Answer:
have you heard of gnoogle?
Explanation:have you heard of goongle?
Answer:
We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).
- Remove all leaves of T. Let the remaining tree be T1.
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Remove all leaves of T1. Let the remaining tree be T2.
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Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
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When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
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If Tk has only one node, that is the center of T. The diameter of T is 2k.
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If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.
Explanation:
We can compute the diameter of the tree T by a pruning procedure, starting at the leaves (external nodes).
- Remove all leaves of T. Let the remaining tree be T1.
-
Remove all leaves of T1. Let the remaining tree be T2.
-
Repeat the "remove" operation as follows: Remove all leaves of Ti. Let remaining tree be Ti+1.
-
When the remaining tree has only one node or two nodes, stop! Suppose now the remaining tree is Tk.
-
If Tk has only one node, that is the center of T. The diameter of T is 2k.
-
If Tk has two nodes, either can be the center of T. The diameter of T is 2k+1.
Answer:
(i) 12 V in series with 18 Ω.
(ii) 0.4 A; 1.92 W
(iii) 1,152 J
(iv) 18Ω — maximum power transfer theorem
Explanation:
<h3>(i)</h3>
As seen by the load, the equivalent source impedance is ...
10 Ω + (24 Ω || 12 Ω) = (10 +(24·12)/(24+12)) Ω = 18 Ω
The open-circuit voltage seen by the load is ...
(36 V)(12/(24 +12)) = 12 V
The Thevenin's equivalent source seen by the load is 12 V in series with 18 Ω.
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<h3>(ii)</h3>
The load current is ...
(12 V)/(18 Ω +12 Ω) = 12/30 A = 0.4 A . . . . load current
The load power is ...
P = I^2·R = (0.4 A)^2·(12 Ω) = 1.92 W . . . . load power
__
<h3>(iii)</h3>
10 minutes is 600 seconds. At the rate of 1.92 J/s, the electrical energy delivered is ...
(600 s)(1.92 J/s) = 1,152 J
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<h3>(iv)</h3>
The load resistance that will draw maximum power is equal to the source resistance: 18 Ω. This is the conclusion of the Maximum Power Transfer theorem.
The power transferred to 18 Ω is ...
((12 V)/(18 Ω +18 Ω))^2·(18 Ω) = 144/72 W = 2 W
Answer:
The radius of a wind turbine is 691.1 ft
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Explanation:
Given;
power generation potential (PGP) = 1000 kW
Wind speed = 5 mph = 2.2352 m/s
Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³
Radius of the wind turbine r = ?
Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²
Wind energy per unit mass of air = 2.517 J/kg
PGP = mass flow rate * energy per unit mass
PGP = ρ*A*V*e
r = 210.64 m = 691.1 ft
Thus, the radius of a wind turbine is 691.1 ft
PGP = CVᵃ
For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)
Let C = 0.4
PGP = Cvᵃ
take log of both sides
ln(PGP) = a*ln(CV)
a = ln(PGP)/ln(CV)
a = ln(1000)/ln(0.4 *2.2352) = 7.73
The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m
Harmonic excitation refers to a sinusoidal external force of a certain frequency applied to a system. ... Resonance occurs when the external excitation has the same frequency as the natural frequency of the system. It leads to large displacements and can cause a system to exceed its elastic range and fail structurally.
