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Fittoniya [83]
3 years ago
10

Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some progr

am and computer B has a clock cycle time of 250 ps and an effective CPI of 1.7 for the same program. Which computer is faster and by how much?
Physics
1 answer:
lesya692 [45]3 years ago
4 0

Answer:

Computer A is 1.41 times faster than the Computer B

Explanation:

Assume that number of instruction in the program is 1

Clock time  of computer A is CT_{A} =200 ps

Clock time  of computer B is CT_{B} =250 ps

Effective CPI of computer A is CPI_{A} =1.5

Effective CPI of computer B isCPI_{B} =1.7

CPU time of A is

CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec

CPU time of B is

CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec

Hence Computer A is Faster by \frac{425}{300} =1.41

Computer A is 1.41 times faster than the Computer B

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Explanation:

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3 years ago
What force is needed to accelerate a 1,000-kilogram car from a stop to 5 m/s/s?
solong [7]
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3 years ago
A frictionless piston–cylinder device contains 5 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin
Arte-miy333 [17]

Answer:

The work input during this process is -742 kJ

Explanation:

Given;

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final temperature of nitrogen T₂ = 450 K

mass of nitrogen, m = 5 kg

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The work input during the process is calculated as;

W = \frac{m*R(T_2-T_1)}{1-n}

where;

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substitute given values in above equation.

W = \frac{m*R(T_2-T_1)}{1-n} = \frac{5*0.2968(450-250)}{1-1.4} = -742 \ kJ

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8 0
4 years ago
A force of 100 N acts upward. Resolve this force into 2 components; one that acts 30º north of west and one that acts 60º north
GrogVix [38]

To resolve these forces we have to make use of the sines and cosines.

To resolve this force in 30 degree north of west, the answer will be

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5 0
3 years ago
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Answer:

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This law had to be combined with the law of conservation of mass when it was determined that mass can be inter-converted with energy.

One can also imagine the energy transformation in a pendulum.  When the ball is at the top of its swing, all of the pendulum’s energy is potential energy.   When the ball is at the bottom of its swing, all of the pendulum’s energy is kinetic energy.   The total energy of the ball stays the same but is continuously exchanged between kinetic and potential forms

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