How do you find output energy given only Power (1500 W) and delta time (173 s) ?

Energy is power * time, so with what's supplied the answer will be in Watt-seconds. Not a typical unit, but a unit of energy nonetheless. You can easily convert to Kilowatt-hours if you like. This may be a unit your familiar with.

One of my supervisors once said that people like numbers between 1 and 10. Ok; 1 and 100; so 10e-9 amps becomes 10 nanoamps etc.

So if the answer in W-s, W-min, Kw-h yields a number between 1-100, that's the one to use, unless your comparing it to something.

Make sense?

umm actually I kind of get what your trying to say there. But my assignment is really based on just using the formula P=E/T and Eg=mgh :S buh thanku so much anyway :)

And P = E/T ---> E = P * T; Right?

Multiply both sides by T

These problems should tell you something. Units are just as important as numbers. So is precision and error bars. But you may not be there yet.

In some cases 1E15 is virtually the same as 3E15 and in some cases it's not - like money.

yea buh I think that would represent or give the input energy! To find output energy my guess was to use the formula q=mct
thanks so much again

q-mcT doesn't apply.

Actually P can be positive or negative. Positive P is energy disapated and negative P is power generated. In the formal sense this is true.

Thus a 16 MW power plant has P = -16MW. A house load of 1 KW is a positive value.

This makes P generated = P disapated

Only the sign changes.

So, if it's energy generated then the sign is negative because P is negative.

We usually don't care because we KNOW that a power plant generates energy, but the governing equations won't solve.

Usually equations like Power(generated)=Power(disapated) are much more difficult in the long run than Power(disapated)+Power(generated)=0

Multiply power and delta time. You will get the answer in J (i.e. Joule).
This is because Watt * second means Joule.