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frez [133]
3 years ago
8

Mechanical (sound) waves to Earth from satellites. How is this possible? aves are unable to travel through a vacuum, such as thr

ough space, but radio waves are transmitted
Physics
1 answer:
Anuta_ua [19.1K]3 years ago
6 0

Sound waves are not able to travel in a vacuum, sound requires a medium such as air or a solid for example.  Satellites are indeed in space, but radio waves are not sound waves they are a form of light.  Light can travel in a vacuum so the messages that satellites beam back to Earth are light waves not sound waves, that is why it is possible.

You might be interested in
What do you meant by varnier constant?
Soloha48 [4]

Answer:

the ratio of the smallest division of main scale to the number of divisions of the vernier scale.

Explanation:

difference between the value of one main scale division and one vernier scale division

3 0
3 years ago
Because of interstellar dust, astronomers can see at most about 5 kpc into the disk of the galaxy at visual wavelengths. What pe
NNADVOKAT [17]

Answer:

96%

Explanation

Let A the total area of the galaxy, is modeled as a disc:

A = πR^2 = π (25 kpc)^2

And let a be the area that astronomers are able to see:

a = πr^2 = π(5 kpc)^2

The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:

P = 100 a/A = (5/25)^2 = 100/25 = 4%

Therefore, the percentage of the galaxy not included, i.e. not seen is:

(100-4)% = 96%

5 0
3 years ago
A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl
Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

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

here, \lambda  is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta  is the angle of scattering.

Given that, the scattering angle is, \theta=157^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8}  } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.

Therfore,

\lambda^{'}-\lambda=4.64 p.m.

Here, the photon's incident wavelength is \lamda=7.33pm

So,

\lambda^{'}=7.33+4.64=11.97 p.m

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

here, \vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therfore,

\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.

This is the de Broglie wavelength of the electron after scattering.

8 0
3 years ago
Briefly explain why arterial injuries are more dangerous than damage to veins
MArishka [77]

Answer:

Injury to a vein increases the risk of forming a blood clot.

Explanation:

hoped this helped

5 0
3 years ago
A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.
Mekhanik [1.2K]

Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

vf^2 = 1900

sqrt(vf^2) = sqrt(1900)

vf = 43.59 m/s

7 0
3 years ago
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