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Anna11 [10]
3 years ago
11

Early LCD backlights use _______________ technology, popular for its low power use, even brightness, and long life.

Physics
1 answer:
Gala2k [10]3 years ago
8 0

Answer:

cold cathode fluorescent lamp

Explanation:

A cold cathode fluorescent lamp (CCFL) is a lighting system that uses two phenomena: electron discharge and fluorescence.

CCFLs are mainly used as light sources for backlights, since they are compact and durable than ordinary fluorescent lamps. They exhibit a wide range of brightness and color (color temperature and chromaticity) that can be achieved by varying the pressure and type of the material injected into the glass tube. The thickness and type of the phosphor used to coat the inner wall of the tube also plays a role in altering the color and brightness.

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Are sinkhole fatalities common?
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Answer:

Sinkhole fatalities are rare, but they do occur sometimes. Also fatalities are more likely where sinkholes are common. For example, sinkholes are more  common in Florida than in California, so you would more likely die from a sinkhole in Florida than you would in California.

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A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in
Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

  • \boxed{u \ or \ v_i = initial \ velocity}
  • \boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}
  • \boxed{a = acceleration \ (constant)}
  • \boxed{d = distance \ travelled}

The formula we follow for this problem are as follows:

\boxed{ \ v^2 = u^2 + 2ad \ }

  • a = acceleration (in m/s²)
  • u = initial velocity  
  • v = final velocity
  • d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

\boxed{ \ 0 = v^2 + 2ad \ }

\boxed{ \ v^2 = -2ad \ }

Both sides are divided by -2d, we get \boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }

Here d' is the stopping distance that we want to look for.

\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }

We crossed out 2 in above and below.

\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }

We multiply both sides by d.

\boxed{ \ v^2 d' = 49.0v^2 d \ }

We crossed out v^2 on both sides.

\boxed{\boxed{ \ d' = 49.0d \ }}

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>
  1. Determine the acceleration of the stuffed bear brainly.com/question/6268248
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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

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I think its most likely a the direction of the impact
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The resistance of the light bulb changed as the voltage (and current) changed. Why does this resistance change occur?
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This resistance current is directly proportional to voltage and inversely proportional to resistance. In other words, as the voltage increases, so does the current. Hope this helps !!
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