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Lady bird [3.3K]
4 years ago
5

Name the five points where product should be scanned during it's path throughout the warehouse.

Engineering
1 answer:
Oxana [17]4 years ago
7 0

Answer and Explanation:

A warehouse can be defined as a place:

is a spot or building, where the articles are kept, put away or in some cases prepared.  

Depending on the sort of use, the checkpoints may fluctuate.  

There can be numerous sorts, for example, pressing, railroad wagon, cool stockpiling, and so on.  

Suppositions are caused in regards to each shape and a few checkpoints for an item to can be summed up.  

  • The point of entry, where it ought to be through into the product house.  
  • Security checkpoint, where the item belonging is to be protected.  
  • Palletized segment, where the item is to be put in a spot and it determines the location of that item in the house.  
  • Quality or bundling segment, where the item must be tried and improved.  
  • Leave segment, through which it must be taken out for further procedures.  

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Think about how could you design, build, and test a light maze. What specific behavior of light will be essential to the success
Mrac [35]

Answer:

Reflection

Explanation:

The specific behavior of light that will be essential to ensure the success of your design is "Reflection". This is because light maze makes use of a mirror and it's the light that is reflected that we see with our eyes. Also, the manner in which light is reflected off objects will affect the colors that are reflected as well.

4 0
3 years ago
A Carnot refrigeration cycle absorbs heat at -12 °C and rejects it at 40 °C. a)-Calculate the coefficient of performance of this
tresset_1 [31]

Answer:

a)COP=5.01

b)W_{in}=2.998 KW

c)COP=6.01

d)Q_R=17.99 KW

Explanation:

Given

T_L= -12°C,T_H=40°C

For refrigeration

  We know that Carnot cycle is an ideal cycle that have all reversible process.

So COP of refrigeration is given as follows

COP=\dfrac{T_L}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{261}{313-261}

a)COP=5.01

Given that refrigeration effect= 15 KW

We know that  COP=\dfrac{RE}{W_{in}}

RE is the refrigeration effect

So

5.01=\dfrac{15}{W_{in}}

b)W_{in}=2.998 KW

For heat pump

So COP of heat pump is given as follows

COP=\dfrac{T_h}{T_H-T_L}  ,T in Kelvin.

COP=\dfrac{313}{313-261}

c)COP=6.01

In heat pump

Heat rejection at high temperature=heat absorb at  low temperature+work in put

Q_R=Q_A+W_{in}

Given that Q_A=15KW

We know that  COP=\dfrac{Q_R}{W_{in}}

COP=\dfrac{Q_R}{Q_R-Q_A}

6.01=\dfrac{Q_R}{Q_R-15}

d)Q_R=17.99 KW

5 0
3 years ago
A large plate is fabricated from a steel alloy that has a plane strain fracture toughness of 82.4 M P a m. If the plate is expos
bekas [8.4K]

Answer:

Answer for the question is given in the attachment.

Explanation:

3 0
3 years ago
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Tryna boost my score for college stuff could you give me the brainiest and a thanks? Hope you find your answer your looking for!
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3 years ago
Luids of viscosities 1  =0.1 N.s/ms and 2  =0.15 N.s/m2 are contained between two plates (each plate 1
Fantom [35]

Answer:

a) 1 / 3 N

b) 5/3 m/s

Explanation:

For constant speed V at the interface of the two fluids the net force is zero.

Hence, F top = F bottom thus the shear stresses at top and bottom must be the same.

t_{top} = t_{bottom}\\where\\t = u.\frac{dU}{dt} \\Hence\\(\frac{V - 1}{h_{2} })*u_{top}  = (\frac{V}{h_{1} })*u_{bottom} \\\\0.3V = 0.5\\\\V = 5/3 m/s

For force of top plate

F_{top} = u_{top} * \frac{V-U}{h_{2} }*A=( 0.15)*(\frac{5/3 -1}{0.3 } )*(1)\\F_{top} =   \frac{1}{3} N

5 0
4 years ago
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