Answer:
E = 2,575 eV
Explanation:
For this exercise we will use the Planck equation and the relationship of the speed of light with the frequency and wavelength
E = h f
c = λ f
Where the Planck constant has a value of 6.63 10⁻³⁴ J s
Let's replace
E = h c / λ
Let's calculate for wavelengths
λ = 4.83 10-7 m (blue)
E = 6.63 10⁻³⁴ 3 10⁸ / 4.83 10⁻⁷
E = 4.12 10-19 J
The transformation from J to eV is 1 eV = 1.6 10⁻¹⁹ J
E = 4.12 10⁻¹⁹ J (1 eV / 1.6 10⁻¹⁹ J)
E = 2,575 eV
Answer:
I've always wanted to be a nurse. growing up my auntie used to show me a lot of stuff about becoming a nurse since she us one.
Answer:
In constructive waves, a <u><em>greater</em></u> amplitude wave is formed. In destructive waves, a wave with a <u><em>smaller</em></u> amplitude is formed. (option A)
Explanation:
Interference is called the superposition or sum of two or more waves. Depending mainly on the wavelengths, amplitudes and the relative distance between them, there are two types of interference: constructive or destructive.
Constructive interference occurs when there are two waves of identical or similar frequency (both have motions equal to an even number of similar wavelengths) and overlap the peak of one with the peak of the other. These effects add together and make a wave of greater amplitude. All of this is possible because the waves were in the same phase in the beginning (in the same position).
Destructive interference occurs in the opposite case to constructive. When the crest of one wave overlaps the valley of the other, they cancel out since they are in different phases when they overlap (they were in different positions). That is, as in the case of constructive waves they were added, in the case of destructive waves they cancel out (subtract).
So, <u><em>In constructive waves, a greater amplitude wave is formed. In destructive waves, a wave with a smaller amplitude is formed. </em></u>
Answer:
The answer is in 3 point
Choose which point you like
Answer:
1.17 m
Explanation:
From the question,
s₁ = vt₁/2................ Equation 1
Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.
Given: v = 343 m/s, t = 0.0115 s
Substitute into equation 1
s₁ = (343×0.0115)/2
s₁ = 1.97 m.
Similarly,
s₂ = vt₂/2.................. Equation 2
Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo
Given: v = 343 m/s, t₂ = 0.0183 s
Substitute into equation 2
s₂ = (343×0.0183)/2
s₂ = 3.14 m
The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁
s₂-s₁ = (3.14-1.97) m = 1.17 m
