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Brut [27]
3 years ago
15

Which list is in order from biggest to smallest? A. Earth Solar system Nebula Galaxy O O B. Universe - Galaxy Solar system Earth

O C. Universe Earth Solar system Nebula O D. Solar system som Galaxy – Nebula Earth​
Physics
2 answers:
gogolik [260]3 years ago
8 0

Answer:

universe, Galaxy, solar system, earth

Explanation:

A P E X

ololo11 [35]3 years ago
3 0

Answer:B, universe, galaxy, solar system, earth

Explanation:

A p e x

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what is the answer for the path that one body in space takes as it revolves around another body is called a .......... a orbit,a
salantis [7]
The answer is orbit, we are orbiting the sun as the moon orbits us
6 0
2 years ago
Apply the concept of density to explain why oil floats on water.
Sauron [17]
Anything less dense than water will float, like oil. Anything more dense than water will sink, like rock.
6 0
3 years ago
A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl
Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known 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=147^{\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-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

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

where,\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}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

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

6 0
3 years ago
How much force is required to accelerate a 800kg car by 15 m/s2
Tom [10]

Answer:

force=12000

Explanation:

F=m*a aka force equals mass times acceleration so 800*15=12000

3 0
2 years ago
Read 2 more answers
What is the speed of a wave with a wavelength of 3 m and a frequency of .1Hz?
Yuri [45]
We know,
Speed = Frequency * Wavelength 
Speed = 3 * 0.1 m/s   [hertz = 1/sec.]
So, your final answer is 0.3 m/s

Hope this helps!!
6 0
3 years ago
Read 2 more answers
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