Answer: 
Explanation:
The Compton Shift 
in wavelength when the photons are scattered is given by the following equation:
(1)
Where:
is a constant whose value is given by 
, being 
the Planck constant, 
the mass of the electron and 
the speed of light in vacuum.
the angle between incident phhoton and the scatered photon.
We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 
:
(2)
(3)
Now, let's find the angle that will produce a fourth of this maximum value found in (3):
(4)
(5)
If we want 
, 
must be equal to 1:
(6)
Finding 
:
Finally:
This is the scattering angle that will produce 
The Swamp
There are many websites that say the rainforest but the rainforest is warmer and gets rain year round and swamps are warm and gets lots of rain but not year round.
That's what scientists and other technical people call the object's "<em>volume</em>".
When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>
On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.
What is Hard stabilization?
- Hard stabilization is the prevention of erosion through the use of artificial barriers.
- Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
- Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
- These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
- Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.
Learn more about the Hard stabilization with the help of the given link:
brainly.com/question/16022736
#SPJ4
Answer:
39.81 N
Explanation:
I attached an image of the free body diagrams I drew of crate #1 and #2.
Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.
∑Fₓ = maₓ
∑Fᵧ = maᵧ
Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:
- ∑Fᵧ = maᵧ
- T₁ - m₂g = m₂aᵧ
Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.
Let's solve for T in the equation...
- T₁ = m₂aᵧ + m₂g
- T₁ = m₂(a + g)
We'll come back to this equation later. Now let's go to the free body diagram for crate #1.
We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.
The normal force is equal to the x-component of the force of gravity.
- (F_n · μ_k) - m₁g sinΘ = m₁aₓ
- (F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
- [m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
- [(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
- [2.539595871] - [-58.0962595] = 6aₓ
- 60.63585537 = 6aₓ
- aₓ = 10.1059759 m/s²
Now let's go back to this equation:
We have 3 known variables and we can solve for the tension force.
- T = 2(10.1059759 + 9.8)
- T = 2(19.9059759)
- T = 39.8119518 N
The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.
