Looking for data about different types of carbon fiber chemical properties

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Research on carbon fibers' chemical properties highlights their composition as primarily carbon, which is generally inert. When combined with plastics or epoxies, these materials enhance strength and resistance to deformation, making them valuable in automotive applications. The chemical and environmental properties of carbon fibers are significantly influenced by the accompanying materials, particularly regarding their interactions with solvents, sunlight, and temperature. For comprehensive data, utilizing databases and research papers focused on carbon fibers is recommended. Accessing resources like Google can provide a wealth of information on this topic.
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TL;DR Summary: data about different types of carbon fibers' chemical properties

hi! is there any databases or informative research papers I can use for my essay? I am focusing on researching carbon fibers' chemical properties. any source regarding carbon fibers' impacts on cars or chemical properties will be useful. thanks.
 
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Carbon fibers are just that, fibers of Carbon. Chemical properties would be the same as Carbon.

Carbon fibers are often mixed with various plastics or epoxies for added strength and resistance to deformation.

As Carbon itself is fairly inert, the plastics or epoxies would be the major concern for chemical and environmental properties (solvents, sunlight, temperature, etc.).

Cheers,
Tom
 
@GoldRemove9500 Welcome to PF.

There is a really good program that can help you here.
It's called Google.
Search; carbon fiber chemical properties
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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