Answer:
a) 
For this case we know the following values:




So then if we replace we got:

b) 
With 
And replacing we have:

And then the scattered wavelength is given by:

And the energy of the scattered photon is given by:

c) 
Explanation
Part a
For this case we can use the Compton shift equation given by:
For this case we know the following values:
So then if we replace we got:
Part b
For this cas we can calculate the wavelength of the phton with this formula:
With
And replacing we have:
And then the scattered wavelength is given by:
And the energy of the scattered photon is given by:
Part c
For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:
Answer: 7.66 m/s
Explanation:
This situation is related to free fall (vertical motion). Hence, this can be considered a one-dimension problem and we can use the following equation:
Where:
is the final velocity of the egg at the asked height
is the initial velocity of the egg
is the acceleration due gravity
is the distance at which the egg is from the nest, when it is
from the ground
Isolating
:
Substituting the known values:
This is the final velocity of the egg
Answer:
Q = ba⁴ * ε₀
Explanation:
From Gauss's Law, we know that
flux Φ = Q / ε₀
where ε₀ = 8.85e-12 C²/N·m²
and also,
Φ = EAcosθ
The field is directed along the x-axis, so that all of the flux passes through the side of the cube at x = a. This means that θ = 0º, and thus
Φ = EAcos0
Φ = EA
E = bx² meanwhile, we are interested in the point where x = a, so we substitute and then
E = ba²
Since A = a² for the cube face, we have
Q / ε₀ = E * A
Q / ε₀ = ba² * a²
so that
Q = ba⁴ * ε₀
Answer:
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Explanation:
The electric field is given by
dE = k dq / r²
In this case as we have a continuous load distribution we can use the concept of linear density
λ= Q / x = dq / dx
dq = λ dx
We substitute in the equation
∫ dE = k ∫ λ dx / x²
We integrate
E = k λ (-1 / x)
We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁
E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)
E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)
We replace the density
E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]
E = k Q 1 / (x₀-x₂) (x₀-x₁)