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1 year ago

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

Physics

1 answer:

son4ous [18]1 year 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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(a) The magnitude of the acceleration of the electron is 5.62 x 10¹³ m/s².

(b) The speed of the electron after the given time is 4.78 x 10⁵ m/s.

<h3>Acceleration of the electron</h3>

The acceleration of the electron is calculated as follows;

F = qE

ma = qE

a = qE/m

a = (1.6 x 10⁻¹⁹ x 320)/(9.11 x 10⁻³¹)

a = 5.62 x 10¹³ m/s²

<h3>Speed of the electron</h3>

v = at

v = 5.62 x 10¹³ m/s² x 8.50 x 10⁻⁹ s

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2 years ago

A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?

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Answer:

L = 2.8 cm

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L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

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2 years ago

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3 years ago

A pair of slits separated by 1 mm, are illuminated with monochromatic light of wavelength 411 nm. The light falls on a screen 1.

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Answer:

$t = 0.192 \mu m$

Explanation:

Path difference due to a transparent slab is given as

$\Delta x = (\mu - 1) t$

here we know that

$\mu = 1.79$

now total shift in the bright fringe is given as

$Shift = \frac{D(\mu - 1)t}{d}$

Also we know that the fringe width of maximum intensity is given as

$\delta x = \frac{\lambda D}{d}$

now we have

$\frac{D}{d} = \frac{\delta x}{\lambda}$

now the shift is given as

$Shift = \frac{(\mu - 1) t \delta x}{\lambda}$

given that the shift is

$Shift = 0.37 \delta x$

here we have

$0.37 \delta x = \frac{(\mu - 1) t \delta x}{\lambda}$

now plug in all values in it

$0.37 = \frac{(1.79 - 1) t}{411 \times 10^{-9}}$

$t = 0.192 \times 10^{-6} m$

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