I think a, im not a 100% sure tho!!!
Answer:
Minimum thickness will be 100 nm
Explanation:
We have given refractive index is n = 1.5
Wavelength of the light incidence
= 600 nm
We have to find the smallest thickness of the film so that there will be minimum light reflect
For minimum thickness of non reflecting film
, here t is thickness,
is wavelength and n is refractive index
Putting all values 
So minimum thickness will be 100 nm
Answer:
No work is performed or required in moving the positive charge from point A to point B.
Explanation:
Lets take
Q= Positive charge which move from point A to point B along
Voltage difference,ΔV =V₁ - V₂
The work done
W = Q . ΔV
Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.
ΔV = 0
So
W = Q . ΔV
W = Q x 0
W= 0 J
So work is zero.
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J