A car owner forgets to turn off the headlights of his car while it is parked in his garage. If the 12.0-V battery in his car is
1 answer:
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
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
Where
So the energy delivered is
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:
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The resistance of a conductive wire is given by:
where
is the material resistivity
is the wire length
is the cross-sectional area of the wire
The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write
Similarly, the radius of the wire is doubled (r'=2r), so the new area is
And if we substitute into the equation, we find that the new resistance of the wire is
Therefore, R=R': this means that the resistance of the wire did not change.
where
The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write
Similarly, the radius of the wire is doubled (r'=2r), so the new area is
And if we substitute into the equation, we find that the new resistance of the wire is
Therefore, R=R': this means that the resistance of the wire did not change.