Answer:
a. true
Explanation:
Firstly, we need to understand what takes places during the compression process in a quasi-equilibrium process. A quasi-equilibrium process is a process in during which the system remains very close to a state of equilibrium at all times. When a compression process is quasi-equilibrium, the work done during the compression is returned to the surroundings during expansion, no exchange of heat, and then the system and the surroundings return to their initial states. Thus a reversible process.
While for a non-quasi equilibrium process, it takes more work to move the piston against this high-pressure region.
Answer:
(N-1) × (L/2R) = (N-1)/2
Explanation:
let L is length of packet
R is rate
N is number of packets
then
first packet arrived with 0 delay
Second packet arrived at = L/R
Third packet arrived at = 2L/R
Nth packet arrived at = (n-1)L/R
Total queuing delay = L/R + 2L/R + ... + (n - 1)L/R = L(n - 1)/2R
Now
L / R = (1000) / (10^6 ) s = 1 ms
L/2R = 0.5 ms
average queuing delay for N packets = (N-1) * (L/2R) = (N-1)/2
the average queuing delay of a packet = 0 ( put N=1)
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