Answer: 22.6 hours
Step-by-step explanation:
The power is the measure of the rate of energy.
In this problem, the 12.0 V battery is rated at 51.0 Ah, which means it delivers 51.0 A of current in a time of t = 1 h = 3600 s. The power delivered by the battery can be written as
![P=IV](https://img.qammunity.org/2021/formulas/physics/high-school/4ofuq13a657if0wubbuuhk1p7xuymrpvco.png)
where
I is the current
V = 12.0 V is the voltage of the battery
So the energy delivered by the battery can be written as
![E=Pt=VIt](https://img.qammunity.org/2021/formulas/physics/high-school/jvv1l4ih1njatfuwkwb6ydd2jsh6kpokwx.png)
Where
![It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s](https://img.qammunity.org/2021/formulas/physics/high-school/rpsnrdd82x9wxh5f395o8oa0w0ruwgd40l.png)
So the energy delivered is
![E=(12.0)(183,600)=2.2\cdot 10^6 J](https://img.qammunity.org/2021/formulas/physics/high-school/d3ish1fu4635budej9iez90qa0q7kq5bac.png)
At the same time, the headlight consumes 27.0 W of power, so 27 Joules of energy per second; Therefore, it will remain on for a time of:
![t=(2.2\cdot 10^6 J)/(27.0 W)=81481 s = 22.6 h](https://img.qammunity.org/2021/formulas/physics/high-school/nmo71xsbsynufpc8velsxni28ub0aufazx.png)