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vodka [1.7K]
2 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]2 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]2 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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What change in entropy occurs when a 0.15 kg ice cube at -18 °C is transformed into steam at 120 °c 4.
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<u>Answer:</u> The change in entropy of the given process is 1324.8 J/K

<u>Explanation:</u>

The processes involved in the given problem are:

1.)H_2O(s)(-18^oC,255K)\rightarrow H_2O(s)(0^oC,273K)\\2.)H_2O(s)(0^oC,273K)\rightarrow H_2O(l)(0^oC,273K)\\3.)H_2O(l)(0^oC,273K)\rightarrow H_2O(l)(100^oC,373K)\\4.)H_2O(l)(100^oC,373K)\rightarrow H_2O(g)(100^oC,373K)\\5.)H_2O(g)(100^oC,373K)\rightarrow H_2O(g)(120^oC,393K)

Pressure is taken as constant.

To calculate the entropy change for same phase at different temperature, we use the equation:

\Delta S=m\times C_{p,m}\times \ln (\frac{T_2}{T_1})      .......(1)

where,

\Delta S = Entropy change

C_{p,m} = specific heat capacity of medium

m = mass of ice = 0.15 kg = 150 g    (Conversion factor: 1 kg = 1000 g)

T_2 = final temperature

T_1 = initial temperature

To calculate the entropy change for different phase at same temperature, we use the equation:

\Delta S=m\times \frac{\Delta H_{f,v}}{T}      .......(2)

where,

\Delta S = Entropy change

m = mass of ice

\Delta H_{f,v} = enthalpy of fusion of vaporization

T = temperature of the system

Calculating the entropy change for each process:

  • <u>For process 1:</u>

We are given:

m=150g\\C_{p,s}=2.06J/gK\\T_1=255K\\T_2=273K

Putting values in equation 1, we get:

\Delta S_1=150g\times 2.06J/g.K\times \ln(\frac{273K}{255K})\\\\\Delta S_1=21.1J/K

  • <u>For process 2:</u>

We are given:

m=150g\\\Delta H_{fusion}=334.16J/g\\T=273K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 334.16J/g}{273K}\\\\\Delta S_2=183.6J/K

  • <u>For process 3:</u>

We are given:

m=150g\\C_{p,l}=4.184J/gK\\T_1=273K\\T_2=373K

Putting values in equation 1, we get:

\Delta S_3=150g\times 4.184J/g.K\times \ln(\frac{373K}{273K})\\\\\Delta S_3=195.9J/K

  • <u>For process 4:</u>

We are given:

m=150g\\\Delta H_{vaporization}=2259J/g\\T=373K

Putting values in equation 2, we get:

\Delta S_2=\frac{150g\times 2259J/g}{373K}\\\\\Delta S_2=908.4J/K

  • <u>For process 5:</u>

We are given:

m=150g\\C_{p,g}=2.02J/gK\\T_1=373K\\T_2=393K

Putting values in equation 1, we get:

\Delta S_5=150g\times 2.02J/g.K\times \ln(\frac{393K}{373K})\\\\\Delta S_5=15.8J/K

Total entropy change for the process = \Delta S_1+\Delta S_2+\Delta S_3+\Delta S_4+\Delta S_5

Total entropy change for the process = [21.1+183.6+195.9+908.4+15.8]J/K=1324.8J/K

Hence, the change in entropy of the given process is 1324.8 J/K

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