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notsponge [240]
3 years ago
8

2.- Si una cámara fotográfica emite un pulso de sonido para enfocar un objeto, determinar

Physics
1 answer:
uranmaximum [27]3 years ago
8 0

Answer:

a. El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

Explanation:

El sonido es un tipo de onda mecánica, que es un tipo de onda que necesita de un medio material para propagarse. En este caso, entendemos que el sonido se propaga a través del aire atmosférico hasta llegar a su destino y devolverse a rapidez constante. Entonces, podemos estimar el tiempo (t), medido en segundos, a partir de la siguiente fórmula:

t = \frac{2\cdot x_{s}}{v_{s}}

Donde:

x_{s} - Distancia entre la cámara fotográfica y el objeto, medida en metros.

v_{s} - Rapidez del sonido en el aire atmosférico, medida en metros por segundo.

A continuación, calculamos el tiempo de recorrido:

a. (x_{s} = 1\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (1\,m)}{340\,\frac{m}{s} }

t = 5.882\times 10^{-3}\,s

El tiempo de recorrido es 5.882\times 10^{-3} segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

b. (x_{s} = 20\,m, v_{s} = 340\,\frac{m}{s})

t = \frac{2\cdot (20\,m)}{340\,\frac{m}{s} }

t = 0.118\,s

El tiempo de recorrido es 0.118 segundos para un objeto localizado a un metro de distancia de la cámara fotográfica.

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Probability of tunneling is 10^{- 1.17\times 10^{32}}

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Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

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The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

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\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

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We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

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P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

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