Answer:
Inductive reactance is 125.7 Ω
Explanation:
It is given that,
Inductance, 
Voltage source, V = 15 volt
Frequency, f = 400 Hz
The inductive reactance of the circuit is equivalent to the impedance. It opposes the flow of electric current throughout the circuit. It is given by :
So, the inductive reactance is 125.7 Ω. Hence, this is the required solution.
Answer:
t = 0.1111 s
Explanation:
Let's reduce the magnitudes to the SI system
d = 120 mm (1m / 1000 mm)
d= 0.120 m
w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)
w= 56.55 rad / s
When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration
W = θ / t
t = θ / w
t = 2π / 56.55
t = 0.1111 s
Answer:
a. (a) grating A has more lines/mm; (b) the first maximum less than 1 meter away from the center
Explanation:
Let n₁ and n₂ be no of lines per unit length of grating A and B respectively.
λ₁ and λ₂ be wave lengths of green and red respectively , D be distance of screen and d₁ and d₂ be distance between two slits of grating A and B ,
Distance of first maxima for green light
= λ₁ D/ d₁
Distance of first maxima for red light
= λ₂ D/ d₂
Given that
λ₁ D/ d₁ = λ₂ D/ d₂
λ₁ / d₁ = λ₂ / d₂
λ₁ / λ₂ = d₁ / d₂
But
λ₁ < λ₂
d₁ < d₂
Therefore no of lines per unit length of grating A will be more because
no of lines per unit length ∝ 1 / d
If grating B is illuminated with green light first maxima will be at distance
λ₁ D/ d₂
As λ₁ < λ₂
λ₁ D/ d₂ < λ₂ D/ d₂
λ₁ D/ d₂ < 1 m
In this case position of first maxima will be less than 1 meter.
Option a is correct .
