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myrzilka [38]
2 years ago
7

During the data transmission there are chances that the data bits in the frame might get corrupted. This will require the sender

to re-transmit the frame and hence it will increase the re-transmission overhead. By considering the scenarios given below, you have to choose whether the packets should be encapsulated in a single frame or multiple frames in order to minimize the re-transmission overhead.
Justify your answer with one valid reason for both the scenarios given below.


Scenario A: Suppose you are using a network which is very prone to errors.


Scenario B: Suppose you are using a network with high reliability and accuracy.
Physics
1 answer:
aalyn [17]2 years ago
6 0

1. Based on Scenario A, multiple frames will minimize re-transmission overhead and should be preferred in the encapsulation of packets.

2. Based on Scenario B, the encapsulation of packets should be in a single frame because of the high level of network reliability and accuracy.

 

<u>Justification:</u>

There will not be further need to re-transmit the packets in a highly reliable and accurate network environment, unlike in an environment that is very prone to errors.  The reliable and accurate network environment makes a single frame economically better.

Encapsulation involves the process of wrapping code and data together within a class so that data is protected and access to code is restricted.

With encapsulation, each layer:

  • provides a service to the layer above it
  • communicates with a corresponding receiving node

Thus, in a reliable and accurate network environment, single frames should be used to enhance transmission and minimize re-transmission overhead.  This is unlike in an environment that is very prone to errors, where multiple frames should rather be used to minimize re-transmission overhead.

Learn more about encapsulation of packets here: brainly.com/question/22471914

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Water drips from the nozzle of a shower onto the floor 190 cm below. The drops fall at regular (equal) intervals of time, the fi
VARVARA [1.3K]

Answer:

Second drop: 1.04 m

First drop: 1.66 m

Explanation:

Assuming the droplets are not affected by aerodynamic drag.

They are in free fall, affected only by gravity.

I set a frame of reference with the origin at the nozzle and the positive X axis pointing down.

We can use the equation for position under constant acceleration.

X(t) = x0 + v0 * t + 1/2 * a *t^2

x0 = 0

a = 9.81 m/s^2

v0 = 0

Then:

X(t) = 4.9 * t^2

The drop will hit the floor when X(t) = 1.9

1.9 = 4.9 * t^2

t^2 = 1.9 / 4.9

t = \sqrt{0.388} = 0.62 s

That is the moment when the 4th drop begins falling.

Assuming they fall at constant interval,

Δt = 0.62 / 3 = 0.2 s (approximately)

The second drop will be at:

X2(0.62) = 4.9 * (0.62 - 1*0.2)^2 = 0.86 m

And the third at:

X3(0.62) = 4.9 * (0.62 - 2*0.2)^2 = 0.24 m

The positions are:

1.9 - 0.86 = 1.04 m

1.9 - 0.24 = 1.66 m

above the floor

8 0
2 years ago
Help please!
JulsSmile [24]

3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:

(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v

where v is the velocity of the combined players. Solve for v :

450 kg•m/s - 320 kg•m/s = (155 kg) v

v = (130 kg•m/s) / (155 kg)

v ≈ 0.84 m/s

4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that

(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v

where v is the new velocity of the 4-kg ball. Solve for v :

30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v

v = (16.4 kg•m/s) / (4 kg)

v = 4.1 m/s

7 0
2 years ago
Rick is moving a wheelbarrow full of bricks out to the curb. The bricks in the wheelbarrow weigh more than Rick is able to carry
USPshnik [31]

Answer is given below

Explanation:

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Sladkaya [172]

Answer:

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Mademuasel [1]
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</span>
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