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2 years ago

List three types of concurrent engineering in manufacturing.

Engineering

1 answer:

klio [65]2 years ago

8 0

Answer:

A famous example of concurrent engineering is the development of the Boeing 777 commercial aircraft. The aircraft was designed and built by geographically distributed companies that worked entirely on a common product database of C A TIA without building physical mock-ups but with digital product definitions.

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Mnsdcbjksdhkjhvdskjbvfdfkjbcv hjb dfkjbkjfvvfebjkhbvefgjdf

Julli [10]

Answer:

ik true eedcggftggggfffff

8 0

3 years ago

Read 2 more answers

A closed system consisting of 4 lb of a gas undergoes a process during which the relation between pressure and volume is pVn 5 c

gayaneshka [121]

Answer:

V1=5ft3

V2=2ft3

n=1.377

Explanation:

PART A:

the volume of each state is obtained by multiplying the mass by the specific volume in each state

V=volume

v=especific volume

m=mass

V=mv

state 1

V1=m.v1

V1=4lb*1.25ft3/lb=5ft3

state 2

V2=m.v2

V2=4lb*0.5ft3/lb= 2ft3

PART B:

since the PV ^ n is constant we can equal the equations of state 1 and state 2

P1V1^n=P2V2^n

P1/P2=(V2/V1)^n

ln(P1/P2)=n . ln (V2/V1)

n=ln(P1/P2)/ ln (V2/V1)

n=ln(15/53)/ ln (2/5)

n=1.377

3 0

3 years ago

AVH

BigorU [14]

Answer: preventive maintenance

Explanation:

3 0

2 years ago

How are industries related to systems

REY [17]

Answer:

The industrial systems of the future are seen as complex systems, composed of vast numbers of devices, interacting with each other and with enterprise systems continuously.

summary:

they are related because they ARE a system. well, a type.

hope this helps!! :)

7 0

2 years ago

The water level in a tank z1, is 16 m above the ground. A hose is connected to the bottom of the tank, and the nozzle at the end

Volgvan

Answer:

40.7 m

Explanation:

Let point 1 represent the surface of the water, point 2 be the top of the water trajectory and the reference be the bottom of the tank. Hence:

$z_1=16\ m,P_1=2\ atm,V_1=0(velocity\ at\ the \ surface \ of\ water\ is\ low)\\V_2=0,P_2=P\\\\Using\ Bernoulli\ equation:\\\\\frac{P_1}{\rho g} +\frac{V_1^2}{2g}+z_1= \frac{P_2}{\rho g} +\frac{V_2^2}{2g}+z_2\\\\Sine\ V_1=0,V_2=0:\\\\\frac{P_1}{\rho g} +z_1=\frac{P}{\rho g} +z_2\\\\z_2=\frac{P_1}{\rho g} -\frac{P}{\rho g} +z_1\\\\z_2=\frac{P_1-P}{\rho g} +z_1\\\\z_2=\frac{P_{1.gage}}{\rho g} +z_1\\\\$

$z_2=\frac{2\ atm}{1000\ kg/m^3*9.81\ m/s^2}(\frac{101325\ N/m^2}{1\ atm} )(\frac{1\ kg.m/s^2}{1 \ N} ) +20\\\\z_2=20.7+20\\\\z_2=40.7\ m$

7 0

3 years ago