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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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An electron moves with a speed of 8.0 × 10^{6} m/s along the +x-axis. It enters a region where there is a magnetic field of 2.5
yanalaym [24]

Answer:

The magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

Explanation:

Given;

speed of the electron, v =  8.0 × 10⁶ m/s

magnetic field strength, B = 2.5 T

angle of inclination of the field, θ = 60°

The magnetic force experienced by the electron in the magnetic field is given as;

F = qvBsinθ

where;

q is charge of electron = 1.6 x 10⁻¹⁹ C

B is strength of magnetic field

v is speed of the electron

Substitute the given values and solve for F

F = (1.6 x 10⁻¹⁹)( 8.0 × 10⁶)(2.5)sin60

F = 2.77 x 10⁻¹² N

Thus, the magnitude of the magnetic force of the electron is 2.77 x 10⁻¹² N

8 0
3 years ago
The volume V of a right circular cylinder of radius r and height h is V=πr2h. (a) How is dVdt related to drdt if h is constant a
Nikolay [14]

Answer:

(a)\frac{dV}{dt}= 2\pi r h  \frac{dr}{dt}

(b)\frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

(c) \frac{dV}{dt} = \pi r^2\frac{dh}{dt}+2\pi r h\frac{dr}{dt}

Explanation:

Differentiating Rules:

  1. \frac{dx^n}{dx}= nx^{n-1}
  2. \frac{dx}{dx}=1
  3. \frac{d}{dx}(mn)= m\frac{dn}{dx}+n\frac{dm}{dx}  [ m and n are the function of x]
  4. \frac{d}{dx}(cn)=c \frac{dn}{dx} [ here c is constant and n is function of x]

Given that,

V= \pi r^2h

(a)

V= \pi r^2h

Differentiating with respect to t

\frac{dV}{dt}= \frac{d}{dt}(\pi r^2h)

\Rightarrow \frac{dV}{dt}= \pi h \frac{d}{dt}(r^2)    [ here \pi h is constant]

\Rightarrow \frac{dV}{dt}= \pi h 2r \frac{dr}{dt}

\Rightarrow \frac{dV}{dt}= 2\pi r h  \frac{dr}{dt}

(b)

V= \pi r^2h

Differentiating with respect to t

\frac{dV}{dt}= \frac{d}{dt}(\pi r^2h)

\Rightarrow \frac{dV}{dt}= \pi r^2 \frac{dh}{dt}

(c)

V= \pi r^2h

Differentiating with respect to t

\frac{dV}{dt}= \frac{d}{dt}(\pi r^2h)

\Rightarrow \frac{dV}{dt} = \pi r^2\frac{dh}{dt}+\pi h\frac{d}{dt}(r^2)

\Rightarrow \frac{dV}{dt} = \pi r^2\frac{dh}{dt}+2\pi r h\frac{dr}{dt}

3 0
4 years ago
A man is at a car dealership, looking for a car to buy. He looks at the sticker on the driver’s window of a car and sees that th
maxonik [38]

Answer:

he will see the sticker because its behind a window bruh and thats a big daddy stack of greens

Explanation:

3 0
3 years ago
What is Hooke's law? what is meant by elastic limit?<br>please answer me​
3241004551 [841]

Answer:

Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit

mark me brainliestt :))

6 0
3 years ago
Find the wavelength (in nm) of a 2.52 eV photon. a. 492.34127 nm.b. 585.88611 nm.c. 541.5754 nm.d. 418.49008 nm.
elena-14-01-66 [18.8K]

Answer:

The correct option is (a).

Explanation:

Given that,

The energy of photon, E = 2.52 eV

We need to find the wavelength of the photon in nm. The formula for the energy of a photon is given by :

E=\dfrac{hc}{\lambda}\\\\\lambda=\dfrac{hc}{E}\\\\\lambda=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{2.52\times 1.6\times 10^{-19}}\\\\=4.93\times 10^{-7}\ m\\\\=493\ nm

The nearest option is a) i.e. 492.34127 nm.

6 0
3 years ago
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