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What is the length of a contention slot in CSMA/CD for (a) a 2-km twin-lead cable (signal propagation speed is 82% of the signal propagation speed in vacuum)

User GProst
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2 Answers

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Final answer:

The propagation speed of the pulse on a string with a tension provided by a spring stretched 2.00 cm and having a spring constant of 100 N/m is approximately 6.73 meters per second, calculated using the tension, the mass per unit length, and the formula v = √(T/μ).

Step-by-step explanation:

Propagation Speed Calculation

To find the propagation speed of a pulse in a string or wire, we can use the following formula which relates the tension (T) in the wire, the mass per unit length (μ), and the propagation speed (v):

v = √(T/μ)

For the case of the string with a mass of 150 g and a length of 3.4 m, and considering that the spring tension is provided by stretching the spring 2.00 cm (0.02 m) with a spring constant (ks) of 100 N/m, the effective tension (T) is calculated using Hooke's law:

T = ks * stretch = 100 N/m * 0.02 m = 2 N

The mass per unit length (μ) for the string is:

μ = mass / length = 150 g / 3.4 m = 0.15 kg / 3.4 m ≈ 0.0441 kg/m

Now, we can calculate the propagation speed (v):

v = √(2 N / 0.0441 kg/m) ≈ √(45.3515 m²/s²) ≈ 6.73 m/s

Thus, the propagation speed of the pulse on the string is approximately 6.73 meters per second.

User RPiAwesomeness
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1 vote

Answer:

1.99*10-4sec

Step-by-step explanation:

Signal propagation speed=0.82∗2.46∗108m/s

d=2000 m

Tp=20000/0.82∗2.46∗108 sec

ContentionPeriod=2Tp=2∗20000/0.82∗2.46∗10^8

= 1.99* 10^-4seconds

User Aleien
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