Recent content by rocky14159

Thanks

in calcluable is not the definition of an imaginary number, to the extent of my knowledge, imaginary numbers are formed when you try to take the square root of a negative number. Is this true?

not necessarily, but if u plot it on a graph, it will end when v=c, so it is incalculatable (for now)

I understand that F=dp/dt is the true law from Newton, but what do the variables stand for?

so in other words, a more accurate form of F=ma would simply be F=(ym)a where y is the Lorentz factor?

Borek, I can't thank you enough! Now this all clicks!

Not that one, the one that HallsofIvy posted in post 19

Hallsofivy, is there a name for the formula which you have provided me? thanks

HallsofIvy, thank you, the formula you have provided shows to me undoubtedly that you cannot go light speed, but very close to it, and i shall use 185999 for my velocity now, and I think i shall gain substantial knowledge and accurate answers from this.

I'm here, but I am in 8th grade and I have never even heard of any of this, but, I also got most of this information from other sources than school

So I take it no one has ever found the ratio between an object's velocity and its relative gained mass, am I correct in assuming this?

I am interested, and I wanted to see around how much U-235 you would have to fission (at 100% efficiency) to propel the space shuttle to light speed (i know its not that simple, but i wanted a round number to see how much you would need) and at the moment all I can do is use F=ma because that's...

There is no explosion after Cadmium-113 receives the extra neutron, this is the precise reason they use them to slow nuclear fission, because it allows the scientists to slowly pull them out, thus slowly increasing the rate of fission

You lost me. Could you explain that in terms that an 8th grader who's only in Geometry could understand?

Thanks, but can you tell me what the variables d, p, and t are? I am new here and not yet familiar with most full formulas