Energy formula mass, velocity.
 
Welp, it seems like: he measured the distance that moved the machine, then cleaning the equations of momentum conservation and acceleration he finds the timeframe in what the energy interchange happened, then back again to momentum conservation to find speed of both the machine and the hand to finally use the typical KE equation.
 
Antoniofer said:
Welp, it seems like: he measured the distance that moved the machine, then cleaning the equations of momentum conservation and acceleration he finds the timeframe in what the energy interchange happened, then back again to momentum conservation to find speed of both the machine and the hand to finally use the typical KE equation.
Ok, So really it is just to find the velocity with using the Force and mass of both the fist and the machine?

I just don't really know what each values represent those factors like would (x)=distance between fist and machine? (v)=acceleration? t=time?
 
He scaled the the distance that covered the machine before stopping via simple scaling (thats x), v = velocity and t = time; with the mass and the velocity one can find KE. Some issues is that he assumed the mass of the machine, and one punch doesn't use the entire mass of the arm.
 
Antoniofer said:
He scaled the the distance that covered the machine before stopping via simple scaling (thats x), v = velocity and t = time; with the mass and the velocity one can find KE. Some issues is that he assumed the mass of the machine, and one punch doesn't use the entire mass of the arm.
Ok, thank you then, I'll just have to apply the momentum convertion and acceleration then.
 
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