Answer: 22.6 hours
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
[tex]P=IV[/tex]
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
[tex]E=Pt=VIt[/tex]
Where
[tex]It=51.0 A\cdot h = 51.0 A \cdot 3600 s/h=183,600 A\cdot s[/tex]
So the energy delivered is
[tex]E=(12.0)(183,600)=2.2\cdot 10^6 J[/tex]
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:
[tex]t=\frac{2.2\cdot 10^6 J}{27.0 W}=81481 s = 22.6 h[/tex]
