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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.

You might be interested in

Which of the following is an example of resourcefulness?

tatiyna

A. Discovering a quick way to handle a new problem

5 0

3 years ago

I don’t have a question because I posted a photoz

Scorpion4ik [409]

Part a:

$Q_{1}$ = 56

$Q_{2}$ = 60

$Q_{3}$ = 63

The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The $Q_{2}$ is the overall medium, $Q_{1}$ is the medium of the first half, and $Q_{3}$ is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between $Q_{1}$ and $Q_{3}$ . In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + $\frac{7}{2}$ ) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0

2 years ago

Which of the following is a small star that has reached the end of its stellar evolution?

Akimi4 [234]

D.) White Dwarf

It is the smallest star whose mass is approximately equal or greater than 1.4M
Here, M = mass of the Sun.

Hope this helps!

3 0

3 years ago

Read 2 more answers

A particle moves along a straight line. Its position at any instant is given by x = 32t− 38t^3/3 where x is in metre and t in se

Rudik [331]

Answer:

The acceleration of the object is -69.78 m/s²

Explanation:

Given;

postion of the particle:

$x = 32t - 38\frac{t^3}{3} \\\\$

The velocity of the particle is calculated as the change in the position of the particle with time;

$v = \frac{dx}{dt} = 32 - 38t^2\\\\when \ the \ particle \ is \ at \ rest, \ v = 0\\\\32-38t^2 = 0\\\\38t^2 = 32\\\\t^2 = \frac{32}{38} \\\\t = \sqrt{\frac{32}{38} } \\\\t = 0.918 \ s$

Acceleration is the change in velocity with time;

$a = \frac{dv}{dt} = -76t\\\\recall , \ t = 0.918 \ s\\\\a = -76(0.918)\\\\a = -69.78 \ m/s^2$

4 0

2 years ago

The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 5.0 rev/s in 8.0 s. At thi

tatiyna

Answer:

50 revolutions

Explanation:

Data provided:

case I: From rest to top spin

The initial angular speed of the washer, ωi = 0 rev /s

Final angular speed of the washer ωf = 5 rev /s

Time taken, t₁ = 8 s

now,

The angular displacement or the number of revolutions taken (θ₁) is calculated as:

θ₁ = ωi t₁ + (1/2)α₁t₁²

where,

α is the angular acceleration

The angular acceleration can be calculated as:

ωf - ωi = α₁t₁

on substituting the values, we get

8α₁ = 5 - 0

α₁ = 0.625 rev/s²

substituting the values in the equation for the number of revolutions, we get

θ₁ = 0 + (1/2) (0.625)(8)²

θ₁ = 20 revolutions

also,

For the case II: From top spin to rest

we have

The initial angular speed, ωi = 5 rev /s

and the final angular speed, ωf = 0 rev /s

Total time taken, t₂ = 12 s

Now, angular acceleration for this case

ωf - ωi = α₂t₂

on substituting the values, we have

12α₂ = 0 - 5

α₂ = - 0.4166 rev/s²

Therefore, the number of revolutions ( i.e angular displacement )

θ₂ = ωit₂ + (1/2)α₂t₂²

on substituting the values, we have

θ₂ = 5 × 12 + (1/2)(-0.4166)(12)²

θ₂ = 30 rev

Hence,

the total number of revolutions made by the washer during the 20s is

θ = θ₁ + θ₂

θ = 20 rev + 30 rev

θ = 50 revolutions

7 0

3 years ago