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Akimi4 [234]
3 years ago
5

Select the correct answer What is the simplest definition of a manufacturing process?

Engineering
2 answers:
anastassius [24]3 years ago
7 0

a process that converts raw materials into a useful product.

lukranit [14]3 years ago
3 0

Answer:

B. A process that mass-produces complex products :)

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A monatomic ideal gas undergoes a quasi-static process that is described by the function p(????)=p1+3(????−????1) , where the st
Alenkasestr [34]

A pure gas made up only of atoms. The noble gases argon, krypton, and xenon are some examples.

Concepts:

Perfect gas law: Work performed on the system: PV = nRT W = -∫PdV

Energy preservation formula: U = Q + W

Reasoning:

W = nRT ln(Vi/Vf) when the process is isothermal.

The temperature is said to be constant, and we are given n, Pfinal, and Vfinal.

Calculation information:

(A) A process that is isothermal has a constant temperature.

PV = nRT, and hence, constant

nRT = PV = 101000 Pa*25*10-3 m3

For a process that is isothermal, W = nRT ln(Vi/Vf).

W/(nRT)=3000 J/(101000 Pa*25*10-3 m3)=-1.19

(The gas produces -W of labor.)

Vi = (25*10-3 m3)/3.28 = 7.62*10-3 m3 = 7.62 L where Vf/Vi = exp(1.19) = 3.28 Vi (b) for a perfect gas PV = nRT. 101000 Pa*25*10-3 m3 = (8.31 J/K) T. T = 303.85 K.

To know more about process click here:

brainly.com/question/29310303

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5 0
1 year ago
A plate clutch has a single friction surface 9-in OD by 7-in ID. The coefficient of friction is 0.2 and the maximum pressure is
Talja [164]

Answer:

the torque capacity is  30316.369 lb-in

Explanation:

Given data

OD = 9 in

ID = 7 in

coefficient of friction = 0.2

maximum pressure = 1.5 in-kip = 1500 lb

To find out

the torque capacity using the uniform-pressure assumption.

Solution

We know the the torque formula for uniform pressure theory is

torque = 2/3 × \pi × coefficient of friction × maximum pressure ( R³ - r³ )    .....................................1

here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in

now put all these value R, r, coefficient of friction and  maximum pressure in equation 1 and we will get here torque

torque = 2/3 × \pi × 0.2 × 1500 ( 4.5³ - 3.5³ )

so the torque =  30316.369 lb-in

3 0
3 years ago
A 50 mm diameter shaft is subjected to a static axial load of 160 kN. If the yield stress of the material is 350 MPa, the ultima
zvonat [6]

In order to develop this problem it is necessary to take into account the concepts related to fatigue and compression effort and Goodman equation, i.e, an equation that can be used to quantify the interaction of mean and alternating stresses on the fatigue life of a materia.

With the given data we can proceed to calculate the compression stress:

\sigma_c = \frac{P}{A}

\sigma_c = \frac{160*10^3}{\pi/4*0.05^2}

\sigma_c = 81.5MPa

Through Goodman's equations the combined effort by fatigue and compression is expressed as:

\frac{\sigma_a}{S_e}+\frac{\sigma_c}{\sigma_u}=\frac{1}{Fs}

Where,

\sigma_a=Fatigue limit for comined alternating and mean stress

S_e =Fatigue Limit

\sigma_c=Mean stress (due to static load)

\sigma_u = Ultimate tensile stress

Fs =Security Factor

We can replace the values and assume a security factor of 1, then

\frac{\sigma_a}{320}+\frac{81.5}{400}=\frac{1}{1}

Re-arrenge for \sigma_a

\sigma_a = 254.8Mpa

We know that the stress is representing as,

\sigma_a = \frac{M_c}{I}

Then,

Where M_c=Max Moment

I= Intertia

The inertia for this object is

I=\frac{\pi d^4}{64}

Then replacing and re-arrenge for M_c

M_c = \frac{\sigma_a*\pi*d^3}{32}

M_c = \frac{260.9*10^6*\pi*0.05^3}{32}

M_c = 3201.7N.m

Thereforethe moment that can be applied to this shaft so that fatigue does not occur is 3.2kNm

5 0
3 years ago
Consider a flat plate that is 25 mm long, 30 mm wide, and 1 mm thick and a 50 mm long cylinder with the same volume as the plate
Blizzard [7]

Answer:

Average heat transfer =42.448w/m^2k

Nud = 13.45978

Explanation:

See attachment for step by step guide

4 0
3 years ago
Is 4/16 equal in measurement to 1/4
Alja [10]

Answer:yes

Explanation:

5 0
3 years ago
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