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Hoochie [10]
3 years ago
13

Consider a single CPU system with an active process A. Explain what happens in the following circumstances including any interru

pts, system calls, etc. and how they are handled until a process is back to running again?
a. Process A forks a new process B
b. Process A is running, and needs to read from a file
c. A timer interrupt occurs while A is running.
Computers and Technology
1 answer:
Anika [276]3 years ago
5 0

Answer:

The answer to this question as follows:

Explanation:

In option a, When we use fork, a new mechanism, that uses fork() method, which is in the parental process, to replicate all sites. It has been installed in a space-differentiating operating system.

In option b, It is the present system cycle that scan and wait for any of the system processor to be installed.

In option c,  The time delay happens when A operates. It is a global that make issues, which are cleared and slowed down when an interrupt happens. In this process, there are not any distractions. It splits into slowly as it heads into the ISR. It helps to understand the code easily.

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Linda wants to change the color of the SmartArt that she has used in her spreadsheet. To do so, she clicks on the shape in the S
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A. True

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3 0
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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
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