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:
PE= m * g *h
work:
PE= 65kg * 9.8 kg *8,800 m
PE=5605600 m/kg
idk the actual units i forgot
The derived unit for voltage is named volt.
According to Newton's 2nd law of motion:
F = m * a where F is the force applied in Newtons, m is the mass of the object in kg, and a is the acceleration of the object in m/

.
Therefore the force applied in this situation is simply:
F = 6 kg * 2.3 m/

= 13.8 N
Hope this helps!