Question

a digital clock reads 8:15:00 when its battery has 4.00 c of charge remaining. if the battery supplies a constant current of 1.00 ma until the charge is depleted, what will the clock read when it stops? 10:33:17 9:55:07 11:44:00 9:21:40

Explanation:

Step1: Calculate the time - interval for the remaining charge

We know that current $I=\frac{Q}{t}$, where $I = 1.00\ mA=1\times10^{- 3}\ A$ and $Q = 4.00\ C$. Rearranging for time $t$, we get $t=\frac{Q}{I}$.
$t=\frac{4.00\ C}{1\times10^{-3}\ A}=4000\ s$.

Step2: Convert the time - interval to hours, minutes and seconds

$4000\ s=1\ h+600\ s$.
$600\ s = 10\ min$. So $t = 1\ h\ 10\ min$.

Step3: Add the time - interval to the initial time

The initial time is $8:15:00$.
Adding $1$ hour and $10$ minutes:
The hour part: $8 + 1=9$.
The minute part: $15+10 = 25$.
The second part remains $0$.
So the time when the clock stops is $9:25:00$. But there is no such option. There may be some misunderstandings in the problem - solving process. Let's calculate in a different way.
We know that $1\ mA=\frac{1\ C}{1000\ s}$. For a charge of $4\ C$ and a current of $1\ mA$, the time is $4000\ s$.
$4000\ s=66\ min\ 40\ s$.
Adding $66\ min\ 40\ s$ to $8:15:00$.
$66\ min=1\ h\ 6\ min$.
Adding $1\ h\ 6\ min\ 40\ s$ to $8:15:00$:
The hour: $8 + 1=9$.
The minute: $15+6 = 21$.
The second: $40$.

Answer:

$9:21:40$