Answer:
The correct answer is option (A) 0.060 uF
Note: Kindly find an attached image of the complete question below
Sources: The complete question was well researched from Quizlet.
Explanation:
Solution
Given that:
C₁ = 0.1 μF
C₂ =0.22 μF
C₃ = 0.47 μF
In this case, C₁, C₂ and C₃ are in series
Thus,
Their equivalent becomes:
1/Ceq = (1/C₁ + 1/C₂ +1/C₃
1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]
1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]
1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)
1/Ceq =[(0.1724)/(0.01034)]
1/Ceq = [(16.67)]
1/Ceq =(1/16.67) = 0.059μf
Ceq = 0.059μf ≈ 0.060μf
Therefore the equivalent capacitance of the three series capacitors is 0.060μf
Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet 
= 96 kPa
the pressure at the exit 
= ???
the pressure 
= 1000 kg/m³
∴
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
Answer:
1.0MG
Explanation:
to solve this problem we use this formula
S₀-S/t = ksx --- (1)
the values have been given as
concentration = S₀ = 250mg
effluent concentration = S= 10mg
value of K = 0.04L/day
x = 3000 mg
when we put these values into this equation,
250-10/t = 0.04x10x3000
240/t = 1200
we cross multiply from this stage
240 = 1200t
t = 240/1200
t = 0.2
remember the question says that 5MGD is required to be treated
so the volume would be
v = 0.2x5
= 1.0 MG
