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lubasha [3.4K]
2 years ago
12

Where do light bulbs get their energy from?(1 point)

Physics
2 answers:
N76 [4]2 years ago
8 0

<u>for instance, steel has a higher thermal conductivity than plastic. Hence, the steel plate gives away heat to the ice block faster than a plastic block does. As a result, ice melts faster on a steel plate than on a plastic one. Faster an object draws heat, the colder it feels.</u>

irina [24]2 years ago
7 0

When a light bulb connects to an electrical power supply, an electrical current flows from one metal contact to the other. As the current travels through the wires and the filament, the filament heats up to the point where it begins to emit photons, which are small packets of visible light.

Essentially, the lightbulb is a very thin filament of hard-to-melt metal – tungsten, usually – encased in a glass bulb filled with inert gases so that the filament doesn't oxidise and disintegrate. The electricity causes the wire to glow and a portion of that energy is turned into light.

Light energy can also be converted into thermal energy when for example the sun heats up your black shirt or a brick wall outside. There are many example we see in our routine life carrying light energy like lightened candle, flash light, fire, Electric bulb, kerosene lamp, stars and other luminous bodies etc.

Why does metal feel colder than plastic when both objects are the exact same room temperature?

It is because the metal conducted heat faster that it feels colder than the wood, which conducted heat slower. They feel different temperatures, but they are actually the same. ... If you stepped on the tiles, it would feel colder because they conduct heat faster.

<h2>▪▪▪Cutest Ghost▪▪▪</h2>
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a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long
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Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

a=\frac{Vf-Vi}{t}

-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}

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-9.8  =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2

To solve for "t" in this quadratic equation, we can factor it out as shown:

0= t(20-\frac{9.8}{2} t)

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s

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~ Ria

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