How much power do a capacitor and inductor dissipate? Assume the capacitor/inductor have no parasitic resistance (no resistor in
1 answer:
You might be interested in
Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)
Now we can use the equation that defines the percentage of increase, in this case for speed
Now we use the equation obtained in the previous step, and replace values
the percent increase in the velocity of air is 25.65%
Answer:
shrinkage ratio = 1.538
Explanation:
given data
water content = 22 %
dry density γ = 82 pcf
required dry density specified γ' = 96 pcf
required to produce = 50,000 yd³ = 50000 × 27 = 1350,000 ft³
solution
we get here first volume of borrow pit that is we know that
dry density ∝
so
v = 1580487.8 ft³
v = 58536.58 yd³
so here
shrinkage ratio will be as
shrinkage ratio =
shrinkage ratio = 1.538
Answer:
Q=33.34 KW/m
Explanation:
Given that
D₁=0.42 m
A₁= π D₁ L
For unit length
A₁= π D₁ = 0.42 π m²
D₂=0.5 m
A₂= 0.5 π m²
ε₁= 1 ,ε₂= 0.55
T₁=950 K ,T₂ = 500 K
F₁₁+F₁₂= 1
F₁₁= 0
So, F₁₂= 1
Q=33.34 KW/m