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3 years ago

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

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:

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3 years ago

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Principals of Construction intro

Nadusha1986 [10]

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:

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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

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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