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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?

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

Cloud [144]

<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>

5 0

3 years ago

simply supported beam is subjected to a linearly varying distributed load ( ) 0 q x x L 5 q with maximum intensity 0 q at B. The

Pavlova-9 [17]

Answer:

q₀ = 350,740.2885 N/m

Explanation:

Given

$q(x)=\frac{x}{L} q_{0}$

σ = 120 MPa = 120*10⁶ Pa

$L=4 m\\w=200 mm=0.2m\\h=300 mm=0.3m\\q_{0}=? \\$

We can see the pic shown in order to understand the question.

We apply

∑MB = 0 (Counterclockwise is the positive rotation direction)

⇒ - Av*L + (q₀*L/2)*(L/3) = 0

⇒ Av = q₀*L/6 (↑)

Then, we apply

$v(x)=\int\limits^L_0 {q(x)} \, dx\\v(x)=-\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6} \\M(x)=\int\limits^L_0 {v(x)} \, dx=-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x$

Then, we can get the maximum bending moment as follows

$M'(x)=0\\ (-\frac{q_{0}}{6L} x^{3}+\frac{q_{0} L}{6}x)'=0\\ -\frac{q_{0}}{2L} x^{2}+\frac{q_{0} L}{6}=0\\x^{2} =\frac{L^{2}}{3}\\ x=\sqrt{\frac{L^{2}}{3}} =\frac{L}{\sqrt{3} }=\frac{4}{\sqrt{3} }m$

then we get

$M(\frac{4}{\sqrt{3} })=-\frac{q_{0}}{6*4} (\frac{4}{\sqrt{3} })^{3}+\frac{q_{0} *4}{6}(\frac{4}{\sqrt{3} })\\ M(\frac{4}{\sqrt{3} })=-\frac{8}{9\sqrt{3} } q_{0} +\frac{8}{3\sqrt{3} } q_{0}=\frac{16}{9\sqrt{3} } q_{0}m^{2}$

We get the inertia as follows

$I=\frac{w*h^{3} }{12} \\ I=\frac{0.2m*(0.3m)^{3} }{12}=4.5*10^{-4}m^{4}$

We use the formula

σ = M*y/I

⇒ M = σ*I/y

where

$y=\frac{h}{2} =\frac{0.3m}{2}=0.15m$

If M = Mmax, we have

$(\frac{16}{9\sqrt{3} }m^{2} ) q_{0}\leq \frac{120*10^{6}Pa*4.5*10^{-4}m^{4} }{0.15m}\\ q_{0}\leq 350,740.2885\frac{N}{m}$

8 0

3 years ago

Are the wooden pillars shown in the image below, a dead load?

Brrunno [24]

Answer:

no

Explanation:

it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls

6 0

3 years ago

All aspects of the Kirby-Bauer test are standardized to assure reliability. What might be the consequence of pouring the plates

EleoNora [17]

Answer:

it would affect the distance the antiantibodies diffuse from the disk

Explanation:

7 0

2 years ago

A civil engineer is studying a left-turn lane that is long enough to hold seven cars. Let X be the number of cars in the line at

BartSMP [9]

Answer:

a) C= 1/120

b) P(X>=5) = 0.333

Explanation:

The attached file contains the explanation for the answers

7 0

3 years ago

Read 2 more answers