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Ad libitum [116K]
2 years ago
15

Optical astronomers need a clear, dark sky to collect good data. Explain why radio astronomers have no problem observing in the

UK where it is often very cloudy.
Physics
1 answer:
gregori [183]2 years ago
3 0

Answer:

In the clarification portion elsewhere here, the definition of the concern is mentioned.

Explanation:

So like optical telescopes capture light waves, introduce it to concentrate, enhance it, as well as make it usable through different instruments via study, so radio telescopes accumulate weak signal light waves, introduce that one to focus, enhance it, as well as make this information available during research. To research naturally produced radio illumination from stars, galaxies, dark matter, as well as other natural phenomena, we utilize telescopes.

Optical telescopes detect space-borne visible light. There are some drawbacks of optical telescopes mostly on the surface:

  • Mostly at night would they have been seen.
  • Unless the weather gets cloudy, bad, or gloomy, they shouldn't be seen.

Although radio telescopes monitor space-coming radio waves. Those other telescopes, when they are already typically very massive as well as costly, have such an improvement surrounded by optical telescopes. They should be included in poor weather and, when they travel through the surrounding air, the radio waves aren't obscured by clouds. Throughout the afternoon and also some at night, radio telescopes are sometimes used.

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A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attach
Olegator [25]

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

3 0
2 years ago
Read 2 more answers
At its maximum speed, a typical snail moves about 4.0 m in 5.0 min.
stiv31 [10]

Answer:

Explanation:

Given

Distance = 4.0m

Time = 5.0 mins = 300secs

Required

Average speed

Average speed = Distance/Time

Average speed = 4.0/300

Average speed = 0.01333m/secs

Hence the average speed of the snail is 0.01333m/s

6 0
3 years ago
Imagine a small child whose legs are half as long as her parent’s legs. If her parent can walk at maximum speed V, at what maxim
AnnZ [28]

Answer:

\boxed{v=\frac {V}{\sqrt {2}}}

Explanation:

We know that speed is given by dividing distance by time or multiplying length and frequency. The speed of the father will be given by Lf where L is the length of the father’s leg ad f is the frequency.

We know that frequency of simple pendulum follows that f=\frac {1}{2\pi} \sqrt {\frac {g}{l}}

Now, the speed of the father will be V=Lf= L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}}) while for the child the speed will be v=\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})

The ratio of the father’s speed to the child’s speed will be

\frac {V}{v}=\frac {\frac {L}{2}\times (\frac {1}{2\pi} \sqrt {\frac {g}{0.5l}})}{ L\times (\frac {1}{2\pi} \sqrt {\frac {g}{l}})}\\\frac {V}{v}=\frac {\sqrt {2}}{2}\\\boxed{v=\frac {V}{\sqrt {2}}}

8 0
3 years ago
How can the frequency with which the direction of a current changes be regulated?
sesenic [268]

Answer:

In an inductive circuit, when frequency increases, the circuit current decreases and vice versa.

Explanation:

5 0
3 years ago
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Answer:

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