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Romashka-Z-Leto [24]
3 years ago
12

calculate how many turns of cable you will need on the drum for your cage and skip to move up and down by 500 m​

Computers and Technology
1 answer:
anyanavicka [17]3 years ago
5 0
Physics. only one, if the drum is 500m around... 50 turns. you will need on the drum for your cage and skip to move up and down by 500m. 500/18.84= 26.5.

Mark me as brainliest pls..... :)
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Of what is famous Ted Nelson?​
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How do you do basic addition and subtraction in binary, octal and hex?
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Explanation:

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2 - The Binary Numbering System

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Assuming you need to represent the number two. Which digit you can use, if there is no number 2 in that system?

We have the following response. In the decimal system, we do not have the number ten and represent the amount of ten using the digit 1 followed by the digit 0. In this case, the number 1 means that we have a group of ten and the digit 0, no drive, which means ten.

In the binary system, do likewise. For you the amount of two, we use the digit 1 followed by the digit 0. The figure 1 means that there is a group of two elements and 0, a group of any unit, thus representing the number two.

The table below helps us to understand the differences between the decimal and binary system, using this rule. The number sequence displayed to the number nine.

Decimal

Binary

0

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3

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5

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6

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9

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2.1 - System Conversion binary to Decimal System

To better understand the conversion we use a decimal number either, for example, 356. This number means:

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6 0
3 years ago
2. Consider a 2 GHz processor where two important programs, A and B, take one second each to execute. Each program has a CPI of
allsm [11]

Answer:

program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

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So, the original processor is better then the new processor for program B.

Explanation:

Finding number of instructions in A and B using time taken by the original processor :

The clock speed of the original processor is 2 GHz.

which means each clock takes, 1/clockspeed

= 1 / 2GH = 0.5ns

Now, the CPI for this processor is 1.5 for both programs A and B. therefore each instruction takes 1.5 clock cycles.

Let's say there are n instructions in each program.

therefore time taken to execute n instructions

= n * CPI * cycletime = n * 1.5 * 0.5ns

from the question, each program takes 1 sec to execute in the original processor.

therefore

n * 1.5 * 0.5ns = 1sec

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So, number of instructions in each program is 1.3333 * 10^9

the new processor :

The cycle time for the new processor is 0.6 ns.

Time taken by program A = time taken to execute n instructions

=  n * CPI * cycletime

= 1.3333 * 10^9 * 1.1 * 0.6ns

= 0.88 sec

Time taken by program B = time taken to execute n instructions

= n * CPI * cycletime

= 1.3333 * 10^9 * 1.4 * 0.6

= 1.12 sec

Now, program A runs in 1 sec in the original processor and 0.88 sec in the new processor.

So, the new processor out-perform the original processor on program A.

Program B runs in 1 sec in the original processor and 1.12 sec in the new processor.

So, the original processor is better then the new processor for program B.

5 0
3 years ago
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