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levacccp [35]
3 years ago
8

What is JIT and why would you advocate it to reduce inventory?

Engineering
1 answer:
bazaltina [42]3 years ago
7 0

Explanation:

Just in time inventory is a system of managing inventory. It is a method to reduce waste, lower cost and increase profit. According to it the raw material orders are directly aligned with the production schedules. In simple words every component of a unit is arriving just in time to be used.

So JIT would be advocated to reduce inventory.

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katrin2010 [14]
Customer service

Since she has reckless driving and her record, anything involving driving of some sorts mechanics as well won’t work well with her!! Hope this helps
8 0
3 years ago
why would it be inappropriate to dimension to a feature on a surface that is not perpendicular to the line of sight?
bazaltina [42]
If you are drawing and dimensioning with a computer program the dimension will be inaccurate... If it is mechanical drawing then the fabricator would not have enough information to accurately measure the component. ie a circle turned a few degrees away from perp. would appear to be an ellipse. and may actually dimension that way
4 0
3 years ago
Why are photopolymers essential for stereolithography?
Leokris [45]

Answer:

They essential for the formation of layers in a 3-D model.

Explanation:

Stereolithography is 3D printing process in which a 3D model can be created. In stereolithography technology, there is vat of the UV curable photopolymer which is placed below in lower compartment of machine.

<u>When the machine begins to construct 3D model by the scanning and the  building of one layer at one time. Each layer in the model is constructed with UV laser and as laser traces next layer of object material hardens on contact.  </u>

Once layer is complete, platform slips down to make room for next layer. This is how, a model is created.

5 0
3 years ago
A cylindrical specimen of a hypothetical metal alloy is stressed in compression. If its original and final diameters are 19.636
luda_lava [24]

Answer:

The original length of the specimen is found to be 76.093 mm.

Explanation:

From the conservation of mass principal, we know that the volume of the specimen must remain constant. Therefore, comparing the volumes of both initial and final state as state 1 and state 2:

Initial Volume = Final Volume

πd1²L1/4 = πd2²L2/4

d1²L1 = d2²L2

L1 = d2²L2/d1²

where,

d1 = initial diameter = 19.636 mm

d2 = final diameter = 19.661 mm

L1 = Initial Length = Original Length = ?

L2 = Final Length = 75.9 mm

Therefore, using values:

L1 = (19.661 mm)²(75.9 mm)/(19.636 mm)²

<u>L1 = 76.093 mm</u>

5 0
3 years ago
In a production facility, 3 cm thick large brass plates (k = 110 W/mC, α = 33.9 × 10-6 m2 /s) that are initially at a uniform
zysi [14]

Answer:

Explanation:

Given.

Thickness of brass plate t = 3 cm

Thermal conductivity of brass k = 110 W/m.°C

Density of brass \rho = 8530 kg/m^3

Specific heat of brass C_p =380J/kg.°C

Thermal diffusivity of brass \alpha = 33.9\times 10^{-6} m^2/s

Temperature of oven T_{\infty} = 700°C

The initial temperature T_i= 25°C

Plate remain in the oven t =10 min  

Heat conduction in the plate is one-dimensional since the plate is large relative to its thickness  and there is thermal symmetry about the center plane.

The thermal properties of the plate are constant.

The heat transfer coefficient is constant and uniform over the entire surface.

The Fourier number is > 0.2 so that the one-term approximate solutions (or the transient  temperature charts) are applicable (this assumption will be verified).

The Biot number for this process Bi = \frac{hL}{k}\\\\Bi=\frac{(80 W/m^2.°C)(0.015 m)}{(110 W/m.°C)}\\=Bi =0.0109

The constants \lambda_1 and A_1 corresponding to this Biot are, from 11-2 tables.

The interpolation method used to find the

\lambda_1=0.1039  and A_1=1.0018&#10;  

The Fourier number \tau=\frac{\alpha t}{L^2}\\\\\tau=\frac{(33.9\times 10^{-6} m^2/s)(10 min \times 60 s/min)}{(0.015m)^2}&#10;\\\\\tau=90.4>0.2

Therefore, the one-term approximate solution (or the transient temperature charts) is applicable.

Then the temperature at the surface of the plates becomes

\theta(L,t)_{wall}=\frac{T(x,t)-T_{\infty}}{(T_i-T_{\infty})}\\\\\theta(L,t)_{wall}=A_1e^{-\lambda_1^2\tau}\cos(\lambda_1L/L)\\\\\theta(L,t)_{wall}=(1.0018)e^{-(0.1039^2(90.4))}\cos(0.1039)\\\\\theta(L,t)_{wall}=0.378\\\\\frac{T(L,t)-700}{25-700}=0.378\\\\T(L,t)=445°C

3 0
3 years ago
Read 2 more answers
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