Answer:
14 rev
Explanation:
= initial angular velocity = 2.5 revs⁻¹
= final angular velocity = 0.8 revs⁻¹
= Angular acceleration = - 0.2 revs⁻²
= Angular displacement
Using the equation

So the number of revolutions are 14
Answer:
0.0133A
Explanation:
Since we have two sections, for the Inductor region there would be a current
. In the case of resistance 2, it will cross a current
Defined this we proceed to obtain our equations,
For
,


For
,


The current in the entire battery is equivalent to,


Our values are,




Replacing in the current for t= 0.4m/s



By raising a position of an object, Its potential energy increases.