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goldfiish [28.3K]
3 years ago
8

Which of the following statements do not correctly describe pull manufacturing? (1). Material flow is determined by the need of

the downstream workstation (2). Control is transferred from the beginning of the line to the end (3). A single piece flow is ideal for a push system and not for a pull system (4). Tries to minimize the number of items pulled at the same time Select one: a. (3) b. (1) and (3) c. (2) and (4) d. (4)
Engineering
1 answer:
weeeeeb [17]3 years ago
5 0

Answer: option B.(1) and (3)

(1). Material flow is determined by the need of the downstream workstation

(3). A single piece flow is ideal for a push system and not for a pull system

Explanation:

A pull system is a Lean manufacturing principle created to reduce waste in the production process. A pull system is to build products based on actual demand and not on forecasts.

One of the most important principles in the Lean Philosophy is the Pull Principle. The Pull principle is based on the assumption that one should only produce what is asked for by the customer. The Pull principle can be applied in more than just production, it can be applied in the office and even in life.

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Which bulb has the lowest total cost of operation? (a) Incandescent (b) Fluorescent (c) LED
Finger [1]

Answer: LED have the lowest cost of operation.

Explanation:

If we ignore the initial procurement cost of the items the operational cost of any device consuming electricity is given by

Cost=Energy\times cost/unit

Among the three item's LED consumes the lowest power to give the same level of brightness as compared to the other 2 item's thus LED's shall have the lowest operational cost.

6 0
3 years ago
A well is located in a 20.1-m thick confined aquifer with a conductivity of 14.9 m/day and a storativity of 0.0051. If the well
ahrayia [7]

Answer:

S = 5.7209 M

Explanation:

Given data:

B = 20.1 m

conductivity ( K ) = 14.9 m/day

Storativity  ( s ) = 0.0051

1 gpm = 5.451 m^3/day

calculate the Transmissibility ( T ) = K * B

                                                       = 14.9 * 20.1 = 299.5  m^2/day

Note :

t = 1

U = ( r^2* S ) / (4*T*<em> t </em>)

  = ( 7^2 * 0.0051 ) / ( 4 * 299.5 * 1 ) = 2.0859 * 10^-4

Applying the thesis method

W(u) = -0.5772 - In(U)

       = 7.9

next we calculate the pumping rate from well ( Q ) in m^3/day

= 500 * 5.451 m^3 /day

= 2725.5 m^3 /day

Finally calculate the drawdown at a distance of 7.0 m form the well after 1 day of pumping

S = \frac{Q}{4\pi T} * W (u)

 where : Q = 2725.5

               T = 299.5

               W(u)  = 7.9

substitute the given values into equation above

S = 5.7209 M

4 0
3 years ago
A horizontal curve of a two-lane undivided highway (12-foot lanes) has a radius of 678 feet to the center line of the roadway. A
OLEGan [10]

Answer:

maximum speed for safe vehicle operation = 55mph

Explanation:

Given data :

radius ( R ) = 678 ft

old building located ( m )= 30 ft

super elevation = 0.06

<u>Determine the maximum speed for safe vehicle operation </u>

firstly calculate the stopping sight distance

m = R ( 1 - cos \frac{28.655*S}{R} )  ----  ( 1 )

R = 678  

m ( horizontal sightline ) = 30 ft

back to equation 1

30 = 678 ( 1 - cos (28.655 *s / 678 ) )

( 1 - cos (28.655 *s / 678 ) )  = 30 / 678 = 0.044

cos \frac{28.65 *s }{678}  = 1.044

hence ; 28.65 * s = 678 * 0.2956

s = 6.99 ≈ 7 ft

next we will calculate the design speed ( u ) using the formula below

S = 1.47 ut  + \frac{u^2}{30(\frac{a}{3.2} )-G1}  ----  ( 2 )

t = reaction time,  a = vehicle acceleration, G1 = grade percentage

assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0

back to equation 2

6.99 = 1.47 * u * 2.5 + \frac{u^2}{30[(11.2/32.2)-0 ]}

3.675 u  + 0.0958 u^2 - 6.99 = 0

u ( 3.675 + 0.0958 u ) = 6.99

5 0
3 years ago
If the car passes point A with a speed of 20 m&gt;s and begins to increase its speed at a constant rate of at = 0.5 m&gt;s 2 , d
sattari [20]

Answer:

a = 1.68m/S^{2}

Explanation:

Please kindly find the attached file for explanations

3 0
3 years ago
A wood pole with a diameter of 10 in. has a moisture content of 5%. The fiber saturation point (FSP) for this wood is 30%. The w
Mekhanik [1.2K]

Answer:

a) Δd(change in wood diameter) = 5%

b) The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) new diameter (D2) = 10.5 in

Explanation:

Wood pole diameter = 10 inches

moisture content = 5%

FSP = 30%

A) The percentage change in the wood's diameter

note : moisture fluctuations from 5% to 30% causes dimensional changes in the wood but above 30% up to 55% causes no change. hence this formula can be used to calculate percentage change in the wood's diameter

Δd/d = 1/5(30 - 5)

Δd/d = 5%  

Δd = 5%

B) would the wood swell or shrink

The wood would swell since the moisture content is increasing which will also led to increase in the wood's diameter

C) The new diameter of the wood

D2 = D + D( \frac{M1}{100} )

D = initial diameter= 10 in , M1 = initial moisture content = 5%

therefore D2 = 10 + 10( 5/100 )

new diameter (D2) = 10.5 in

5 0
3 years ago
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