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Debora [2.8K]
1 year ago
8

Jeffrey works with a huge database of spreadsheet records each day. To organize and identify these spreadsheets, he wants to ass

ign some keywords to each spreadsheet record. Which option will help him do so?
Computers and Technology
1 answer:
mariarad [96]1 year ago
8 0

Answer:

A. Document Properties

B. Permission

C. Document Options

D. File Details

The answer to this question is "File Details " . This will help Jeffrey works efficiently with a huge database of spreadsheet records each day. He can assign names to these File Details which represents each spreadsheet records. This will help his report more organized and easy to identity.

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Consider a single-platter disk with the following parameters: rotation speed: 7200 rpm; number of tracks on one side ofplatter:
horsena [70]

Answer:

Given Data:

Rotation Speed = 7200 rpm

No. of tracks on one side of platter = 30000

No. of sectors per track = 600

Seek time for every 100 track traversed = 1 ms

To find:

Average Seek Time.

Average Rotational Latency.

Transfer time for a sector.

Total Average time to satisfy a request.

Explanation:

a) As given, the disk head starts at track 0. At this point the seek time is 0.

Seek time is time to traverse from 0 to 29999 tracks (it makes 30000)

Average Seek Time is the time taken by the head to move from one track to another/2

29999 / 2 = 14999.5 ms

As the seek time is one ms for every hundred tracks traversed.  So the seek time for 29,999 tracks traversed is

14999.5 / 100 = 149.995 ms

b) The rotations per minute are 7200

1 min = 60 sec

7200 / 60 = 120 rotations / sec

Rotational delay is the inverses of this. So

1 / 120 = 0.00833 sec

          = 0.00833 * 100

          = 0.833 ms

So there is  1 rotation is at every 0.833 ms

Average Rotational latency is one half the amount of time taken by disk to make one revolution or complete 1 rotation.

So average rotational latency is: 1 / 2r

8.333 / 2 = 4.165 ms

c) No. of sectors per track = 600

Time for one disk rotation = 0.833 ms

So transfer time for a sector is: one disk revolution time / number of sectors

8.333 / 600 = 0.01388 ms = 13.88 μs

d)  Total average time to satisfy a request is calculated as :

Average seek time + Average rotational latency + Transfer time for a sector

= 149.99 ms + 4.165 ms + 0.01388 ms

= 154.168 ms

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Acceleration is the rate at which an object changes its velocity. It is typically represented by symbol a and measured in m/s2 (
Natalija [7]

Answer:

<u>Pseudocode:</u>

INPUT velocity

INPUT time

SET velocity = 0.44704 * velocity

SET acceleration = velocity / time

SET acceleration = round(acceleration, 1)

PRINT acceleration

<u>Code:</u>

velocity = float(input("Enter a velocity in miles per hour: "))

time = float(input("Enter a time in seconds: "))

velocity = 0.44704 * velocity

acceleration = velocity / time

acceleration = round(acceleration, 1)

print("The acceleration is {:.2f}".format(acceleration))

Explanation:

*The code is in Python.

Ask the user to enter the velocity and time

Convert the miles per hour to meters per second

Calculate the acceleration using the formula

Round the acceleration to one decimal using round() method

Print the acceleration with two decimal digits

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