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mario62 [17]
3 years ago
12

The Manufacturing sector is the only sector where Lean manufacturing philosophy can be applied. a)- True b)- False

Engineering
1 answer:
AlladinOne [14]3 years ago
5 0

Answer:

b). false

Explanation:

<u>Lean manufacturing</u>

Lean manufacturing, a philosophy developed by Toyota Production System are means to eliminate wastes. They are defined as the techniques or management activities in eliminating wastes and increasing the efficiency inside an organisation.

    According to the concept of lean manufacturing, mainly seven types of wastes are identified. They are :

1. Transportation waste

2. Inventory waste

3. Over production

4. Waiting

5. Defects

6. Motion waste

7. Non utilized talent

All these waste affect greatly to the efficiency of an organisation and devalue its services.  Lean manufacturing advises to prevent  all these waste in order to increase the productivity.

       All the management activities and techniques used in lean manufacturing may be different according to the business application but they are all based on the same basic principle of removing wastes and errors and increase efficiency.

The different sectors that are benefiting from lean manufacturing methodology are

Healthcare

Hospitality

Food and Beverage

Government

Manufacturing

Lean manufacturing can be used in different sectors.

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6.Identification of Material ParametersThe principal in-plane stresses and associated strains in a plate of material areσ1= 50 k
ad-work [718]

Answer:

See attached image for diagrams and solution

6 0
2 years ago
Why does the auto industry prefer uniform (national) standards for automobile emissions as opposed to regionally varying standar
WINSTONCH [101]

Answer:

Explanation:

For automobile emission, a uniform standard is preferred, because no unnecessary advantage is given by it to any company that is located in particular states where the regional standards are less severe.

Since pollution has its impact across the states and in the whole of the USA, then there should be uniform standards across all the states. It will also invalidate the impact of regional standards as a factor in the selection of plant locations for the automobile company. It means that a state offering less valid emission standards, will attract more companies to herself and it will be against the other states who care more about the natural environment. It can make more states to opt for the permissive emission standards, that will be more harmful to the USA as a country, than the good. So, a uniform standard is preferred to eliminate it as a factor in plant location decisions.

Yes, uniform standards are beneficial to everyone, because it will bring effective control upon the pollution level because there will be no state where the culprit firm can hide. Besides, it is more effective as efforts done towards environment conservation.

3 0
3 years ago
Select three types of engineering that involve work in inaccessible environments.
Pavel [41]

Answer:

Chemical Engineer,Geological Engineer,Aerospace Engineer

Explanation:

8 0
2 years ago
Read 2 more answers
Six forces act on a beam that forms part of a building's
IrinaVladis [17]

Answer:

<h2> FA = 13 kN </h2><h2>FG = 15.3 kN</h2>

Explanation:

write each force in terms of magnitude and directions  

Fx = F sin Ф

Fy = F cos Ф

where Ф is to be measured from x axis.

∑F at y = o

FAy + FBy + FCy + FDy + FEy + FGy = 0

∑F at x = o

FAx + FBx + FCx + FDx + FEx + FGx = 0

Let  

FA = FA sin (110)   +   FA cos (110)

FB = 20 sin (270)  +  20 cos (270)

FC = 16 sin (140)    +  16 cos (140)

FD = 9 sin (40)       +  9 cos (40)

FE = 20 sin (270)    +  20 cos (270)

FG = FG sin (50)     +  FG cos (50)

add x and y forces:

FAx + FBx + FCx + FDx + FEx + FGx = 0

FAy + FBy + FCy + FDy + FEy + FGy = 0

FA sin (110)  + 0  + 16 sin (140)  + 9 sin (40)  + 0   + FG sin (50) = 0

FA cos (110) - 20 + 16 cos (140) + 9 cos (40) - 20 + FG cos (50 = 0

FA sin (110)  + 0  + 10.285  + 5.785  + 0   + FG sin (50) = 0

FA cos (110) - 20 - 12.257 + 6.894 - 20 + FG cos (50) = 0

FA sin (110)  + 16.070 + FG sin (50) = 0        

FA cos (110) - 45.363 + FG cos (50) = 0

solving for FA, and FG

FA = 13 kN

FG = 15.3 kN

7 0
3 years ago
The current in a 20 mH inductor is known to be: 푖푖=40푚푚푚푚푡푡≤0푖푖=푚푚1푒푒−10,000푡푡+푚푚2푒푒−40,000푡푡푚푚푡푡≥0The voltage across the induct
Anni [7]

Answer:

a) The expression for electrical current: i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

The expression for voltage: v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) For t<=0 the inductor is storing energy and for t > 0 the inductor is delivering energy.

Explanation:

The question text is corrupted. I found the complete question on the web and it goes as follow:

The current in a 20 mH inductor is known to be: i = 40 mA at t<=0 and i = A1*e^(-10,000*t) + A2*e^(-40,000*t) A at t>0. The voltage across the inductor (passive sign convention) is -68 V at t = 0.

a. Find the numerical expressions for i and v for t>0.

b. Specify the time intervals when the inductor is storing energy and is delivering energy.

A inductor stores energy in the form of a magnetic field, it behaves in a way that oposes sudden changes in the electric current that flows through it, therefore at moment just after t = 0, that for convenience we'll call t = 0+, the current should be the same as t=0, so:

i = A1*e^(-10,000*(0)) + A2*e^(-40,000*(0))

40*10^(-3) = A1*e^(-10,000*0) + A2*e^(-40,000*0)

40*10^(-3) = (A1)*1 + (A2)*1

40*10^(-3) = A1 + A2

A1 + A2 = 40*10^(-3)

Since we have two variables (A1 and A2) we need another equation to be able to solve for both. For that reason we will use the voltage expression for a inductor, that is:

V = L*di/dt

We have the voltage drop across the inductor at t=0 and we know that the current at t=0 and the following moments after that should be equal, so we can use the current equation for t > 0 to find the derivative on that point, so:

di/dt = d(A1*e^(-10,000*t) + A2*e^(-40,000*t))/dt

di/dt = [d(-10,000*t)/dt]*A1*e^(-10,000*t) + [d(-40,000*t)/dt]*A2*e^(-40,000*t)

di/dt = -10,000*A1*e^(-10,000*t) -40,000*A2*e^(-40,000*t)

By applying t = 0 to this expression we have:

di/dt (at t = 0) = -10,000*A1*e^(-10,000*0) - 40,000*A2*e^(-40,000*0)

di/dt (at t = 0) = -10,000*A1*e^0 - 40,000*A2*e^0

di/dt (at t = 0) = -10,000*A1- 40,000*A2

We can now use the voltage equation for the inductor at t=0, that is:

v = L di/dt (at t=0)

68 = [20*10^(-3)]*(-10,000*A1 - 40,000*A2)

68 = -400*A1 -800*A2

-400*A1 - 800*A2 = 68

We now have a system with two equations and two variable, therefore we can solve it for both:

A1 + A2 = 40*10^(-3)

-400*A1 - 800*A2 = 68

Using the first equation we have:

A1 = 40*10^(-3) - A2

We can apply this to the second equation to solve for A2:

-400*[40*10^(-3) - A2] - 800*A2 = 68

-1.6 + 400*A2 - 800*A2 = 68

-1.6 -400*A2 = 68

-400*A2 = 68 + 1.6

A2 = 69.6/400 = 0.174

We use this value of A2 to calculate A1:

A1 = 40*10^(-3) - 0.174 = -0.134

Applying these values on the expression we have the equations for both the current and tension on the inductor:

i = -0.134*e^(-10,000*t) + 0.174*e^(-40,000*t) A

v = [20*10^(-3)]*[-10,000*(-0.134)*e^(-10,000*t) -40,000*(0.174)*e^(-40,000*t)]

v = [20*10^(-3)]*[1340*e^(-10,000*t) - 6960*e^(-40,000*t)]

v = 26.8*e^(-10,000*t) - 139.2*e^(-40,000*t) V

b) The question states that the current for the inductor at t > 0 is a exponential powered by negative numbers it is expected that its current will reach 0 at t = infinity. So, from t =0 to t = infinity the inductor is delivering energy. Since at time t = 0 the inductor already has a current flow of 40 mA and a voltage, we can assume it already had energy stored, therefore for t<0 it is storing energy.

8 0
3 years ago
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