Answer:
the average kinetic energy of water particles - C.
105 miles because you have to use the gif arable
Answer:

and

Explanation:
Given:
- first charge,

- second charge,

- position of first charge,

- position of second charge,

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.
<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

- since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.





Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

and

Answer:
<h2>135,000 J</h2>
Explanation:
The work done by an object can be found by using the formula
workdone = force × distance
From the question we have
workdone = 900 × 150
We have the final answer as
<h3>135,000 J</h3>
Hope this helps you