Answer:
Answer for the question:
Let Deterministic Quicksort be the non-randomized Quicksort which takes the first element as a pivot, using the partition routine that we covered in class on the quicksort slides. Consider another almost-best case for quicksort, in which the pivot always splits the arrays 1/3: 2/3, i.e., one third is on the left, and two thirds are on the right, for all recursive calls of Deterministic Quicksort. (a) Give the runtime recurrence for this almost-best case. (b) Use the recursion tree to argue why the runtime recurrence solves to Theta (n log n). You do not need to do big-Oh induction. (c) Give a sequence of 4 distinct numbers and a sequence of 13 distinct numbers that cause this almost-best case behavior. (Assume that for 4 numbers the array is split into 1 element on the left side, the pivot, and two elements on the right side. Similarly, for 13 numbers it is split with 4 elements on the left, the pivot, and 8 elements on the right side.)
is given in the attachment.
Explanation:
Answer:
<em>minimum required diameter of the steel linkage is 3.57 mm</em>
<em></em>
Explanation:
original length of linkage l = 10 m
force to be transmitted f = 2 kN = 2000 N
extension e = 5 mm= 0.005 m
maximum stress σ = 200 N/mm^2 = 
maximum stress allowed on material σ = force/area
imputing values,
200 = 2000/area
area = 2000/(
) = 
m^2
recall that area = 
= 
= 
= 
= 
m = 3.57 mm
<em>maximum diameter of the steel linkage d = 3.57 mm</em>
Answer:
Zero 1 = -1
Zero 2 = -3
Pole 1 = 0
Pole 2 = -2
Pole 3 = -4
Pole 4 = -6
Gain = 4
Explanation:
For any given transfer function, the general form is given as
T.F = k [N(s)] ÷ [D(s)]
where k = gain of the transfer function
N(s) is the numerator polynomial of the transfer function whose roots are the zeros of the transfer function.
D(s) is the denominator polynomial of the transfer function whose roots are the poles of the transfer function.
k [N(s)] = 4s² + 16s + 12 = 4[s² + 4s + 3]
it is evident that
Gain = k = 4
N(s) = (s² + 4s + 3) = (s² + s + 3s + 3)
= s(s + 1) + 3 (s + 1) = (s + 1)(s + 3)
The zeros are -1 and -3
D(s) = s⁴ + 12s³ + 44s² + 48s
= s(s³ + 12s² + 44s + 48)
= s(s + 2)(s + 4)(s + 6)
The roots are then, 0, -2, -4 and -6.
Hope this Helps!!!
Answer:
Ps=19.62N
Explanation:
The detailed explanation of answer is given in attached files.
