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DedPeter [7]
3 years ago
8

In addition to National Electrical Code requirements, some areas have ___ requirements for energy conservation that also affect

luminaire selection.
Physics
1 answer:
Alex777 [14]3 years ago
4 0

Answer:

building code

Explanation:

National Electrical Code establishes the minimum conditions that must be met by electrical installations to preserve the safety of people and property, as well as ensure the reliability of their operation. It applies to installations in buildings destined for homes, shops, offices and for facilities in premises where similar functions are fulfilled, including temporary or provisional ones.

The construction code is something extra to help the conservation of energy of the building affecting the provision of energy even its luminaries. This type of code is chosen to increase the sustainability of said building .

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You push a 2.3 kg block against a horizontal spring, compressing the spring by 17 cm. then you release the block, and the spring
horsena [70]
Some dogs may inherit a susceptibility to epilepsy.

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3 years ago
Ocean waves pass through two small openings, 20.0 m apart, in a breakwater. You're in a boat 70.0 m from the breakwater and init
Klio2033 [76]

Answer:

λ = 5.65m

Explanation:

The Path Difference Condition is given as:

δ=(m+\frac{1}{2})\frac{lamda}{n}  ;

where lamda is represent by the symbol (λ) and is the wavelength we are meant to calculate.

m = no of openings which is 2

∴δ= \frac{3*lamda}{2}

n is the index of refraction of the medium in which the wave is traveling

To find δ we have;

δ= \sqrt{70^2+(33+\frac{20}{2})^2 }-\sqrt{70^2+(33-\frac{20}{2})^2 }

δ= \sqrt{4900+(\frac{66+20}{2})^2}-\sqrt{4900+(\frac{66-20}{2})^2}

δ= \sqrt{4900+(\frac{86}{2})^2 }-\sqrt{4900+(\frac{46}{2})^2 }

δ= \sqrt{4900+43^2}-\sqrt{4900+23^2}

δ= \sqrt{4900+1849}-\sqrt{4900+529}

δ= \sqrt{6749}-\sqrt{5429}

δ=  82.15 -73.68

δ= 8.47

Again remember; to calculate the wavelength of the ocean waves; we have:

δ= \frac{3*lamda}{2}

δ= 8.47

8.47 = \frac{3*lamda}{2}

λ = \frac{8.47*2}{3}

λ = 5.65m

3 0
4 years ago
What are the magnitude and direction of the acceleration of an electron at a point where the electric field has magnitude 7400 N
bezimeni [28]

Answer: a = 1.32 * 10^18m/s² due north

Explanation: The magnitude of the force required to move the electron is given as

F = ma

The force exerted on the charge by the electric field of intensity (E) is given by

F = Eq

Thus

Eq = ma

a = E * q/ m

Where a = acceleration of charge

E = strength of electric field = 7400N/c

q = magnitude of electronic charge = 1.609 * 10^-6c

m = mass of an electronic charge = 9.109 * 10^-31kg

a = 7400 * 1.609 * 10^-16/ 9.109 * 10^-31

a = 11906.6 * 10^-16 / 9.019 * 10^-31

a = 1.19 * 10^-12 / 9.019 * 10^-31

a = 0.132 * 10^19

a = 1.32 * 10^18m/s²

As stated in the question, the direction of the electric field is due north hence, the direction of it force will also be north thus making the electron experience a force due north ( according to Newton second law of motion)

5 0
3 years ago
An elastic band is hung on a hook and a mass is hung on the lower end of the band. When the mass is pulled downward and then rel
pshichka [43]

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

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a(t) = v′(t) = −9cos(t) −9sin(t)

5 0
3 years ago
Explain the process of a nuclear fusion reaction using hydrogen. Include the particles that are used to start and maintain the c
Zielflug [23.3K]

Answer it might be this

Explanation:

8 0
3 years ago
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