Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = 
- d fi / dt
= B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A 
Dt = - A 
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
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
Explanation:
c. if the vector is oriented at 0° from the X -axis.
If you were given distance & period of time, you would be able to calculate the speed.
Hope this helps!
