A capacitor with plates separated by distance d is charged to a potential difference ΔVC. All wires and batteries are disconnect
ed, then the two plates are pulled apart (with insulated handles) to a new separation of distance 2d. Part A Does the capacitor charge Q change as the separation increases? If so, by what factor? If not, why not?
2 answers:
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Answer:
Explanation:
Given height of lamp from the ceiling = 2.6m
mass of the lamp = 3.8kg
acceleration due to gravity = 9.81m/s²
As the body falls to the ground, it falls under the influence of gravity.
Gravitational potential energy = mass*acc due to gravity * height
Gravitational potential energy = 3.8*2.6*9.81
Gravitational potential energy = 96.923 Joules
b) Kinetic energy = 1/2 mv²
m = mass of the body (in kg)
v = velocity of the body (in m/s²)
To get the velocity v, we will use the equation of motion
Since mass = 3.8kg
c) To know how fast the lamp is moving when it hits the ground, we will use the formula. When the body hits the ground, the height covered will be 0m. this means that the body is not moving once it hits the ground. It stays in one position. The energy possessed by the body at this point is potential energy. The correct answer is therefore 0 m/s