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RUDIKE [14]
3 years ago
7

Demuestre que la función de onda: y(x, t) = A e^−i(kx−ωt+φ) satisface la ecuación de onda lineal. (donde i =√−1).

Physics
1 answer:
Alekssandra [29.7K]3 years ago
5 0

Answer:

La ecuación de onda lineal para una función f(x, t) se escribe como:

\frac{d^2f}{dt^2}  = c^2*\frac{d^2f}{dx^2}

Donde c es una constante que depende de la velocidad de propagación de la onda.

En este caso, tenemos:

f(x) = A*e^{-i*(k*x - w*t + \phi)}

Recordar que la derivada de la exponente es tal que:

k(x,y) = A*e^{c*x + n*y}\\dk/dx = c*A*e^{c*x + n*y}

Entonces las derivadas de f van a ser:

\frac{df}{dt} =  (i*w)*f(x)

\frac{d^2f}{dt^2} = (i*w)^2*f(x) = -(w)^2*f(x)

\frac{df}{dx} = (-i*k)*f(x)

\frac{d^2f}{dx^2} = (-i*k)^2*f(x) = -k^2*f(x)

Entonces podemos reescribir la ecuación de onda como:

\frac{d^2f}{dt^2}  = c^2*\frac{d^2f}{dx^2}

-(w)^2*f(x) = c^2*(-k^2*f(x))

Ahora podemos simplemente definir c^2 = (w/k)^2 y vemos que la ecuación de onda se cumple.

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Identify and define Four causes of infections diseases
Anastasy [175]

Answer:

Bacteria

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5 0
3 years ago
A point charge of 6.0 nC is placed at the center of a hollow spherical conductor (inner radius = 1.0 cm, outer radius = 2.0 cm)
egoroff_w [7]

Explanation:

The given data is as follows.

             q = 6.0 nC = 6 \times 10^{-9} C

         inner radius (r) = 1.0 cm = 0.01 m   (as 1 cm = 100 m)

So, there will be same charge on the inner surface as the charge enclosed with an opposite sign.

Formula to calculate the charge density is as follows.

            \sigma = \frac{q_{in}}{A} .......... (1)

Since, area of the sphere is as follows.

               A = 4 \pi r^{2} ........... (2)

Hence, substituting equation (2) in equation (1) as follows.

      \sigma = \frac{q_{in}}{4 \pi r^{2}}

                   = \frac{6 \times 10^{-9} C}{4 \times 3.14 \times (0.01)^{2}}            

                   = 0.477 \times 10^{-5}

or,               = 4.77 \mu C/m^{2}

Thus, we can conclude that the resulting charge density on the inner surface of the conducting sphere is 4.77 \mu C/m^{2}.

5 0
3 years ago
A ring with an 18mm diameter falls off a scientist's finger into the solenoid in the lab. The solenoid is 25 cm long, 5.0 cm in
Troyanec [42]

Answer:

The value is  \epsilon =  3.84 *10^{-5} \  V

Explanation:

From the question we are told that

  The diameter of the ring is  d =  18 \ mm  =  0.018 \  m

   The length of the solenoid is l = 25 \ cm  =  0.25 \ m

   The diameter of the solenoid is  D = 5.0 \ cm  = 0.05 \ m

    The number of turns is  N = 1500

   The change in  current in the solenoid is   \Delta  I   = 20 \ A

   The time taken is  \Delta  t  = 1 \ s

Generally the radius of the ring is  

     r = \frac{d}{2}

=>  r = \frac{0.018 }{2}

=>  r = 0.009 \ m

Generally the area of the ring is mathematically represented as  

      A = \pi r^2

=>   A = 3.142 *  0.009^2    

=>   A = 2.545 *10^{-4}\ m^2

Generally the induced emf is mathematically represented as

       \epsilon  =  A * \frac{dB}{dt}

Here    

         \frac{dB }{dt} =  \mu_o * \frac{N}{l} *\frac{ \Delta I }{\Delta t}

Here  \mu_o is the permeability of free space with value  

         \mu_o =  4\pi *10^{-7} \ N/A^2

So  

     \frac{dB }{dt} =   4\pi * 10^{-7} * \frac{1500}{0.25} *\frac{20 }{1}

=>  \frac{dB }{dt} =   0.150816\  T/s

So

     \epsilon =   0.150816 *  2.545 *10^{-4}

=>   \epsilon =  3.84 *10^{-5} \  V

3 0
3 years ago
Calculate the amount of air in a room 6m long, 5m wide and 3mm high.​
sattari [20]

Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

The amount of air is equal to the volume.

6 0
2 years ago
Read 2 more answers
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