Because dark line spectra result from passing white light through ionized gasses and plasmas, which is what the atmosphere of stars are made of. These frequencies are scattered by the star's atmosphere as it leaves the surface (photosphere) of the star, and don't make it to earth.
As we know that the formula of kinetic energy will be

now here we know that
m = 2 kg
v = 1 m/s
so from the above equation we have


The answer is 59.4 degrees.
5m/s
Explanation:
Given parameters:
Mass of ball = 0.1kg
Force on the ball = 5N
time taken = 0.1s
Unknown:
final speed of the ball = ?
Solution:
According to newton's second law "the net force on a body is the product of its mass and acceleration".
Force = mass x acceleration equation 1
Acceleration =
V is the final velocity
U is the initial velocity
T is the time taken
U = O since it is a stationary body;
a = 
Input "a" into equation 1
F = m x 
5 = 0.1 x 
V = 5m/s
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Answer:
- The energy that must be added to the electron to move it to the third excited state is -1.153 eV
- The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
Explanation:
Given;
Energy of electron in ground state (n = 1 ) = 1.23 eV
E₁ = 1.23 eV
Eₙ = E₁ /n²
where;
E₁ is the energy of the electron in ground state
n is the energy level,
For third excited state, n = 4
E₄ = E₁ /4²
E₄ = (1.23 eV) / 16
E₄ = 0.077 eV
Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV
The energy that must be added to the electron to move it to the third excited state is -1.153 eV
For fourth excited state, n = 5
E₅ = E₁ /5²
E₄ = (1.23 eV) / 25
E₄ = 0.049 eV
Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV
The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV