Answer:
Range will become 4 times of initial range
Explanation:
Let the velocity of projection is u
And angle at which projectile is projected is
And acceleration due to gravity is
So range of projectile is equal to ........eqn 1
Now in second case it is given that velocity of launching is doubled
So new velocity
So new range will be equal to .....eqn 2
Now dividing eqn 2 by eqn 1
So if we double the initial launch speed then range will become 4 times
Answer:
The center of mass move with the velocity of -3.487 m/s.
Explanation:
Given values of block A.
Mass of block A, (M1) = 4 kg
Speed of block A, (V1) = 2 m/s
Given values of block B.
Mass of block B, (M2) = 8.4 kg
Speed of block B, (V2) = -6.1 m/s
Below is the formula to find the velocity of center of mass.
Answer:
the Zeeman effect without spin is three lines
Explanation:
The Zeeman effect is the result of the interaction of the magnetic field with the orbital angular momentum of the electrons, if we do not take the spin into account it is called the Normal Zeeman effect.
Therefore we only take into account the orbital moments (m_l) of the transition, from the selection rules of the refreshed harmonics, only the transition with
= 0, ± 1
ΔE
Let's analyze for the case of the Hydrogen atom
For a transition between levels n = 1 and n = 2 the values of m_l are n_f = 1 m_l = 0 and for n₀ = 2 m_l = 0, 1
so we only have two lines.
For transition n_f = 2 and n₀ = 3
n_f = 2 m_l = 0, 1
n₀ = 3 m_l = -1, 0, 1
There are only two lines plus the central line, so there are three spectral lines
for n_f = 3 and n_o = 4
n_f = 3 ml = -1, 0, 1
n₀= 4 ml = -2, -1, 0, 1, 2
Only transitions with Δm_l = ±1 are allowed, so there are only two transitions plus the central transition (Δm_l = 0), so there are only 3 spectral lines.
In summary, due to the selection rule of spherical harmonics, the greatest number of lines in the Zeeman effect without spin is three lines.