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In a high-quality coaxial cable, the power drops by a factor of 10 approximately every 2.75 km. If the original signal power is 0.45 W (= 4.5 x 10-1), how far will a signal be transmitted before the power is attenuated to 45 μW?

Include a table showing the signal power (in W) vs distance in 2.75 km intervals (like Table 1.2 in Section 1.4.4) as part of your answer.

If optical fibre is used instead of the coaxial cable, briefly explain how you would expect the above calculated distance value to change (see Section 1.4.4). You are not required to include another table for this last part.

Respuesta :

The distance up to the signal being transmitted before the power is attenuated to 45 μW will be 8.25 km.

What is electric power?

Electric power is the product of the voltage and current. Its unit is the watt. It is the rate of the electric work done.

Given condition;

Power drops by a factor of 10 approximately every 2.75 km

0.45 W -> 0 km

0.045 W  -> 2.75 km

0.0045 W  -> 5.5 km

0.00045 W -> 8.25 km

0.00045W = 0.45 mW = 45 uW  -> 8.25 km

Hence, the distance up to the signal being transmitted before the power is attenuated to 45 μW will be 8.25 km.

To learn more about the electric power refer to the link;

https://brainly.com/question/12316834

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Answer:

The distance of the transmitted signal is 11 km before it is attenuated to 45uW

Explanation:

First work out 45uW into scientific notation

1/10e+6=0.000001 =1uW

So 45/10e+6=0.000045 =45uW

Scientific notation = 4.5 x 10^-5

The signal power of the cable is 0.45 (4.5 x 10^-1 at 0 distance

Table

0 = 10^-1

2.75 = 10^-2

5.5 = 10^-3

8.25 = 10^-4

11 = 10^-5

Answer 11 km before it is attenuated to 45uW

Optical fibre is very great compared to high quality coaxial cable. Optical fibre drops by a factor of 10 approximately every 30 km so it will transmit 120 km before it is attenuated to 45uW

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