A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.
<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>
Answer:
reduce the velocity of collision
This question is incomplete, but I can do it for you, considering the equation to be *In its most famous form*:
A+B⇒C+D
A and B here are the reactants, while C and D are the products.
The reactants are generally the input materials in the beginning of any chemical reactions and they usually, if not always, are on the left hand side of the chemical equation. While the products are on the right hand side and are the final output of the chemical reaction.
Hope this helps.
