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2 years ago

The specific grounding and bonding requirements for an information technology room are located in which article of the nec?

1 answer:

6 0

The specific grounding and bonding requirements for an information technology room are located in Article 645.

<h3>What does NEC stand for?</h3>

The National Electrical Code (NEC) is a collection of requirements for the secure installation of electrical wiring in the US that are routinely updated.

Bonding is the joining of conductors that do not convey current, such as enclosures and buildings. Bonded systems are attached to the earth by grounding. To protect persons and property from electric risks, both are required.

Article 645 compliance is optional. The two main pardons provided by Article 645 are. First, underneath a raised floor, non-plenum-rated cables are acceptable. Second, cables, boxes, and similar items for the mentioned IT equipment are exempt from the need for security. sted IT equipment cables, boxes, and the like are not required to be secured in place.

To learn more about NEC refer to:

brainly.com/question/20348027

#SPJ4

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If a structure can withstand seismic stress, what is it prepared for?.

mars1129 [50]

Answer:

Steel and wood

Explanation:

For a material to resist stress and vibration, it must have high ductility, which is the ability to undergo large deformations and tension. Modern buildings are often constructed with structural steel, a component that comes in a variety of shapes and allows buildings to bend without breaking.

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2 years ago

Print a message telling a user to press the letterToQuit key numPresses times to quit. End with newline. Ex: If letterToQuit = '

ella [17]

Answer:

Vb.Net

msgbox ("Press "q" twice to quit", msgboxstyle.information)

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3 years ago

KVL holds for the supermesh, so we can write a KVL equation to generate the second equation we need to solve for the two unknown

kaheart [24]

Answer:

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

Explanation:

As the complete question is not given the complete question is found online and is attached herewith.

By applying KCL at node 1

$i_x+50mA=i_y\\i_x-i_y=0.05A$

Also

$V_{\Delta}=1K*i_y$

Now applying KVL on loop 1 as indicated in the attached figure

$1K*i_y+5K(i_y-i_z)+3K*i_x=0\\3i_x+6 i_y-5i_z=0$

Similarly for loop 2

$2V_{\Delta}+5K(i_z-i_y)=0\\2*1K*i_y+5K(i_z-i_y)=0\\2K*i_y+5K(i_z-i_y)=0\\3i_y-5i_z=0$

So the system of equations become

$i_x-i_y+0i_z=0.05\\3i_x+6i_y-5i_z=0\\0i_x-3i_y+5i_z=0$

Solving these give the values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA. Also the value of voltage is given as

$V_{\Delta}=1K*i_y\\V_{\Delta}=1K*-25 mA\\V_{\Delta}=-25 V$

The values of i_x,i_y and i_z as 25 mA, -25 mA and 15 mA while that of V_Δ is -25 V

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4 years ago

El tiempo hasta que falle un sistema informático sigue una distribución Exponencial con media de 600hs. (Utilice 3 decimales par

Lesechka [4]

Answer:

La probabilidad pedida es $0.717$

Explanation:

Primero comencemos definiendo la variable aleatoria. Para nuestro problema, la variable aleatoria es la siguiente :

$X:$ '' El tiempo (en horas) hasta que falle un sistema informático ''

La variable aleatoria $X$ será entonces una variable aleatoria continua.

Sabemos que sigue una distribución exponencial con una media de 600 hs.

Esto se escribe :

$X$ ~ ε ( λ ) (I)

En donde λ es igual a la inversa de la media. Esto se escribe :

λ $=\frac{1}{E(X)}$

En donde $E(X)$ es la media de la variable. Por ende, si reemplazamos los datos del ejercicio obtenemos ⇒

λ $=\frac{1}{E(X)}$ ⇒ λ $=\frac{1}{600}$

Si reemplazamos el valor de λ en (I) obtenemos :

$X$ ~ ε $(\frac{1}{600})$

La función de distribución de $X$ (por ser una variable aleatoria exponencial) es :

$F_{X}(x)=P(X\leq x)=$ 1 - e ^ ( - λx) con x > 0 y $F_{X}(x)=0$ en caso contrario.

Si reemplazamos el valor de λ en la función de distribución de $X$ obtenemos :

$F_{X}(x)=P(X\leq x)=1-e^{-\frac{x}{600}}$

Dado que la variable aleatoria $X$ se distribuye de manera exponencial, el hecho de saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos aporta ninguna información sobre lo que ocurrirá después. Esta característica se conoce como propiedad de perdida de memoria de la variable aleatoria exponencial. Entonces, la probabilidad pedida se reduce a calcular :

$P(X>200)$

Dado que saber que el sistema ha estado funcionando sin fallas durante 400 hs no nos dice nada sobre lo que ocurrirá instantes posteriores a esas 400 hs.

Calculamos entonces la probabilidad pedida :

$P(X>200)=1-P(X\leq 200)=1-F_{X}(200)=1-(1-e^{-\frac{200}{600}})=e^{-\frac{1}{3}}=0.717$

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3 years ago

What is the purpose of an engineer?

amid [387]

Answer:

Their purpose is to invent, design, analyze, build and test machines, complex systems, structures, gadgets and materials to fulfill functional objectives and requirements while considering the limitations imposed by practicality, regulation, safety and cost.

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3 years ago

Read 2 more answers