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.
-
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.
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:a
a) Vo/Vi = - 3.4
b) Vo/Vi = - 14.8
c) Vo/Vi = - 1000
Explanation:
a)
R1 = 17kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
(Va - Vi)/5kΩ + (Va -Vo)/17kΩ = 0
sin we know Va≈Vb=0
so
-Vi/5kΩ + -Vo/17kΩ = 0
Vo/Vi = - 17k/5k
Vo/Vi = -3.4
║Vo/Vi ║ = 3.4 ( negative sign phase inversion)
b)
R2 = 74kΩ
for ideal op-amp
Va≈Vb=0 so Va=0
so
(Va-Vi)/5kΩ + (Va-Vo)74kΩ = 0
-Vi/5kΩ + -Vo/74kΩ = 0
Vo/Vi = - 74kΩ/5kΩ
Vo/Vi = - 14.8
║Vo/Vi ║ = 14.8 ( negative sign phase inversion)
c)
Also for ideal op-amp
Va≈Vb=0 so Va=0
Now for position 3 we apply nodal analysis we got at position 1
(Va - Vi)/5kΩ + (Va - Vo)/5000kΩ = 0 ( 5MΩ = 5000kΩ )
so
-Vi/5kΩ + -Vo/5000kΩ = 0
Vo/Vi = - 5000kΩ/5kΩ
Vo/Vi = - 1000
║Vo/Vi ║ = 1000 ( negative sign phase inversion)
Because of the skin depth effect, the current at high frequency tends to flow at very low depth from radius. Then at high frequency the effective cross section of the wire is narrower than at DC.
Fro example skin depth at 100 kHz is 0.206 mm (0.008”), a wire more thicker than AWG26 could be a waste of copper, better use a bunch of thin wire (Litz wire) to rise the Q factor.
Answer:

Explanation:
The mass inside the rigid tank before the high pressure stream enters is:



The final mass inside the rigid tank is:



The supplied air mass is:



Answer:
The break force that must be applied to hold the plane stationary is 12597.4 N
Explanation:
p₁ = p₂, T₁ = T₂


The heat supplied =
× Heating value of jet fuel
The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s
The heat supplied =
·
= 20 kg/s
The heat supplied = 20*
= 21,350 kJ/s
= 1.15 kJ/kg
T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K
p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa
p₃ = p₂ = 855 kPa
T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K
T₄ = 1413.3 - 204.88 = 1208.42 K

T₅ = 1208.42*(2/2.333) = 1035.94 K
= √(1.333*287.3*1035.94) = 629.87 m/s
The total thrust =
×
= 20*629.87 = 12597.4 N
Therefore;
The break force that must be applied to hold the plane stationary = 12597.4 N.