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Oliga [24]
3 years ago
12

Estimate the daily carbon utilization to remove chlorobenzene from 1.0 MGD of ground water saturated with chlorobenzene. Assume

a chlorobenzene concentration of 5 mg/L is acceptable for discharge to a POTW.
Contaminant = chlorobenzene [C_6H_5CI]
Final concentration of chlorobcnzene (C_f) = 5 mg/L
Saturation concentration = 500 mg/L
Density = 1.107 at 20°
Initial concentration = 500 mg/L
Freundlich isotherm:q or C_s = 91*Cf^0.99
Engineering
1 answer:
Setler [38]3 years ago
5 0

Answer:

4.6398 kg/day

Explanation:

Please find attached file for complete answer solution.

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A binary star system consists of two stars of masses m1m1m_1 and m2m2m_2. The stars, which gravitationally attract each other, r
d1i1m1o1n [39]

Answer:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

Explanation:

The question is: <em>Find the magnitude of the centripetal acceleration of the star with mass m₂</em>

The <em>centripetal acceleration</em> is the quotient of the centripetal force and the mass.

                a_c=\dfrac{F_c}{m}

Thus, you can write the equations for each star:

     

       a_c_1=\dfrac{F_c_1}{m_1}

       a_c_2=\dfrac{F_c_2}{m_2}

As per Newton's third law, the centripetal forces are equal in magnitude. Then:

       a_c_1\times m_1=a_c_2\times m_2

Now you can clear a_c_2:

          a_c_2=\dfrac{a_c_1\times m_1}{m_2}

6 0
4 years ago
A ten-station transfer machine has an ideal cycle time of 30 sec. The frequency of line stops is 0.075 stops per cycle. When a l
goldfiish [28.3K]

Answer:

62.5%

Explanation:

We are given that

A ten-station transfer machine has ideal cycle time=30 sec=\frac{30}{60}=0.5 min

Frequency of line=0.075 stops per cycle

Average time=4 min

We have to determine the line efficiency

T_p=0.5+0.075(4)=0.5+0.3=0.8

Line efficiency=\frac{0.5}{0.8}\times 100=62.5%

Hence, the line efficiency=62.5%

8 0
4 years ago
Suppose there are 76 packets entering a queue at the same time. Each packet is of size 5 MiB. The link transmission rate is 2.1
tia_tia [17]

Answer:

938.7 milliseconds

Explanation:

Since the transmission rate is in bits, we will need to convert the packet size to Bits.

1 bytes = 8 bits

1 MiB = 2^20 bytes = 8 × 2^20 bits

5 MiB = 5 × 8 × 2^20 bits.

The formula for queueing delay of <em>n-th</em> packet is :  (n - 1) × L/R

where L :  packet size = 5 × 8 × 2^20 bits, n: packet number = 48 and R : transmission rate =  2.1 Gbps = 2.1 × 10^9 bits per second.

Therefore queueing delay for 48th packet = ( (48-1) ×5 × 8 × 2^20)/2.1 × 10^9

queueing delay for 48th packet = (47 ×40× 2^20)/2.1 × 10^9

queueing delay for 48th packet = 0.938725181 seconds

queueing delay for 48th packet = 938.725181 milliseconds = 938.7 milliseconds

4 0
3 years ago
Limited time only for christmas give yourself free 100 points YES YES muhahahahhaha
Setler79 [48]
Answer:

Thank you so much and may god bless you.
6 0
2 years ago
Read 2 more answers
What are the two types of chatter (vibration) in metal cutting operations?
ioda

Answer:

Forced vibrations and self-generated vibrations

Explanation:

Forced vibration is caused due to the motion of the cutting operation itself it is common in operations that have interrupted cutting like milling.

The size of the chips are related to self-generated vibrations. If larger chips are formed this means higher depth of cut is given. If smaller chips are formed this means lower depth of cut is given. When we give excess depth of cut then self-generated vibrations are produced.

5 0
4 years ago
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