The 10 foot wide circle quarter gate AB is articulated at A. Determine the contact force between the gate and the smooth surface
1 answer:
Answer:
Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)
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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:
diameter of the sprue at the bottom is 1.603 cm
Explanation:
Given data;
Flow rate, Q = 400 cm³/s
cross section of sprue: Round
Diameter of sprue at the top = 3.4 cm
Height of sprue, h = 20 cm = 0.2 m
acceleration due to gravity g = 9.81 m/s²
Calculate the velocity at the sprue base
= √2gh
we substitute
= √(2 × 9.81 m/s² × 0.2 m )
= 1.98091 m/s
= 198.091 cm/s
diameter of the sprue at the bottom will be;
Q = AV = (π/4) ×
= √(4Q/π
)
we substitute our values into the equation;
= √(4(400 cm³/s) / (π×198.091 cm/s))
= 1.603 cm
Therefore, diameter of the sprue at the bottom is 1.603 cm