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4 years ago

What is pixel's intensity ?

Engineering

1 answer:

bonufazy [111]4 years ago

7 0

Answer:

Small points of colored light arranged in a grid, each is formed from three colored lights: red,green,and blue. (RGB), nothing is absorbed,nothing is reflected, just see pure colored lights.

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Air at 80 kPa and 10Â°C enters an adiabatic diffuser steadily with a velocity of 150 m/s and leaves with a low velocity at a pre

il63 [147K]

Answer:

The exit temperature is 293.74 K.

Explanation:

Given that

At inlet condition(1)

P =80 KPa

V=150 m/s

T=10 C

Exit area is 5 times the inlet area

Now

$A_2=5A_1$

If consider that density of air is not changing from inlet to exit then by using continuity equation

$A_1V_1=A_2V_2$

So $A_1\times 150=5A_1V_2$

$V_2=30$ m/s

Now from first law for open system

$h_1+\dfrac{V_1^2}{2}+Q=h_2+\dfrac{V_2^2}{2}+w$

Here Q=0 and w=0

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

When air is treating as ideal gas

$h=C_pT$

Noe by putting the values

$h_1+\dfrac{V_1^2}{2}=h_2+\dfrac{V_2^2}{2}$

$1.005\times 283+\dfrac{150^2}{2000}=1.005\times T_2+\dfrac{30^2}{2000}$

$T_2=293.74K$

So the exit temperature is 293.74 K.

7 0

3 years ago

Which condition allows the instrument air reserve header cyclinders to be stored with the instrument air compressor

goblinko [34]

Answer:

C) The only motor-driven machinery in the room is for the instrument air equipment

Explanation:

Quizlet just practiced flashcards :)

8 0

2 years ago

C&A's potato chip filling process has a lower specification limit of 9.5 oz. and an upper specification limit of 10.5 oz. Th

Law Incorporation [45]

Answer:

Process capability index = (USL - LSL)/6Sigma = 1/6(0.3)= 0.56

Answer (a) 0.56

Explanation:

3 0

3 years ago

Aggregate blend composed of 65% coarse aggregate (SG 2.701), 35% fine aggregate (SG 2.625)

dimaraw [331]

Answer:

2.674
2.428
91.695%
2.592
3.305%
11.786%
78.1%

Explanation:

coarse aggregate (ca) = 65%, SG = 2.701

Fine aggregate = 35%, SG = 2.625

A) Bulk specific gravity of aggregate

= $\frac{65*2.701 + 35*2.625}{100} = 2.674$

B) Wm = 1257.9 g { weight in air }

Ww = 740 g { submerged weight }

therefore Bulk specific gravity of compacted specimen

= $\frac{Wm}{Wm-Ww}$ = $\frac{1257.9}{1257.9 - 740 }$ = 2.428

Theoretical specific gravity = 2.511

Percent stone

= 100 - asphalt content - Vv

= 100 - 5 - 3.305 = 91.695%

c) percent of void

= $\frac{9.511-2.428}{2.511} * 100$ Vv = 3.305%

d) let effective specific gravity in stone

= $\frac{91.695*unstone+ 5 *1.030 }{96.695} = 2.511$

= Instone = 2.592 effective specific gravity of stone

e) Vv = 3.305%

f ) volume filled with asphalt (Vb) = $\frac{\frac{Wb}{lnb} }{\frac{Wm}{Inm} } * 100$

Vb = $\frac{5 * 2.428}{1.030 * 100} * 100$

Vb = 11.786 %

Volume of mineral aggregate = Vb + Vv

VMA = 11.786 + 3.305 = 15.091

g) percent void filled with alphalt

= Vb / VMA * 100

VMA = 11.786 + 3.305 = 15.091

percent void filled with alphalt

= Vb / VMA * 100 = (11.786 / 15.091) * 100 = 78.1%

3 0

3 years ago

You are asked by your college crew to estimate the skin friction drag in their eight-seat racing shell. The hull of the shell ma

Fynjy0 [20]

Answer:

The total skin friction drag on the hull under these conditions is 276N

Explanation:

In this question, we are asked to determine the total skin friction drag on the hull under the given conditions.

Please check attachment for complete solution and step by step explanation

7 0

3 years ago

What is pixel's intensity ?​

What is pixel's intensity ?