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Elan Coil [88]
3 years ago
14

Which parts of LED bulb needs to stay cool as well as which part must transfer energy away from the diode???

Chemistry
1 answer:
storchak [24]3 years ago
6 0

Answer:

every method of removing heat from LED's should be considered. Conduction, convection, and radiation are the three means of heat transfer. Typically, LED's are encapsulated in a transparent resin, which is a poor thermal conductor. Nearly all heat produced is conducted through the back side of the chip. Heat is generated from the PN junction by electrical energy that was not converted to useful light, and conducted to outside ambiance through a long path, from junction to solder point, solder point to board, and board to the heat sink and then to the atmosphere. A typical LED side view and its thermal model are shown in the figures.

Explanation:

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Please help , science tho. by the way i have to do 5
kap26 [50]

Answer:

850

Explanation:

It's between 840 and 860

8 0
2 years ago
Why is it important for scientists to use a classification
Kobotan [32]
So they can tell what exact species it is. 
7 0
3 years ago
Which of these solutes raises the boiling point of water the most?
sergey [27]

Answer:

D. ionic sodium phosphate (Na3PO4)

Explanation:

Molecule for molecule, the solute that raises the boiling point of water the most is the one that makes the most particles in the solution. Lithium chloride breaks up into two ions (Li+ and Cl-). So does sodium chloride (Na+ and Cl-). Molecular molecules don't break up at all, so sucrose has only 1 particle per molecule. Sodium phosphate makes 4 total particles (3 Na+ ions and 1 PO4^3-). And magnesium bromide would make 3 particles (1 Mg2+ and 2 Br-). So the most is 4.

6 0
3 years ago
The alkali metals cesium (Cs) and rubidium (Rb) were discovered based on their characteristic flame colors. Cesium is named afte
Dmitry_Shevchenko [17]

well Cs=blue/violet and Rb= red

5 0
3 years ago
On the basis of the information above, what is the approximate percent ionization of HNO2 in a 1.0 M HNO2 (aq) solution?
enot [183]

Answer:

The answer is "2%"

Explanation:

Equation:

HNO_2\ (aq) \leftrightharpoons  H^{+} \ (aq) + NO_2^{-}\ (aq) \\\\\  K_a = 4.0\times \ 10^{-4}

H^{+}=?

Formula:

Ka = \frac{[H^{+}][NO_2^{-}]}{[HNO_2]}

Let

[H^{+}] = [NO_2^{-}] = x at equilibrium

x^2 = (4.0\times 10^{-4})\times 1.0\\\\x = ((4.0\times 10^{-4})\times 1.0)^{0.5} = 2.0 \times 10^{-2} \  M\\\\

therefore,

[H^{+}] = 2.0\times 10^{-2} \ M = 0.02 \ M

Calculating the % ionization:

= \frac{([H^{+}]}{[HNO_2])} \times 100 \\\\= \frac{0.02}{1}\times 100 \\\\= 2\%\\\\

6 0
3 years ago
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