Calculate the peak voltage of a generator that rotates its 200-turn, 0.100 m diameter coil at 3600 rpm in a 0.800 T field.
1 answer:
Answer:
The peak voltage is 473.86 V
Explanation:
Given;
number of turns of generator, N = 200 turn
diameter of the coil, d = 0.1 m
radius of the coil, r = 0.05 m
Magnitude of the magnetic field, B = 0.8 T
angular frequency, ω = 3600 rpm
angular frequency, ω (rad/s) =
The peak voltage is given by;
E = NBAω
Where;
A is the area of the coil = πr² = π (0.05)² = 7.855 x 10⁻³ m²
E = (200)(0.8)(7.855 x 10⁻³)(377.04)
E = 473.86 V
Therefore, the peak voltage is 473.86 V
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Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
Answer:
The force constant is
The energy stored in the spring is
Explanation:
From the question we are told that
The mass of the object is
The period is
The period of the spring oscillation is mathematically represented as
where k is the force constant
So making k the subject
substituting values
The energy stored in the spring is mathematically represented as
Where x is the spring displacement which is given as
substituting values