A. I believe, lmk if I’m right
Answer:
The given grammar is :
S = T V ;
V = C X
X = , V | ε
T = float | double
C = z | w
1.
Nullable variables are the variables which generate ε ( epsilon ) after one or more steps.
From the given grammar,
Nullable variable is X as it generates ε ( epsilon ) in the production rule : X -> ε.
No other variables generate variable X or ε.
So, only variable X is nullable.
2.
First of nullable variable X is First (X ) = , and ε (epsilon).
L.H.S.
The first of other varibles are :
First (S) = {float, double }
First (T) = {float, double }
First (V) = {z, w}
First (C) = {z, w}
R.H.S.
First (T V ; ) = {float, double }
First ( C X ) = {z, w}
First (, V) = ,
First ( ε ) = ε
First (float) = float
First (double) = double
First (z) = z
First (w) = w
3.
Follow of nullable variable X is Follow (V).
Follow (S) = $
Follow (T) = {z, w}
Follow (V) = ;
Follow (X) = Follow (V) = ;
Follow (C) = , and ;
Explanation:
Answer:
0.024 m = 24.07 mm
Explanation:
1) Notation
= tensile stress = 200 Mpa
= plane strain fracture toughness= 55 Mpa
= length of a surface crack (Variable of interest)
2) Definition and Formulas
The Tensile strength is the ability of a material to withstand a pulling force. It is customarily measured in units (F/A), like the pressure. Is an important concept in engineering, especially in the fields of materials and structural engineering.
By definition we have the following formula for the tensile stress:
(1)
We are interested on the minimum length of a surface that will lead to a fracture, so we need to solve for
Multiplying both sides of equation (1) by
(2)
Sequaring both sides of equation (2):
(3)
Dividing both sides by we got:
(4)
Replacing the values into equation (4) we got:
3) Final solution
So the minimum length of a surface crack that will lead to fracture, would be 24.07 mm or more.
Answer:
X_cp = c/2
Explanation:
We are given;
Chord = c
Angle of attack = α
p u (s) = c 1
p1(s)=c2,
and c2 > c1
First of all, we need to find the resultant normal force on the plate and the total moment about leading edge.
I've attached the solution