The pump draws water at the rate of 500 kg/min, which is not the same as 500 kg. You are then multiplying 500 kg/min by v2 in (m/s)2. I don't think the end result of this calculation is Joules. Ditto for the calculation of PE. And I'm stuck on part b, here is what...
K.E. = 2 Joules / sec (2 Watts) ? 4) thanks for the link - but the explanation on the Yahoo answers, highlights the point I made in my original post. Using the simple Ohms law relation V/R = I if V is fixed at 4.8V, and you use a very low resistance load - it would ap...