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Lelu [443]
3 years ago
13

- How long does it take a packet of length 1,000 bytes to propagate over a link of distance 2,500 km, propagation speed 2.5 *10^

8 m/s, and transmission rate 2 Mbps? More generally, how long does it take a packet of length L to propagate over a link of distance d, propagation speed s, and transmission rate R bps? Does this delay depend on packet length? Does this delay depend on transmission rate?
Physics
1 answer:
Sunny_sXe [5.5K]3 years ago
3 0

Answer:

It will take 0.01 s or 10 ms

Solution:

As per the question:

Length of the packet, L = 1,000 bytes = 1000\times 8 = 8000 bits

Distance, d = 2500 km = 2.5\times 10^{6}\ m

Speed of propagation, s = 2.5\times 10^{8}\ m/s

Transmission rate, R = 2 Mbps

Now,

Propagation time, t can be calculated as:

t = \frac{d}{s} = \frac{2.5\times 10^{6}}{2.5\times 10^{8}} = 0.01\ s

t = 10 ms

  • In general, propagation time, t is given by:

       t = \frac{link\ distance}{Propagation\ speed}

  • No, this delay is independent of the length of the packet.
  • No, this delay is independent of the rate of transmission.

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Answer:

La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h

Explanation:

La velocidad es una magnitud que expresa el desplazamiento que realiza un objeto en una unidad determinada de tiempo, esto es, relaciona el cambio de posición (o desplazamiento) con el tiempo.

Siendo la velocidad es el espacio recorrido en un período de tiempo determinado, entonces 90 km/h indica que en 1 hora la motocicleta recorre 90 km. Entonces, siendo 1 h= 3600 segundos (1 h=60 minutos y 1 minuto=60 segundos) podes aplicar la siguiente regla de tres: si en 3600 segundos (1 hora) la motocicleta recorre 90 km, entonces en 1 segundo ¿cuánta distancia recorrerá?

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Por otro lado, aplicas la siguiente regla de tres: si 1 km es igual a 1,000 metros, ¿0.025 km cuántos metros son?

distancia=\frac{0.025 km*1,000 metros}{1 km}

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<u><em>La motocicleta recorre 25 metros en 1 segundo si circula a una velocidad de 90 km/h</em></u>

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