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IRINA_888 [86]
2 years ago
5

Hi, everyone I'm a high school student in Texas. My engineering teacher is asking us to find an active engineer to complete a li

st of questions, but I don't know any engineers. so If anyone can help I would be appreciated. Thanks

Engineering
1 answer:
jenyasd209 [6]2 years ago
6 0

Answer:ill look it up

Explanation:

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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
Alexxandr [17]

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

6 0
3 years ago
What was the purpose of the vasa ship
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The main purpose was for power. The vessel has come to symbolize Sweden's Great Power Period, when the nation became a major European power and controlled much of the Baltic.
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3 years ago
What is the answer of this question please tell me <br><br>dadada please please​
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3 years ago
What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?
Yakvenalex [24]

Answer:

Activation\ Energy=2.5\times 10^{-19}\ J

Explanation:

Using the expression shown below as:

N_v=N\times e^{-\frac {Q_v}{k\times T}

Where,

N_v is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = 1.38\times 10^{-23}\ J/K

{Q_v} is the activation energy

T is the temperature

Given that:

N_v=2.3\times 10^{13}

N = 10 moles

1 mole = 6.023\times 10^{23}

So,

N = 10\times 6.023\times 10^{23}=6.023\times 10^{24}

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (425 + 273.15) K = 698.15 K  

T = 698.15 K

Applying the values as:

2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}

Q_v=2.5\times 10^{-19}\ J

4 0
3 years ago
An electric motor is to be supported by four identical mounts. Each mount can be treated as a linear prevent problems due requir
Artyom0805 [142]

GIVEN:

Amplitude, A = 0.1mm

Force, F =1 N

mass of motor, m = 120 kg

operating speed, N = 720 rpm

\frac{A}{F} =  \frac{0.1\times 10^{-3}}{1} = 0.1\times 10^{-3}

Formula Used:

A = \frac{F}{\sqrt{(K_{t} - m\omega ^{2}) +(\zeta \omega ^{2})}}

Solution:

Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K

We know that:

\omega = \frac{2 \pi\times N}{60}

so,

\omega = \frac{2 \pi\times 720}{60} = 75.39 rad/s

Using the given formula:

Damping is negligible, so, \zeta = 0

\frac{A}{F} will give the tranfer function

Therefore,

\frac{A}{F} = \frac{1}{\sqrt{(4K - 120\ ^{2})}}

0.1\times 10^{-3} =  \frac{1}{\sqrt{(4K - 120\ ^{2})}}

Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm

8 0
3 years ago
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