4000 seconds
Explanation:
speed = distance / time
0.0004m/s = 1.6m / time
Subject time
time = 1.6 / 0.0004
time = 4000 seconds.
Hope this helps. Mark as brainliest if possible. tks
Answer:
The answer is based on the conservation of energy law; something you should really understand by now.
For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.
So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.
It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.
It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.
But there's more.
Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.
And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.
Answer:
a. 
b. 
c. 
d. 
e. 
Explanation:
<u>Work and Kinetic Energy
</u>
When an object moves at a certain velocity v0 and changes it to v1, a change in its kinetic energy is achieved:

Knowing that

We have

The work done by the force who caused the change of velocity (acceleration) is

If we know the distance x traveled by the object, the work can also be calculated by

Being F the force responsible for the change of velocity
The 75 kg baseball player has an initial velocity of 6 m/s, then he slides and stops
a. Before the slide, his initial kinetic energy is



b. Once he reaches the base, the player is at rest, thus his final kinetic energy is


c. The change of kinetic energy is


d. The work done by friction to stop the player is


e. We compute the force of friction by using

and solving for x



The negative sign indicates the force is against movement