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myrzilka [38]
3 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]3 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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Which equation below represents a generic equation suggested by a graph showing a hyperbola?
ruslelena [56]

Answer:

y = k/x

Explanation:

y = k/x is a graph of a hyperbola that has been rotated about the origin.

8 0
3 years ago
Assume you are in the car and the car is moving at a certain speed to
Setler79 [48]
You are at rest with respect to the car.
You are in motion with respect to the School.

6 0
3 years ago
It is 5.00 km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0 k
8_murik_8 [283]

Answer:

a. Walking burns up more energy.

b. 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

Explanation:

a. We know energy W = Pt where P = power and t = time.

Now for walking, t = d/v where d = distance = 5.00 km and v = speed = 3.00 km/hr and P = 290 W

So, t = d/v = 5.00 km/3.00 km/hr = 5/3 hr = 5/3 × 3600 s = 6000 s

W = Pt = 290 W × 6000 s = 1740000 = 1740 kJ

Now for running, t = d/v where d = distance = 5.00 km and v = speed = 10.00 km/hr

So, t = d/v = 5.00 km/10.00 km/hr = 0.5 hr = 0.5 × 3600 s = 1800 s and P = 700 W

W = Pt = 700 W × 1800 s = 1260000 = 1260 kJ

Since walking burns up 1740 kJ and running burns up 1260 kJ, walking burns up more energy.

b. It burns up 1740 kJ

c. This is because more intense exercise releases a lot of energy in a short period of time, whereas, less intense energy releases it energy gradually over a long period of time.

4 0
3 years ago
Select the correct answer.
Stels [109]

Answer:

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of the jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?

A.

We’ll see the bell move, but we won’t hear it ring.

B.

We won’t see the bell move, but we’ll hear it ring.

C.

We’ll see the bell move and hear it ring.

D.

We won’t see the bell move or hear it ring.

E.

We’ll see the sound waves exit the vacuum pump.

Explanation:

so, the answer to the question is

A.

We'll see the bell move, but we won’t hear it ring.

5 0
3 years ago
Read 2 more answers
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.
34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

Q = \dot Q \cdot \Delta t (1)

If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

Q' \approx 4340.589\,\frac{W}{m^{2}}

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0
3 years ago
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