A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
1 answer:
Answer:
the capacitive reactance is reduced by a factor 2
Explanation:
The capacitive reactance is given by:
where
f is the frequency of the source
C is the capacitance
In this case, the frequency of the source is doubled:
f' = 2f
while the capacitance does not change. So the new capacitive reactance will be
so the capacitive reactance is reduced by a factor 2.
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Answer:
a)
b)
c)
d) or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:
Part b
For this case first we can convert the spring constant to N/m like this:
And the work donde by the spring on this case is given by:
Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:
And if we solve for the initial velocity we got:
Part d
Let d1 represent the new maximum distance, in order to find it we know that :
And replacing we got:
And we can put the terms like this:
If we multiply all the equation by 2 we got:
Now we can replace the values and we got:
And solving the quadratic equation we got that the solution for or 18.3 cm because the negative solution not make sense.
