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blsea [12.9K]
3 years ago
15

Which type of engineering design uses an already existing design

Engineering
1 answer:
Snezhnost [94]3 years ago
3 0

Answer:

Reverse engineering

Explanation:

Reverse Engineering is the remaking of already made products following the deconstruction and examination of the product to make known the product design, code and architecture features, gain knowledge of the composition and construction in a scientific research approach

Reverse engineering is also known as back engineering and consists of three main stages

1) Recovery implementation

2) Design recovery

3) Recovery analysis.

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A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027
Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = \frac{d}{2} = 3 inches

density,  \rho} = 493 lbm/ ft^{3}

S.G = 1.0027

g = 9.8 m/ m^{2} = 386.22 inch/ s^{2}

Solution:

Using the formula for terminal velocity,

v_{T} = \sqrt{\frac{2V\rho  g}{A \rho C_{d}}}              (1)

(Since, m = V\times \rho)

where,

V = volume of sphere

C_{d} = drag coefficient

Now,

Surface area of sphere, A = 4\pi r^{2}

Volume of sphere, V = \frac{4}{3} \pi r^{3}

Using the above formulae in eqn (1):

v_{T} = \sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho  g}{4\pi r^{2} \rho C_{d}}}

v_{T} = \sqrt{\frac{2gr}{3C_{d}}}  

v_{T} = \sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}

Therefore, terminal velcity is given by:

v_{T} = \frac{27.79}{\sqrt{C_d}} inch/sec

3 0
3 years ago
State three active materials of a lead acid cell​
igomit [66]

Answer:

lead dioxide,sulfate and lead acid

6 0
3 years ago
Read 2 more answers
Do not answer pls thank you
astra-53 [7]

The answer is answered! Explanation:

4 0
3 years ago
Read 2 more answers
Calculate the convective heat-transfer coefficient for water flowing in a round pipe with an inner diameter of 3.0 cm. The water
olasank [31]

Answer:

h = 10,349.06 W/m^2 K

Explanation:

Given data:

Inner diameter = 3.0 cm

flow rate  = 2 L/s

water temperature 30 degree celcius

Q = A\times V

2\times 10^{-3} m^3 = \frac{\pi}{4} \times (3\times 10^{-2})^2 \times velocity

V = \frac{20\times 4}{9\times \pi} = 2.83 m/s

Re = \frac{\rho\times V\times D}{\mu}

at 30 degree celcius = \mu = 0.798\times 10^{-3}Pa-s , K  = 0.6154

Re = \frac{10^3\times 2.83\times 3\times 10^{-2}}{0.798\times 10^{-3}}

Re = 106390

So ,this is turbulent flow

Nu = \frac{hL}{k} = 0.0029\times Re^{0.8}\times Pr^{0.3}

Pr= \frac{\mu Cp}{K} = \frac{0.798\times 10^{-3} \times 4180}{0.615} = 5.419

\frac{h\times 0.03}{0.615}  = 0.0029\times (1.061\times 10^5)^{0.8}\times 5.419^{0.3}

SOLVING FOR H

WE GET

h = 10,349.06 W/m^2 K

6 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
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