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Evgen [1.6K]
3 years ago
13

Electrical equipment is required to be adequately protected against short-circuit current conditions per NEC 110.10 and several

other complementary NEC requirements. OSHA requires equipment short-circuit current protection in 1910.303(B)(viii)(5).
Physics
1 answer:
Anna11 [10]3 years ago
8 0

ANSWER:

The statment is true.

STEP-BY-STEP EXPLANATION:

Per Osha 1910.303 b (5):

The overcurrent protective devices, the total impedance, the component short-circuit current ratings, and other characteristics of the circuit to be protected shall be selected and coordinated to permit the circuit protective devices used to clear a fault to do so without the occurrence of extensive damage to the electrical components of the circuit. This fault shall be assumed to be either between two or more of the circuit conductors, or between any circuit conductor and the grounding conductor or enclosing metal raceway.

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a street light is mounted at the top of a 15 foot pole. A man 6 ft tall walks away from the pole wit a speed of 7 ft/s along a s
Mariulka [41]

Answer:

16.3 ft/s

Explanation:

Let d=distance

and

x = length of shadow.

Therfore,

x=(d + x)

 = 6/15

So,

    15x = 6x + 6d

     9x = 6d.

x = (2/3)d.

As we know that:

dx=dt

   = (2/3) (d/dt) 

Also,

Given:

d(d)=dt

     = 7 ft/s

Thus,

d(d + x)=dt

           = (7/3)d (d/dt)

Substitute, d= 7  

d(d + x) = 49/3 ft/s.

Hence,

d(d + x) = 16.3 ft/s.

4 0
3 years ago
an electric current of 285.0 ma flows 674. milliseconds. Calculate the amount of electric charge transported. Be sure your answe
yulyashka [42]

Answer:

<em>The amount of electric charge transported = 0.192 C</em>

Explanation:

Electric Charge: This is defined as the product of electric current and time in an electric circuit, The S.I unit of electric charge is Coulombs (C)

Q = It..................... Equation 1

Where Q = Electric charge, I = electric current, t = time.

<em>Given:</em> I = 285 mA, t = 674 milliseconds.

<em>Conversion: (i) Convert from 285 mA to A = (285/1000) A = 0.285 A</em>

<em>       (ii) convert from 674 milliseconds to seconds = (674/1000) s = 0.674 s          </em>

Substituting these values into equation 1

Q = 0.285 × 0.674

<em>Q = 0.192 C</em>

<em>Therefore the amount of electric charge transported = 0.192 C</em>

<em></em>

<em></em>

5 0
2 years ago
Which would most likely cause the cylinder head temperature and engine oil temperature gauges to exceed their normal operating r
dusya [7]

Answer: Using fuel that has a lower-than-specified fuel rating.

Explanation:

Most likely, what causes the cylinder head temperature and engine oil temperature gauges to exceed their normal operating ranges is using fuel that has a lower-than-specified fuel rating. This can lead to detonation of the engine which is the tendency for the fuel to pre-ignite or auto-ignite in an engine's combustion chamber.The cylinder head and the engine oil are part of the automobile systems that helps in fuel combustion.

3 0
3 years ago
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Vlad1618 [11]
C. thalamus
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3 years ago
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e force acting between two charged particles A and B is 5.2 × 10-5 newtons. Charges A and B are 2.4 × 10-2 meters apart. If the
anastassius [24]
The force acting between the particles is

F=k \frac{Q_{1}Q_{2}}{r^2}
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Q_{2}= \frac{5.2 \times 10^-^5 \times 0.024^2}{ 9.0 \times 10^9 7.2 \times 10^-^8} =4.622 \times 10^-^1^1C




7 0
3 years ago
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