a load of 400 Newton is lifted by a first class lever in which the load is at the distance of 20 cm and the effort is at the dis
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Answer:
The electric field strength inside the capacitor is 49880.77 N/C.
Explanation:
Given:
Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m
Separation between the plates (d) = 0.407 mm =
Energy stored in the capacitor (U) =
Assuming the medium to be air.
So, permittivity of space (ε) =
Area of the square plates is given as:
Capacitance of the capacitor is given as:
Now, we know that, the energy stored in a parallel plate capacitor is given as:
Rewriting in terms of 'E', we get:
Now, plug in the given values and solve for 'E'. This gives,
Therefore, the electric field strength inside the capacitor is 49880.77 N/C
Explanation:
Given that,
The mass of rock, m = 2.35-kg
It was released from rest at a height of 21.4 m.
(a) The kinetic energy is given by :
As the rock was at rest initially, it means, its kinetic energy is equal to 0.
(b) The gravitational potential energy is given by :
It can be calculated as :
(c) The mechanical energy is equal to the sum of kinetic and potential energy such that,
M = 0 J + 492.84 J
M = 492.84 J
Hence, this is the required solution.