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.
Because, they are all required to configure to it to be recognized by an operating system.
Explanation:
It is the Operating System Software that instructs the hardware and puts them together to work well. When the manufacturer does not configure the device to be recognized by an operating system, then it will not work with other components.
Example.
If an Apple machine's sound card is being put in an HP machine which uses Microsoft designed operating system, won't work due to the operating system that the sound card has been designed for.
Answer:
D. nothing, as the alkyne would not react to an appreciable extent.
Explanation:
Nothing, as the alkyne would not react to an appreciable extent.
Answer:
19.71 ms
Explanation:
The disk rotates at 3600 rpm, hence the time for one revolution = 60 / 3600 rpm = 16.67 ms.
Hence time to read or write on a sector = time for one revolution / number of sectors per track = 16.67 ms / 32 sectors = 0.52 ms
Head movement time from track 8 to track 9 = seek time = 2 ms
rotation time to head up sector 1 on track 8 to sector 1 on track 9 = 16.67 * 31/32 = 16.15 ms
The total time = sector read time +head movement time + rotational delay + sector write time = 0.52 ms + 2 ms + 16.15 ms + 0.52 ms = 19.19 ms
