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3 years ago

14

Which option identifies the most likely scaling factor in the above drawing of the Mercury capsule and booster unit? 1:1,000 1:1

0,000 1,000:1 10,000:1

1 answer:

Elina [12.6K]3 years ago

3 0

Answer:

I would say 10000;1 but thats just me

Explanation:

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The Ethernet (CSMA/CD) alternates between contention intervals and successful transmissions. Assume a 100 Mbps Ethernet over 1 k

Vesnalui [34]

<h3><u>CSMA/CD Protocol: </u></h3>

Carrier sensing can transmit the data at anytime only the condition is before sending the data sense carrier if the carrier is free then send the data.

But the problem is the standing at one end of channel, we can’t send the entire carrier. Because of this 2 stations can transmit the data (use the channel) at the same time resulting in collisions.

There are no acknowledgement to detect collisions, It's stations responsibility to detect whether its data is falling into collisions or not.

<u>Example: </u>

$T_{P}=1 H r$ , at time t = 10.00 AM, A starts, 10:59:59 AM B starts at time 11:00 AM collision starts.

12:00 AM A will see collisions

Pocket Size to detect the collision.

$\begin{aligned}&T_{t} \geq 2 T_{P}\\&\frac{L}{B} \geq 2 T_{P}\\&L \geq 2 \times T_{P} \times B\end{aligned}$

CSMA/CD is widely used in Ethernet.

<u>Efficiency of CSMA/CD:</u>

In the previous example we have seen that in worst case $2 T_{P}$ time require to detect a collision.

There could be many collisions may happen before a successful completion of transmission of a packet.

We are given number of collisions (contentions slots)=4.

$\text { Propagation day }=\frac{\text {distance}}{\text {speed}}$

Distance = 1km = 1000m

$\begin{aligned}&\text { Speed }=2 \times 10^{8} \mathrm{m} / \mathrm{sec}\\ &T_{P}=\frac{1000}{2 \times 10^{8}}=(0.5) \times 10^{-5}=5 \times 10^{-6}\\ &T_{t}=5 \mu \mathrm{sec}\end{aligned}$

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3 years ago

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VashaNatasha [74]

Answer:

Explanation:

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3 years ago

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$2.

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$5 - $3 = $2

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2 years ago

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Answer:

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Explanation:

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