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velikii [3]
3 years ago
10

4. In the Hyatt Regency walkway case study, it is reported that Jack Gillum stamps the 42 shop drawings, including the revised S

hop Drawing 30 and Erection Drawing E-3 denoting the revised hanger connection detail, with his engineering review seal, authorizing construction. When stamping these drawings, what is Jack Gillum accepting? A one-word response is sufficient.
Engineering
1 answer:
mario62 [17]3 years ago
3 0

Answer:

Responsibility

Explanation:

By stamping the drawings that he was looking over, Jack Gillum conveys the fact that he is accepting responsibility for this work. The purpose of Gillum's stamp is to explain that such work has been under engineering review, and that it has fulfilled all the requirements that he watches our for. By putting his stamp in this work, Gillum accepts responsibility in case an error or a discrepancy is found in the drawings.

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Tin atoms are introduced into an FCC copper ,producing an alloy with a lattice parameter of 4.7589×10-8cm and a density of 8.772
slega [8]

Answer:

atomic percentage = 143 %

Explanation:

Let  x be the number of tin atoms and there are 4 atoms / cell in the FCC structure , then 4 -x  be the number of copper atoms . Therefore, the value of x can be determined by using the density equation as shown below:

\mathbf{density (\rho) = \dfrac{(no \ of \ atoms/cell)(atomic \ mass )}{(lattice \ parameter )^3(6.022*10^{23} atoms/ mol)} }

where;

the lattice parameter is given as : 4.7589 × 10⁻⁸ cm

The atomic mass of tin is 118.69 g/mol

The atomic mass of copper is 63.54 g/mol

The density is 8.772 g/cm³

\mathbf{8.772 g/cm^3 = \dfrac{(x)(118.69 \ g/mol) +(4-x)(63.54 \ g/mol)}{(4.7589*10^{-8} cm )^3(6.022*10^{23} atoms/ mol)} }

569.32 = 118.69x + 254.16-63.54x

569.32 - 254.16 = 118.69x - 63.54 x

315.16 = 55.15x

x = 315.16/55.15

x = 5.72 atoms/cell

As there are four atoms per cell in FCC structure for the metal, thus, the atomic percentage of the tin is  calculated as follows :

atomic % = \frac{no \ of \ atoms \ per  \ cell \ in \ tin }{no \ of \ atoms \ per  \ cell \ in \ the \ metal}*100

atomic % = \frac{5.72 \ atoms / cell}{4 \ atoms/ cell} *100

atomic % = 143 %

7 0
3 years ago
Mathew is manufacturing overhead transmission wires for the working of electric trains. On what does the insulation of these wir
elena-14-01-66 [18.8K]

Answer:

B

Explanation:

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3 years ago
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Technician A says when tightening lug or wheel nuts a torque wrench must be used. Technician B says that an air wrench can be us
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3 years ago
A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
3 years ago
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