As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?
1 answer:
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Work done can be computed using the formula:
Where:
W = work (J)
F = Force (N)
d = Distance (d)
Looking at the given, you know that you do not have a value for force, so you will have to solve for it.
Where:
F = Force
m = mass
a = acceleration
Because the object is being lifted, the acceleration will rely on gravity. Acceleration due to gravity is a constant 9.8 m/s^2. Let's list our given first:
F = ?
m = 100kg
a = 9.8m/s^2
Put that into our equation and solve:
Our force is then 980 N.
Now that we have force we can solve for Work. The given for work is as follows:
F= 980N
d = 1.4m
Put that into our formula and solve:
The work done is 1,372J.
Where:
W = work (J)
F = Force (N)
d = Distance (d)
Looking at the given, you know that you do not have a value for force, so you will have to solve for it.
Where:
F = Force
m = mass
a = acceleration
Because the object is being lifted, the acceleration will rely on gravity. Acceleration due to gravity is a constant 9.8 m/s^2. Let's list our given first:
F = ?
m = 100kg
a = 9.8m/s^2
Put that into our equation and solve:
Our force is then 980 N.
Now that we have force we can solve for Work. The given for work is as follows:
F= 980N
d = 1.4m
Put that into our formula and solve:
The work done is 1,372J.