Answer:
A
Explanation:
because thats what I put and got it right
Answer:
If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
Explanation:
Option A is incorrect because, given this case, it is easier to calculate the field.
Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated
Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.
Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."
Let
be the average acceleration over the first 2.46 seconds, and
the average acceleration over the next 6.79 seconds.
At the start, the car has velocity 30.0 m/s, and at the end of the total 9.25 second interval it has velocity 15.2 m/s. Let
be the velocity of the car after the first 2.46 seconds.
By definition of average acceleration, we have


and we're also told that

(or possibly the other way around; I'll consider that case later). We can solve for
in the ratio equation and substitute it into the first average acceleration equation, and in turn we end up with an equation independent of the accelerations:


Now we can solve for
. We find that

In the case that the ratio of accelerations is actually

we would instead have

in which case we would get a velocity of
