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vodka [1.7K]
3 years ago
7

IBM has a computer it calls the Blue Gene/L that can do 136.8 teracalculations per second. How many calculations can it do in a

microsecond?
The answer has to be in four significant digits.
Physics
2 answers:
nikdorinn [45]3 years ago
7 0

IBM has a computer it calls the Blue Gene/L that can do  136.8* 10^{6} calculations in a microsecond or  136.8 megacalculations in a microsecond

<h3>Further explanation </h3>

Conversion of units is the conversion between different units of measurement for the same quantity, typically through multiplicative conversion factors.

There are methods for converting values with multiple units

  • Write down your problem.
  • Find the conversion for one unit
  • Multiply your number by the conversion fraction.
  • Cancel out your units.
  • Multiply with another conversion fraction the same way.
  • Cancel units.
  • Repeat until the conversion is done

The seven base quantities and their corresponding units are:

  • length (metre)
  • mass (kilogram)
  • time (second)
  • electric current (ampere)
  • thermodynamic temperature (kelvin)
  • amount of substance (mole)
  • luminous intensity (candela)

IBM has a computer it calls the Blue Gene/L that can do 136.8 teracalculations per second. How many calculations can it do in a microsecond?

prefixes : tera = 10^{12}

micro = 10^-6, there are  10^6  microseconds \mu s in 1 second

136.8* 10^{12} \frac{calculations}{second} = 136.8* 10^{12} \frac{calculations}{10^6 \mu s}

So, it can do 136.8* 10^{6} calculations in a microsecond or  136.8 megacalculations in a microsecond. The answer is in four significant digits

<h3>Learn more</h3>
  1. Learn more about physics microsecond brainly.com/question/4942348

<h3>Answer details</h3>

Grade:  9

Subject:  physics

Chapter:  microsecond

Keywords:  microsecond

Norma-Jean [14]3 years ago
4 0
There are 1,000,000 micro seconds in one second so multiple 136.8 by 1000000 and you'll get 136,800,000 Tera calculations per second.
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\frac{1}{10}M

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As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒I_{1} \times W_{1} =I_{2} \times W_{2}

  • I_{1} is the moment of interia of earth before impact
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  • W_{2} is the angular velocity of earth and asteroid system about the same axis

let  W_{1}=W

since \text{Time period of rotation}∝\frac{1}{\text{Angular velocity}}

⇒ if time period is to increase by 25%, which is \frac{5}{4} times, the angular velocity decreases 25% which is \frac{4}{5}  times

therefore W_{1} = \frac{4}{5} \times W_{1}

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where M is mass of earth

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I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where M_{1} is mass of asteroid

⇒ \frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})

\frac{1}{2} \times M\times R^{2}= (\frac{2}{5} \times M\times R^{2}+ M_{1}\times R^{2})

M_{1}\times R^{2}= \frac{1}{10} \times M\times R^{2}

⇒M_{1}=}\frac{1}{10} \times M

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