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Fofino [41]
2 years ago
5

How many conditions does the NEC list whereby conductors shall be considered to be outside of a building or other structure?

Physics
1 answer:
Likurg_2 [28]2 years ago
4 0

Answer:

4 conditions

Explanation:

The National Electrical Code (NEC) provides safety guidelines for the installation of electrical wiring and electrical equipment in the United States. The purpose behind NEC is to standardize the safe electrical installation practices.

There is always a confusion among people to distinguish between inside or outside of a building or structure. Therefore, NEC has listed 4 conditions in the Article 230.6 where it has mentioned conductors to be considered outside a building or structure.

Conductors are considered outside a building when they are installed:

1. Under not less than 2 inches of concrete beneath a building or structure.

2. Within a building or structure in a raceway that is encased in no less than 2 inches thick of concrete or brick.

3. Installed in a vault that meets the construction requirements of Article 450, Part III.

4. In conduit under not less than 18 inches of earth beneath a building or structure.

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The nucleus of an atom has all of the following characteristics except that it
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THE ANSWER IS: contains nearly all of the atom's volume.

Explanation:

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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

y = A cos(\omega t)

here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

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now we have

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6 0
2 years ago
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A physics student with a stopwatch drops a rock into a very deep well, and measures the time between when he drops the rocks and
tankabanditka [31]

Answer:

h= 161.06 m

Explanation:

Given that

Speed of the sound ,C= 343 m/s

Total time ,t= 6.2 s

lets take the depth of the well = h

The time taken by stone before striking the water = t₁

we know that

h=\dfrac{1}{2}gt_1^2

t_1=\sqrt{\dfrac{2h}{g}}

The time taken by sound =t₂

t_2=\dfrac{h}{343}

The total time

t = t₁+ t₂

6.2 = \sqrt{\dfrac{2h}{g}}+\dfrac{h}{343}

6.2 = \sqrt{\dfrac{2h}{9.81}}+\dfrac{h}{343}

Now by solving the above equation we get

h= 161.06 m

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6 0
2 years ago
The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo
irina1246 [14]

Answer:

p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0
2 years ago
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