Answer:
B. G = 333 mS, B = j250 mS
Explanation:
impedance of a circuit element is Z = (3 + j4) Ω
The general equation for impedance
Z = (R + jX) Ω
where
R = resistance in ohm
X = reactance
R = 3Ω X = 4Ω
Conductance = 1/R while Susceptance = 1/X
Conductance = 1/3 = 0.333S
= 333 mS
Susceptance = 1/4 = 0.25S
= 250mS
The right option is B. G = 333 mS, B = j250 mS
Answer:
![X_B = 1.8 \times 10^{-3} m = 1.8 mm](https://tex.z-dn.net/?f=X_B%20%3D%201.8%20%5Ctimes%2010%5E%7B-3%7D%20m%20%3D%201.8%20mm)
Explanation:
Given data:
Diffusion constant for nitrogen is ![= 1.85\times 10^{-10} m^2/s](https://tex.z-dn.net/?f=%3D%201.85%5Ctimes%2010%5E%7B-10%7D%20m%5E2%2Fs)
Diffusion flux ![= 1.0\times 10^{-7} kg/m^2-s](https://tex.z-dn.net/?f=%3D%201.0%5Ctimes%2010%5E%7B-7%7D%20kg%2Fm%5E2-s)
concentration of nitrogen at high presuure = 2 kg/m^3
location on which nitrogen concentration is 0.5 kg/m^3 ......?
from fick's first law
![J = D \frac{C_A C_B}{X_A X_B}](https://tex.z-dn.net/?f=J%20%3D%20D%20%5Cfrac%7BC_A%20C_B%7D%7BX_A%20X_B%7D)
Take C_A as point on which nitrogen concentration is 2 kg/m^3
![x_B = X_A + D\frac{C_A -C_B}{J}](https://tex.z-dn.net/?f=x_B%20%3D%20X_A%20%2B%20D%5Cfrac%7BC_A%20-C_B%7D%7BJ%7D)
Assume X_A is zero at the surface
![X_B = 0 + ( 12\times 10^{-11} ) \frac{2-0.5}{1\times 10^{-7}}](https://tex.z-dn.net/?f=X_B%20%3D%200%20%2B%20%28%2012%5Ctimes%2010%5E%7B-11%7D%20%29%20%5Cfrac%7B2-0.5%7D%7B1%5Ctimes%2010%5E%7B-7%7D%7D)
![X_B = 1.8 \times 10^{-3} m = 1.8 mm](https://tex.z-dn.net/?f=X_B%20%3D%201.8%20%5Ctimes%2010%5E%7B-3%7D%20m%20%3D%201.8%20mm)
Answer:
True
Explanation:
Nikola Tesla defeated Thomas Edison in the AC/DC battle of electric current.
Answer:
a)We know that acceleration a=dv/dt
So dv/dt=kt^2
dv=kt^2dt
Integrating we get
v(t)=kt^3/3+C
Puttin t=0
-8=C
Putting t=2
8=8k/3-8
k=48/8
k=6
Answer:
a) Ef = 0.755
b) length of specimen( Lf )= 72.26mm
diameter at fracture = 9.598 mm
c) max load ( Fmax ) = 52223.24 N
d) Ft = 51874.67 N
Explanation:
a) Determine the true strain at maximum load and true strain at fracture
True strain at maximum load
Df = 9.598 mm
True strain at fracture
Ef = 0.755
b) determine the length of specimen at maximum load and diameter at fracture
Length of specimen at max load
Lf = 72.26 mm
Diameter at fracture
= 9.598 mm
c) Determine max load force
Fmax = 52223.24 N
d) Determine Load ( F ) on the specimen when a true strain et = 0.25 is applied during tension test
F = 51874.67 N
attached below is a detailed solution of the question above