Answer:
well my bff is trans so yes I guess xd
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:
Not knowing the units the tolerance is 0.02. I would presume mm but hopefully your question has more detail.
Explanation:
The tolerance is the portion after the main dimension (+/- 0.02). In our case we have bilateral tolerance since there is tolerance in both directions (positive and negative). If you were building a part the acceptable range would be 2.98 to 3.02 based on the tolerance provided.
Answer:
μ=0.329, 2.671 turns.
Explanation:
(a) ln(T2/T1)=μβ β=angle of contact in radians
take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.
T2=5000 lb and T1=80 lb
we have two full turns which makes total angle of contact=4π radians
μ=ln(T2/T1)/β=(ln(5000/80))/4π
μ=0.329
(b) using the same relation as above we will now compute the angle of contact.
take greater tension as T2 and smaller as T1.
T2=20000 lb T1=80 lb μ=0.329
β=ln(20000/80)/0.329=16.7825 radians
divide the angle of contact by 2π to obtain number of turns.
16.7825/2π =2.671 turns
