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lukranit [14]
3 years ago
14

What measures can be taken to mitigate or prevent the formation of stress corrosion cracks?

Engineering
1 answer:
sasho [114]3 years ago
7 0

Answer:

a) Selection and control of material

b) Control of stress

c) Control of the environment

Explanation:

The measures that can be taken to control and prevent the formation of strss corrosion cracks are:

a) Selection and control of material

It deals with the selection of the best quality material that can resist the stress corrosion cracks to a high extent.

b) Control of stress

It is to control or reduce the intensity of the major stress that causes the stress corrosion cracks.

c) Control of the environment

It includes the prevention of the material from the corrosive environment by removing the component from the region or controlling the environment condition around the object.

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1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu
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Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

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3 years ago
An escalator handles a steady load of 26 people per minute in elevating them from the first to the second floor through a vertic
photoshop1234 [79]

Answer:

\eta = 70.711\,\%

Explanation:

The power needed to make the escalator working is obtained by means of the Work-Energy Theorem:

\dot W  = \dot U_{g}

\dot W = \dot n \cdot m_{p}\cdot g \cdot \Delta y

\dot W = \left(26\,\frac{persons}{min}\right)\cdot (124\,lbm)\cdot \left(32.174\,\frac{ft}{s^{2}}\right)\cdot \left(\frac{1\,lbf}{32.174\,\frac{lbm\cdot ft}{s^{2}} } \right)\cdot (27.5\,ft)

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The mechanical efficiency of the escalator is:

\eta = \frac{2.687\,hp}{3.8\,hp}\times 100\,\%

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3 0
3 years ago
A 4KJ of energy is supplied to a machine used for lifting a mass. The force required is 800N. If the machine has an efficiency o
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The height at which the mass will be lifted is; 3 meters

<h3>How to utilize efficiency of a machine?</h3>

Formula for efficiency is;

η = useful output energy/input energy

We are given

η = 60% = 0.6

Input energy = 4 KJ = 4000 J

Thus;

0.6 = useful output energy/4000

useful output energy = 0.6 * 4000

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Work done in lifting mass(useful output energy) = force * distance moved

Useful output energy = 800 * h

where h is height to lift mass

Thus;

800h = 2400

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Read more about Machine Efficiency at; brainly.com/question/3617034

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