Answer and Explanation:
Given data:
Distance (D) = 40 KM
Speed of light in the fiber =Distance/ speed of light in the fiber
a) Delay (P) = Distance/ speed of light in the fiber
= (40,000 Meters/2×108 m/s)
=( 40×103 Meters/2×108 m/s)
Propagation delay (P) = 0.0002 seconds or 200 microseconds
b)
if propagation delay is 0.0002 Seconds roundup trip time (RTT) will be 0.0004 Seconds or 400 micro Seconds
Essentially since transmission times and returning ACKs are insignificant all we really need is a value slightly greater than 0.0004 seconds for our timeout value.
c)
The obvious reasons would be if the data frame was lost, or if the ACK was lost. Other possibilities include extremely slow processing on the receive side (late ACK).
Or Extremely Slow Processing of the ACK after it is received back at the send side.
1. true
2. pixel
3. raster images are built with pixels
4. false
5. image size
6. rollover
7. sharp with clear details
8. 78px
9. 24in
10. scalability
C. The viewing area of a web browser.
The viewport is the user's visible area of a web page.
Answer:
Your answer is <em>I think </em>A) A loop
Explanation:
In computer programming, a loop is a sequence of instruction s that is continually repeated until a certain condition is reached.
Use an adapter, costs money though.