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3 years ago

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

2 answers:

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

You might be interested in

The fulcrum is the

olga_2 [115]

Answer:

point of support on which a lever rotates.

Explanation:

The fulcrum is the point of support on which a lever rotates. Fulcrum is a pivotal part of simple machines.

The fulcrum provides the platform for a lever to torque.

The force that opposes motion by the applied force is termed the frictional force.
Friction is a force that opposes motion.
The stored energy of an object is its potential energy.
The potential energy is the energy due to the position of a body.
The distance an object moves when doing work is termed its displacement.

6 0

3 years ago

A plane drops a rubber raft to the survivors of a shipwreck. The plane is flying at a height of 1960 m and with a speed of 109 m

Korolek [52]

In this case the rubber raft has horizontal and vertical motion.

Considering vertical motion first.

We have displacement $s = ut +\frac{1}{2}at^2$ , u = Initial velocity, t = time taken, a = acceleration.

In vertical motion

s = 1960 m, u = 0 m/s, a = 9.81 $m/s^2$

$1960 = 0*t+\frac{1}{2} *9.81*t^2\\ \\ t = 20 seconds$

So raft will take 20 seconds to reach ground.

Now considering horizontal motion of raft

u = 109 m/s, t = 20 s, a = 0 $m/s^2$

So $s = 109*20+\frac{1}{2} *0*20^2 = 2180 m$

So shipwreck was 2180 meter far away from the plane when the raft was dropped.

8 0

3 years ago

For this discussion, you will work in groups to answer the questions. In a video game, airplanes move from left to right along t

Mariulka [41]

Answer:

When fired from (1,3) the rocket will hit the target at (4,0)

When fired from (2, 2.5) the rocket will hit the target at (12,0)

When fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

When fired from (4,2.25) the rocket will hit the target at (40,0)

Explanation:

All of the parts of the problem are solved in the same way, so let's start with the first point (1,3).

Let's assume that the rocket's trajectory will be a straight line, so what we need to do here is to find the equation of the line tangent to the trajectory of the airplane and then find the x-intercept of such a line.

In order to find the line tangent to the graph of the trajectory of the airplane, we need to start by finding the derivative of such a function:

$y=2+\frac{1}{x}$

$y=2+x^{-1}$

$y'=-x^{-2}$

$y'=-\frac{1}{x^{2}}$

so, we can substitute the x-value of the given point into the derivative, in this case x=1, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(1)^{2}}$

m=y'=-1

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-3=-1(x-1})$

$y-3=-1x+1$

$y=-x+1+3$

$y=-x+4$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-x+4=0$

and solve for x

x=4

so, when fired from (1,3) the rocket will hit the target at (4,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2, 2.5)

so, we can substitute the x-value of the given point into the derivative, in this case x=2, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2)^{2}}$

$m=y'=-\frac{1}{4}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.5=-\frac{1}{4}(x-2})$

$y-2.5=-\frac{1}{4}x+\frac{1}{2}$

$y=-\frac{1}{4}x+\frac{1}{2}+\frac{5}{2}$

$y=-\frac{1}{4}x+3$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{4}x+3=0$

and solve for x

x=12

so, when fired from (2, 2.5) the rocket will hit the target at (12,0)

Now, let's calculate the coordinates where the rocket will hit the target if fired from (2.5, 2.4)

so, we can substitute the x-value of the given point into the derivative, in this case x=2.5, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(2.5)^{2}}$

$m=y'=-\frac{4}{25}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.4=-\frac{4}{25}(x-2.5})$

$y-2.4=-\frac{4}{25}x+\frac{2}{5}$

$y=-\frac{4}{25}x+\frac{2}{5}+2.4$

$y=-\frac{4}{25}x+\frac{14}{5}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{4}{25}x+\frac{14}{5}=0$

and solve for x

$x=\frac{35}{20}$

so, when fired from (2.5, 2.4) the rocket will hit the target at $(\frac{35}{2},0)$

Now, let's calculate the coordinates where the rocket will hit the target if fired from (4, 2.25)

so, we can substitute the x-value of the given point into the derivative, in this case x=4, so:

$y'=-\frac{1}{x^{2}}$

$y'=-\frac{1}{(4)^{2}}$

$m=y'=-\frac{1}{16}$

so we can now use this slope and the point-slope form of the line to find the equation of the line tangent to the trajectory of the airplane so we get:

$y-y_{1}=m(x-x_{1})$

$y-2.25=-\frac{1}{16}(x-4})$

$y-2.25=-\frac{1}{16}x+\frac{1}{4}$

$y=-\frac{1}{16}x+\frac{1}{4}+2.25$

$y=-\frac{1}{16}x+\frac{5}{2}$

So we can now set y=0 so find the x-coordinate where the rocket hits the x-axis.

$-\frac{1}{16}x+\frac{5}{2}=0$

and solve for x

$x=40$

so, when fired from (4,2.25) the rocket will hit the target at (40,0)

I uploaded a graph that represents each case.

8 0

3 years ago

Which statements describe Earth? Select three options. It is an outer planet. It is a terrestrial planet. It was formed from gas

AVprozaik [17]

The formation of Earth is supported by the statements like, it is a terrestrial planet. it was formed from gas and dust. it was formed in a debris disk from colliding planetesimals.

The given problem is based on the correct statements entitling the description of Earth. Earth is amongst 8 planets in our solar system, that happen to orbit around the sun.

The statements given to describe the Earth are as follows:

Earth is not an Outer planet. It is an inner planet that lies closest to the sun after Mercury and Venus.
Earth is a terrestrial planet because it is having a compact and rocky surface. Also, it is known to be an as largest terrestrial planet in the solar system with extensive regions of liquids and water.
The abundant rocky surfaces have evolved from the cloud of dust and gas, during the post-Big Bang Era. So, it is somewhere true to say that Earth is formed from gas and dust.
The earlier atmosphere of Earth was known for having proportional layers of Hydrogen and Helium. Hence is quite true to say that the Earth is having an atmosphere of Hydrogen and helium gases, but it is not as thick as the like sun.
Majority of terrestrial planets are formed from the collision of planetesimals in a debris disk. With Earth being one of them, it is quite correct to consider the given statement.

Thus, we can conclude that the formation of Earth is supported by the statements like: it is a terrestrial planet, it was formed from gas and dust and it was formed in a debris disk from colliding planetesimals.

Learn more about the planet Earth here:

brainly.com/question/24878669

3 0

2 years ago

Boyle's Law mainly involves _______.

goblinko [34]

Your answer is B, gases

6 0

3 years ago