Answer:
The answer to your question is a = -1.85 m/s² the acceleration is negative because it is coming to stop.
Explanation:
Data
vo = 25 m/s
t = 13.5 s
a= ?
vf = 0 m/s
Formula
vf = vo + at
solve for a
a = (vf - vo)/t
Substitution
a = (0 - 25) / 13.5
Simplification
a = -25/13.5
Result
a = -1.85 m/s²
Answer:
C) No work is required to move the negative charge from point A to point B.
Explanation:
An equipotential surface is defined as a surface connecting all the points at the same potential.
Therefore, when a charge moves along an equipotential surface, it moves between points at same potential.
The work done when moving a charge is given by

where
q is the charge
is the potential difference between the initial and final point of motion of the charge
However, the charge in this problem moves along an equipotential surface: this means that the potential does not change, so

And so, the work done is also zero.
Answer:![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Explanation:
Given
Three charges of magnitude q is placed at three corners and fourth charge is placed at last corner with -q charge
Force due to the charge placed at diagonally opposite end on -q charge

where
Distance between the two charges

negative sign indicates that it is an attraction force
Now remaining two charges will apply the same amount of force as they are equally spaced from -q charge

The magnitude of force by both the charge is same but at an angle of 
thus combination of two forces at 2 and 3 will be

Now it will add with force due to 1 charge
Thus net force will be
![F_{net}=\frac{kq^2}{(L)^2}\left [ \frac{1}{2}+\sqrt{2}\right ]](https://tex.z-dn.net/?f=F_%7Bnet%7D%3D%5Cfrac%7Bkq%5E2%7D%7B%28L%29%5E2%7D%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B2%7D%2B%5Csqrt%7B2%7D%5Cright%20%5D)
Average speed = distance traveled / time
average speed = (126.5 m * 3.5 laps) / (4.17 min)
= 106.2 m/min
Answer:
The driver was not telling the truth because it is not possible for a car to hit another car from behind and generate a force to the sides that deflects it from its path.
Explanation:
First, we analyze the driver's statement.
The driver when arriving at the curve, is collided from behind by another car and deviates from his path and crashes into a tree. For the car to go to the tree there must be a force towards the tree.
The net force that causes the car to deviate must be formed by the sum of the motion vector of the first car plus the force that is directed towards the tree.
Here we verify that a car hitting from behind will not generate a force to the sides, but will generate a force in the same direction that the car moves, forward.