Answer:
attached below
Explanation:
a) G(s) = 1 / s( s+2)(s + 4 )
Bode asymptotic magnitude and asymptotic phase plots
attached below
b) G(s) = (s+5)/(s+2)(s+4)
phase angles = tan^-1 w/s , -tan^-1 w/s , tan^-1 w/4
attached below
c) G(s)= (s+3)(s+5)/s(s+2)(s+4)
solution attached below
Answer:
Explanation:
Pie charts generally should have no more than eight segments.
Answer:
Q= 4.6 × 10⁻³ m³/s
actual velocity will be equal to 8.39 m/s
Explanation:
density of fluid = 900 kg/m³
d₁ = 0.025 m
d₂ = 0.05 m
Δ P = -40 k N/m²
C v = 0.89
using energy equation
![\dfrac{P_1}{\gamma}+\dfrac{v_1^2}{2g} = \dfrac{P_2}{\gamma}+\dfrac{v_2^2}{2g}\\\dfrac{P_1-P_2}{\gamma}=\dfrac{v_2^2-v_1^2}{2g}\\\dfrac{-40\times 10^3\times 2}{900}=v_2^2-v_1^2](https://tex.z-dn.net/?f=%5Cdfrac%7BP_1%7D%7B%5Cgamma%7D%2B%5Cdfrac%7Bv_1%5E2%7D%7B2g%7D%20%3D%20%5Cdfrac%7BP_2%7D%7B%5Cgamma%7D%2B%5Cdfrac%7Bv_2%5E2%7D%7B2g%7D%5C%5C%5Cdfrac%7BP_1-P_2%7D%7B%5Cgamma%7D%3D%5Cdfrac%7Bv_2%5E2-v_1%5E2%7D%7B2g%7D%5C%5C%5Cdfrac%7B-40%5Ctimes%2010%5E3%5Ctimes%202%7D%7B900%7D%3Dv_2%5E2-v_1%5E2)
under ideal condition v₁² = 0
v₂² = 88.88
v₂ = 9.43 m/s
hence discharge at downstream will be
Q = Av
Q =
Q =
Q= 4.6 × 10⁻³ m³/s
we know that
![C_v =\dfrac{actual\ velocity}{theoretical\ velocity }\\0.89 =\dfrac{actual\ velocity}{9.43}\\actual\ velocity = 8.39m/s](https://tex.z-dn.net/?f=C_v%20%3D%5Cdfrac%7Bactual%5C%20velocity%7D%7Btheoretical%5C%20velocity%20%7D%5C%5C0.89%20%3D%5Cdfrac%7Bactual%5C%20velocity%7D%7B9.43%7D%5C%5Cactual%5C%20velocity%20%3D%208.39m%2Fs)
hence , actual velocity will be equal to 8.39 m/s
Answer:
True
Explanation:
Tensile testing which is also referred to as tension testing is a process which materials are subjected to so as to know how well it can be stretched before it reaches breaking point. Hence, the statement in the question is true