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sineoko [7]
3 years ago
9

How much charge does a 9.0 v battery transfer from the negative to the positive terminal while doing 39 j of work?

Physics
1 answer:
sdas [7]3 years ago
3 0
The work done by the battery is equal to the charge transferred during the process times the potential difference between the two terminals of the battery:
W=q \Delta V
where q is the charge and \Delta V is the potential difference.

In our problem, the work done is W=39 J while the potential difference of the battery is \Delta V = 9.0 V, so we can find the charge transferred by the battery:
q= \frac{W}{\Delta V}= \frac{39 J}{9.0 V}=4.33 C
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Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

F \propto \frac{1}{r}

F = \frac{k}{r} (Eq. 1)

Where:

F - Force, measured in newtons.

k - Proportionality ratio, measured in newton-meters.

r - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

F_{A}\cdot r_{A} = F_{B}\cdot r_{B}

F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}(Eq. 2)

Where:

F_{A}, F_{B} - Initial and final forces, measured in newtons.

r_{A}, r_{B} - Initial and final distances, measured in meters.

If we know that F_{A} = 5\,N, r_{A} = 0.9\,m and r_{B} = 0.35\,m, then final force is:

F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)

F_{B} = 12.857\,N

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