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.
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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.
-
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:
Using python
num_boys = int(input("Enter number of boys :"))
num_girls = int(input("Enter number of girls :"))
budget = int(input("Enter the number of dollars spent per school year :"))
try:
dollarperstudent = budget/(num_boys+num_girls)
print("Dollar spent per student : "+str(dollarperstudent))#final result
except ZeroDivisionError:
print("unavailable")
Answer:
k = 4.21 * 10⁻³(L/(mol.s))
Explanation:
We know that
k = Ae
------------------- euqation (1)
K= rate constant;
A = frequency factor = 4.36 10^11 M⁻¹s⁻¹;
E = activation energy = 93.1kJ/mol;
R= ideal gas constant = 8.314 J/mol.K;
T= temperature = 332 K;
Put values in equation 1.
k = 4.36*10¹¹(M⁻¹s⁻¹)e![^{[(-93.1*10^3)(J/mol)]/[(8.314)(J/mol.K)(332K)}](https://tex.z-dn.net/?f=%5E%7B%5B%28-93.1%2A10%5E3%29%28J%2Fmol%29%5D%2F%5B%288.314%29%28J%2Fmol.K%29%28332K%29%7D)
k = 4.2154 * 10⁻³(M⁻¹s⁻¹)
here M =mol/L
k = 4.21 * 10⁻³((mol/L)⁻¹s⁻¹)
or
k = 4.21 * 10⁻³((L/mol)s⁻¹)
or
k = 4.21 * 10⁻³(L/(mol.s))
Answer:
The fluid level difference in the manometer arm = 22.56 ft.
Explanation:
Assumption: The fluid in the manometer is incompressible, that is, its density is constant.
The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.
And P(gage) = ρgh
ρ = density of the manometer fluid = 60 lbm/ft³
g = acceleration due to gravity = 32.2 ft/s²
ρg = 60 × 32.2 = 1932 lbm/ft²s²
ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³
h = fluid level difference between the two arms of the manometer = ?
P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²
1353.6 = ρg × h = 60 lbf/ft³ × h
h = 1353.6/60 = 22.56 ft
A diagrammatic representation of this setup is presented in the attached image.
Hope this helps!
