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Otrada [13]
3 years ago
12

Time complexity of quick short

Engineering
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:

The time complexity will be "O(n log n)".

Explanation:

  • Many realistic Quick sort implementations choose a randomized special edition. The time complexity variable O(n Logn) was predicted in the randomized edition.
  • Throughout the randomized version, probably the most disgusting case is also conceivable, but by far the worst scenario for something like a given pattern does not exist as well as randomized Quick sort performs well throughout the practice.
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A desktop computer is to be cooled by a fan whose flow rate is 0.34 m3 /min. Determine the mass flow rate of air through the fan
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m = 0.238  <u>kg</u>

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r = 0.063m

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The heavier an atomic nucleus gets, the less energy the star is able to extract from it through nuclear fusion. When it gets to
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Iron

Explanation:

Once a star starts fusing iron in its core it gives off very large amounts of energy. Also helium, hydrogen, carbon, oxygen, and silicon still exist in the star in different shells while fusin is taking place at different parts of the star. Right at the surface, hydrogen continues to fuse to helium, as e go further down helium is fusing to carbon and oxygen; right inside the core we have silicon that is fusing with iron. At this point, Iron being of stable atomic structure and quite heavier will no longer be fused into anything because of the quite large amounts of energy and force required to fuse iron atoms. The process comes to a standstill at this point.

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Show What You Know: Creativity
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Two grenades, A and B, are thrown horizontally with different speeds from the top of a cliff 70 m high. The speed of A is 2.50 m
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3 years ago
A body is moving with simple harmonic motion. It's velocity is recorded as being 3.5m/s when it is at 150mm from the mid-positio
natima [27]

Answer:

1) A=282.6 mm

2)a_{max}=60.35\ m/s^2

3)T=0.42 sec

4)f= 2.24 Hz

Explanation:

Given that

V=3.5 m/s at x=150 mm     ------------1

V=2.5 m/s at x=225 mm   ------------2

Where x measured  from mid position.

We know that velocity in simple harmonic given as

V=\omega \sqrt{A^2-x^2}

Where A is the amplitude and ω is the natural frequency of simple harmonic motion.

From equation 1 and 2

3.5=\omega \sqrt{A^2-0.15^2}    ------3

2.5=\omega \sqrt{A^2-0.225^2}   --------4

Now by dividing equation 3 by 4

\dfrac{3.5}{2.5}=\dfrac {\sqrt{A^2-0.15^2}}{\sqrt{A^2-0.225^2}}

1.96=\dfrac {{A^2-0.15^2}}{{A^2-0.225^2}}

So    A=0.2826 m

A=282.6 mm

Now by putting the values of A in the equation 3

3.5=\omega \sqrt{A^2-0.15^2}

3.5=\omega \sqrt{0.2826^2-0.15^2}

ω=14.609 rad/s

Frequency

ω= 2πf

14.609= 2 x π x f

f= 2.24 Hz

Maximum acceleration

a_{max}=\omega ^2A

a_{max}=14.61 ^2\times 0.2826\ m/s^2

a_{max}=60.35\ m/s^2

Time period T

T=\dfrac{2\pi}{\omega}

T=\dfrac{2\pi}{14.609}

T=0.42 sec

8 0
4 years ago
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