Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
The only correct statement on that list is <em>choice-C</em>: If a positively charged rod is brought close to a positively charged object, the two objects will repel.
"The path difference between the two waves should be one-quarter of a wavelength" is the statement among the choices given in the question that describes the <span>path difference between the two waves. The correct option among all the options that are given in the question is the fifth statement or the penultimate statement.</span>
Beta emission is occurring in the given nuclear reaction.
Answer: Option B
<u>Explanation:</u>
In this equation, the reactant is the Thorium atom, which is reduced to palladium. As the atomic number get decreased by one, so an electron will be emitted. This process of emission of electrons by radiation or decaying the reactant nuclei to form a new product nuclei is termed as beta emission.
So, the electrons are generally termed as beta particles while the positrons are termed as positive beta particles. So this is a kind of radioactive reactions where the reactant changes to new element by releasing an electron and thus there is a change in the atomic number of the product by one.
Answer:
The direction in which a positive charge would move.
Explanation:
The direction of an electric current is by convention the direction in which a positive charge would move. Thus, the current in the external circuit is directed away from the positive terminal and toward the negative terminal of the battery. Electrons would actually move through the wires in the opposite direction.
