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3 years ago

List three types of concurrent engineering in manufacturing.

Engineering

1 answer:

klio [65]3 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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Create a separate function file fieldtovar.m that receives a single structure as an input and assigns each of the field values t

Soloha48 [4]

Answer:

Explanation gives the answer

Explanation:

% Using MATLAB,

% Matlab file : fieldtovar.m

function varargout = fieldtovar(S)

% function that accepts single structure as input, assigning each

% of the field values to user-defined variables

fields = fieldnames(S); % get the field names of the input structure

% check if number of user-defined variables and number of fields in

% structure are equal

if nargout == length(fields)

% if equal assign each value of structure to user-defined varable

for i=1:nargout

varargout{i} = getfield(S,fields{i});

end

else

% if not equal display an error message

error('The number of output variables does not equal the number of fields');

end

%This brings an end to the program

4 0

3 years ago

A mass of 5 kg of saturated water vapor at 100 kPa is heated at constant pressure until the temperature reaches 200°C.

Alex73 [517]

Answer: you can watch a video on how to solve this question on you tube

6 0

3 years ago

Can someone help me with this maze shown below.

Gnoma [55]

We can’t see the maze

3 0

3 years ago

Plis 3 conclusiones de este video

vazorg [7]

No hay videos? de cual video estás hablando?

6 0

3 years ago

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a

Virty [35]

Answer:

a) The additional time required for the truck to stop is <u>8.5 seconds</u>

b) The additional distance traveled by the truck is <u>230.05 ft</u>

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

$v^2 - u^2 = 2*a*s$

$54.12^2 - 108.24^2 = 2*a*690$

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

$v^2 - u^2 = 2*a*s$

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

5 0

3 years ago