Answer:
Attenuators are electrical components designed to reduce the amplitude of a signal passing through the component, without significantly degrading the integrity of that signal. They are used in RF and optical applications
Explanation:
Answer:
D center punch
Explanation:
it has a fine point so it is good with accuracy
Answer:
8.5 days
Explanation:
Given data :
Flow ( Q ) = 2.6 MGD = 11819.834 m^3/day
BOD = 131 mg/L
BOD loading rate = 35 Ibs/1000 ft^3 per day = 0.5606 kg/m^3/day
<u>Calculate the sludge age of the facility </u>
Given the BOD applied to the aeration tank = 11819.834 m^3/day * 131mg/l
= 1548.398 kg/day
first calculate the volume of the aeration tank
V = BOD applied / BOD loading rate
V = 1548.398 / 0.5606 = 2762.03 m^3
Hydraulic Detention time = V / Q
= 2762.03 / 11819.834 = 0.2336 day = 5.6 hour
next : determine the mass rate of the waste
= 44% * 0.5606 kg/m^3/day
= 0.2466 kg/m^2/day
finally determine the sludge age
= V * Xt / ∅w * R
= ( 2762.03 m^3 * 2100 * 10^-3 ) / ( 0.2466 * 2762.03 kg/day )
= 8.5 days
A.an internationally standard used to measure a physical quality
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
