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sukhopar [10]
2 years ago
15

One type of illumination system consists of rows of strip fluorescents and a ceiling that will transmit light. For this system t

o be efficient, the surfaces above the luminous ceiling must be ____.
Engineering
1 answer:
marshall27 [118]2 years ago
3 0

Answer:

For this system to be efficient, the surfaces above the luminous ceiling must be white, so that all the light been transmitted is reflected.

Explanation:

White surfaces reflect light at almost 100% efficiency, this ensures that non of the light been transmitted is absorbed.

Thus, if one type of illumination system consists of rows of strip fluorescents and a ceiling that will transmit light. For this system to be efficient, the surfaces above the luminous ceiling must be white so that all the light been transmitted is reflected.

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Which one of the following activities is not an example of incident coordination
Lady bird [3.3K]
Directing, ordering, or controlling
7 0
2 years ago
Two metallic terminals protrude from a device. The terminal on the left is the positive reference for a voltage called vx (the o
tresset_1 [31]

Answer:

Vx = 6.242 x 10raised to power 15

Vy = -6.242 x 10raised to power 15

Explanation:

from E = IVt

but V = IR from ohm's law and Q = It from faraday's first law

I = Q/t

E = Q/t x V x t = QV

hence, E =QV

V = E/Q

3 0
2 years ago
Read 2 more answers
A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (From Sample Prob. 12.7 is the definition of rat
forsale [732]

Answer:

tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962

\theta = tan^{-1} (0.962) = 43.89

Explanation:

If the question is: Determine the banking angle θ

We have the forces involved on the figure attached.

For this case we know that the weight is given by:

W = mg

And for this case the centripetal acceleration would be given by:

a=\frac{v^2}{r}

If we analyze the sum of forces on x and y we have:

\sum F_x = m a_x

F + W sin \theta = ma cos theta

And if we solve for the force we got:

F = ma cos \theta - mg sin \theta = \frac{mv^2}{r} cos \theta - mg sin \theta

\sum F_y = m a_y

N - W cos \theta = ma sin \theta

If we solve for the normal force we got:

N =W cos \theta + ma sin \theta = \frac{mv^2}{r} sin \theta + mg cos \theta

In order to find the banking angle we use the fact that F =0

0 = \frac{mv^2}{r} cos \theta - mg sin \theta

tan \theta= \frac{v^2}{rg}

The velocity on this case is 120 mi/h if we convert this into ft/ s we got:

120 mi/h * \frac{5280 ft}{1mi} *\frac{1hr}{3600 s}= 176 ft/s

And then we have this:

tan \theta = \frac{(176ft/s)^2}{1000 ft 32.2 ft/s^2}= 0.962

\theta = tan^{-1} (0.962) = 43.89

5 0
2 years ago
The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr
Rashid [163]

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below;  if we take an integral look at the component of force in the direction of the displacement; we have

F_x= Fcos 40°

F_x= 100 (cos 40°)

F_x= 76.60 N

Workdone by the friction force can now be determined as:

W = F_x × displacement

W = 76.60 × 2

W = 153.2 J

∴  the work done by the friction force = 153.2 J

7 0
2 years ago
The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting
Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

    AC = \frac{5L}{3}T, CD = \frac{4L}{3} C

B) AB = AD = BC = BD = 0

   AC = \frac{5L}{3} T, CD = \frac{4L}{3} C

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = \frac{5L}{3} T,  CD = \frac{4L}{3}C

7 0
2 years ago
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