Solution :
Given :
k = 0.5 per day


Volume, V 
Now, input rate = output rate + KCV ------------- (1)
Input rate 


The output rate 
= ( 40 + 0.5 ) x C x 1000

Decay rate = KCV
∴
= 1.16 C mg/s
Substituting all values in (1)

C = 4.93 mg/L
Answer:
A) max factored load ( pv = 1.4 * 18 ) = 25.2 kips
B) max load factored load = ( Pa = 18 + 2 ) = 20 kips
Explanation:
service dead load = 18 kips
service live load = 2 kips
A) Determine the maximum factored load and controlling AISC load combination
max factored load ( pv = 1.4 * 18 ) = 25.2 kips
DL = 18 kips
LL = 2 kips
B) Determine the max load and controlling AISC load combination
max load factored load = ( Pa = 18 + 2 ) = 20 kips
attached below
The answer is most likely Biological because insects and other organisms thrive in stagnant water.
Answer:
C = 59.17 nF
Q = 2.6
Explanation:
given data
frequencies = 40k Hz
frequencies = 90k Hz
solution
we take here R, L C take in series
so cut off frequency is express as
Wc1 =
= 40000
wc2 =
= 90000
so here
wc2 - wc1 will be
wc2 - wc1 = 90000 - 40000 = 50000
so
= 50000
we consider here R is 500
so L =
L = 10 m H
and here total cut off frequency is
total cut off frequency = 40000 + 90000 = 130000
so capacitance will be
capacitance C is =
so C = 59.17 nF
quality factor Q will be
Q =
Q = 2.6
Fast cracking sound is the answer