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3 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 ____.

1 answer:

marshall27 [118]3 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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A distillation column is initially designed to separate a mixture of toluene and xylene at around ambient temperature (say, 100°

weeeeeb [17]

Answer:

Purchase cost= 87056

Bar module cost= 292725

Explanation:

solution is attached below

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4 years ago

Determine the following parameters for the water having quality x=0.7 at 200 kPa:

ra1l [238]

Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

Temperature = 120.21°C

c). Specific volume

$v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

$v_x=v_f+x(v_g-v_f)$

$=0.001061+0.7(0.88578-0.001061)$

$=0.62036 \ m^3/kg$

d). Specific energy ( $u_x$ )

$u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

$u_x=504.5 + 0.7(2024.6)$

$=1921.72 \ kJ/kg$

e). Specific enthalpy $(h_x)$

At $h_f = 504.71, \ \ h_{fg} = 2201.6$

$h_x=504.71+(0.7\times 2201.6)$

$= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

$H=mh_x$

$= 0.5 \times 2045.83$

= 1022.91 kJ

7 0

3 years ago

15 POINTS! Help.

IRISSAK [1]

Answer: it would overload

Explanation:

4 0

3 years ago

Read 2 more answers

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago

214Bi83 --> 214Po84 + eBismuth-214 undergoes first-order radioactive decay to polonium-214 by the release of a beta particle,

Zolol [24]

Answer:

(C) ln [Bi]

Explanation:

Radioactive materials will usually decay based on their specific half lives. In radioactivity, the plot of the natural logarithm of the original radioactive material against time will give a straight-line curve. This is mostly used to estimate the decay constant that is equivalent to the negative of the slope. Thus, the answer is option C.

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3 years ago

Read 2 more answers