Answer:
The cassette player will operate at normal power for 3333.33 seconds.
Explanation:
The first step is to identify the operating voltage and the operating current with the purpose to determine the power that the cassette player consumes. Remember that power equals the voltage multiplied by the current [tex]P=V\times I[/tex], where P is in Watts (W), V is in Volts (V) and I is in Amperes (A).
The problem says that four batteries are connected in series to provide a voltage of 6V to the player circuit. So, the operating voltage is 6V, [tex]V=6V[/tex]
Then, the problem says that cassette player draws a constant current of 10mA from the battery pack. So, the operating current is 10mA or 0.01A, [tex]I=0.01A[/tex]
From previous, it could be said that the cassette player consumes 0.06W.
[tex]P=V\times I=(6V)\times (0.01A)=0.06W[/tex]
Now, the idea is to calculate how long the cassette will operate at 0.06W.
The problem says that the battery pack stores [tex]200\, W\cdot s[/tex], it means that the battery pack could provide 200W in a second; after a second, the battery pack will not work properly. However, the battery pack just have to provide 0.06W so, it will last more time. For calculating that, you must divide the total power per time the cell can provide by the power that the cassette player needs to work.
[tex]t=\frac{200W\cdot s}{0.06W}=3333.33s[/tex]
As you can see, the W units are canceled and second remains.
Thus, the cassette player will operate at normal power for 3333.33 seconds.
