1) Length of the wire.
2) Thickness of the wire.
3) Temperature.
4) Type of metal.
Hope this helps!
-Payshence
Answer:
nuclear energy.............
Answer: the link isnt loading
Answer:

Explanation:
Angular acceleration is defined by 
Angular velocity is related to the period by 
Putting all together:

Taking our initial (i) point now and our final (f) point one year later, we would have:



So for our values we have:

Where the minus sign indicates it is decelerating.
We know that arc length (x(t)) is given with the following formula:

Where r is the radius of the barrel. We must keep in mind that as barrel rolls its radius decreases because less and less tape is left on it.
If we say that the thickness of the tape is D then with every full circle our radius shrinks by d. We can write this down mathematically:

When we plug this back into the first equation we get:

We must solve this quadratic equation.
The final solution is:

It is rather complicated solution. If we asume that the tape has no thickness we get simply: