78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
Answer:
<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>
The number of atoms of lead required is 1.73x10²³.
Explanation:
To find the number of atoms of lead we need to find first the volume of the plate:

<u>Where</u>:
A: is the surface area = 160
t: is the thickness = 0.002
<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>
Now, using the density we can find the mass:

Finally, with the Avogadros number (
) and with the atomic mass (A) we can find the number of atoms (N):
Hence, the number of atoms of lead required is 1.73x10²³.
I hope it helps you!
Answer:
Stress corrosion cracking
Explanation:
This occurs when susceptible materials subjected to an environment that causes cracking effect by the production of folds and tensile stress. This also depends upon the nature of the corrosive environment.
Factors like high-temperature water, along with Carbonization and chlorination, static stress, and material properties.
Answer:
Tension in cable BE= 196.2 N
Reactions A and D both are 73.575 N
Explanation:
The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence
hence

Therefore, tension in the cable, 
Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then



Similarly,


Therefore, both reactions at A and D are 73.575 N
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