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3 years ago

What are some of the challenges that you might face if you wanted to observe and study stars from Earth?

2 answers:

7 0

<span>The earth moves, which is unstoppable and if the earth moves the telescope won't be able to see it clearly because the telescope needs to be able to move at the same pace as earth to keep up to the objects</span>

AlexFokin [52]3 years ago

4 0

Another main issue with astronomical sighting is light pollution. The lights from our cities and such drown out the light of the stars. You could also say clouds, humidity, fog, or anything of the kind.

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Particles that attract each other are

photoshop1234 [79]

B. I belive :)
Hopes this helps

3 0

2 years ago

Read 2 more answers

1.How far is Object Z from the origin at t = 3 seconds?

GenaCL600 [577]

Answer:

Please find the answer in the explanation

Explanation:

1.) How far is Object Z from the origin at t = 3 seconds

The distance of the object Z from the origin will be the slope of the graph.

Slope = 4/2 = 2m

2.) Which object takes the least time to reach a position 4 meters from the origin ?

According to the graph given to the question above, object Z has the list time which is 2 seconds since object X does not start from the origin.

3.) Which object is farthest from the origin at t = 2 seconds?

The correct answer is still object Z because it has the highest slope.

4 0

2 years ago

Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the

Anit [1.1K]

Answer:

C. Quadruple

Explanation:

¨Drivers who are talking on the phone, even on a hands-free device, are up to four times more likely to be involved in a crash.¨

I hope this helps! Have a great day!

3 0

2 years ago

Two ships of equal mass are 110 m apart. What is the acceleration of either ship due to the gravitational attraction of the othe

dusya [7]

Answer:

Acceleration of the ship, $a=2.14\times 10^{-7}\ m/s^2$

Explanation:

It is given that,

Mass of both ships, $m=39000\ metric\ tons=39\times 10^6\ kg$

Distance between two ships, d = 110 m

The gravitational force between two ships is given by :

$F=G\dfrac{m^2}{d^2}$

$F=6.67\times 10^{-11}\ Nm^2/kg^2\times \dfrac{(39\times 10^6\ kg)^2}{(110\ m)^2}$

F = 8.38 N

Let a is the acceleration. Now, using second law of motion as :

$a=\dfrac{F}{m}$

$a=\dfrac{8.38\ N}{39\times 10^6\ kg}$

$a=2.14\times 10^{-7}\ m/s^2$

So, the acceleration of either ship due to the gravitational attraction of the other is $2.14\times 10^{-7}\ m/s^2$ . Hence, this is the required solution.

7 0

2 years ago

If the second ball has a mass of 2.4kg and a constant acceleration of a⃗ 2= 2.8m/s2 j^ , what must the mass of the first ball be

kompoz [17]

Hope this helps you!

8 0

2 years ago