Answer:
-8.04 m/s2
Explanation:
To find the answer to this, you have to use the 4th kinematic equation:
You plug into the equation to get:
solve for a to get
-8.04 m/s2
If the electric field of an electromagnetic wave is in the x-direction and the magnetic field of the wave is in the y-direction,
1 answer:
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Answer:
tex]\lambda_{Be}[/tex] = 22.78 nm
Explanation:
Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)
= k e² / 2a₀ (1 /n²)
ao = h'² / k m e² h' = h/2πi
For another atom with a single electron in the last layer
a₀ ’= h’² / k m (Ze)²
a₀ ’= a₀ / Z²
Therefore, when replacing in the equation
= - Z² Eo/n²
E₀ = 13,606 eV
The transition occurs when the electron stops from one level to another
-
= Z² E₀ (1 / n² - 1 / m²) = Z² ΔE
Let's relate this expression to the wavelength
c = λ f
E = h f
E = h c /λ
h c / λ = Z² ΔE
λ = 1 / Z² (hc / ΔE)
λ = 1 / Z² λ_hydrogen
Let's apply this last equation to our case
Lithium Z = 3
= - 9 Eo / n²
40.5 10-9 = 1/9 λ_hydrogen
Beryllium Z = 4
λ = 1/16 λ_hydrogen
Let's write our two equations is and solve
40.5 10-9 = 1/9 λ_hydrogen
tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen
40.5 10⁻⁹ = 1/9 (16 )
tex]\lambda_{Be}[/tex] = 40.5 9/16
tex]\lambda_{Be}[/tex] = 22.78 nm
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) =
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx