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3 years ago

9

A sample of municipal sewage is diluted to 1% by volume prior to running a BOD5 analysis. After 5 days the oxygen consumption is

determined to be 2.00 mg · L−1. What is the BOD5 of the sewage

1 answer:

liberstina [14]3 years ago

4 0

Answer:

BOD5 = 200 mg/L

Explanation:

given data

diluted = 1% = 0.01

time = 5 day

oxygen consumption = 2.00 mg · L−1

solution

we get here BOD5 that is BOD after 5 day

and here total volume is 100% = 1

so dilution factor is $\frac{100}{1}$ = 100

so BOD5 is

BOD5 = oxygen consumption × dilution factor

BOD5 = 2 × 100

BOD5 = 200 mg/L

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3 years ago

1: A baseball is hit 4 feet above the ground leaves the bat with an initial speed of 98 ft/sec at an angle of 0 45 is caught by

zvonat [6]

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$To \ find \ the \ distance \ of \ the \ ball \ from \ the \ home \ plate. \\ \\ From \ the \ given \ information:$

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So;

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3 years ago

If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less

liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

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A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

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Where

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E = 11.68*8.85x10¹⁴ f/cm

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The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

$Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm$

7 0

3 years ago