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lisabon 2012 [21]
3 years ago
9

What are the important factors needed to be considered while selecting a brake or clutch?

Engineering
1 answer:
AysviL [449]3 years ago
4 0

Answer:

The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:

-Engagement

-Friction

-Electromagnetic

-Mechanical

-Actuation

-Electric

-Fluid power

-Self-actuation

-Key concepts

-Application notes

-Selection criteria

Explanation:

Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.

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Explanation:

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A gas stream contains 4.0 mol % NH3 and its ammonia content is reduced to 0.5 mol % in a packed absorption tower at 293 K and 10
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Answer:

Explanation:

Step by step solved solution is given in the attached document.

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3 years ago
(True/False) Unix is written in the C language. *<br> True<br> O False
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Answer:

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Misha Larkins [42]

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If we have silicon at 300K with 10 microns of p-type doping of 4.48*10^18/cc and 10 microns of n-type doping at 1000 times less
liq [111]

Answer:

The resistance is 24.9 Ω

Explanation:

The resistivity is equal to:

R=\frac{1}{N_{o}*u*V } =\frac{1}{4.48x10^{15}*1500*106x10^{-19}  } =0.93ohm*cm

The area is:

A = 60 * 60 = 3600 um² = 0.36x10⁻⁴cm²

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{A} }+\frac{1}{N_{D} })

If NA is greater, then, the term 1/NA can be neglected, thus the equation:

w=\sqrt{\frac{2E(V_{o}-V) }{p}(\frac{1}{N_{D} })

Where

V = 0.44 V

E = 11.68*8.85x10¹⁴ f/cm

V_{o} =\frac{KT}{p} ln(\frac{N_{A}*N_{D}}{n_{i}^{2}  } , if n_{i}=1.5x10^{10}cm^{-3}  \\V_{o}=0.02585ln(\frac{4.48x10^{18}*4.48x10^{15}  }{(1.5x10^{10})^{2}  } )=0.83V

w=\sqrt{\frac{2*11.68*8.85x10^{-14}*(0.83-0.44) }{1.6x10^{-19}*4.48x10^{15}  } } =3.35x10^{-5} cm=0.335um

The length is:

L = 10 - 0.335 = 9.665 um

The resistance is:

Re=\frac{pL}{A} =\frac{0.93*9.665x10^{-4} }{0.36x10^{-4} } =24.9ohm

7 0
3 years ago
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