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grigory [225]
3 years ago
11

A dairy in Fresno County discharges its wastewater to a circular treatment pond. The dairy wants to expand its operations, which

will increase the influent flow but not the influent concentration. What is the maximum influent flow that can be handled by the existing pond in the future without exceeding the legally allowed effluent concentration of 30 mg/L.
Data on the existing facilities and flows are shown below. The soil is sandy and there is considerable seepage. The wastewater pollutants are particulate in form and degrade with a rate constant of 0.25 d-1.

Current conditions

influent Flow = 3200 m3/d

Influent concentration = 220 g/m3

Pond effluent concentration = 22 mg/L

Pond volume = 120,000 m3

Evaporation negligible

Seepage = 1500 m3/d
Engineering
1 answer:
Strike441 [17]3 years ago
4 0

Answer:

420.69 mg

Explanation:

i just did that last week

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Answer:

The operating system

Explanation:

The job of the operating system is to manage system resources allowing the abstraction of the hardware, providing a simple user interface for the user.  The operating system is also responsible for handling application's access to system resources.

For this purpose, the operating system allows a user to run applications on their computing device.

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3 years ago
3. Which of the following is an example of qualitative data?
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Answer:

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3 years ago
Consider a resistor made of pure silicon with a cross-sectional area pf 0.5 μm2, and a length of 50 μm. What is the resistance o
lukranit [14]

Answer: 24 pA

Explanation:

As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.

Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵  Ω  cm.

The resistance R of a given resistor, is expressed by the following formula:

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Replacing by the values for resistivity, L and A, we have

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R = 2.1. 10¹¹ Ω

Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:

I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA

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Airida [17]
I think downwards as that's how most saw's work.
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