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Fofino [41]
2 years ago
6

Add the following two vectors via the component method:

Physics
1 answer:
NeX [460]2 years ago
8 0

A quantity which have both magnitude and direction is know as vector quantity.

Resultant vector is 6.81 m south east.

Triangle law of vector addition is when two vectors are represented as two sides of the triangle with the order of magnitude and direction, then the third side of the triangle represents the magnitude and direction of the resultant vector.

Since, vector K is 3.0 m north and vector L is 8.6 m southeast

So, angle between vectors K and L is,

                                                                 90+45 = 135 degree

Resultant vector,  R=\sqrt{K^{2}+L^{2}=2*K*L*cosx  \\\\

                             R=\sqrt{9+73.96-36.49} \\\\R=\sqrt{46.47} \\\\R=6.81 m

So, Resultant vector R is 6.81 m with direction of southeast.

Learn more:

brainly.com/question/19791435

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Answer:

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Explanation:

4 0
3 years ago
What is the speed of a wave if the wavelength is 3 meters and the frequency is 200hz
RideAnS [48]
Wave speed = frequency * wavelength

Input the numbers into this equation :

Wave speed = 200 * 3

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6 0
3 years ago
Calculate the speed of a gamma ray with a frequency of 3.0 x 10^19 Hz and a wavelength of 1.0 x 10^-11 m.
Ulleksa [173]

Answer:

Speed of gamma rays = 3 x 10⁸ m/s

Explanation:

Given:

Frequency of gamma ray = 3 x 10¹⁹ Hz

Wavelength of gamma rays = 1 x 10⁻¹¹ meter

Find:

Speed of gamma rays

Computation:

Velocity = Frequency x wavelength

Speed of gamma rays = Frequency of gamma ray x Wavelength of gamma rays

Speed of gamma rays = [3 x 10¹⁹][1 x 10⁻¹¹]

Speed of gamma rays = 3 x [10¹⁹⁻¹¹]

Speed of gamma rays = 3 x [10⁸]

Speed of gamma rays = 3 x 10⁸ m/s

6 0
3 years ago
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

7 0
3 years ago
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