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.
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.