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stira [4]
3 years ago
14

Which of these would best define sound intensity? A) how fast the wave is traveling B) the frequency of the sound wave the power

of the sound wave per unit area the mass of the wave divided by its volume
Physics
2 answers:
faust18 [17]3 years ago
5 0

Which of these would best define sound intensity?

A) how fast the wave is travelling

<u>B) the frequency of the sound wave the power of the sound wave per unit area the mass of the wave divided by its volume</u>

STatiana [176]3 years ago
4 0

The correct answer is B. Hope this helps! :)

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A 20 KeV electron emits two bremsstrahlung photons as it is being brought to rest in two successive decelerations. The wavelengt
Degger [83]

Answer:

λ₁ = 87.5 10⁻¹² m ,  λ₂ =  2.175 10⁻¹⁰ m,    E₂ = 5.8 10³ eV

Explanation:

In this case you can use the law of conservation of energy, all the energy of the electron is converted into energized emitted photons

Let's reduce to the SI system

          E₀ = 20 10³ eV (1.6 10⁻¹⁹ J / 1eV) = 3.2 10⁻¹⁵ J

          Δλ = 1.30 A = 0.13 nm = 0.13 10⁻⁹ m

          Ef = E₁ + E₂

         E₀ = Ef

         E₀ = E₁ + E₂

The energy can be found with the Planck equation

          E = h f

          c = λ f

          f = c / λ

          E = hc / λ

They indicate that the wavelength of the second photon is

 

           λ₂ =  λ₁ +0.130 10⁻⁹

We replace

           E₀ = hv / λ₁ + hc / ( λ₁ + 0.130 10⁺⁹)

           E₀ / hv = 1 / λ₁ + 1 / ( λ₁ + 0.13 10⁻⁹)

          3.2 10⁻¹⁵ / (6.63 10⁻³⁴ 3 10⁸) = ( λ₁ + 0.13 10⁻⁹ +  λ₁) /  λ₁ ( λ₁ + 0.13 10⁻⁹)

          1.6 10¹⁰ ( λ₁² +0.13 10⁻⁹  λ₁) = 2  λ₁ + 0.13 10⁻⁹

           λ₁² + 0.13 10⁻⁹  λ₁ = 1.25 10⁻¹⁰  λ₁ + 8.125 10⁻²¹

            λ₁² + 0.005 10⁻⁹  λ₁ = 8.125 10⁻²¹

            λ₁² + 5 10⁻¹²  λ₁ - 8.125 10⁻²¹ = 0

Let's solve the second degree equation

            λ₁ = [-5 10⁻¹² ±√((5 10⁻¹²)² + 4 8.125 10⁻²¹)] / 2

    λ₁ = [-5 10⁻¹² ±√(25 10⁻²⁴ +32.5 10⁻²¹)] / 2 = [-5 10⁻¹² ±√ (32525 10⁻²⁴)] / 2

             λ₁ = [-5 10⁻¹² ± 180 10⁻¹²] / 2

            λ₁ = 87.5 10⁻¹² m

             λ₂ = -92.5 10⁻¹² m

We take the positive wavelength

The wavelength of the photons is

            λ₁ = 87.5 10⁻¹² m

            λ₂ =  λ₁ + 0.13 10⁻⁹

             λ₂ = 87.5 10⁻¹² + 0.13 10⁻⁹

             λ₂ = 0.2175 10⁻⁹ m = 2.175 10⁻¹⁰ m

The energy after the first deceleration is

            E₂ = E₀ –E₁

            E₂ = E₀ –hc / λ₁

            E₂ = 3.2 10⁻¹⁵ - 6.63 10⁺³⁴ 3 10⁸ / 87.5 10⁻¹²

            E₂ = 3.2 10⁻¹⁵ - 2.27 10⁻¹⁵

             E₂ = 0.93 10⁻¹⁵ J

             E₂ = 0.93 10⁻¹⁵ J (1 eV / 1.6 10⁻¹⁹ J)

             E₂ = 5.8 10³ eV

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