A.Ensure all circuits are de-energized before beginning work
Answer:
See the attachment
Verification shown in explanation
Explanation:
wc= 1/(RC)
The value of R and C shown in the attachment can be calculated as follows:
R=100 ohms
1000= 1/(100×C)
C= 10 microfarads
Gain= R/√(R²+Xc²)
at f=500, wc= 3142.59 rad/s
Xc= 1/(2πfC)
Xc=31.83
Gain= 100/√(100²+31.83²)
Gain= 0.953
at fc= 100, wc= 628.32 rad/s
Xc= 159.15
gain= 100/√(100²+ 159.15²)
gain= 0.532
As see form calculations that gain at low frequencies is smaller and gain at very high frequencies is near to 1. This means that at high frequencies Vin≈ Vout
Hence this filter allows high frequencies to pass but blocks low frequencies by lowering the gai
Answer:
36.0 kpsi2.
Explanation:
From the question
Sut=110 kpsi
Se’=0.5(110)=55 kpsi
For surface factor ka,
a=2.70, b= -0.265, ka=a(Sut)b=2.70(110)-0.265=0.777
Assuming the worst case for size factor kb, and since 0.11 less than or equal to d less than or equal to 2in,
Therefore:
kb=0.879d-0.107 = 0.879(1.5)-0.107=0.842
Hence, the endurance strength is Se= ka kb Se’ = 36.0 kpsi2.