All categories

2 years ago

Select four items that an industrial engineer must obtain in order to practice in the field.

Engineering

2 answers:

WARRIOR [948]2 years ago

5 0

Professional engineering license:-

This is mandatory to become engineer
You may be arrested if you do anything without licence

Bachelor s degree

You must have a degree that certifies your engineer profession

Computer science classes

Computer is a helpful thing
Using program he can. create designs

English/communication classes

It's helpful while communcating with workers staffs

alex41 [277]2 years ago

3 0

Answer:

Professional engineering license

Bachelor's degree

Computer science classes

job recommendations

You might be interested in

The north-south streets of a grid have block lengths of 250 m and the east-west streets have block lengths of 200 m. Desired spe

antoniya [11.8K]

Answer:

See attached pictures.

Explanation:

See attached pictures for detailed explanation.

7 0

3 years ago

A 225 MPa conducted in which the mean stress was 50 MPa and the stress amplitude was (a) Compute the maximum and (b) Compute the

tamaranim1 [39]

Answer:

Explanation:

Given data in question

mean stress = 50 MPa

amplitude stress = 225 MPa

to find out

maximum stress, stress ratio, magnitude of the stress range.

solution

we will find first maximum stress and minimum stress

and stress will be sum of (maximum +minimum stress) / 2

so for stress 50 MPa and 225 MPa

$\sigma _{m}$ = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2

50 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........1

and

225 = $\sigma _{maximum}$ + $\sigma _{minimum}$ / 2 ...........2

from eqution 1 and 2 we get maximum and minimum stress

$\sigma _{maximum}$ = 275 MPa ............3

and $\sigma _{minimum}$ = -175 MPa ............4

In 2nd part we stress ratio is will compute by ratio of equation 3 and 4

we get ratio = $\sigma _{minimum}$ / $\sigma _{maximum}$

ratio = -175 / 227

ratio = -0.64

now in 3rd part magnitude will calculate by subtracting maximum stress - minimum stress i.e.

magnitude = $\sigma _{maximum}$ - $\sigma _{minimum}$

magnitude = 275 - (-175) = 450 MPa

3 0

3 years ago

A farmer wants to buy a 10 kg bag of fertilizer (organic soil). They have the choice between two merchants. Merchant A sells the

sergejj [24]

Answer:

Since the farmer wants to buy a 10 kg bag of fertilizer, he should buy it from merchant A. However, Merchant A and B are selling at the same price for a unit value. In other words, Both Merchant A and B are selling 1kg of dry fertilizer for $1.

Explanation:

Which merchant has the better deal means which merchant offers the farmer a better deal.

For Merchant A, 10 kg bag = $10

meaning it contains a real 10 kg bag of dry fertilizer which the farmer can use without losing any Kg to drying.

While for Merchant B, 10 kg bag = $8

where the 10kg = 80% dry fertilizer + 20% water content

But the farmer can only use the solid constituents of the bag which means,

Merchant B is giving 80/100 x 10Kg of dry fertilizer for $8

That is, 8kg for $8

5 0

3 years ago

Is it possible to have an iron-carbon alloy for which the mass fractions of total ferrite and proeutectoid cementite are 0.846 a

MAXImum [283]

Answer:

Yes it is possible.

Explanation:

This problem is about to possibility to have alloy of iron-carbon for which mass fraction of ferrite, $W_{\alpha} = 0.846$ and proeutectoid cementite, $W_{Fe_3C}=0.049$

An alloy formation is possible when the composition values of the two alloy are equal.

Now writing the expression for the mass fraction of total ferrite, we have

$W_{\alpha}=\frac{C_{Fe_3C}-C_0}{C_{Fe_3C}-C_{\alpha}}$

$0.846}=\frac{6.70-C_0}{6.70-0.022}$

$5.649588 = 6.70 - C_0$

$\therefore C_0 = 1.05$ wt. % of C

Now write the expression for the mass fraction of the proeutectoid cementite :

$W_{Fe_3C}=\frac{C_1-0.76}{5.94}$

$0.049=\frac{C_1-0.76}{5.94}$

$C_1 = 1.05$ % wt. C

Since, $C_0 =C_1$ , it is possible to have an alloy of iron - carbon.

5 0

3 years ago

Suppose the beam is carrying a known shear load of RD = 28 kN . In this particular situation, the resistance factor is ϕ=0.9 for

andreyandreev [35.5K]

Answer:

49.5 kN

Explanation:

From the information given:

$R_D = 28 \ kN$ $\delta _D = 1.4; \ \ \ \delta _L = 1.6$

$\sigma_n = 204 \ MPa; \ \ \ A_w = 6.45 \ cm^2 = 645 \ mm^2$

Thus ; $P_n = \dfrac{\sigma_n}{\frac{1}{A}} \\ \\ = \ {\sigma_n}*{A} \\ \\ = 645 *204 \\ \\ = 131.58 \ kN$

From the given inequality; maximum live load (in addition to RD) that can be supported in shear by this beam is calculated by using the relation;

$\phi P_n \geq \sum \delta_i R_i \\ \\ \geq \delta_DR_D + \delta_L R_L \\ \\ 0.9*131.58 \geqq [1.4*28+1.6*R_L ] \\ \\ 118.4 \geq 39.2+ 16 R_L \\ \\ 118.4 - 39.2 \geq 16R_L \\ \\ 79.2 \geq 16R_L\\ \\ R_L \leq 49.5 \ kN$

4 0

3 years ago