Answer:
a. Using a straight forward scheme, the RC4 algorithm stores 2064 bits in the interval state.
b. 1700 bits
Explanation:
See RC4 Algorithm attached with details
Answer:
operates at a substantially constant load for an indefinitely long time. ... Short-time duty: operates at a substantially constant load for a time that is definite, short, and specified. Varying duty: the loads and intervals of operation change.
Explanation:
<h2>
˜”*°•.˜”*°• Question •°*”˜.•°*”˜
</h2>
<em>What Member function places a new node at the end of the linked list?
</em>
<h2>
˜”*°•.˜”*°• Answer •°*”˜.•°*”˜</h2>
appendNode
<h2>˜”*°•.˜”*°• Explanation •°*”˜.•°*”˜</h2>
The appendNode() member function places a new node at the end of the linked list. The appendNode() requires an integer representing the current data of the node.
<h2>˜”*°•.˜”*°• Details •°*”˜.•°*”˜</h2>
Subject: Coding (?)
Grade: College
Keywords: Function, linked list, appendNode, integer
Hope this helped. <3
Answer:
A)
D = 158.42 kmol/h
B = 191.578 kmol/h
B) Rmin = 1.3095
Explanation:
<u>a) Determine the distillate and bottoms flow rates ( D and B ) </u>
F = D + B ----- ( 1 )
<em>Given data :</em>
F = 350 kmol/j
Xf = 0.45 mole
yD ( distillate comp ) = 0.97
yB ( bottom comp ) = 0.02
back to equation 1
350(0.45) = 0.97 D + 0.02 B ----- ( 2 )
where; B = F - D
Equation 2 becomes
350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3
solving equation 3
D = 158.42 kmol/h
resolving equation 2
B = 191.578 kmol/h
<u>B) Determine the minimum reflux ratio Rmin</u>
The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve
first we calculate the value of the enriching line
Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )
q - line ; y = ( 9 / 9-1 ) x - xf/9-1
therefore ; x = 0.45
Finally Rmin
= (( 0.97 / (Rm + 1 )) = 0.42
0.42 ( Rm + 1 ) = 0.97
∴ Rmin = 1.3095
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Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
