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polet [3.4K]
3 years ago
10

The Hubble Space Telescope orbits Earth and sends images of distant objects. Why does Hubble form better images than the optical

telescopes used on land?
a. It’s larger than land-based telescopes.
b. It uses more powerful lenses than telescopes on land.
c. It orbits beyond Earth’s atmosphere to avoid scattering of light.
d. It revolves around the outer edge of Earth.
Physics
2 answers:
postnew [5]3 years ago
7 0

Answer: option c: It orbits beyond the Earth's atmosphere to avoid scattering of light.

Explanation:

Hubble space telescope orbits Earth and sends images of distant objects. The images formed by Hubble are better than the optical telescopes used on land. This is because the Hubble telescope is a space telescope. Light from the distant objects when reaches the land telescopes transmits through atmosphere, where scattering occurs. Some the light rays bounce back. This is avoided by the space telescope Hubble.

Nataly [62]3 years ago
6 0

the awnser is c I hope it helps :)

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1. What is the potential energy of a 4-kilogram potted plant that is on a 1 meter-high plant
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Read 2 more answers
Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2
Alinara [238K]

Answer:

a)E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

b)E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

E=K\dfrac{q}{r^2}

E_1=K\dfrac{q_1}{r_1^2}

E_2=K\dfrac{q_2}{r_2^2}

Now by putting the va;ues

a)

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C

E_1=72.11\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C

E_2=21.58\times 10^{6}\ N/C

The net electric field

E=E_1-E_2

E=50.53\times 10^{6}\ N/C

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

b)

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C

E_1=120.3\times 10^{6}\ N/C

E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C

E_2=147.92\times 10^{6}\ N/C

The net electric field

E=E_1+E_2

E=268.22\times 10^{6}\ N/C

The direction will be positive direction.

   

7 0
3 years ago
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