Answer:
Explanation:
Width of central diffraction peak is given by the following expression
Width of central diffraction peak= 2 λ D/ d₁
where d₁ is width of slit and D is screen distance and λ is wave length.
Width of other fringes become half , that is each of secondary diffraction fringe is equal to
λ D/ d₁
Width of central interference peak is given by the following expression
Width of each of bright fringe = λ D/ d₂
where d₂ is width of slit and D is screen distance and λ is wave length.
Now given that the central diffraction peak contains 13 interference fringes
so ( 2 λ D/ d₁) / λ D/ d₂ = 13
then ( λ D/ d₁) / λ D/ d₂ = 13 / 2
= 6.5
no of fringes contained within each secondary diffraction peak = 6.5
Actually Welcome to the Concept of the Wave Optics.
Whenever the waves are out of phase they form a DESTRUCTIVE interference.
thus,
Destructive Interference occurs when two waves are out of phase.
Answer:
27,000 m
450 m/s
Explanation:
Assuming the initial velocity is 0 m/s:
v₀ = 0 m/s
a = 15 m/s²
t = 60 s
A) Find: Δy
Δy = v₀ t + ½ at²
Δy = (0 m/s) (60 s) + ½ (15 m/s²) (60 s)²
Δy = 27,000 m
B) Find: v_avg
v_avg = Δy / t
v_avg = 27,000 m / 60 s
v_avg = 450 m/s
Ω₀ = the initial angular velocity (from rest)
t = 0.9 s, time for a revolution
θ = 2π rad, the angular distance traveled
Let
α = the angular acceleration
ω = the final angular velocity
The angular rotation obeys the equation
(1/2)*(α rad/s²)*(0.9 s)² = (2π rad)
α = 15.514 rad/s²
The final angular velocity is
ω = (15.514 rad/s²)*(0.9 s) = 13.963 rad/s
If the thrower's arm is r meters long, the tangential velocity of release will be
v = 13.963r m/s
Answer: 13.963 rad/s
Two terminal device which can maintain a fixed voltage. Hope this help I need more detail in your question this is all I can provide. :)
