1) Current: 4.5 A
2) Time taken: 4.7 s
Explanation:
1)
The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:
![I=\frac{q}{t}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bq%7D%7Bt%7D)
where
I is the current
q is the amount of charge passing a given point in a time t
For the wire in this problem, we have
q = 9.0 C is the amount of charge
t = 2.0 s is the time interval
Solving for I, we find the current:
![I=\frac{9.0}{2.0}=4.5 A](https://tex.z-dn.net/?f=I%3D%5Cfrac%7B9.0%7D%7B2.0%7D%3D4.5%20A)
2)
To solve this problem, we can use again the same formula
![I=\frac{q}{t}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7Bq%7D%7Bt%7D)
where
I is the current
q is the amount of charge passing a given point in a time t
In this problem, we have:
I = 3.0 A (current)
q = 14.0 C (charge)
Therefore, the time taken for the charge to move past a particular spot in the wire is
![t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s](https://tex.z-dn.net/?f=t%3D%5Cfrac%7Bq%7D%7BI%7D%3D%5Cfrac%7B14.0%7D%7B3.0%7D%3D4.7%20s)
Learn more about electric current:
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