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Serhud [2]
3 years ago
10

Can light cause plastic to melt? If yes does it matter what kind of light?

Physics
2 answers:
ad-work [718]3 years ago
6 0

Answer:

plastics are susceptible to degradation due to ultraviolet (UV) light from the sun. With enough exposure, UV light can cause a chemical reaction in the plastic, which results in scission, or severing, of those big polymer molecules.

Explanation:

harkovskaia [24]3 years ago
4 0

Answer:

the short answer is yes and sun or strong uv lights

Explanation:

LEDs cannot melt plastic fixtures as they just do not get that hot, not even at the base These new bulbs are based on field-induced polymer electroluminescent (FIPEL) technology, with a twist. Without heat as a continual stressor, the polymer should remain stable for years. Any kind of light bulbs, from fluorescent to incandescent to halogen, can cause fires if they are not used correctly. <u>As more heat is put into the plastic, more bonds are broken and the material becomes more like a liquid and less like a solid. Semi-crystalline polymers have a melting temperature which is the point at which the secondary bonds in the crystalline regions break. At this point the mateiral is completely liquid. There is melt and there is soften. Certainly they soften and become weak in sunlight both by the heat and by the UV degredation of the plastic. Unfortunately, the plastics used for garbage bags have a low melting point but by design, they should not melt in strong sunlight.</u> the underlined sentences are the most important

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Answer:

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Explanation:

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3 years ago
A satellite of Mars, called Phobos, has an orbital radius of 9.4 ✕ 106 m and a period of 2.8 ✕ 104 s. Assuming the orbit is circ
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Answer:

6.27*10^{23}kg

Explanation:

assume

M= mass of Mars

m=mass of phobos

r=orbital radius

T=period

we can apply F=ma to this orbital motion (considering the cricular motion laws)

where,

F=\frac{GMm}{r^{2} }  and a=rω^2

where ω=\frac{2\pi }{T} and G is the universal gravitational constant.

G = 6.67 x 10-11 N m2 / kg2

F=ma\\\frac{GMm}{r^{2} }=mr(\frac{2\pi }{T} )^{2}\\  M=\frac{r^{3}}{G}  (\frac{2\pi }{T} )^{2}\\M=\frac{(9.4*10x^{6} )^{3}*(2\pi )^{2} }{(2.8*10^{4}) ^{2} *6.67*10^{-11} } \\M=6.27*10^{23}kg

6 0
3 years ago
Replacing an object attached to a spring with an object having 14 the original mass will change the frequency of oscillation of
Pachacha [2.7K]

Answer:

<em>The frequency changes by a factor of  0.27.</em>

<em></em>

Explanation:

The frequency of an object with mass m attached to a spring is given as

f = \frac{1}{2\pi } \sqrt{\frac{k}{m} }

where f is the frequency

k is the spring constant of the spring

m is the mass of the substance on the spring.

If the mass of the system is increased by 14 means the new frequency becomes

f_{n} = \frac{1}{2\pi } \sqrt{\frac{k}{14m} }

simplifying, we have

f_{n} = \frac{1}{2\pi \sqrt{14} } \sqrt{\frac{k}{m} }

f_{n} = \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }

if we divide this final frequency by the original frequency, we'll have

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  ÷  \frac{1}{2\pi } \sqrt{\frac{k}{m} }

==> \frac{1}{3.742*2\pi  } \sqrt{\frac{k}{m} }  x  2\pi \sqrt{\frac{m}{k} }

==> 1/3.742 = <em>0.27</em>

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