Answer:
The energy stored in the capacitor quadruples its original value.
Explanation:
The energy stored in a capacitor is given by the equation
where
C is the capacitance
V is the voltage across the plates
The capacitance, C, depends only on the properties of the capacitor, so it does not change when the voltage applied is changed.
Instead, in this problem the voltage applied is doubled:
V' = 2V
So the new energy stored is
so, the energy stored has quadrupled.
To develop this problem we will apply the concept of energy conservation. For which the work carried out must be equivalent to the potential energy stored on the capacitor. We will start by finding the capacitance to later be able to calculate the energy and therefore the work in the capacitor
Here,
C = Capacitance
V = Potential difference between the plates
Q = Charge between the capacitor plates
At the same time the energy stored in the capacitor can be defined as,
We will start by finding the value of the capacitance, so we will have to,
Finally using the expression for the energy we have that,
Therefore the minimum amount of work that must be done in charging this capacitor is
Answer:
Part 1) Time of travel equals 61 seconds
Part 2) Maximum speed equals 39.66 m/s.
Explanation:
The final speed of the train when it completes half of it's journey is given by third equation of kinematics as
where
'v' is the final speed
'u' is initial speed
'a' is acceleration of the body
's' is the distance covered
Applying the given values we get
Now the time taken to attain the above velocity can be calculated by the first equation of kinematics as
Since the deceleration is same as acceleration hence the time to stop in the same distance shall be equal to the time taken to accelerate the first half of distance
Thus total time of journey equals
Part b)
the maximum speed is reached at the point when the train ends it's acceleration thus the maximum speed reached by the train equals