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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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serious [3.7K]
The answer is true a convex lens can produce a real image but not a viral image
5 0
4 years ago
Calculate the power provided by a 12 volt car battery with a rating of 350 amperes.
mamaluj [8]

Answer:

Option D

4200 W

Explanation:

Power, P is also given as the product of voltage and current, expressed as P=VI

Here, P is power, V is voltage in the xircuit and I is current theough voltage.

Taking 12 V for voltage across and 350A for current across circuit then power will be

P=350*12=4200 W

Therefore, option D is correct.

8 0
3 years ago
What is the amount of thermal energy needed to make 5 kg of ice at - 10 °C to
agasfer [191]

Answer:

The amount of thermal energy needed is 15167500 joules.

Explanation:

By First Law of Thermodynamics, we see that amount of thermal energy (Q), in joules, is equal to the change in internal energy. From statement we understand that change in internal energy consisting in two latent components (U_{l,ice}, U_{l,steam}), in joules, and two sensible component (U_{s,w}), in joules, that is:

Q = U_{l,ice} + U_{s, w} + U_{s,ice} + U_{l,steam} (1)

By definitions of Sensible and Latent Heat, we expanded the formula:

Q = m\cdot (h_{f,w}+h_{v,w}+c_{ice}\cdot \Delta T_{ice}+c_{w}\cdot \Delta T_{w}) (2)

Where:

m - Mass, in kilograms.

h_{f,w} - Latent heat of fussion of water, in joules per kilogram.

h_{v,w} - Latent heat of vaporization of water, in joules per kilogram.

c_{ice} - Specific heat of ice, in joules per kilogram per degree Celsius.

c_{w} - Specific heat of water, in joules per kilogram per degree Celsius.

\Delta T_{ice} - Change in temperature of ice, measured in degrees Celsius.

\Delta T_{w} - Change in temperature of water, measured in degrees Celsius.

If we know that m = 5\,kg, h_{f,w} = 3.34\times 10^{5}\,\frac{J}{kg}, h_{v,w} = 2.26\times 10^{6}\,\frac{J}{kg}, c_{ice} = 2.090\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, c_{w} = 4.186\times 10^{3}\,\frac{J}{kg\cdot ^{\circ}C}, \Delta T_{ice} = 10\,^{\circ}C and \Delta T_{w} = 100\,^{\circ}C, then the amount of thermal energy is:

Q = 15167500\,J

The amount of thermal energy needed is 15167500 joules.

7 0
3 years ago
A research submarine can withstand an external pressure of 62 megapascals (million pascals) all the while maintaining a comforta
Alex

Answer:

<em>The depth will be equal to</em> <em>6141.96 m</em>

<em></em>

Explanation:

pressure on the submarine P_{sea} = 62 MPa = 62 x 10^6 Pa

we also know that P_{sea} = ρgh

where

ρ is the density of sea water = 1029 kg/m^3

g is acceleration due to gravity = 9.81 m/s^2

h is the depth below the water that this pressure acts

substituting values, we have

P_{sea} = 1029 x 9.81 x h = 10094.49h

The gauge pressure within the submarine P_{g} = 101 kPa =  101000 Pa

this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.

Equating the pressure P_{sea}, we have

62 x 10^6 = 10094.49h

depth h = <em>6141.96 m</em>

7 0
3 years ago
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Nikitich [7]
Uhhhhhhhhh just tryna get a point so I can ask a question so eh I’m using ur question heheheheheh
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