Gravity is pulling on you downwards with a force which we call your weight. The reason why you aren't accelerating downwards is
1 answer:
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Answer:
λ =365.4 nm
Explanation:
Boh's atomic model of the Hydrogen atom the energy of each level is
Eₙ = - 13.606 / n²
where the synergy is in electonvotes and the value of E₀ = 13.606 eV is the energy of the base state of hydrogen.
An atomic transition occurs when an electron goes from an excited state and joins everything of lower energy.
ED = 13.606 (1 / n₀² - 1 /)
we are going to apply this relationship to answer slash.
At the beginning of the studies of atomic transitions, each group did not consider having a different name
name Initial state
Lymman 1
Balmer 2
the final state is any other state sta the continuum that corresponds to n = inf
Let's look for the highest energy of the Balmer series
ΔE = 13.606 (1/2² - 1 /∞)
ΔE = 3.4015 eV
Let's use the Planck relation for the energy
E = h f = h c /λ
λ = h c / E
Let's reduce the energy to J
E = 3.4015 eV (1.6 10⁻¹⁹ J / 1 eV) = 5.4424 10⁻¹⁹
λ = 6.63 10⁻³⁴ 3 10⁸ / 5.4424 10⁻¹⁹
λ = 3.654 10⁻⁷ m
λ = 3,654 10⁻⁷ m (10⁹ nm / 1m)
λ =365.4 nm
this eta radiation in the ultraviolet range
<u>If the disk turns with constant angular velocity, the following statements about it are true </u>
- The linear acceleration of Q is twice as great as the linear acceleration of P
- is moving twice Q as fast as P.
Answer: Options D and E
<u>Explanation: </u>
Let us consider that R is the radius of the circular disc. So as Q is on the rim, so the distance of Q from the centre of the disc is R and as P is the midpoint between centre and rim of the disk, so the distance of P from the centre is R/2.
As we know that the angular velocity of the circular disk will be equal to the ratio of distance covered by that point to the time taken. So the angular velocity at point Q will be
As R is the distance of point Q from the centre of the disc.
Similarly ,
So if we equate v with v’ we obtain that
Therefore, the point Q will be moving twice as fast as P. As the velocity of Q is more than O, the linear acceleration of point Q will also be twice as great as the linear acceleration of P.
This is because acceleration is directly proportional to the rate of change in velocity. So if velocity increases in the factor of 2, the acceleration of point Q will also increase twice with respect to point P.