Another solution to minimize resonance is to _______

Which of these tools can best be used as a self assessment for career planning purposes? a personality test an asset analysis a

Ivahew [28]

Answer:

a goal development checklist

Explanation:

8 0

2 years ago

A company requires an Ethernet connection from the north end of its office building to the south end. The distance between conne

BlackZzzverrR [31]

Answer:

A company requires an Ethernet connection from the north end of its office building to the south end. The distance between connections is 161 meters and will need to provide speeds of up to 10 Gbps in full duplex mode. Which of the following cable combinations will meet these requirements?

ANSWER: Use multi-mode fiber optic cable

Explanation:

MULTI-MODE FIBER OPTIC CABLE

For Ethernet distances up to 100 meters, the Copper CATX cable will do just fine. In the question, we are dealing with a distance of up to 161 meters, so we need an Ethernet extension. LAN extension over fiber optic cable with media converter can be used to convert the Ethernet cable runs from copper to fiber. Multi-mode fiber has a range of 550 meters for 10/100/1000 Ethernet links.

Multi-mode optical fiber is commonly used for communication over short distances, like within a building or on a campus. They are capable of data rate of up to 100 Gbps, which surpasses the requirements in this question. They are more economical in this case as they are not as expensive a the single-mode optical fiber cable. Fiber optic cable has the advantage of being immune to electromagnetic interferences, spikes, ground loops, and surges. This makes it more suited for this purpose.

7 0

2 years ago

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

4 0

2 years ago

Stan’s assignment is to print a three-dimensional image on a piece of paper. Which printing technique should he use?

Zigmanuir [339]

Answer:

to print a three-dimensional image on a piece of paper he should use autocad

8 0

2 years ago

Read 2 more answers

Write the 8-bit signed-magnitude, two's complement, and ones' complement representations for each decimal number: +25, + 120, +

pshichka [43]

Answer:

Let's convert the decimals into signed 8-bit binary numbers.

As we need to find the 8-bit magnitude, so write the powers at each bit.

Sign -bit 64 32 16 8 4 2 1

+25 - 0 0 0 1 1 0 0 1

+120- 0 1 1 1 1 0 0 0

+82 - 0 1 0 1 0 0 1 0

-42 - 1 0 1 0 1 0 1 0

-111 - 1 1 1 0 1 1 1 1

One’s Complements:

+25 (00011001) – 11100110

+120(01111000) - 10000111

+82(01010010) - 10101101

-42(10101010) - 01010101

-111(11101111)- 00010000

Two’s Complements:

+25 (00011001) – 11100110+1 = 11100111

+120(01111000) – 10000111+1 = 10001000

+82(01010010) – 10101101+1= 10101110

-42(10101010) – 01010101+1= 01010110

-111(11101111)- 00010000+1= 00010001

Explanation:

To find the 8-bit signed magnitude follow this process:

For +120

put 0 at Sign-bit as there is plus sign before 120.

Put 1 at the largest power of 2 near to 120 and less than 120, so put 1 at 64.

Subtract 64 from 120, i.e. 120-64 = 56.

Then put 1 at 32, as it is the nearest power of 2 of 56. Then 56-32=24.

Then put 1 at 16 and 24-16 = 8.

Now put 1 at 8. 8-8 = 0, so put 0 at all rest places.

To find one’s complement of a number 00011001, find 11111111 – 00011001 or put 0 in place each 1 and 1 in place of each 0., i.e., 11100110.

Now to find Two’s complement of a number, just do binary addition of the number with 1.

6 0

2 years ago

Another solution to minimize resonance is to ________ the bridge.