Answer:
The question is incomplete, below is the complete question
"The real power delivered by a source to two impedance, Z1=4+j5Ω and Z2=10Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."
answer:
a. 615W, 384.4W
b. 17.4A
Explanation:
To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.
recall that the symbol for admittance is Y and express as
Hence for each we have,
for the second impedance we have
we also determine the voltage cross the impedance,
P=V^2(Y1 +Y2)
The real power in the impedance is calculated as
for the second impedance
b. We determine the equivalent admittance
We convert the equivalent admittance back into the polar form
the source current flows is
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:
(B) FALSE
Explanation:
view factor 
depends on the surface emissivity and the surface of geometry view factor is the term used in radiative heat transfer. View factor is depends upon the radiation which leave the surface and strike the surface.View factor is also called shape factor configuration factor it is denoted by
Answer:
Explanation:
Sorry I'm new and I need points ty
