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2 years ago

Which telescopes must be placed in orbit around earth in order to observe short-wavelength radiation?.

Physics

1 answer:

son4ous [18]2 years ago

6 0

Space telescopes must be placed in orbit around earth in order to observe short-wavelength radiation.

<h3>What is telescope?</h3>

A telescope is an optical instrument that uses lenses, curved mirrors, or a combination of both to watch distant objects.

When atoms in a gas reach this temperature, they travel so quickly that when they collide, they release X-ray photons with wavelengths smaller than 10 nanometers.

Because the Earth's atmosphere prevents all X-rays from space, these wavelengths must be seen using space telescopes.

To study short-wavelength radiation, space telescopes must be put in orbit around the Earth.

Hence, space telescope is the correct answer.

To learn more about the telescope, refer:

brainly.com/question/556195

#SPJ1

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The heart damaging chemical found in the blood that association with long term depression is

adoni [48]

The answer is homocysteine hehe hope this helps

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4 years ago

Jogger A has a mass m and a speed v, jogger B has a mass m/2 and a speed 3v, jogger C has a mass 3m and a speed v/2, and jogger

finlep [7]

Answer: B>A=D>C

Explanation:

Kinetic Energy is the product of mass and square of the velocity

For Jogger A

$K.E._a=\frac{1}{2}mv^2$

For Jogger B

$K.E._b=\frac{1}{2}\times\frac{m}{2}\times(3v)^2=\frac{9}{4}mv^2\\K.E._b=2.25mv^2$

For Jogger C

$K.E._c=\frac{1}{2}\times3m\times(\frac{v}{2})^2=\frac{3}{8}mv^2\\K.E._c=0.375mv^2$

For Jogger D

$K.E._d=\frac{1}{2}\times4m\times(\frac{v}{2})^2=\frac{1}{2}mv^2$

Kinetic Energy of Joggers in increasing order

B>A=D>C

8 0

3 years ago

A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa

andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

$\omega=\frac{v}{r}$

where

$\omega$ is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s$

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

$\omega=\frac{v}{r}$

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

$\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s$

(c) 5550 m

The maximum playing time of the CD is

$t =74.0 min \cdot 60 s/min = 4,440 s$

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

$d=vt=(1.25 m/s)(4,440 s)=5,550 m$

(d) $-6.4\cdot 10^{-3} rad/s^2$

The angular acceleration of the CD is given by

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 21.6 rad/s$ is the final angular speed (when the CD is scanned at the outermost part)

$\omega_i = 50.0 rad/s$ is the initial angular speed (when the CD is scanned at the innermost part)

$t=4440 s$ is the time elapsed

Substituting into the equation, we find

$\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2$

5 0

4 years ago

Read 2 more answers

A force of magnitude 33.73 lb directed toward the right is exerted on an object. What other force must be applied to the object

vova2212 [387]

Answer:

33.73 lb to the left

Explanation:

You need to exert a force with the same magnitude, but opposite direction. You can visualize it in this way: When you push an object, the object will follow your path, but if there is another person opposing the force you are exerting, the object will just not move. If the force that the other person exerts were higher, then the object would move in the opposite direction. So, you need them to have the same magnitude.

7 0

3 years ago

During an oil spill, oil coats the surface of the water, endangering sea life. Why doesn’t oil dissolve in ocean water?

Natasha2012 [34]

The properties of oil and water do not mix

5 0

3 years ago