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:

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:

2)
To solve this problem, we can use again the same formula

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

Learn more about electric current:
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