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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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Suppose a person pushes thumbtack that is 1/5 centimeter long into a bulletin board, and the force (in dynes) exerted when the d
mario62 [17]

Answer:

W = 290.7 dynes*cm

Explanation:

d = 1/5 cm = 0.2 cm

The force is in function of the depth x:

F(x) = 1000 * (1 + 2*x)^2

We can expand that as:

F(x) = 1000 * (1 + 4*x + 4x^2)

F(x) = 1000 + 4000*x + 4000*x^2

Work is defined as

W = F * d

Since we have non constant force we integrate

W = \int\limits^{0.2}_{0} {(1000 + 4000*x + 4000*x^2)} \, dx

W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2

W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3

W = 200 + 80 + 10.7 = 290.7 dynes*cm

3 0
4 years ago
A particle moves in a straight line with the velocity function v ( t ) = sin ( w t ) cos 3 ( w t ) . find its position function
Sunny_sXe [5.5K]

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

<h3>How to get the position equation of the particle?</h3>

Let the velocity of the particle is:

$v(t)=\sin (\omega t) * \cos ^2(\omega t)

To get the position equation we just need to integrate the above equation:

$f(t)=\int \sin (\omega t) * \cos ^2(\omega t) d t

$\mathrm{u}=\cos (\omega \mathrm{t})

Then:

$d u=-\sin (\omega t) d t

\Rightarrow d t=-d u / \sin (\omega t)

Replacing that in our integral we get:

$\int \sin (\omega t) * \cos ^2(\omega t) d t$

$-\int \frac{\sin (\omega t) * u^2 d u}{\sin (\omega t)}-\int u^2 d t=-\frac{u^3}{3}+c$

Where C is a constant of integration.

Now we remember that $u=\cos (\omega t)$

Then we have:

$f(t)=\frac{\cos ^3(\omega t)}{3}+C

To find the value of C, we use the fact that f(0) = 0.

$f(t)=\frac{\cos ^3(\omega * 0)}{3}+C=\frac{1}{3}+C=0

C = -1 / 3

Then the position function is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

Integrating the velocity equation, we will see that the position equation is:

$f(t)=\frac{\cos ^3(\omega t)-1}{3}

To learn more about motion equations, refer to:

brainly.com/question/19365526

#SPJ4

4 0
2 years ago
A bicyclist in the Tour de France crests a mountain pass as he moves at 18 km/h. At the bottom, 4.0 km farther, his speed is at
Allisa [31]

We are given:

v0 = initial velocity = 18 km/h

d = distance = 4 km

v = final velocity = 75 km/h

a =?

<span>
We can solve this problem by using the formula:</span>

v^2 = v0^2 + 2 a d

 

75^2 = 18^2 + 2 (a) * 4

5625 = 324 + 8a

<span>a = 662.625 km/h^2</span>

6 0
3 years ago
Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
Nana76 [90]

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

4 0
3 years ago
Read 2 more answers
An electrical device in which a long wire is wound into a succession of closely spaced loops around a n insulating core is calle
Iteru [2.4K]

Answer:

Solenoid I think.

Explanation:

6 0
3 years ago
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