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fiasKO [112]
3 years ago
14

The heliocentric theory suggests blank

Physics
1 answer:
astra-53 [7]3 years ago
4 0

Answer:

Heliocentrism is the astronomical model in which the Earth and planets revolve around the Sun at the center of the Universe.

Explanation:

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It transfers and changes into different types of energy, this is why the ground feels hot when something moves fast over it.
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Which of the following is not an example of work?
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The Earth orbits around the sun because the gravitational force that the sun
kotykmax [81]
<h3>Question -:</h3>

The Earth orbits around the sun because the gravitational force that the sun

exerts on the Earth:

O A. causes Earth's acceleration toward the sun.

O B. is very small because the sun is so far from the Earth.

O c. is smaller than the force the Earth exerts on the sun.

O D. pushes the Earth away from the sun.

<h3>Answer -:</h3>

O A. causes Earth's acceleration toward the sun.

<em>I </em><em>hope </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em><em>have </em><em>a </em><em>nice </em><em>time </em><em>ahead!</em>

5 0
3 years ago
The maximum Compton shift in wavelength occurs when a photon isscattered through 180^\circ .
vlabodo [156]

Answer: 90\°

Explanation:

The Compton Shift \Delta \lambda in wavelength when the photons are scattered is given by the following equation:

\Delta \lambda=\lambda_{c}(1-cos\theta)     (1)

Where:

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}c}, being h the Planck constant, m_{e} the mass of the electron and c the speed of light in vacuum.

\theta) the angle between incident phhoton and the scatered photon.

We are told the maximum Compton shift in wavelength occurs when a photon isscattered through 180\°:

\Delta \lambda_{max}=\lambda_{c}(1-cos(180\°))     (2)

\Delta \lambda_{max}=\lambda_{c}(1-(-1))    

\Delta \lambda_{max}=2\lambda_{c}     (3)

Now, let's find the angle that will produce a fourth of this maximum value found in (3):

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{4}2\lambda_{c}(1-cos\theta)      (4)

\frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}(1-cos\theta)      (5)

If we want \frac{1}{4}\Delta \lambda_{max}=\frac{1}{2}\lambda_{c}, 1-cos\theta   must be equal to 1:

1-cos\theta=1   (6)

Finding \theta:

1-1=cos\theta

0=cos\theta  

\theta=cos^{-1} (0)  

Finally:

\theta=90\°    This is the scattering angle that will produce \frac{1}{4}\Delta \lambda_{max}      

7 0
3 years ago
Which is not a device for reproducing sound?
Alona [7]
Phonograph. Hope that helps
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