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3 years ago

Which of these is the BEST description of

Engineering

1 answer:

strojnjashka [21]3 years ago

3 0

Answer:

i would say C but i may be wrong have a great day

Explanation:

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Go online and search for information about companies that have been harmed or bankrupted by a disaster. Choose one such company

tester [92]

Answer:

See explaination

Explanation:

In recent times and in the past, so many companies have been dealt with or bankrupted by disaster.

This disasters are in the form of floods, earthquake and tsunami. This has led to many companies across world to have faced problem. As Japan is particularly vulnerable to natural disaster because of its climate and topography let me take example of japan .The tsunami that struck Japan in March this year which has lead to a debt of worth nearly $8 billion due to failed businesses. A total of 341 firms in japan with a combined 6,376 employees has got affected by this disaster.

One of company that hit hard by these disaster is Toyota. Toyota is a Japanese multinational automotive manufacturer headquartered in Toyota, Aichi, Japan. Actually due to flood in thailand toyota has faced a big loss. As flood hit thailand the parts of vehicle which is produced in thailand has stopped . It leads to interruption in the supply chain of some Thailand made components.

This lead to suspension of production as thai made component was unavailable.It has been reported until the situation of thailand will be good toyota plants in Indiana, Kentucky and Ontario, Canada, will be shut down . And production rate at plant in North America will get slow down .This flood leads to a loss of 37500 vehicles .Due to this reason toyota was forced to open its plants in Southeast asian country just to enhance the production rate.Sales of vehicle has also got reduced in thailand . Hence toyota faced a total loss of 1.6 billion dollar.

In Nigeria, due to the insurgency of the Boko Haram, many companies have entirely stopped operations in some regions, this harsh realities leads to massive loss of revenue for such companies.

3 0

3 years ago

Turbine blades mounted to a rotating disc in a gas turbine engine are exposed to a gas stream that is at [infinity] = 1100°C and

Nataly_w [17]

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment

∴ Calculation maybe wrong but method is correct

Download docx

3 0

3 years ago

A rigid 10-L vessel initially contains a mixture of liquid and vapor water at 100 °C, with a quality factor of 0.123. The mixtur

masya89 [10]

Answer:

$Q_{in} = 46.454\,kJ$

Explanation:

The vessel is modelled after the First Law of Thermodynamics. Let suppose the inexistence of mass interaction at boundary between vessel and surroundings, changes in potential and kinectic energy are negligible and vessel is a rigid recipient.

$Q_{in} = U_{2} - U_{1}$

Properties of water at initial and final state are:

State 1 - (Liquid-Vapor Mixture)

$P = 101.42\,kPa$

$T = 100\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 675.761\,\frac{kJ}{kg}$

$x = 0.123$

State 2 - (Liquid-Vapor Mixture)

$P = 476.16\,kPa$

$T = 150\,^{\textdegree}C$

$\nu = 0.2066\,\frac{m^{3}}{kg}$

$u = 1643.545\,\frac{kJ}{kg}$

$x = 0.525$

The mass stored in the vessel is:

$m = \frac{V}{\nu}$

$m = \frac{10\times 10^{-3}\,m^{3}}{0.2066\,\frac{m^{3}}{kg} }$

$m = 0.048\,kg$

The heat transfer require to the process is:

$Q_{in} = m\cdot (u_{2}-u_{1})$

$Q_{in} = (0.048\,kg)\cdot (1643.545\,\frac{kJ}{kg} - 675.761\,\frac{kJ}{kg} )$

$Q_{in} = 46.454\,kJ$

3 0

3 years ago

Please help is due tonight

Kipish [7]

Answer:

tHE answer is b

Explanation:

7 0

3 years ago

Read 2 more answers

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago