Answer:
Approximate linear dimension is 2 light years.
Explanation:
Radius of the spiral galaxy r = 62000 LY
Thickness of the galaxy h = 700 LY
Volume of the galaxy = πr²h
= (3.14)(62000)²(700)
= (3.14)(62)²(7)(10)⁸
= 84568×10⁸
=
(LY)³
Since galaxy contains number of stars = 1078 billion stars ≈ 
Now volume covered by each star of the galaxy = 
= 
= 7.839 Light Years
Now the linear dimension across the volume
= 
= 
= 1.99 LY
≈ 2 Light Years
Therefore, approximate linear dimension is 2 light years.
Answer:
<em>At constant mass, the acceleration of an object varies (</em><em>directly</em><em>) with the net external force applied. That is to say, that an object's acceleration increases as the force applied is (</em><em>increased</em><em>), but its acceleration decreases if the force applied is (</em><em>decreased</em><em>).</em>
Explanation:
<u>Mechanical Force
</u>
According to the second Newton's law, the acceleration of an object varies directly proportional to the external net force applied and inversely proportional to the mass of the object.
If the mass is constant, then the acceleration will vary in the same way as the force does.
Completing the sentences:
At constant mass, the acceleration of an object varies (directly) with the net external force applied. That is to say, that an object's acceleration increases as the force applied is (increased), but its acceleration decreases if the force applied is (decreased).
Power is work over time.

We can solve for work by using F*d

Now solve for power:
If a mass m is attached to an ideal massless spring and has a period of t, then the period of the system when the mass is 2m is
.
Calculation:
Step-1:
It is given that a mass m is attached to an ideal massless spring and the period of the system is t. It is required to find the period when the mass is doubled.
The time it takes an object to complete one oscillation and return to its initial position is measured in terms of a period, or T.
It is known that the period is calculated as,

Here m is the mass of the object, and k is the spring constant.
Step-2:
Thus the period of the system with the first mass is,

The period of the system with the second mass is,

Then the period of the system with the second mass is
times more than the period of the system with the first mass.
Learn more about period of a spring-mass system here,
brainly.com/question/16077243
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