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natulia [17]
3 years ago
6

The Hubble Telescope The Hubble Telescope is a large telescope that orbits Earth. It was named after Edwin Hubble. Hubble was an

astronomer. The Hubble was launched in 1990 by NASA, the United States' space program. It uses a 94- inch mirror to collect light. Data are sent back to Earth by radio. Since it is above Earth's atmosphere, it gives us clearer pictures of space than telescopes on Earth can. It can also see farther. With this powerful tool, we've learned a lot about the universe. What Scientists Have Learned from the Hubble The Hubble has allowed us to watch stars form. It also has let us see into distant galaxies. We've been able to test our ideas. For example, before the Hubble, black holes were just a theory. We didn't know for sure that they existed. The Hubble has shown us evidence of black holes in space. The telescope has helped us find places to land data-gathering robots on Mars. We've used it to track the paths of comets. We've also been able to see volcanoes erupt on 10, one of Jupiter's moons. Telescopes that are larger than the Hubble are now being created But the Hubble is the first of its kind.
Why does being in outer space make Hubble a better telescope?
Physics
1 answer:
Marrrta [24]3 years ago
3 0

Quoting from the article itself:

"Since it is above Earth's atmosphere, it gives us clearer pictures of space than telescopes on Earth can."

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A planet has been observed orbiting a nearby star. This star has a mass of 3.5 solar masses, and the planet is 4.2 AU from the s
kondaur [170]

Answer:

4.6 years

Explanation:

This is solved using Kepler's third law which says:

T^2=\frac{4\pi ^2}{GM} a^2

Where

T = Orbital period of the planet (in seconds)

a = Distance from the star (in meters)

G = Gravitational constant

M = Mass of the parent star (in kg)

From the information given

M = 3.5M_{sun} = 6.96*10^{30} kg

a=4.2AU = 6.28*10^{11} meters

G = 6.67*10^{-11}m^3kg^{-1}s^{-2}

We put this into Kepler's law and get:

T=\sqrt{\frac{4\pi ^2}{6.67*10^{-11}*6.96*10^{30}} (6.28*10^{11})^3}=145,128,196 seconds.

This when converted to years is 4.6 years.

4 1
3 years ago
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A person consumes a snack containing 14 food calories (14kcal). what is the power this food produces if it is to be "burned off"
Inessa05 [86]

Answer:

B) 2.7W

Explanation:

Converting Cal to Joule

        1 cal = 4.186J

        14 kcal = 14 x 1000 x 4.186

                     = 58604 J

Converting hour to seconds

             6 hours = 6 x 60 x 60 seconds

                           = 21600 seconds

Power is the time rate of doing work.

Power = Work/Time

P = (58604) / (21600)

P = 2.7W

8 0
3 years ago
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If my cylinder of air lasts 60 minutes while I am at the surface breathing normally, assuming all else is the same, how long wil
AfilCa [17]

Answer:

Explanation:

Let the volume of air be V. at atmospheric pressure, that is 10⁵  Pa

At 20 m below surface pressure will be

atmospheric pressure + hdg

10⁵ + 20 x 9.8 x 1000 = 2.96 x 10⁵Pa

At this pressure volume V becomes V/ 2.96  

This volume will last 1/2.96 times  time that is 60/2.96 = 20.27 minutes.

8 0
3 years ago
A heat engine uses thermal energy that flows from a heat source with a temperature of 325 K to a cold sink, which has a temperat
Ilya [14]
The answer is "58.5"
5 0
3 years ago
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To meet a U.S. Postal Service requirement, employees' footwear must have a coefficient of static friction of 0.5 or more on a sp
motikmotik

Answer:

0.79 s

Explanation:

We have to calculate the employee acceleration, in order to know the minimum time. According to Newton's second law:

\sum F_x:f_{max}=ma_x\\\sum F_y:N-mg=0

The frictional force is maximum since the employee has to apply a maximum force to spend the minimum time. In y axis the employee's acceleration is zero, so the net force is zero. Recall that f_{max}=\mu N

Now, we find the acceleration:

\mu N=ma_x\\\mu mg=ma_x\\a_x=\mu g\\a_x=0.83(9.8\frac{m}{s^2})\\a_x=8.134\frac{m}{s^2}

Finally, using an uniformly accelerated motion formula, we can calculate the minimum time. The employee starts at rest, thus his initial speed is zero:

x=v_0t+\frac{1}{2}a_xt^2\\2x=a_xt^2\\t=\sqrt{\frac{2x}{a}}\\t=\sqrt{\frac{2(3.2m)}{8.134\frac{m}{s^2}}}\\t=0.79 s

8 0
3 years ago
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