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ra1l [238]
3 years ago
8

How do information systems support the activities in a supply chain?

Engineering
1 answer:
Degger [83]3 years ago
6 0

Answer:

the no. of activities supply in a cahin like in the figuration wise they supply the chain

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A method that uses low temperature heat-treating that imparts toughness without reduction in hardness is called:_______
lbvjy [14]

Answer:

"Tempering Process" seems to be the appropriate choice.

Explanation:

  • Tempering seems to be a method of heat preparation which is mostly used in completely hard materials to increase consistency, strength, durability, and also some decreasing brittleness.
  • The tempering method is used to examine good functionality as well as flexural by reducing stiffness again after the substance has indeed been quenched towards its toughest state.
5 0
3 years ago
Name the main classes of polymer and define their characteristic properties
Svetlanka [38]

Answer:

Polymers are the naturally occurring or synthetic macromolecules that are composed of repeating subunits, called monomers.

The three main classes of polymers are: thermoplastic, thermosetting, and the elastomers.

Thermoplastic polymers have linear bonding. These polymers can be melted again and thus can recycled.

Thermosetting polymers have cross-linked bonding. These polymers decompose when heated and thus can not be remelted and recycled.

Elastomers have linear bonding with some cross-linking. These polymers extreme elastic extensibility and thus can revert back to its original shape after deformation, without causing any permanent damage.

8 0
4 years ago
Electric Resistance Heating. A house that is losing heat at a rate of 50,000 kJ/h when the outside temperature drops to 4 0C is
ryzh [129]

Answer:

a) \dot W = 0.978\,kW, b) I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW

Explanation:

a) The ideal Coefficient of Performance for the heat pump is:

COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}

COP_{HP} = \frac{298.15\,K}{298.15\,K - 277.15\,K}

COP_{HP} = 14.198

The reversible work input is:

\dot W = \frac{\dot Q_{H}}{COP_{HP}}

\dot W = \left(\frac{50000\,\frac{kJ}{h} }{14.198} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}  \right)

\dot W = 0.978\,kW

b) The irreversibility is given by the difference between real work and ideal work inputs:

I = \dot W_{real} - \dot W_{ideal}

I = \left(50000\,\frac{kJ}{h} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)\cdot \left(\frac{1}{COP_{real}} \right) - 0.978\,kW

7 0
4 years ago
What is the function of a fixed resistor?
agasfer [191]

Answer:

  add resistance to a circuit

Explanation:

It depends on the design in which it is incorporated. A fixed resistor has many uses, including, but not limited to ...

  • dropping voltage
  • limiting current
  • contributing to a time delay
  • adjusting frequency response
  • eliminating (or creating) signal reflections
  • acting as a fuse
  • calibrating or trimming a response
  • providing protection against electrical shock or ESD
  • acting as a reference when measuring variable resistors
8 0
4 years ago
An uninsulated, thin-walled pipe of 100-mm diameter is used to transport water to equipment that operates outdoors and uses the
Viefleur [7K]

Answer:

4.6 mm

Explanation:

Given data includes:

thin-walled pipe diameter = 100-mm =0.1 m

Temperature of pipe T_p = -15° C = (-15 +273)K =258 K

Temperature of water T_w = 3° C = (3 + 273)K = 276 K

Temperature of ice T_i = 0° C = (0 +273)K =273 K

Thermal conductivity (k) from the ice table = 1.94 W/m.K  ;  R = 0.05

convection coefficient Lh_l =2000 W/m².K

The energy balance can be expressed as:

q_{conduction} =q_{convention}

where;

q_{conduction} = \frac{2\pi LK(T_i-T_p)}{In(R/r)}       -------------   equation (1)

q_{convention} = \pi DLh_l(T_w-T_i)  ------------ equation(2)

Equating both equation (1) and (2); we have;

\frac{2\pi LK(T_i-T_p)}{In(R/r)} = \pi DLh_l(T_w-T_i)

Replacing the given data; we have:

\frac{2\pi (1)(1.94)(273-258)}{In(0.05/r)} = \pi (0.1)*2000(276-273)

\frac{182.84}{In(\frac{0.05}{r}) } = 1884.96

In(\frac{0.05}{r})*1884.96 = 182.84

In(\frac{0.05}{r}) = \frac{182.84}{1884.96}

In(\frac{0.05}{r}) =0.0970

\frac {0.05}{r} =e^{0.0970}

\frac {0.05}{r} =1.102

r=\frac{0.05}{1.102}

r = 0.0454

The thickness (t) of the ice layer can now be calculated as:

t = (R - r)

t = (0.05 - 0.0454)

t = 0.0046 m

t = 4.6 mm

6 0
4 years ago
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