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)
Didactic apparatus is a method of teaching in which scientific approach is follow in order to present the information to the student. This method effectively teaches the student with the required theoretical knowledge .
Answer:
14,700 J
Explanation:
PE = Mgh = (75 kg)(9.8 m/s²)(20 m) = 14,700 J
Answer:
The shaft work generated per kilogram is 
Explanation:
Given:
Temperature 
K
Initial Pressure 
MPa
Final pressure 
MPa
From the table superheated,
and 
Work done by shaft is,
But here efficiency is 0.56,
So work generated per kg is,
Work = 
Work = 
Therefore, the shaft work generated per kilogram is
Answer:
Explanation:
R1 = 1 k; R2 = 1k; R3 = 3k; R4 = 6k; R5 = 5k; R6 = 6k and R7 = 2k
Vt = 428V
The series and parallel circuit combination is as follow:
(R6║R7 + R5) + R4 + R3║R2 + R1
(6*2/6 + 2) + 5 = 13/ k
(13/2*6/13/6 + 6) = 78/31k
78/3 + 3 = 171/3 = 57k
57k║R2 = 57k║1k = 57/58k + R1 = (57/58 + 1)k = 115/58k = 2k
It = Vt/2k = 428/2000 = 0.2A
∴ I1 = 0.2A
I1 = I2 + I3
Using current divider rules to obtain I2 and I3
∴ I2 = I1 X (I2/I2 + I3) = 0.2 X ( 1/4) = 0.05A
and I3 = I1 X (I3/I2 + I3) = 0.2 X (3/4) = 0.15A
I3 = I4 + I5, using current divider
I4 = I3 X (I4/I4 + I5) = 0.15 X (6/6 + 5) = 0.08
I5 = 0.15 - 0.08 = 0.07A
I5 = I6 + I7, using current divider
I6 = I5 X (I6/I6 + I7) = 0.07 X (6/6 + 2) = 0.05A
I7 = 0.07 - 0.05 = 0.02A
