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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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Answer:

Average speed = 10,000 m/s

Explanation:

Given the following data;

Distance = 2m

Time = 0.0002secs

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Average speed = distance/time

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In an RC circuit, what fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for
4vir4ik [10]

Answer:

The  fraction fraction of the final energy is stored in an initially uncharged capacitor after it has been charging for 3.0 time constants is  

      k  = 0.903

Explanation:

From the question we are told that

     The time  constant  \tau  =  3

The potential across the capacitor can be mathematically represented as

     V  =  V_o  (1 -  e^{- \tau})

Where V_o is the voltage of the capacitor when it is fully charged

    So   at  \tau  =  3

     V  =  V_o  (1 -  e^{- 3})

     V  =  0.950213 V_o

   Generally energy stored in a capacitor is mathematically represented as

             E = \frac{1}{2 } * C  * V ^2

In this equation the energy stored is directly proportional to the the square of the potential across the capacitor

Now  since capacitance is  constant  at  \tau  =  3

        The  energy stored can be evaluated at as

         V^2 =  (0.950213 V_o )^2

       V^2 =  0.903  V_o ^2

Hence the fraction of the energy stored in an initially uncharged capacitor is  

      k  = 0.903

4 0
3 years ago
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