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nignag [31]
3 years ago
9

A sample of wastewater is diluted 10 times. The diluted solution has an ultimate biochemical oxygen demand (BOD), Lo, of 30 mg/L

and a reaction rate constant, k, of 0.20/d. The values of ultimate BOD and the reaction rate constant of the original wastewater are most nearly _____.
Engineering
1 answer:
zzz [600]3 years ago
3 0

Answer:

474.59 mg/L

Explanation:

Given that

BOD = 30 mg/L

Original BOD  = 30 mg/L × dilution factor

Original BOD  = 30 mg/L  × 10 = 300 mg/L

L_o = \frac{BOD}{1-e^{-5t}}

here L_o is the ultimate BOD ; BOD is the  biochemical oxygen demand ;  t = 0.20 /day

L_o = \frac{300}{1-e^{-5(0.20)}}

L_o = 474.59 \ mg/L

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In the 5 Code of Federal Regulations (C.F.R.), it is recommended that an individual has security awareness training before s/he
GarryVolchara [31]

Answer: A it is true.

Explanation: It is true under Subpart C—Information Security Responsibilities for Employees who Manage or Use Federal Information Systems

Number 3 and 4

(3) Program and functional managers must receive training in information security basics; management and implementation level training in security planning and system/application security management; and management and implementation level training in system/application life cycle management, risk management, and contingency planning.

(4) Chief Information Officers (CIOs), IT security program managers, auditors, and other security-oriented personnel (e.g., system and network administrators, and system/application security officers) must receive training in information security basics and broad training in security planning, system and application security management, system/application life cycle management, risk management, and contingency planning.

6 0
3 years ago
Read 2 more answers
You leave your house at 5:02 PM and run 20 yards down the street. You don't realize that you forgot your Wallet back at home and
Sidana [21]

Answer:

The average velocity is 0.203 m/s

Explanation:

Given;

initial displacement, x₁ = 20 yards = 18.288 m

final displacement, x₂ = ¹/₃ x 18.288 = 6.096 m

change in time between 5:02 PM and 5:03 PM, Δt = 3 mins - 2 mins = 1 min = 60 s

The average velocity is given by;

V = change in displacement / change in time

V = (x₂ - x₁) / Δt

V = (18.288 - 6.096) / 60

V = 0.203 m/s

Therefore, the average velocity is 0.203 m/s

7 0
3 years ago
So in my settings i set it to send notifications too my email so ik when smtn happens but it doesn't even send stuff too it.....
attashe74 [19]

Answer:

Did you make sure you did it and your notifications are on

3 0
3 years ago
Consider tests of an unswept wing that spans the wind tunnel and whose airfoil section is NACA 23012. Since the wing model spans
Dominik [7]

Answer:

Check the explanation

Explanation:

to know the lift per unit span (N/m) that is expected to be measured when the wing attack angle is 4°

as well as the corresponding section lift coefficient and die moment coefficient .

Kindly check the attached image below to see the step by step explanation to the above question.

3 0
3 years ago
To provide some perspective on the dimensions of atomic defects, consider a metal specimen that has a dislocation density of 105
GenaCL600 [577]

Answer:

62.14\ \text{miles}

6213727.37\ \text{miles}

Explanation:

The distance of the chain would be the product of the dislocation density and the volume of the metal.

Dislocation density = 10^5\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^5\times 1000=10^8\ \text{mm}\\ =10^5\ \text{m}

1\ \text{mile}=1609.34\ \text{m}

\dfrac{10^5}{1609.34}=62.14\ \text{miles}

The chain would extend 62.14\ \text{miles}

Dislocation density = 10^{10}\ \text{mm}^{-2}

Volume of the metal = 1000\ \text{mm}^3

10^{10}\times 1000=10^{13}\ \text{mm}\\ =10^{10}\ \text{m}

\dfrac{10^{10}}{1609.34}=6213727.37\ \text{miles}

The chain would extend 6213727.37\ \text{miles}

3 0
3 years ago
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