Answer:
The distance is 
Explanation:
From the question we are told that
The coefficient of static friction is 
The initial speed of the train is 
For the crate not to slide the friction force must be equal to the force acting on the train i.e

The negative sign shows that the two forces are acting in opposite direction
=> 
=> 
=> 
=> 
From equation of motion

Here v = 0 m/s since it came to a stop
=> 
=> 
=> 
Answer:
False
Explanation:
Please see the attached file
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:
B. There are no forces acting on the ball.
Explanation:
There are no forces acting on the ball.
(30, 5)
(10, 1)
change of y / change of x
= (30 - 10) / (5 - 1)
= 20 /4
= 5