Originally posted by sonhouseWatts are power, not energy.
What do you think I defined? E. One megawatt. Jeez.
Getting close to c is what the fitzgerald contraction formula is all about.
Did you read my PM's?
Of course you add kinetic energy to a body under acceleration, call it a battery, so what? Ion rockets do that by accelerating at a constant but low acceleration, the total kinetic energy always goes up. That's just the way the cookie crumbles!
Originally posted by joe shmoYou only have one power of t on top and t^2 on the bottom. They do not cancel out.
This is explicitly solvable as stated
"one megawatt applied for one second can give you how much mass the thing can accelerate for one second(sonhouse)" (You also stated how they were the same thing, that however is incorrect)
none of the others (from you OP and including up to the above statment) had solutions as stated.
Here is the solution to yo ...[text shortened]...
P*t = 1/2 *m*(a*t)^2
m= (2*P*t)/(a*t)^2
using your numbers I come up with 20,825 kg
EDIT - Let me check:
Pt = K
(m d^2/t^3)(t) = (m d^2/2t^2)
(m d^2/t^3)(t) = 1/2 m(at)^2
2(m d^2/t^3)(t)/(at)^2 = m
2P/ta^2 = m
See? The mass depends on the time given constant power and acceleration.
Originally posted by AThousandYoungit shows you that
Those are all correct, but I don't see the relevance.
Power*time does indeed = energy
and the kinetic Energy is 1/2 *mass*velocity squared
since we know that the velocity is defined as acceleration* time
you can sub it in and see that Energy still = Energy
so if you solve the equation for mass you will indeed get a mass.
Originally posted by joe shmoI agree with all of that, except the last sentence. The last sentence does not follow.
it shows you that
Power*time does indeed = energy
and the kinetic Energy is 1/2 *mass*velocity squared
since we know that the velocity is defined as acceleration* time
you can sub it in and see that Energy still = Energy
so if you solve the equation for mass you will indeed get a mass.