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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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A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur
morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\

\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] =  4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\

Substitute the given parameters and solve for the angular speed;

\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\  \ +\  \ 0.35(0.02^2\  + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

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2 years ago
How much force is required to accelerate a 50 kg mass at 2 m/s
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Answer:

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Explanation:

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You have a string with a mass of 0.0127 kg. You stretch the string with a force of 9.33 N, giving it a length of 1.93 m. Then, y
melomori [17]

Answer:

wavelength = 0.968 m

frequency = 39.02 Hz

Explanation:

given data

mass = 0.0127 kg

force = 9.33 N

length = 1.93 m

to find out

wavelength and Frequency

solution

we know here linear density that is

linear density = \frac{mass}{length}   .........1

linear density = \frac{0.0127}{1.93}

linear density = 6.5803 × 10^{-3} kg/m

so

wavelength will be here

wavelength = \frac{2L}{n}   ..............2

here n = 4 for forth harmonic

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and

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frequency = \frac{4}{2L} \sqrt{\frac{tension}{linear\ density} }    ..........3

frequency = \frac{4}{2*1.93} \sqrt{\frac{9.33}{6.5803*10^{-3}} }

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20 point question B)
frez [133]

Description:

As we know that the sun offers what most of the world has to go examples could be energy, air. As we also know that heating causes liquid and water that is frozen to evaporate to water vapour gas. Meaning the sun provides the energy necessary for evaporation.

Answer:  B

Please mark brainliest

<em><u>Hope this helps.</u></em>

4 0
3 years ago
Read 2 more answers
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