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Vilka [71]
3 years ago
8

Why might a telescope be built high in the mountains above weather sys

Physics
2 answers:
Debora [2.8K]3 years ago
6 0

Large telescopes aren't built on mountain tops for the purpose of watching
weather systems.  The ones that are built at high altitudes are intended to be
used to observe celestial objects ... planets, stars, galaxies, comets, nebulae,
quasars, novae, and the space around black holes. 

The way a telescope does that is:  It collects visible light and radiation with
other electromagnetic wavelengths, and people then analyze the radiation
that the telescope has collected.

When we use the telescope to do that, we want anything it collects to be
as close as possible to the radiation that actually left the star.  The problem
is that anything the telescope collects must come down through AIR. The trip
through air changes the radiation before you have a chance to collect it, so
you can never see exactly what left the star.

The solution: 

==> Build your telescope in a place where the light goes through less air
before it reaches the telescope. 

==> Or ... if you can work it out somehow ... through NO air.

That means: 

==> Build your telescope at high altitude, on a mountaintop, where
most of the Earth's air is BELOW you.

==> Or put your telescope in a spacecraft.  Put the spacecraft in orbit
around the Earth, where there is almost NO air, and let the telescope
send its pictures and other data to you by radio. 

Schach [20]3 years ago
6 0

you could look down on the weather systems AND look up into the troposphere to see every weather system or whatever is happening in the troposphere
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8 0
1 year ago
Initially (at time t = 0) a particle is moving vertically at 7.5 m/s and horizontally at 0 m/s. Its horizontal acceleration is 1
Lelu [443]

Answer:

t = 0.657 s

Explanation:

given,

initial vertical velocity = 7.5 m/s

initial horizontal velocity = 0 m/s

angle = 49◦

using kinetic equation

final velocity in vertical direction

v sinθ    = u_y   - gt ........................(1)

final velocity in horizontal direction

v cosθ = u_x   + a_x × t

here   u_x = 0.0 m/s

v cosθ = a_x×t ......................(2)

Dividing equation (1) / (2)

tan \theta =\dfrac{u_y - gt}{a_x\times t}

solving for time t

t = \dfrac{u_y}{tan \theta \times a_x + g}

u_y   =   initial velocity along x direction

acceleration along a_x = 1.4 m/s²

g = acceleration due to gravity   = 9.8 m/s²

θ = 43° ,   u_y   =  7.5 m/s    

t = \dfrac{7.5}{tan 49^0\times 1.4+ 9.8}

t = 0.657 s

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6 0
2 years ago
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