What kind of bonding is found in graphite?
Consider the dark pressed powder inside of the pencil. When you apply pressure on the pencil to a paper a mark is left behind. What really is happening there is if we zoom in to the nanoscopic size of the tip of the pencil we can see a stack of graphene (1 layer of hexagonally bonded carbon). The graphene sheets shear away from neighboring sheets one by one releasing the weak out-of-plane bond that was present in graphite. This type of weak bonding is often referred to as van der Waals bonding. In-plane bonding of graphene is the other type we must consider as well. The graphene sheet looks like an infinite array of hexagons that are aligned similarly to a soccer ball's pattern on its surface. More accurately, it shares infinitely bonded hexagonally packed carbon atoms. This type of bonding is quite strong since the sheets are not readily shearable in-plane. This degree of strength is typical of a covalent bond.
Given that f(x) = x^2 + 2xy +4, what is the derivative of f(x) with respect to y? Now, differentiate f(x) with respect to x.
First thing you need to realize in this problem is that you have 2 major steps to overcome. 1. Recognize your task: Take the derivative of something. 2. What is the something you have to derivate by?: y d f(x)/ dy = taking the derivative of f(x) with respect to y f(x) = x^2 + 2xy +4 d f(x)/ dy = 0 + 2x +0 So, what happened here? We looked at f(x) and saw that only 1 term had "y" in it, meaning we could differentiate the whole expression simply by considering x to be a constant. A derivative of a constant is just 0. So, with the same logic, lets differentiate with respect to x. f(x) = x^2 + 2xy +4 d f(x)/ dx= 2x + 2x +0 d f(x)/ dx= 4x Still you see the same behavior. 2 terms have x this time and require a derivative step by step. the derivative of x^2 is just the exponent times the variable raised to the power of the original exponent minus 1. So, d (x^2)/ dx = 2 x^(2-1) = 2x. We can repeat what we did before about considering x a constant in the previous problem, except we now consider that constant to be y!. This leaves us with the same type of derivative and with simplification brings the derivative's expression to 4x.
Using the common steel phase diagram, where does the eutectic point lie and what does it signify?
The eutectic point is the temperature and compositional point on a phase diagram where a melt is cooled from the liquid phase directly into binary phase. This means that as the melt cools down, 2 phases will precipitate out as alpha and beta. This point lies at 727 celcius at about 0.76% carbon.