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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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A vector A has a magnitude of 5 units and points in the −y-direction, while a vector B has triple the magnitude of A and points
Harman [31]

Answer:

A+B; 5√5 units, 341.57°

A-B; 5√5 units, 198.43°

B-A; 5√5 units, 18.43°

Explanation:

Given A = 5 units

By vector notation and the axis of A, it is represented as -5j

B = 3 × 5 = 15 units

Using the vector notations and the axis, B is +15i. The following vectors ate taking as the coordinates of A and B

(a) A + B = -5j + 15i

A+B = 15i -5j

|A+B| = √(15)²+(5)²

= 5√5 units

∆ = arctan(5/15) = 18.43°

The angle ∆ is generally used in the diagrams

∆= 18.43°

The direction of A+B is 341.57° based in the condition given (see attachment for diagrams

(b) A - B = -5j -15i

A-B = -15i -5j

|A-B|= √(15)²+(-5)²

|A-B| = √125

|A-B| = 5√5 units

The direction is 180+18.43°= 198.43°

See attachment for diagrams

(c) B-A = 15i -( -5j) = 15i + 5j

|B-A| = 5√5 units

The direction is 18.43°

See attachment for diagram

5 0
3 years ago
Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded goalp
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<span>The answer is -0.8 m/s. We know acceleration is the average of final minus initial velocity over time (a = (vf-v0)/t). We also know that Force is equal to Mass times Acceleration (F = ma). Using our force equation, we know that the acceleration we get is negative 8.8 (-8.8). The force is acting in the opposite direction of the rugby player, hence the negative sign. From there, plug in that number for a in the velocity equation, and solve for vf, as v0 and t are known. We get 0.8 m/s in the opposite direction that the player was running.</span>
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How was Mendeleev's table different from Meyer's table? A. Mendeleev did not arrange according to chemical properties. B. Meyer
olya-2409 [2.1K]

Answer: C. Meyer did not leave gaps while Mendeleev did.

Explanation: Mendeleev's table was table different from Meyer's table because Meyer did not leave gaps while Mendeleev did.

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How can the rate of a reaction be increased?
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