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tensa zangetsu [6.8K]

3 years ago

8

Name the two spectral windows through which electromagnetic radiation easily reaches the surface of earth and describe the large

st-aperture telescope currently in use for each window.

1 answer:

Anestetic [448]3 years ago

4 0

Answer: • Visible Light, Radio Waves

• Radio - 305-m , at Arecibo, Puerto Rico

Visible Light - 10.4m Mirror, Canary Island

Explanation:

The spectral window is simply the range of frequencies that are correctly measured. It should be noted that the signals that are outside the spectral window are folded when they show up in spectrum.

The two spectral windows through which electromagnetic radiation easily reaches the surface of earth are the visible light and the radio waves.

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The James Webb Space Telescope is positioned around 1.5 million kilometres from the Earth on the side facing away from the Sun.

Bad White [126]

The angular velocity depends on the length of the orbit and the orbital

speed of the telescope.

Response:

First question:

The angular velocity of the telescope is approximately <u>0.199 rad/s</u>

Second question:

The telescope should accelerates away by approximately F = <u>0.0005·m </u>

Third question:

<u>The pulling force between the Earth and the satellite</u>

<h3>What equations can be used to calculate the velocity and forces acting on the telescope?</h3>

The distance of the James Webb telescope from the Sun = 1.5 million kilometers from Earth on the side facing away from the Sun

The orbital velocity of the telescope = The Earth's orbital velocity

First question:

$Angular \ velocity = \mathbf{\dfrac{Angle \ turned}{Time \ taken}}$

The orbital velocity of the Earth = 29.8 km/s

The distance between the Earth and the Sun = 148.27 million km

The radius of the orbit of the telescope = 148.27 + 1.5 = 149.77

Radius of the orbit, r = 149.77 million kilometer from the Sun

The length of the orbit of the James Webb telescope = 2 × π × r

Which gives;

r = 2 × π × 149.77 million kilometers ≈ 941.03 million kilometers

Therefore;

$Angular \ velocity = \dfrac{29.8}{941.03}\times 2 \times \pi \approx 0.199$

The angular velocity of the telescope, ω ≈ <u>0.199 rad/s</u>

Second question:

Centrifugal force force, $F_{\omega}$ = m·ω²·r

Which gives;

$F_{\omega} = m \cdot \dfrac{28,500^2 \, m^2/s^2}{149.77 \times 10^9 \, m} \approx 0.0054233 \cdot m$

$Gravitational \ force, F_G = \mathbf{G \cdot \dfrac{m_{1} \cdot m_{2}}{r^{2}}}$

Universal gravitational constant, G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Mass of the Sun = 1.989 × 10³⁰ kg

Which gives;

$F_G = 6.67408 \times 10^{-11} \times \dfrac{1.989 \times 10^{30} \times m}{149.77 \times 10^9} \approx 0.00592 \cdot m$

Which gives;

$F_{\omega}$ < $F_G$ , therefore, the James Webb telescope has to accelerate away from the Sun

F = $\mathbf{F_{\omega}}$ - $\mathbf{F_G}$

The amount by which the telescope accelerates away is approximately 0.00592·m - 0.0054233·m ≈ <u>0.0005·m (away from the Sun)</u>

Third part:

Other forces include;

<u>The force of attraction between the Earth and the telescope </u>which can contribute to the the telescope having a stable orbit at the given speed.

Learn more about orbital motion here:

brainly.com/question/11069817

3 0

3 years ago

What causes an aurora?

geniusboy [140]

Answer:

When charged particles from the sun strike atoms in Earth's atmosphere, they cause electrons in the atoms to move to a higher-energy state. When the electrons drop back to a lower energy state, they release a photon: light. This process creates the beautiful aurora, or northern lights.

Explanation:

6 0

4 years ago

What is work in science

mr_godi [17]

Answer:

The scientific defenision of work reveals its relationship to energy; whenever work is done, energy is transfered.

4 0

3 years ago

Find the frequency (in hertz) of the first overtone for a 1.75 m pipe that is closed at only one end. Use 350 m/s for the speed

serg [7]

Answer:

The first overtone frequency is 100 Hz.

Solution:

According to the question:

Length of pipe, l = 1.75 m

Speed of sound in air, v_{sa} = 350 m/s

Frequency of first overtone, $f_{1}$ is given by:

$f_{1} = \frac{v_{sa}}{2l}$

$f_{1} = \frac{350}{2\times 1.75} = 100 Hz$

Since, the frequency, as clear from the formula depends only on the speed

and the length. It is independent of the air temperature.

Thus there will be no effect of air temperature on the frequency.

7 0

3 years ago

A ball is spun around in circular motion such that it completes 50 rotations in 25 s. What is the frequency of its rotation? 2.

Elis [28]

Answer:

1. $f = 2 Hz$

2. $f = 0.011 Hz$

3. $f = 0.067 Hz$

4. $t = 4 s$

Explanation:

1. The frequency of rotation is given by:

$f = \frac{\omega}{2\pi}$

Where:

ω: is the angular speed = 50 rotations (revolutions) in 25 s.

We need to convert the units of ω.

$\omega = \frac{50 rev}{25 s}*\frac{2\pi rad}{1 rev} = 4\pi rad/s$

Now, the frequency is:

$f = \frac{4\pi rad/s}{2\pi} = 2 Hz$

2. The frequency is:

We know:

5 laps = 5 revolutions

t: time = 450 s

$f = \frac{\omega}{2\pi} = \frac{\frac{5 rev}{450 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.011 Hz$

3. The frequency of the pendulum is:

$f = \frac{\omega}{2\pi} = \frac{\frac{1 rev}{15 s}*\frac{2\pi rad}{1 rev}}{2\pi} = 0.067 Hz$

4. We have:

θ: number of revolutions = 48 rev

f = 12 Hz

t =?

The time can be calculated as follows:

$f = \frac{\omega}{2\pi} = \frac{\theta}{2\pi t}$

$t = \frac{\theta}{2\pi f} = \frac{48 rev*\frac{2\pi rad}{1 rev}}{2\pi*12 Hz} = 4 s$

I hope it helps you!

4 0

3 years ago