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Sonja [21]
3 years ago
6

A disk rotates at a rate of 7200 revolutions per minute. Seek operations (i.e., moving the access head to a desired track) take

on average 20 milli-seconds. Once the access head is on the appropriate track, it takes an average of one half revolution of the disk to get to the beginning of a requested sector. Each track contains 128 sectors and each sector contains 512 bytes of data. How long on average does it take to get the access head to the beginning of a randomly selected sector on a randomly selected track? Express your answer in micro-seconds (not seconds or milli-seconds).
Computers and Technology
1 answer:
Leno4ka [110]3 years ago
8 0

Answer:

24.167 micro seconds.

Explanation:

The given rotation rate = 7200 rpm = 7200 rounds per minute

Rotational latency is the average time taken for the head to reach starting of sector .

Rotational latency (in micro seconds) = (1 / (RPM / 60)) * 0.5 * 1000

(1/(7200/60))* 0.5 * 1000 = 4.167 ms

Thus, rotational latency = 4.167 ms.

Seek time = 20 ms

The average time taken for the access head to get to the beginning of randomly selected sector will be equal to the average time to first reach the random track plus the average time taken to reach random sector .

= 20 ms + 4.167  ms  = 24.167 micro seconds.

Thus, it would take 24.167 micro seconds to get the access head to the beginning of a randomly selected sector on a randomly selected track.

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How many binary digits does a single hexadecimal digit represent?
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3 years ago
1. Write a high level algorithm for cooking a cheeseburger.
pishuonlain [190]

Answer:

1.  A high level algorithm for cooking a cheeseburger could be:

  1. Heat fry pan
  2. Cook one side of the hamburger
  3. Wait
  4. Turn hamburger upside down
  5. Put cheese over hamburger
  6. Wait
  7. Cut hamburger bread in half
  8. Put cooked hamburger inside bread
  9. End (eat)

2. A detailed algorithm for cooking a cheeseburger could be:

  1. Place fry pan over the stove heater
  2. Turn on heater (max temp)
  3. IF fry pan not hot: wait, else continue
  4. Place raw hamburger on fry pan
  5. IF hamburger not half cooked: Wait X time then go to line 5, else continue
  6. Turn hamburger upside down
  7. Put N slices of cheese over hamburger
  8. IF hamburger not fully cooked: Wait X time then go to line 8, else continue
  9. Turn off heater
  10. Cut hamburger bread in half horizontally
  11. Put cooked hamburger on one of the bread halves.
  12. Put second bread half on top of hamburger
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Explanation:

An algorithm is simply a list of steps to perform a defined action.

On 1, we described the most relevant steps to cook a simple cheeseburger.

Then on point 2, the same steps were taken and expanded with more detailed steps and conditions required to continue executing the following steps.

In computational terms, we used pseudo-code for the algorithm, since this is a list of actions not specific to any programming language.

Also we can say this is a structured programming example due to the sequential nature of the cooking process.

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When creating a presentation in Libre Office Impress, where does the editing of slides take place?
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Given public class Fishing { byte b1 = 4; int i1 = 123456; long L1 = (long) i1; //Line A short s2 = (short) i1; //Line B byte b2
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Answer:

Line E.

Explanation:

The  given program is as follows:

public class Fishing {

   byte b1 = 4; int i1 = 123456; long L1 = (long) i1;   //Line A

   short s2 = (short) i1;     //Line B

   byte b2 = (byte) i1;     //Line C

   int i2 = (int)123.456;    //Line D

   byte b3 = b1 + 7;     //Line E

   }

In the above code Line E will not compile and give following compilation error:

error: incompatible types: possible lossy conversion from int to byte

This error is coming because in Java b1 + 7 will be interpreted as int variable expression. Therefore in order to make code of Line E work the expression should be type casted as byte.  

The correct code will be as follows:

public class Fishing {

   byte b1 = 4; int i1 = 123456; long L1 = (long) i1;   //Line A

   short s2 = (short) i1;     //Line B

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   int i2 = (int)123.456;    //Line D

   byte b3 = (byte)(b1 + 7);     //Line E

   }

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