Answer:
Flow energy is defined as, flow energy is the energy needed to push fluids into control volume and it is the amount of work done required to push the entire fluid. It is also known as flow work. Flow energy is not the fundamental quantities like potential and kinetic energy.
Fluid at state of rest do not possess any flow energy. It is mostly converted into internal energy as, rising in the fluid temperature.
Carbonation is more of a healer to the engine
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 )
Answer: (a) 9.00 Mega Newtons or 9.00 * 10^6 N
(b) 17.1 m
Explanation: The length of wall under the surface can be given by
The average pressure on the surface of the wall is the pressure at the centeroid of the equilateral triangular block which can be then be calculated by multiplying it with the Plate Area which will provide us with the Resultant force.
![F(resultant) = Pavg ( A) = (Patm + \rho g h c)*A \\= [100000 N/m^2 + (1000 kg/m^3 * 9.81 m/s^2 * 25m/2)]* (140*25m/sin60)\\= 8.997*10^8 N \\= 9.0*10^8 N](https://tex.z-dn.net/?f=F%28resultant%29%20%3D%20Pavg%20%28%20A%29%20%3D%20%28Patm%20%2B%20%20%5Crho%20g%20h%20c%29%2AA%20%5C%5C%3D%20%5B100000%20N%2Fm%5E2%20%2B%20%281000%20kg%2Fm%5E3%20%2A%209.81%20m%2Fs%5E2%20%2A%2025m%2F2%29%5D%2A%20%28140%2A25m%2Fsin60%29%5C%5C%3D%208.997%2A10%5E8%20N%20%5C%5C%3D%209.0%2A10%5E8%20N)
Noting from the Bernoulli equation that
From the second image attached the distance of the pressure center from the free surface of the water along the surface of the wall is given by:
Substituting the values gives us the the distance of the surface to be equal to = 17.1 m
