Answer:
Both come from the sun
Both are reusable sources
and both don't cause pollution
Explanation:
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)
Answer:
A). Dry unit weight = 1657.08Kg/m3
B). Porosity = 0.37
C). Void ratio = 0.593
D). 0.712
Explanation:
Total unit weight, Y = 120pcf =1922.2 Kg/m3
Specific gravity of solids, Gs = 2.64
Water content, w = 16%
A). Dry unit weight
Yd = Y/(1+w)
= 1922.2/(1+0.16) = 1657.08Kg/m3
B). Porosity
However void ratio, e = Gs×Yw/Yd, where Yw = 1000Kg/m3
Void ratio = 2.64×1000/1657.08 = 0.593
And porosity = e/(1+e) =0.593/(1+0.593) = 0.37
C). void ratio, e = 0.593
D). Degree of saturation, S = m×Gs/e where m =water content
S = 0.16×2.64/0.593 = 0.712
Answer:
for a) F= 744.97 N
for a) F= 167.85 N
for a) F= 764.57 N
Explanation:
the pressure developed by the piston should be higher than the saturated vapor pressure of water for boiling point at T=120 to ensure boiling.
Then from steam tables
T= 120°C → P required=Pr= 198.67 kPa
then the pressure developed by the piston is
P = (m*g + F)/A
where m= mass of the piston ,g= gravity F= force required and A= area of the piston
then
Pr = P = (m*g + F)/A
F = Pr*A-m*g
since A= π/4*D²
F =π/4* Pr*D²-m*g
replacing values
F =π/4* Pr*D²-m*g = π/4*198.67 *10³Pa*(0.07m)² -2kg* 9.8m/s²
F= 744.97 N
b) for T₂=80°C → Pr₂=47.41 kPa
F₂ =π/4* Pr₂*D²-m*g = π/4*47.41*10³Pa*(0.07m)² -2kg* 9.8m/s²
F₂= 167.85 N
c) for m=0 (mass of the piston neglected) ,the force required is
F₃ =π/4*Pr*D² = π/4*198.67 *10³Pa*(0.07m)²= 764.57 N
F₃ =764.57 N
Answer:
im so sorry I rlly need these points
Explanation:
