The power developed is 500 W ( to the nearest Watt)
Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.


Therefore,

Substitute the given values of force , displacement and time


Thus the Power can be rounded off to the nearest value of 500 W
Answer:
(a) 6650246.305 N/C
(b) 24150268.34 N/C
(c) 6408227.848 N/C
(d) 665024.6305 N/C
Explanation:
Given:
Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]
Total charge of the ring (Q) = 75.0 μC =
[1 μC = 10⁻⁶ C]
Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

Plug in the given values for each point and solve.
(a)
Given:
, 
Electric field is given as:

(b)
Given:
, 
Electric field is given as:

(c)
Given:
, 
Electric field is given as:

(d)
Given:
, 
Electric field is given as:
The angular momentum of an object is equal to the product of its moment of inertia and angular velocity.
L = Iω
I = 1/2 MR²
I = 1/2 x 13 x (0.2)
I = 1.3
ω = 2π/t
ω = 2π/0.3
ω = 20.9
L = 1.3 x 20.9
= 27.2 kgm²/s
Answer:

Explanation:
According to Coulomb's law, the magnitude of the electric force between two point charges is directly proportional to the product of the magnitude of both charges and inversely proportional to the square of the distance that separates them:

Here k is the Coulomb constant. In this case, we have
,
and
. Replacing the values:

The negative sign indicates that it is an attractive force. So, the magnitude of the electric force is:
