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myrzilka [38]
1 year ago
10

Suppose a packet is 10K bits long, the channel transmission rate connecting a sender and receiver is 10 Mbps, and the round-trip

propagation delay is 10 ms. What is the maximum channel utilization of a stop-and-wait protocol for this channel
Computers and Technology
1 answer:
jarptica [38.1K]1 year ago
6 0

Answer:

The Answer to the question is given with proper step by step solution.

Given that : the round-trip-time (RTT in abbreviation) = 20ms = 0.02 micro seconds

L = 2000

Byte = 2000*8

bits = 16000 bits

R = 500Mbps = 500*106 bps

Now,calculate L/R = 16000 / 500*106

= 32 / 106

= 32 micro seconds

The Utilization (U) = fraction when the time sender is busy sending

= (L/R) / (L/R + RTT)

= (32) / (32 + 0.02)

= 32 / 32.02

= 0.99937539 seconds

= 999375.39 micro seconds

= 999375.4 (nearest tenth)

Explanation:

um I dont know what to put for the explanation so yea

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3 years ago
Which of the following internet protocols is used to request and send pages and files on the World Wide Web
Marianna [84]

it is . org Explanation:

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(iii) ............ characters can be stored in memo field. (a) 50 (b) 64000 (c) 255 (d) 200 ​
Svetlanka [38]

Answer:

(b) 64000

Explanation:

Assuming that the "Memo" being mentioned is a Microsoft Access memo field, it can hold up to 64,000 characters, even in the more recent version of the application where the "Memo" is now know as the "Long Text" field.

According to microsoft.com, "In earlier versions of Access, we used the Memo data type to store large amounts of text... the Long Text field works the same as the Memo field of old... [it] can only display the first 64,000 characters."

<em>Please put "Brainliest" on my answer if it helped you out the most!</em>

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7 0
2 years ago
four quantum numbers that could represent the last electron added (using the Aufbau principle) to the Argon atom. A n = 2, l =0,
marshall27 [118]

Answer:

  • n = 3
  • l  = 1
  • m_l = 1
  • m_s=+1/2

Explanation:

Argon atom has atomic number 18. Then, it has 18 protons and 18 electrons.

To determine the quantum numbers you must do the electron configuration.

Aufbau's principle is a mnemonic rule to remember the rank of the orbitals in increasing order of energy.

The rank of energy is:

1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7d

You must fill the orbitals in order until you have 18 electrons:

  • 1s² 2s² 2p⁶ 3s² 3p⁶   : 2 + 2 + 6 + 2 + 6 = 18 electrons.

The last electron is in the 3p orbital.

The quantum numbers associated with the 3p orbitals are:

  • n = 3

  • l = 1 (orbitals s correspond to l = 0, orbitals p correspond to l  = 1, orbitals d, correspond to l  = 2 , and orbitals f correspond to  l = 3)

  • m_l can be -1, 0, or 1 (from - l  to + l )

  • the fourth quantum number, the spin can be +1/2 or -1/2

Thus, the six possibilities for the last six electrons are:

  • (3, 1, -1 +1/2)
  • (3, 1, -1, -1/2)
  • (3, 1, 0, +1/2)
  • (3, 1, 0, -1/2)
  • (3, 1, 1, +1/2)
  • (3, 1, 1, -1/2)

Hence, the correct choice is:

  • n = 3
  • l  = 1
  • m_l = 1
  • m_s=+1/2
5 0
3 years ago
When light hits an opaque object it will be _______________.
nikklg [1K]
None of it passes through. Most of the light is either reflected by the object or absorbed and converted to heat. Materials such as wood, stone, and metals are opaque to visible light.

Good luck
7 0
2 years ago
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