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Andru [333]
3 years ago
11

Which of the following would be addressed by an employer completing an EAP template?

Engineering
1 answer:
zloy xaker [14]3 years ago
4 0
I can help you, but what are the options that were given to you?
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A long rod of 60-mm diameter and thermophysical properties rho=8000 kg/m^3, c=500J/kgK, and k=50 W/mK is initally at a uniform t
Monica [59]

Answer:

Tc = 424.85 K

Explanation:

Given that,

D = 60 mm = 0.06 m

\rho = 8000 kg/m^3

k = 50 w/m . kc = 500 j/kg.k

h_{\infty} = 1000 w/m^2t_{\infity} = 750 kt_w = 500 K

surface area = As = \pi dL \\\frac{As}{L} = \pi D = \pi \timeS 0.06

HEAT FLOW Q  is

Q = h_{\infty} As (T_[\infty} - Tw)  = 1000 \pi\times 0.06 (750-500)

 = 47123.88 w per unit length of rod

volumetric heat rate

q = \frac{Q}{LAs}

= \frac{47123.88}{\frac{\pi}{4} D^2 \times 1}

q = 1.66\times 10^{7} w/m^3

Tc = \frac{- qR^2}{4K} + Tw

= \frac{ - 1.67\times 10^7 \times (\frac{0.06}{2})^2}{4\times 50} +  500

  = 424.85 K

3 0
3 years ago
What is the first step of the engineering design process?
scoundrel [369]

Answer:

The first step  is to identify the need and constraints

Explanation:

5 0
3 years ago
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Principals of Construction intro
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Answer:

<em>The</em><em> </em><em>answer</em><em> </em><em>is</em><em> </em><em>safety</em><em> </em><em>documents</em><em> </em>

Explanation:

Every company is responsible

For people documents

Is very important for every company

8 0
3 years ago
The electrical panel schedules are located on EWR Plan number ___.
Stells [14]
A8 is the answer because yea and because I am a teacher
5 0
3 years ago
A saturated 1.5 ft3 clay sample has a natural water content of 25%, shrinkage limit (SL) of 12% and a specific gravity (GS) of 2
Svetllana [295]

79 f t^{3} is the volume of the sample when the water content is 10%.

<u>Explanation:</u>

Given Data:

V_{1}=100\ \mathrm{ft}^{3}

First has a natural water content of 25% = \frac{25}{100} = 0.25

Shrinkage limit, w_{1}=12 \%=\frac{12}{100}=0.12

G_{s}=2.70

We need to determine the volume of the sample when the water content is 10% (0.10). As we know,

V \propto[1+e]

\frac{V_{2}}{V_{1}}=\frac{1+e_{2}}{1+e_{1}}  ------> eq 1

e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}

The above equation is at S_{r}=1,

e_{1}=w_{1} \times G_{s}

Applying the given values, we get

e_{1}=0.25 \times 2.70=0.675

Shrinkage limit is lowest water content

e_{2}=w_{2} \times G_{s}

Applying the given values, we get

e_{2}=0.12 \times 2.70=0.324

Applying the found values in eq 1, we get

\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904

V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}

7 0
3 years ago
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