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:
Distribution factor P = =38.33
V = 7.826 ml
Explanation:
given details:
BOD =230 mg/l
DO inital = 8.0mg/l
DO final = 2.0mg/l
we know
BOD = [DO inital -DO final] * distribution factor
230 = [8 - 2] D.F
Distribution factor P
Distribution factor P = =38.33
THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is
distribution factor
V = 7.826 ml
Answer:
7.615 kW
Explanation:
Solution in pen paper form in the attachment section
Answer:
There are 2 expected readings greater than 2.70 V
Solution:
As per the question:
Total no. of readings, n = 60 V
Mean of the voltage,
standard deviation,
Now, to find the no. of readings greater than 2.70 V, we find:
The probability of the readings less than 2.70 V, :
Now, from the Probability table of standard normal distribution:
Now,
Now, for the expected no. of readings greater than 2.70 V:
No. of readings expected to be greater than 2.70 V =
No. of readings expected to be greater than 2.70 V = ≈ 2