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3 years ago

What is meant in astronomy by the phrase "adaptive optics?

1 answer:

7 0

<h2>Answer: a.The mirrors and eyepiece of a large telescope are spring-loaded to allow them to return quickly to a known position. </h2>

Explanation:

Adaptive optics is a method used in several astronomical observatories to counteract in real time the effects of the Earth's atmosphere on the formation of astronomical images.

This is done through the insertion into the optical path of the telescope of sophisticated deformable mirrors supported by a set of computationally controlled actuators. Thus obtaining clear images despite the effects of atmospheric turbulence that cause the unwanted distortion.

It should be noted that with this technique it is also necessary to have a moderately bright reference star that is very close to the object to be observed and studied. However, it is not always possible to find such stars, so a powerful laser beam is used to point towards the Earth's upper atmosphere and create artificial stars.

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Will is a scientist. He’s designing a spacecraft that would allow people to land on Mars. Will’s mass on Earth is 75 kilograms.

Vesnalui [34]

Mass will remain unchanged, always. His weight, which is the gravitational force acting on that mass will be less in this case.

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3 years ago

Read 2 more answers

(a) Calculate the magnitude of the gravitational force exerted on a 445-kg satellite that is a distance of 1.77 earth radii from

victus00 [196]

Answer:

a)1396.52 N

b)1396.52 N

c) $a_{satellite}= 3.13 m/sec^2$

d) $a_{earth}=2.32\times10^{-22} m/s^2$

Explanation:

The force experienced by the satelite is giveb by

$F= \frac{Gm_{satellite}m_{earth}}{r^2} \\$

m_{satellite}= 445 Kg

m_{earth}= 6×10^24 Kg

radius r= 1.77Re= 1.77×6.38×10^6 m

now putting values we get

$F= \frac{6.67\times10^{-11}(445)(6\times10^24)}{(1.77\times6.38\times10^6)^2}$

⇒F= 1396.52 N

now,

$a_{satellite}= \frac{F}{m_{satellite}}$

$a_{satellite}= \frac{1396.52}{445}$

$a_{satellite}= 3.13 m/sec^2$

also,

$a_{earth}= \frac{F}{m_{earth}}$

$a_{earth}= \frac{1396.52}{(6\times10^24)}$

$a_{earth}=2.32\times10^{-22} m/s^2$

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3 years ago

State advantages of ultrasonic sound in determining the depth of the ocean

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Answer:

It has <u>greater accuracy than other nondestructive methods in determining the depth of internal flaws and the thickness of parts with parallel surfaces.</u>

Explanation:

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3 years ago