A 1.50-V battery supplies 0.204 W of power to a small flashlight. If the battery moves 8.33 1020 electrons between its terminals during the time the flashlight is in operation, how long was the flashlight used

981.41 secs

Explanation:

Parameters given:

Voltage, V = 1.5V

Power, P = 0.204W

Number of electrons, n = 8.33 * 10^20

First, we calculate the current:

P = I*V

I = P/V

I = 0.204/1.5 = 0.136A

The total charge of 8.33 * 10^20 electrons is:

Q = 8.33 * 10^20 * 1.6023 * 10^(-19)

Q = 133.47 C

Current, I, is given as:

I = Q/t

=> t = Q/I

t = 133.47/0.136 = 981.41 secs

false br

neither accurate nor precise.