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Vinvika [58]
3 years ago
7

The older computer had a CPU that ran at 266MHz compared to a current CPU that runs 3.7GHz. How many times faster is the new CPU

than the older CPU? [1T]
266MHz=0.266GHz
Computers and Technology
1 answer:
sergij07 [2.7K]3 years ago
3 0

Answer:

Number of times new computer faster than old computer = 13.90 times (Approx)

Explanation:

Given:

Clock speed of old computer = 266 MHz

Clock speed of new computer = 3.7 GHz

266MHz = 0.266GHz

Find:

Number of times new computer faster than old computer

Computation:

Number of times new computer faster than old computer = Clock speed of new computer / Clock speed of old computer

Number of times new computer faster than old computer = 3.7 / 266

Number of times new computer faster than old computer = 3.7 / 0.266

Number of times new computer faster than old computer = 13.90 times (Approx)

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Verizon [17]

Answer:

megapixel refers to the unit of resolution i.e. one million

Explanation:

Interestingly the higher the pixels does not mean higher quality of image, it's more about the camera and it's sensor.

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5 0
3 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
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3 years ago
Prove that any amount of postage greater than or equal to 64 cents can be obtained using only 5-cent and 17-cent stamps?
elixir [45]
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6 0
3 years ago
Why is web arnea of so much commetition today
Sonja [21]

Answer:

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Explanation:

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Answer:

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7 0
3 years ago
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