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Masja [62]
3 years ago
15

Galileo’s most important contribution to astronomy was his

Physics
2 answers:
olchik [2.2K]3 years ago
6 0

The answer is D I just took the test.

ikadub [295]3 years ago
3 0

When Galileo got ahold of a little telescope ... no more than
a kid's toy today ... he soon turned it towards the night sky. 
The story says that he was the first one to do that.  Apparently
everyone else who had these things used them to look at their
neighbors' windows.  Anyway, When Galileo looked at Jupiter,
he saw four tiny dots.  As he watched them night after night, he
saw that the four dots always stayed with Jupiter, but got nearer
and farther from it.  He eventually realized that they were small
bodies in orbit around Jupiter.  And then ... because he was Galileo ...
he made the big intellectual jump:  He reasoned that if there were
smaller bodies in orbit around Jupiter, then Jupiter and the other
planets might also be in orbit around something, and not necessarily
around the Earth.  He understood this as evidence that the sun is
the center of the solar system.
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In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

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Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

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Explain the relationship between brightness and distance. Explain by relating the brightness of the light to the energy of the l
Iteru [2.4K]

Answer:

What part of the light wave is related to the brightness of the light?

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