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4 years ago

Which object absorbs the most visible light?

Physics

2 answers:

klio [65]4 years ago

4 0

Answer:

C. a black t-shirt

Explanation:

wariber [46]4 years ago

3 0

<h3><u>Answer;</u></h3>

<em>A black T-shirt </em>

<h3><u>Explanation;</u></h3>

<em><u>Materials can be either reflectors and absorbers of waves such as light waves, sound waves etc.</u></em> Reflectors are those materials that reflects waves, such that the waves bounces back on hitting the surface of those materials.
Examples of such materials are those <em><u>with smooth and shiny surfaces, while those with rough and opaque surfaces are poor reflectors of light. </u></em>Additionally, these materials that are good reflectors are poor absorbers while those that are poor reflectors are good absorbers of light.
<u><em>Dull surfaces are the best emitters and absorbers of light but the worst reflectors of light waves.</em></u> White surfaces or polished surfaces are bad and worst emitters and absorbers of light waves respectively

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How large amount of energy is produced during the fission of uranium

vova2212 [387]

Answer:

answer is a very large amount of energy is produced from a very small mass

Explanation:

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3 years ago

Given that oxygen- 16 and oxygen -18 both have an atomic nmber of 8, how many electrons, protons, and neutrons do these oxygen a

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3 years ago

We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes

Ronch [10]

Answer:

292.3254055 W/m²

469.26267 V/m

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = $\frac{d}{2}=\frac{7}{2}=3.5\ cm$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

The intensity is given by

$I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2$

5% of this energy goes to the visible light so the intensity is

$I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2$

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

$u=\frac{1}{2}\epsilon_0E^2$

Energy density is also given by

$\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m$