Recent content by ozma

i did average V / R = (1.5 / 2) / 5 = 0.15 I = dQ / dt --> 0.15 = dQ / (2 x 60 x 60) --> dQ = 1080 C and it's correct! finally! thanks for your help.

I don't understand why the average V would be 1.5 / 2. Shouldn't I divide by the time in seconds? 2 hours = 2 x 60 x 60 seconds, that's why I used this number as the denominator. Q_{total}=\frac{mt^2}{2R}|_{t_i}^{t_F} What numbers do I put in this? Is m the slope of the graph? Would...

So I tried this: average of the voltage (1.5 V) over 2 hours = 1.5 / (2 x 60 x 60) = 2.1 x 10^-4 average I = average V / R = 2.1 x 10^-4 / 5 = 4.2 x 10^-5 average I = dQ / dt --> 4.2 x 10^-5 = dQ / (60 x 2 x 2) --> dQ = 0.01 but this answer is still incorrect. Any ideas? Thanks.

total charge lifted by the "charge escalator" Homework Statement A 1.5 V flashlight battery is connected to a wire with a resistance of 5.00 ohms. The figure shows the battery's potential difference as a function of time. The slope is linear going from (0, 1.5 V) to (2 hrs, 0)...