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arsen [322]
3 years ago
5

Sunglasses that reduce glare take advantage of which kind of wave

Physics
2 answers:
marysya [2.9K]3 years ago
6 0

Answer:

It should be option B polarization

masya89 [10]3 years ago
5 0
B polarización the answer
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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

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:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on
sammy [17]

Answer: First, we determine the circumference of the Mars by the equation below.  

                                     C = 2πr

Substituting the known values,

                                     C = 2(π)(3,397 km) = 6794π km

To determine the tangential speed, we divide the circumference calculated above by the time it takes for Mars to complete one rotation and that is,

                   tangential speed = 6794π km / 24.6 hours = 867.64 km/h

6 0
3 years ago
Read 2 more answers
How to eat butter . I want to know how you all eat the butter.
g100num [7]
Many people never even taste butter, instead use oleo or margerine...fake stuff. Real butter is the best topping for popcorn! And crab legs are good dipped in it, as well as hot corn on the cobb. This is how I eat butter.
5 0
3 years ago
A current of 2.0A flows through a light bulb. What is the amount of charge flowing through the light bulb in 40 seconds?
lidiya [134]

Answer:

Quantity of charge = 80 Coulombs

Explanation:

Given the following data;

Current = 2 A

Time = 40 seconds

To find the amount of charge flowing through the light bulb;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge = current * time

Substituting into the formula, we have;

Quantity of charge = 2 * 40

Quantity of charge = 80 Coulombs

5 0
3 years ago
When energy is added to a wave, how can the wave change?.
nadya68 [22]

Answer:

When waves overlap in-phase (crest meets crest or trough meets trough) the waves energy is additive and the amplitude increases.

Explanation:

When waves overlap out-of-phase (crest meets trough) the waves cancel and the amplitude (energy) decreases. When two interfering waves cancel each other out.

6 0
2 years ago
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