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AnnZ [28]
3 years ago
5

How many LEDs must be combined into one light source to give a total of 3.8 W of visible-light output (comparable to the light o

utput of a 100 W incandescent bulb)?
Physics
1 answer:
notka56 [123]3 years ago
4 0

Answer:

This is an incomplete question. The complete question is --

An individual white LED (light-emitting diode) has an efficiency of 20% and uses 1.0 W of electric power.

How many LEDs must be combined into one light source to give a total of 3.8W of visible-light output (comparable to the light output of a 100W incandescent bulb)?

And the answer is --

19 LEDs

Explanation:

The full form of LED is Light emitting diode.

It is given that the efficiency of the LED bulb is 20 %

1 LED uses power = 1 W

So the output power of 1 LED = 0.2 W

We need to find the power required to give a 3.8 W light.

Power required for 3.8 W = Number of LEDs required  =   (total required power  /  power required for 1 LED )

                                                                                         = 3.8 / 0.2

                                                                                         = 19

Therefore, the number of LEDs required is 19.

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velikii [3]

Answer:

Approximately 4.2\; {\rm s} (assuming that the projectile was launched at angle of 35^{\circ} above the horizon.)

Explanation:

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\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}.

Since there is no drag on this projectile, the vertical acceleration of this projectile would be g. In other words, a = g = -9.81\; {\rm m\cdot s^{-2}}.

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\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}.

Hence, this projectile would be in the air for approximately 4.2\; {\rm s}.

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