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prohojiy [21]
3 years ago
5

Which sentence explains why optical land telescopes are not able to take images as clear as those taken by the Hubble Space Tele

scope?
Hubble takes pictures of planets, stars, and galaxies. The mixture of gases that surround a planet is called its atmosphere. Earth’s atmosphere changes and blocks some of the light that comes from space. Hubble orbits high above Earth and its atmosphere. So, Hubble can see space more clearly than telescopes on Earth can. Hubble is not the kind of telescope that you look through with your eye. Hubble uses a digital camera to take pictures. Then, Hubble uses radio waves to send the pictures through the air back to Earth.
Physics
2 answers:
Murrr4er [49]3 years ago
8 0

Sorry I'm a good year or so late, but its "Earth’s atmosphere changes and blocks some of the light that comes from space." !

ad-work [718]3 years ago
7 0

he Hubble Space Telescope has beamed hundreds of thousands of images back to Earth over the past two decades. One might call it the most skilled paparazzo, snapping countless images of the stars.

<span>Thanks to these images, scientists have been able to determine the age of the universe and shed light on the existence of dark energy. These extraordinary advancements have been possible because the Hubble images surpass those taken by Earth-based telescopes.</span>

<span>While ground-based observatories are usually located in highly elevated areas with minimal light pollution, they must contend with atmospheric turbulence, which limits the sharpness of images taken from this vantage point. (The effects of atmospheric turbulence are clear to anyone looking at the stars ? this is why they appear to twinkle.)</span>

 

In space, however, telescopes are able to get a clearer shot of everything from exploding stars to other galaxies.

Another disadvantage for ground-based telescopes is that the Earth's atmosphere absorbs much of the infrared and ultraviolet light that passes through it. Space telescopes can detect these waves.

Newer ground-based telescopes are using technological advances such as adaptive optics to try to correct or limit atmospheric distortion, but there's no way to see the wavelengths that the atmosphere blocks from reaching Earth, according to the Space Telescope Science Institute (STScI), which manages the Hubble research program.

<span>One downside to space telescopes like the Hubble is that they are extremely difficult to maintain and upgrade. The Hubble is the first telescope specifically designed to be repaired in space by astronauts, while other space telescopes cannot be serviced at all.</span>

<span>NASA scientists estimate that the telescope will only be able to keep taking pictures for five more years.</span>

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(2 points) A pair of students measure the intensity of light from a desklamp in Sec. 4.5 of the experiment. When the detector is
Novosadov [1.4K]

Answer:

Intensity at 45 cm will be 0.2283 volt

Explanation:

We have given that intensity of signal measured by the detector is 0.74 volt at a distance of 25 cm

We know that intensity of signal is given by I=\frac{s}{r^2} , here s is strength of light and r is distance

So 0.74=\frac{s}{{25}^2}

s=462.5

In second case

Intensity is given by I=\frac{s}{r^2}

So I=\frac{462.5}{45^2}=0.2283volt

4 0
3 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
4 years ago
4. How much mass is required to exert a force of 25 Newtons, accelerating at 5 m/s?
lions [1.4K]
F = ma
F/a = m
(25 N) / (5 m/s2) = m
4 kg = m
8 0
3 years ago
Please help !!! HELP
kozerog [31]

Answer:

The answer is C

I just took the quiz

5 0
3 years ago
Water flows horizontally through a garden hose with an inner diameter of .012 m at a speed of 7.8 m/s. it exits out a small nozz
tigry1 [53]

<span>Answer: 110.12 m/s </span>

We will use the formula A1V1 = A2V2 where 7.8 m/s is divided with 0.0085 m then multiply to 0.12 m, the result will be 110.117 or 110.12 m/s. This is related to the continuity of fluid flow in which as liquid moves horizontally, the same amount of liquid goes out as it comes in or the liquid itself do not change as it moves but the speed does when the diameter changes.

 

3 0
3 years ago
Read 2 more answers
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