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Gre4nikov [31]
3 years ago
5

Which of these parts of a telescope is used to make the light focused from distant stars visible to the human eye?

Physics
1 answer:
Zina [86]3 years ago
4 0
The correct option is A.
A telescope is an optical instrument that is used to observe celestial bodies in order to gather information about them. There are different types of telescopes but all of them have some common parts. One of the common parts is the eye piece. The eye piece is the part of the telescope that one looks through and is the part that makes the focused light visible to the eye. The eye piece is made up of magnifying lenses.<span />
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When the pressure on a gas increases,what does the volume do?
kupik [55]

Answer:

As pressure goes up, volume goes down.

Explanation:

Pressure and volume of a gas are inversely proportional.  This means that as pressure goes up, volume goes down.  And as volume goes up, pressure goes down.

Cheers.

4 0
3 years ago
Read 2 more answers
Write a hypothesis about the use of an object's physical
KengaRu [80]

If the mass of the object and the volume of the object is determined;

Then, the density of the object is determined by taking the ratio of the mass and volume.

<h3>What is density of an object?</h3>

The density of an object is the ratio of the mass and volume of that object.

Mathematically;

  • Density = mass/volume

To determine the density of an object therefore, the physical characteristics of mass and the volume of the object are measured.

The mass of the object is obtained using a scale or a balance.

The volume of the object if a solid is obtained using a displacement bottle. If it is a liquid, a measuring cylinder is used.

The density of the object is then obtained by taking the ratio of the mass and the volume of the object.

In conclusion, the density of an object is determined from the volume and mass ratio.

Learn more about density at: brainly.com/question/1354972

#SPJ1

3 0
2 years ago
A marble rolls 269cm across the floor with a constant speed of in 44.1cm/s.
Marrrta [24]

Answer:

t = 6.09 seconds

Explanation:

Given that,

Speed, v = 44.1 cm/s

Distance, d = 269 cm

We need to find the time interval of the marble. Speed is distance per unit time.

v=\dfrac{d}{t}\\\\\implies t=\dfrac{d}{v}\\\\t=\dfrac{269\ \text{cm}}{44.1\ \text{cm/s}}\\\\t=6.09\ s

Hence, the time interval of the marble is 6.09 seconds.

6 0
4 years ago
How is momentum conserved in a newtons cradle <br><br> when one steel ball hits the other
arsen [322]

In Newton's Cradle experiment we know that all cradles of same mass and identical to each other

Now we know that when two identical objects collide elastically then they interchange their velocity

So here we have same illustration

When Newton pulls up a cradle and release it will move hit another cradle which is at rest

Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.

So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.

So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.

So here we can say that in horizontal direction when all cradles are colliding each other there is no external force on the system so momentum is conserved and they all will move off with same speed and hence we observe the above condition.

8 0
3 years ago
The masses of the two moons are determined to be 2M2M for Moon AA and MM for Moon BB . It is observed that the distance between
seraphim [82]

Answer:

 F_A = 8 F_B

Explanation:

The force exerted by the planet on each moon is given by the law of universal gravitation

        F = G \frac{m M}{r^{2} }

where M is the mass of the planet, m the mass of the moon and r the distance between its centers

let's apply this equation to our case

Moon A

the distance between the planet and the moon A is r and the mass of the moon is 2m

        F_A = G \frac{2m M}{r^{2} }

Moon B

        F_B = G \frac{m M}{(2r)^{2} }

         F_B = G \frac{m M}{4 r^{2} }

the relationship between these forces is

         F_B / F_A = \frac{1}{2 \ 4 } = 1/8

         F_A = 8 F_B

7 0
3 years ago
Read 2 more answers
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