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2 years ago

Hey answer quick for 20 points

Engineering

2 answers:

SSSSS [86.1K]2 years ago

4 0

Thank you!!!!!!!!!!!!!!!!

creativ13 [48]2 years ago

3 0

Answer:

<h2>Hey.....Thanks but FYI this is only 8 points....If u want to put 20....try putting 40...</h2>

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QUESTÃO 13. Explique o uso das aspas no trecho "Darei a cada uma de vocês

lesya [120]

Answer: speaks Portuguese

Eu disse a todos a tradução para que possam te ajudar

Explanation: Y’all can help I have no idea

QUESTION 13. Explain the use of quotation marks in the excerpt "I will give each of you

seed. The one who will bring me the most beautiful flower within six months will be chosen but

wife and the future empress of China. ".

QUESTION 14. The palace servant considered the idea of her daughter attending the celeb

organized by the prince of the region, a foolish idea, a madness. This is an OP

Do you agree with this opinion of the character? Justify your answer.

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3 years ago

Wireless technology has made communicating

emmainna [20.7K]

Answer:

Taking as a basis of calculation 100 mol of gas leaving the conversion reactor, draw andcompletely label a flowchart of this process. Then calculate the moles of fresh methanol feed,formaldehyde product solution, recycled methanol, and absorber off-gas, the kg of steamgenerated in the waste-heat boiler, and the kg of cooling water fed to the heat exchangerbetween the waste-heat boiler and the absorber. Finally, calculate the heat (kJ) that must beremoved in the distillation column overhead condenser, assuming that methanol enters as asaturated vapor at 1 atm and leaves as a saturated liquid at the same pressure.

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Explanation:

6 0

3 years ago

The state of plane strain on an element is:

balu736 [363]

Answer:

a. ε₁=-0.000317

ε₂=0.000017

θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

$\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2} \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2} \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6} \pm 1.67 \times 10^{-4}$

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

$tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5$

θ= -13.28° and 76.72°

To determine the direction of ε₁ and ε₂

$\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2} + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2} + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56) + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\$

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and θ₂=76.72°

b. maximum in-plane shear strain

$\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}$

=3.335 *10^-4

$\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )$

ε(avg) =150 *10^-6

orientation of γmax

$tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}$

θ = 31.71 or -58.29

To determine the direction of γmax

$\gamma _{x'y' }= - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }= - \frac{-300*10^{-6} - \ 0}{2} sin(63.42) + \frac{150*10^{-6}}{2}cos(63.42)$

= 1.67 *10^-4

4 0

3 years ago

Two wastewater treatment plant workers (one male and one female) are exposed to hydrogen sulfide in confined spaces in the treat

Vlada [557]

Answer:

Go to explaination for the details of the answer.

Explanation:

In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:

CDI = Chronic Daily Intake

C= Chemical concentration

CR= Contact Rate

EFD= Exposure Frequency and Distribution

BW= Body Weight

AT = Average Time.

Having names our variables lets create the equations that will be used to derive our answers.

Please kindly check attachment for details of the answer.

5 0

3 years ago

A number 12 copper wire has a diameter of 2.053 mm. Calculate the resistance of a 37.0 m long piece of such wire.

Alinara [238K]

Answer:

R=1923Ω

Explanation:

Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.

Coil length(L) of the wire=37.0m

Cross-sectional area of the conductor or wire (A) = πr^2

A= π * (2.053/1000)/2=3.31*10^-6

To calculate for the resistance (R):

R=ρ*L/A

R=(1.72*10^8)*(37.0)/(3.31*10^-6)

R=1922.65Ω

Approximately, R=1923Ω

5 0

3 years ago