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:
(a)27.12 N-m (b) 69.96 
(c) 1.27 mm/sec
Explanation:
We have to convert
(a) 20 lb-ft to N-m
We know that 1 lb = 4.45 N
So 20 lb = 20×4.45 = 89 N
1 feet = 0.3048 m
So 
(b) 450 
to KN/
We know that 1 lb = 4.45 N = 0.0044 KN
So 
(c) 15 ft/hr to mm/sec
We know that 1 feet = 0.3048 m = 304.8 mm
And 1 hour = 60×60=3600 sec
So 
Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
<u>Surface Charge Density = 26.55 x 10^-12 C/m²</u>
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
Answer:
rate of makeup water added is 367.75 kg/s
Explanation:
given data
output power = 600 MW
efficiency = 36 %
waste heat = 15%
heat taken away in cooling tower = 85%
to find out
At what rate must 15ºC makeup water be provided from the river to offset the water lost in the cooling tower
solution
input power = 
input power = 
input power = 1666.67 MW
total heat = input - output
total heat = 1666.67 - 600
total heat = 1066.67 MW
and we know
15% of waste heat is released to atmosphere
so Q atm = 0.15 × 1066.667
Q atm = 160 MW
and
and 85% of heat is taken away in the cooling water so
Q cooling tower = 0.85 × 1066.667
Q cooling tower = 906.66 MW
so
we know at 15ºC from saturated water tables hfg will be
hfg = 2465.4 kJ/kg
so rate of water lost in cooling tower added by means of make up water that is
Q cooling tower = m × hfg
906.66 × 1000 = m × 2465.4
m = 367.75 kg/s
so rate of makeup water added is 367.75 kg/s
