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:
See explanation
Explanation:
The magnetic force is
F = qvB sin θ
We see that sin θ = 1, since the angle between the velocity and the direction of the field is 90º. Entering the other given quantities yields
F
=
(
20
×
10
−
9
C
)
(
10
m/s
)
(
5
×
10
−
5
T
)
=
1
×
10
−
11
(
C
⋅
m/s
)
(
N
C
⋅
m/s
)
=
1
×
10
−
11
N
Answer:
The white lined paper
Explanation:
The teacher is most likely putting the while line paper in jeopardy because of the detail process involved in taking care of the paper prior to the submission of the home work.
The fact that a mistake must not be visible due to the instruction of every erasures being thorough and clean. this can cause jeopardy to the paper.
Answer:
Welding, carpentry, masonry, construction worker, barber
Explanation:
The rate of gain for the high reservoir would be 780 kj/s.
A. η = 35%
W = 
W = 420 kj/s
Q2 = Q1-W
= 1200-420
= 780 kJ/S
<h3>What is the workdone by this engine?</h3>
B. W = 420 kj/s
= 420x1000 w
= 4.2x10⁵W
The work done is 4.2x10⁵W
c. 780/308 - 1200/1000
= 2.532 - 1.2
= 1.332kj
The total enthropy gain is 1.332kj
D. Q1 = 1200
T1 = 1000
<h3>Cournot efficiency = W/Q1</h3>
= 1200 - 369.6/1200
= 69.2 percent
change in s is zero for the reversible heat engine.
Read more on enthropy here: brainly.com/question/6364271
