A.the composition of the inner and outer planets, current observations of star formation, and the motion of the solar system I hope this helps
The electric force between two charge objects is calculated through the Coulomb's law.
F = kq₁q₂/d²
The value of k is 9.0 x 10^9 Nm²/C² and the charge of proton is 1.602 x10^-19 C. Substituting the known values from the given,
2.30x10^-26 = (9.0 x 10^9 Nm²/C²)(1.602 x10^-19C)²/d²
The value of d is equal to 0.10 m.
As we know that KE and PE is same at a given position
so we will have as a function of position given as

also the PE is given as function of position as

now it is given that
KE = PE
now we will have




so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula

now to find the fraction of kinetic energy



now since total energy is sum of KE and PE
so fraction of PE at the same position will be


Answer:
B: False
Explanation:
The second law of thermodynamics states that: the entropy of an isolated system will never decrease because isolated systems always tend to evolve towards thermodynamic equilibrium which is a state with maximum entropy.
Thus, it means that the entropy change will always be positive.
Therefore, the given statement in the question is false.
As per the question the wavelength of spectral line of the neutral hydrogen atom is 21 cm.
Spectral line of hydrogen atom will be observed when an electron will jump from higher excited to the ground state. If E is the energy of the ground state and E' is the energy of the higher excited state,the energy emitted when electron will jump from higher excited to ground state is E'-E.
As per Planck's quantum theory -

Where f is the frequency of emitted radiation and h is the Planck's constant.Actually this energy emitted is also electromagnetic in nature. We know that all the electromagnetic radiation will move with the same velocity which is equal to the velocity of light.
The velocity of light c= 3×10^8 m/s.Hence the velocity of wave corresponding to this spectral line is 3×10^8 m/s.
We know that velocity of a wave is the product of frequency and wavelength of that corresponding wave. Mathematically we can write it as-

[
is the wavelength of the wave ]
It is given that wavelength =21 cm=0.21 m
The frequency is calculated as -




Hence the antenna will be tuned to a frequency of 14.2857×10^8 Hz
Here Hertz[ Hz] is the unit of frequency.