C. The average acceleration is positive and smaller in magnitude than the initial acceleration.
<h3>
What is average velocity?</h3>
Average velocity is defined as the change in position or displacement (∆x) divided by the time intervals (∆t) in which the displacement occurs.
a = Δv/Δt
where;
- Δv is change in velocity
- Δt is change in time
a = (250 - 248) / (70 - 40)
a = 0.067 m/s²
Thus, we can conclude that the average acceleration is positive and smaller in magnitude than the initial acceleration.
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Answer:
that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
Explanation:
Let us carefully analyze the situation, when the bar is facing the index post a load of equal magnitude, but opposite sign on its surface, these two charges are in balance; When the hand touches the pole, it creates a path to the ground where the charges that were induced on the pole can be balanced with the charge coming from the ground, leaving a zero charge on the pole.
Now if the hand is removed, there can be no exchange of charges with the earth. When the bar is removed, the induced loads are redistributed in the post, but the excess loads that came from the earth that have the same value and are of a sign opposite to the induced ones remain, you want to sign that they are of the same sign as the charges of the bar.
In summary, after the process, the post has a load of equal magnitude and sign (negative) that of the bar.
If we assume that initially the weather vane was at rest, by this load that remained on the pole it would begin to move.
It can be both flat or it can be when you have new eyeglasses on and you look down it makes you think the ground looks like that but its not
6 . . . . . a crest
7 . . . . . the amplitude
8 . . . . . the wavelength
9 . . . . . a trough
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
