Answer:
The peak emf of the generator is 40.94 V.
Explanation:
Given that,
Number of turns in primary coil= 11
Number of turns in secondary coil= 18
Peak voltage = 67 V
We nee to calculate the peak emf
Using relation of number of turns and emf


Where, N₁ = Number of turns in primary coil
N₂ = Number of turns in secondary coil
E₂ = emf across secondary coil
Put the value into the formula


Hence, The peak emf of the generator is 40.94 V.
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Newton's law of conservation states that energy of an isolated system remains a constant. It can neither be created nor destroyed but can be transformed from one form to the other.
Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.
Mathematically also potential energy is represented as
Potential energy= mgh
Where m is the mass of the pendulum.
g is the acceleration due to gravity
h is the height from the bottom z the ground.
At the bottom of the swing,the height is zero, hence the potential energy is also zero.
The kinetic energy is represented mathematically as
Kinetic energy= 1/2 mv^2
Where m is the mass of the pendulum
v is the velocity of the pendulum
At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.
Energy can neither be created e destroyed. It can only be transferred from one form to another. Implying this law and the above explainations we conclude that at the bottom of the pendulum,the potential energy=0 and the kinetic energy=294J as the entire potential energy is converted to kinetic energy at the bottom.
Answer:

Explanation:
Given:
- Three identical charges q.
- Two charges on x - axis separated by distance a about origin
- One on y-axis
- All three charges are vertices
Find:
- Find an expression for the electric field at points on the y-axis above the uppermost charge.
- Show that the working reduces to point charge when y >> a.
Solution
- Take a variable distance y above the top most charge.
- Then compute the distance from charges on the axis to the variable distance y:

- Then compute the angle that Force makes with the y axis:
cos(Q) = sqrt(3)*a / 2*r
- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:
F_1,2 = 2*F_x*cos(Q)
- The total net force would be:
F_net = F_1,2 + kq / y^2
- Hence,

- Now for the limit y >>a:

- Insert limit i.e a/y = 0

Hence the Electric Field is off a point charge of magnitude 3q.