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3 years ago

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A BOD test is to be run on a sample of wastewater that has a five-day BOD of 230 mg/L. If the initial DO of a mix of distilled w

ater and wastewater is 8.0 mg/L, and the test requires a decrease in DO of at least 2.0 mg/L with at least 2.0 mg/L of DO remaining at the end of the five days, what range of dilution factors (P) would produce acceptable results? In 300 mL bottles, what range of wastewater volumes could be used?

1 answer:

Alexxandr [17]3 years ago

6 0

Answer:

Distribution factor P = =38.33

V = 7.826 ml

Explanation:

given details:

BOD =230 mg/l

DO inital = 8.0mg/l

DO final = 2.0mg/l

we know

BOD = [DO inital -DO final] * distribution factor

230 = [8 - 2] D.F

Distribution factor P $= \frac{230}{6}$

Distribution factor P = =38.33

THE RANGE OF WASTE WATER VOLUME IN 300 ml bottle is

distribution factor $= \frac{300}{V}$

$V = \frac{300}{38.33}$

V = 7.826 ml

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Answer:

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Explanation:

Original diameter $d_{o}$ = 30 mm

Final diameter $d_{1}$ = 30.04 mm

Change in diameter Δd = 0.04 mm

Final length $l_{1}$ = 105.20 mm

Elastic modulus E = 65.5 G pa = 65.5 × $10^{3}$ M pa

Shear modulus G = 25.4 G pa = 25.4 × $10^{3}$ M pa

We know that the relation between the shear modulus & elastic modulus is given by

$G = \frac{E}{2(1 + \mu)}$

$25.5 = \frac{65.5}{2 (1 + \mu)}$

$\mu = 0.28$

This is the value of possion's ratio.

We know that the possion's ratio is given by

$\mu = \frac{\frac{0.04}{30} }{\frac{change \ in \ length}{l_{o} } }$

${\frac{change \ in \ length}{l_{o} } = \frac{\frac{0.04}{30} }{0.28}$

${\frac{change \ in \ length}{l_{o} } =$ 0.00476

$\frac{l_{1} - l_{o} }{l_{o} } = 0.00476$

$\frac{l_{1} }{l_{o} } = 1.00476$

Final length $l_{o}$ = 105.2 m

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$l_{o} = 104.7 mm$

This is the original length of the specimen.

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3 years ago

You are in charge of ordering the concrete for a basement wall concrete pour. The wall forms are all set up and ready. The wall

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Answer:

189.15cy

Explanation:

To understand this problem we need to understand as well the form.

It is clear that there is four wall, two short and two long.

The two long are $\rightarrow 120ft5in+2(10ft)$

The two long are $\rightarrow 122ft1in=122.08ft$

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The height and the thickness are 14ft and 0.83ft respectively.

So we only calculate the Quantity of concrete,

$Q_c = [(2*122.08)+(2*86-375)]*14*0.833\\Q_c=4864.02ft^3$

That in cubic yards is equal to $180.15 (1cy=27ft^3)$

Hence, we need order 5% plus that represent with the quantity

$Q_{ordered}=1.05*180.15=189.15cy$

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2 years ago

A 3.52 kg steel ball is tossed upward from a height of 6.93 meters above the floor with a vertical velocity of 2.99 m/s. What is

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Answer : The final velocity of the ball is, 12.03 m/s

Explanation :

By the 3rd equation of motion,

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u = initial velocity = 2.99 m/s

v = final velocity = ?

a = acceleration = $9.8m/s^2$

Now put all the given values in the above equation, we get the final velocity of the ball.

$v^2-(2.99m/s)^2=2\times (9.8m/s^2)\times (6.93m)$

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Thus, the final velocity of the ball is, 12.03 m/s

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A double-threaded Acme stub screw of 2-in. major diameter is used in a jack having a plain thrust collar of 2.5-in. mean diamete

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This is the answer for the question

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2 years ago

A motor car shaft consists of a steel tube 30 mm internal diameter and 4 mm thick. The engine develops 10 kW at 2000 r.p.m. Find

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The maximum shear stress in the tube when the power is transmitted through a 4: 1 gearing is 28.98 MPa.

<h3>What is power?</h3>

Power is the energy transferred per unit time.

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Ts =Ps x 60/(2π x 500)

Ts =190.96 N.m

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Substitute the values into the equation, we get

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Thus, the maximum shear stress in the tube is 28.98 MPa.

Learn more about power.

brainly.com/question/13385520

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