An uncharged capacitor is connected to the terminals of a 4.0 V battery, and 9.0 μC flows to the positive plate. The 4.0 V batte
1 answer:
Answer:
Explanation:
At the beginning, we have:
V = 4.0 V potential difference across the capacitor
charge stored on the capacitor
Therefore, we can calculate the capacitance of the capacitor:
Later, the battery is replaced with another battery whose voltage is
V = 5.0 V
Since the capacitance of the capacitor does not change, we can calculate the new charge stored:
Since the capacitor has been connected exactly as before, we have that the charge on the positive plate has increased from to
. Therefore, the additional charge that moved to the positive plate is
You might be interested in
Answer:
Part 1. Pendulum clock is slower.
Part 2. Spring-loaded clock remains the same.
Explanation:
The period of a simple pendulum is given by
where = length of pendulum and
= acceleration due to gravity.
It is seen that the period is inversely proportional to the square root of the gravitational acceleration. So if gravity increases, period decreases and vice versa.
on the moon is about one-fifth that of the Earth. Hence, the pendulum will have a larger period, about twice (
). A larger period means it takes longer to finish an oscillation, so the pendulum clock is slower.
The period of a loaded spring is given by
where = the spring constant and
= mass of load on the spring.
It is seen that this relation does not depend on gravity nor does it have any parameter that depends on gravity: k is a constant of the spring that does not change while mass is independent of location.
Hence, the spring-loaded clock will remain the same.
As a note, one might assume that gravity affects the loaded spring because the load is 'pushed' down by gravity. In fact, only the equilibrium position is affected by gravity; it only determines where the oscillation starts from, not how long it takes.