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nydimaria [60]
2 years ago
9

LED lightbulbs are the most energy efficient way to light your home, as they use _____ of the energy of a comparable compact flu

orescent bulb.
Physics
1 answer:
Solnce55 [7]2 years ago
8 0

Light - emitting diode lightbulbs known as LED lightbulbs are energy saving lightbulbs. They consume very small amount of energy compare to other lightbulbs. Therefore, LED lightbulbs are the most energy efficient way to light your home, as they use about ten percent (10%) of the energy of a comparable compact fluorescent bulb.

There are varieties of different light bulbs. some of them are  incandescent, compact fluorescent (CFL), halogen, and light-emitting diode (LED) to mention a few.

Before the advent of light - emitting diode (LED) bulbs, some of the bulbs consume a lot of energy. For instance, we have incandescent bulbs of 20 watts, 50 watts, 100 watts and even 200 watts.

But with the advent of light - emitting diode (LED) bulbs, consumers can save energy because light - emitting diode (LED) bulbs save a lot of energy and they are very bright.

We can therefore conclude that light - emitting diode lightbulbs known as LED lightbulbs are the most energy efficient way to light your home, as they use about ten percent (10%) of the energy of a comparable compact fluorescent bulb.

Learn more here: brainly.com/question/13423221

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Answer:

the force will decrease to 3/4 of its original value.

Explanation:

The initial electric force between the two charges is:

F = k \frac{q\cdot q}{r^2}

where

k is the Coulomb's constant

q is the magnitude of each charge

r is their separation

Later, half of one charge is transferred to the other charge; this means that one charge will have a charge of

q+\frac{q}{2}=\frac{3}{2}q

while the other charge will be

q-\frac{q}{2}=\frac{q}{2}

So, the new force will be

F' = k \frac{(\frac{q}{2})\cdot (\frac{3}{2}q)}{r^2}=\frac{3}{4} (k\frac{q\cdot q}{r^2})=\frac{3}{4}F

So, the force will decrease to 3/4 of its original value.

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Answer:

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