Answer:
ω=314.15 rad/s.
0.02 s.
Explanation:
Given that
Motor speed ,N= 3000 revolutions per minute
N= 3000 RPM
The speed of the motor in rad/s given as

Now by putting the values in the above equation

ω=314.15 rad/s
Therefore the speed in rad/s will be 314.15 rad/s.
The speed in rev/sec given as

ω= 50 rev/s
It take 1 sec to cover 50 revolutions
That is why to cover 1 revolution it take

Answer:
a) it is periodic
N = (20/3)k = 20 { for K =3}
b) it is Non-Periodic.
N = ∞
c) x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
We know that In Discrete time system, complex exponentials and sinusoidal signals are periodic only when ( 2π/w₀) ratio is a rational number.
then the period of the signal is given as
N = ( 2π/w₀)K
k is least integer for which N is also integer
Now, if x(n) = x1(n) + x2(n) and if x1(n) and x2(n) are periodic then x(n) will also be periodic; given N = LCM of N1 and N2
now
a) cos(2π(0.15)n)
w₀ = 2π(0.15)
Now, 2π/w₀ = 2π/2π(0.15) = 1/(0.15) = 1×20 / ( 0.15×20) = 20/3
so, it is periodic
N = (20/3)k = 20 { for K =3}
b) cos(2n);
w₀ = 2
Now, 2π/w₀ = 2π/2) = π
so, it is Non-Periodic.
N = ∞
c) cos(π0.3n) + cos(π0.4n)
x(n) = x1(n) + x2(n)
x1(n) = cos(π0.3n)
x2(n) = cos(π0.4n)
so
w₀ = π0.3
2π/w₀ = 2π/π0.3 = 2/0.3 = ( 2×10)/(0.3×10) = 20/3
∴ N1 = 20
AND
w₀ = π0.4
2π/w₀ = 2π/π0. = 2/0.4 = ( 2×10)/(0.4×10) = 20/4 = 5
∴ N² = 5
so, x(n) is periodic
N = LCM ( 5, 20 )
Explanation:
Superheater has two types of parts which are:
- The primary super-heater
- The secondary super-heater
Primary super-heater is first heater which is passed by the steam after steam comes out of steam drum.
After steam is heated on super primary heater, then the steam is passed on secondary super-heater so to be heated again. Thus, on secondary super-heater, the steam formed is hottest steam among others.
Steam from secondary super-heater which becomes the superheated steam, flow to rotate the High-Pressure Turbine.
Independent variable if I’m not mistaken