Answer:
See the explanation below.
Explanation:
Attached is a picture of a chocolate bar that can be easily found on the market. In this picture, we see the table of contents with the amount of energy that depends on the mass in grams of the chocolate bar.
We see that for the bar size of 100G, the energy value is 483 [kcal].
We can see that the energy depends on the size of the chocolate bar.
Atomic structure applies to just that, atoms in isolation. Witness looking at atomic spectra using laboratory discharge tubes to ionise low pressure gases of atoms of the elements you want to examine for energy level structure (sodium and the yellow D line "doublet").Crystal structure applies to vast collections of atoms/molecules/ions in complicated geometrical arrays which form lattices. Simple cubic, body centred cubic, face centred cubic being a few "simple" examples.
Answer:
A. Kinetic energy is converted to electric potential energy, and the proton moves more slowly.
Explanation:
When a moving proton is brought close to a stationary one, the kinetic energy of the moving one is converted to electric potential and the proton moves more slowly.
Kinetic energy is the energy due to the motion of a body. A moving proton will possess this form of energy.
Two protons according to coulombs law will repel each other with an electrostatic force because they both have similar charges. This will increase their electric potential energy of both of them.
Potential energy is the energy at rest of a body. As it increases, the motion of a body will be slower and it will tend towards being stationary.
Answer:
2.33 nC, 4.67 nC
Explanation:
when the two spheres are connected through the wire, the total charge (Q=7.00 nC) re-distribute to the two sphere in such a way that the two spheres are at same potential:
(1)
Keeping in mind the relationship between charge, voltage and capacitance:

we can re-write (1) as
(2)
where:
Q1, Q2 are the charges on the two spheres
C1, C2 are the capacitances of the two spheres
The capacitance of a sphere is given by

where R is the radius of the sphere. Substituting this into (2), we find
(3)
we also know that sphere 2 has twice the diameter of sphere 1, so the radius of sphere 2 is twice the radius of sphere 1:

So the eq.(3) becomes

And re-arranging it we find:

And since we know that the total charge is

we find
