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n200080 [17]
4 years ago
7

An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati

ve to the frequency of the incident photon? it is lower.
Physics
1 answer:
Viefleur [7K]4 years ago
8 0

The frequency of the scattered photon decreases or it will be lower compare to the frequency of incident photon. An x-ray photon scatters in one direction after a collision and some energy is transferred to the electron as it recoils in another direction resulting to have less energy in the scattered photon. In addition, the frequencies will also depend on the differences of the angle at which the scattered photon leaves the collision and this incident is called Compton Effect.

You might be interested in
If a bicycle is going 8 m/s for 29 seconds how far does it go?
VladimirAG [237]

Answer:

232

Explanation:

D=S*T

(By distance formula)

8*29=

232

4 0
3 years ago
An object of mass 3.07 kg, moving with an initial velocity of 5.07 m/s, collides with and sticks to an object of mass 2.52 kg wi
Alenkasestr [34]

Answer:

This is an inelastic collision. This means, unfortunately, that KE cannot save you, at least in the problem's current form.  

Let's see what conservation of momentum in both directions does ya:

Conservation in the x direction:

Only 1 object here has a momentum in the x direction initally.  

m1v1i + 0 = (m1 + m2)(vx)

3.09(5.10) = (3.09 + 2.52)Vx

Vx = 2.81 m/s

Explanation:

Conservation in the y direction:

Again, only 1 object here has initial velocity in the y:

0 + m2v2i = (m1 +m2)Vy

(2.52)(-3.36) = (2.52 + 3.09)Vy

Vy = -1.51 m/s

++++++++++++++++++++

Now that you have Vx and Vy of the composite object, you can find the final velocity by doing Vf = √Vx^2 + Vy^2)

Vf = √(2.81)^2 + (-1.51)^2

Vf = 3.19 m/s

3 0
4 years ago
A 2.00-kg, frictionless block is attached to an ideal spring with force constant 300 N/mN/m. At t=0 t=0 the block has velocity -
slega [8]

Answer:

The amplitude of the spring is 32.6 cm.

Explanation:

It is given that,

Mass of the block, m = 2 kg

Force constant of the spring, k = 300 N/m

At t = 0, the velocity of the block, v = -4 m/s

Displacement of the block, x = 0.2 mm = 0.0002 m

We need to find the amplitude of the spring. We know that the velocity in terms of amplitude and the angular velocity is given by :

v=\omega\sqrt{A^2-x^2}

\omega=\sqrt{\dfrac{k}{m}}

\omega=\sqrt{\dfrac{300}{2}}    

\omega=12.24\ rad/s

So, \dfrac{v^2}{\omega^2}+x^2=A^2

\dfrac{(-4)^2}{(12.24)^2}+(0.0002)^2=A^2            

A = 0.326 m

or

A = 32.6 cm

So, the amplitude of the spring is 32.6 cm. Hence, this is the required solution.

4 0
3 years ago
N=-D(n2-n1)/(x2-x1) D is diffastion.What are the dimensions of D.
MrMuchimi

We don't have much to go on.

The dimensions of D depend on the dimensions of N, n, and x, and we don't know what any of those stand for.

It might help if we had ever heard of 'diffastion', but we're striking out there too.

8 0
4 years ago
Thirty five soccer players were challenged to do as many sit-ups possible in two minutes
Ghella [55]

Answer:

Explanation:

What is the question?

7 0
3 years ago
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