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:
Circle because it’s round and we all love round things
Answer:
Q=0.95 W/m
Explanation:
Given that
Outer diameter = 0.3 m
Thermal conductivity of material
![K= 0.055(1+2.8\times 10^{-3}T)\frac{W}{mK}](https://tex.z-dn.net/?f=K%3D%200.055%281%2B2.8%5Ctimes%2010%5E%7B-3%7DT%29%5Cfrac%7BW%7D%7BmK%7D)
So the mean conductivity
![K_m=0.055\left ( 1+2.8\times 10^{-3}T_m \right )](https://tex.z-dn.net/?f=K_m%3D0.055%5Cleft%20%28%201%2B2.8%5Ctimes%2010%5E%7B-3%7DT_m%20%5Cright%20%29)
![T_m=\dfrac{160+273+40+273}{2}](https://tex.z-dn.net/?f=T_m%3D%5Cdfrac%7B160%2B273%2B40%2B273%7D%7B2%7D)
![T_m=373 K](https://tex.z-dn.net/?f=T_m%3D373%20K)
![K_m=0.055\left ( 1+2.8\times 10^{-3}\times 373 \right )](https://tex.z-dn.net/?f=K_m%3D0.055%5Cleft%20%28%201%2B2.8%5Ctimes%2010%5E%7B-3%7D%5Ctimes%20373%20%5Cright%20%29)
![K_m=0.112 \frac{W}{mK}](https://tex.z-dn.net/?f=K_m%3D0.112%20%5Cfrac%7BW%7D%7BmK%7D)
So heat conduction through cylinder
![Q=kA\dfrac{\Delta T}{L}](https://tex.z-dn.net/?f=Q%3DkA%5Cdfrac%7B%5CDelta%20T%7D%7BL%7D)
![Q=0.112\times \pi \times 0.15^2\times 120](https://tex.z-dn.net/?f=Q%3D0.112%5Ctimes%20%5Cpi%20%5Ctimes%200.15%5E2%5Ctimes%20120)
Q=0.95 W/m
Answer:
Vc2= V(l+e) ^2/4
Vg2= V(l-e^2)/4
Explanation:
Conservation momentum, when ball A strikes Ball B
Where,
M= Mass
V= Velocity
Ma(VA)1+ Mg(Vg)2= Ma(Va)2+ Ma(Vg)2
MV + 0= MVg2
Coefficient of restitution =
e= (Vg)2- (Va)2/(Va)1- (Vg)1
e= (Vg)2- (Va)2/ V-0
Solving equation 1 and 2 yield
(Va)2= V(l-e) /2
(Vg)2= V(l+e)/2
Conservative momentum when ball b strikes c
Mg(Vg)2+Mc(Vc)1 = Mg(Vg)3+Mc(Vc)2
=> M[V(l+e) /2] + 0 = M(Vg)3 + M(Vc) 2
Coefficient of Restitution,
e= (Vc)2 - (Vg)2/(Vg)2- (Vc)1
=> e= (Vc)2 - (Vg)2/V(l+e) /2
Solving equation 3 and 4,
Vc2= V(l+e) ^2/4
Vg2= V(l-e^2)/4