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Vanyuwa [196]
3 years ago
5

An industrial system is rated at 600 v the system includes a partially exposed terminal block thats mounted on a bulkhead a new

electrical apprentice who isnt considered to be qualified according to OSHA's standards approches the energized block the closest the apprentice should come to the block is?
Engineering
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

10 ft or 3.048 m (for unqualified person)

1.09 ft or 0.33 m (for qualified person)

Explanation:

OSHA has prescribed "Minimum Approach Distance - MAD" to ensure safety of personnel working with electrical power lines.

For a unqualified person:

The minimum approach distance is 10 ft or 3.048 m

For a qualified person:

For a voltage in the range of 300V to 750V (Phase to Ground or Phase to Phase) the minimum approach distance is 1.09 ft or 0.33 m

Since 600V lies in the above range hence it is applicable on 600V.

OSHA has developed a full list of minimum approach distance according to the level of voltages (50V to 72000V)

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Never operate electric tools outdoors or in wet conditions unless circuit is protected by what?.
Andreyy89

Never operate electric tools outdoors or in wet conditions unless the circuit is protected by a ground fault circuit interrupter (GFCI).

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When a number of electric entities are connected through various conductors is called a circuit.

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A ground fault circuit interrupter (GFCI) helps to avoid the repercussion caused due to electric shocks. An individual receives a shock, the GFCI perceives this and cuts off the current before the individual can get any damage. it is usually installed where electrical circuits come into reference with water.

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2 years ago
What is the increase of entropy principle?
Sonja [21]

Answer:

Explanation:

For an isolated system (one that does not exchange matter of energy with its surroundings) entropy tends to raise.

The entire universe might be considered as an isolated system, and thus the entropy of the universe tends to rise and it never lowers.

Note that this is for isolated systems, not any systems.

A certain system might have its entropy go down, but for that it must exchange energy with its surroundings, and the entropy of the surroundings will raise. The change of entropy of the surroundings plus the change of entropy of the system (remember their signs) will always be positive.

5 0
3 years ago
A high compression ratio may result in;
Delvig [45]

A high compression ratio may result in compressor failure.

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A compressor refers to a mechanical device that is designed and developed to provide power to refrigerators, especially by increasing the pressure on air or other applicable gases.

According to heating, ventilation, and air conditioning (HVAC) information, a high compression ratio of 8: 1 or higher is most likely to result in compressor failure.

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4 0
2 years ago
The aluminum rod (E1 = 68 GPa) is reinforced with the firmly bonded steel tube (E2 = 201 GPa). The diameter of the aluminum rod
Vsevolod [243]

Answer:

Explanation:

From the information given:

E_1 = 68 \ GPa \\ \\ E_2 = 201 \ GPa  \\ \\ d = 25 \ mm \  \\ \\ D = 45 \ mm \ \\ \\ L   = 761 \ mm  \\ \\ P = -88 kN

The total load is distributed across both the rod and tube:

P = P_1+P_2 --- (1)

Since this is a composite column; the elongation of both aluminum rod & steel tube is equal.

\delta_1=\delta_2

\dfrac{P_1L}{A_1E_1}= \dfrac{P_2L}{A_2E_2}

\dfrac{P_1 \times 0.761}{(\dfrac{\pi}{4}\times .0025^2 ) \times 68\times 10^4}= \dfrac{P_2\times 0.761}{(\dfrac{\pi}{4}\times (0.045^2-0.025^2))\times 201 \times 10^9}

P_1(2.27984775\times 10^{-8}) = P_2(3.44326686\times 10^{-9})

P_2 = \dfrac{ (2.27984775\times 10^{-8}) P_1}{(3.44326686\times 10^{-9})}

P_2 = 6.6212 \ P_1

Replace P_2 into equation (1)

P= P_1 + 6.6212 \ P_1\\ \\ P= 7.6212\ P_1 \\ \\  -88 = 7.6212 \ P_1  \\ \\ P_1 = \dfrac{-88}{7.6212} \\ \\  P_1 = -11.547 \ kN

Finally, to determine the normal stress in aluminum rod:

\sigma _1 = \dfrac{P_1}{A_1} \\ \\  \sigma _1 = \dfrac{-11.547 \times 10^3}{\dfrac{\pi}{4} \times 25^2}

\sigma_1 = - 23.523 \ MPa}

Thus, the normal stress = 23.523 MPa in compression.

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