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Ilia_Sergeevich [38]
3 years ago
6

What are the two most important things to remember when at the end of your interview?

Engineering
1 answer:
Cloud [144]3 years ago
5 0
<h3>Answer:</h3><h3><em>1. Ask questions</em></h3><h3><em>2. Thank the interviewer for their time </em></h3><h3>Explanation:</h3>

1<em>. When the interviewer asked if you have any questions at the end of the interview don't say no. You should always say yes your interviewer is expecting you to ask a few good questions before ending the interview. </em>

<h3><em /></h3>

<em>2. Always thank the interviewer for their time and effort to interview you. This would look very good for you and its a nice way to help wrap up the interview. </em>

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The costs of mining and transporting coal are roughly independent of the heating value of the coal. Consider:
Paul [167]

Answer:

2

Explanation:

5 0
3 years ago
Find the value of L
KonstantinChe [14]

Answer:

the value is 356732 Volt

6 0
2 years ago
5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of
marta [7]

Answer:

(a) 3.185*10^{21} cells

(b) 6.37*10^{21} atoms

Explanation:

(a)

Volume, V of unit cell

V=(2.866*10^{-8})^{3}=2.354*10^{-23}

Number of unit cells, N

N=\frac {W_{mat}}{V\rho_{mat}} Where W_{mat} is weight of material and \rho_{mat} is density of material

N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron=3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms

8 0
3 years ago
Teresa is emotionally volatile, particularly with friends and boyfriends. She is extremely dramatic about even the smallest disa
Ilia_Sergeevich [38]

Answer: Borderline personality Disorder.

Explanation: This is a type of mental disorder which could affects mood, behavior and relationships.

Its symptoms includes unstable emotions, sense of insecurity, worthlessness, and impulsivity.

This condition can not be cured, but treatments such as therapies, medication (in some cases) could help.

4 0
3 years ago
Air enters the combustor of a jet engine at p1=10 atm, T1=1000°R, and M1=0.2. Fuel is injected and burned, with a fuel/air mass
snow_lady [41]

Answer:

M2 = 0.06404

P2 = 2.273

T2 = 5806.45°R

Explanation:

Given that p1 = 10atm, T1 = 1000R, M1 = 0.2.

Therefore from Steam Table, Po1 = (1.028)*(10) = 10.28 atm,

To1 = (1.008)*(1000) = 1008 ºR

R = 1716 ft-lb/slug-ºR cp= 6006 ft-lb/slug-ºR fuel-air ratio (by mass)

F/A =???? = FA slugf/slugaq = 4.5 x 108ft-lb/slugfx FA slugf/sluga = (4.5 x 108)FA ft-lb/sluga

For the air q = cp(To2– To1)

(Exit flow – inlet flow) – choked flow is assumed For M1= 0.2

Table A.3 of steam table gives P/P* = 2.273,

T/T* = 0.2066,

To/To* = 0.1736 To* = To2= To/0.1736 = 1008/0.1736 = 5806.45 ºR Gives q = cp(To* - To) = (6006 ft-lb/sluga-ºR)*(5806.45 – 1008)ºR = 28819500 ft-lb/slugaSetting equal to equation 1 above gives 28819500 ft-lb/sluga= FA*(4.5 x 108) ft-lb/slugaFA =

F/A = 0.06404 slugf/slugaor less to prevent choked flow at the exit

5 0
3 years ago
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