Reeeeeeeeeeeeeeeeeeeeeeeeeeee
Answer:
<u>True</u>
<u>Explanation:</u>
We can make this conclusion because the term Computer graphics usually refers not to a computer language, but to a technology found in computers that allows the display of images on the computer.
This technology is made possible by certain computer hardware and software but <em>not </em>the "Graphics language electronic processing."
Answer:
0.2 kcal/mol is the value of
for this reaction.
Explanation:
The formula used for is:
![\Delta G_{rxn}=\Delta G^o+RT\ln Q](https://tex.z-dn.net/?f=%5CDelta%20G_%7Brxn%7D%3D%5CDelta%20G%5Eo%2BRT%5Cln%20Q)
![\Delta G^o=-RT\ln K](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D-RT%5Cln%20K)
where,
= Gibbs free energy for the reaction
= standard Gibbs free energy
R =Universal gas constant
T = temperature
Q = reaction quotient
k = Equilibrium constant
We have :
Reaction quotient of the reaction = Q = 46
Equilibrium constant of reaction = K = 35
Temperature of reaction = T = 25°C = 25 + 273 K = 298 K
R = 1.987 cal/K mol
![\Delta G_{rxn}=-RT\ln K+RT\ln Q](https://tex.z-dn.net/?f=%5CDelta%20G_%7Brxn%7D%3D-RT%5Cln%20K%2BRT%5Cln%20Q)
![=-1.987 cal/K mol\times 298 K\ln [35]+1.987 cal/K mol\times 298K\times \ln [46]](https://tex.z-dn.net/?f=%3D-1.987%20cal%2FK%20mol%5Ctimes%20298%20K%5Cln%20%5B35%5D%2B1.987%20cal%2FK%20mol%5Ctimes%20298K%5Ctimes%20%5Cln%20%5B46%5D)
![=-2,105.21 cal/mol+2,267.04 cal/mol=161.82 cal/mol=0.16182 kcal/mol\approx 0.2 kcal/mol](https://tex.z-dn.net/?f=%3D-2%2C105.21%20cal%2Fmol%2B2%2C267.04%20cal%2Fmol%3D161.82%20cal%2Fmol%3D0.16182%20kcal%2Fmol%5Capprox%200.2%20kcal%2Fmol)
1 cal = 0.001 kcal
0.2 kcal/mol is the value of
for this reaction.
Answer:
See attachment for complete answer.
Explanation:
Given that:Suppose that all the dislocations in 2000 mm3 of crystal were somehow removed and linked end to end. Given 1 m =0.0006214 mile, how far (in miles) would this chain extend for dislocation densities of
(a) 103 mm-2 (undeformed metal)
(b) 1010 mm-2 (cold-worked metal)
See attchment for completed solvings
1) ![a_x=4.287+2.772x\\a_y=-5.579+2.772y](https://tex.z-dn.net/?f=a_x%3D4.287%2B2.772x%5C%5Ca_y%3D-5.579%2B2.772y)
2) 8.418
Explanation:
1)
The two components of the velocity field in x and y for the field in this problem are:
![u=1.85+2.05x+0.656y](https://tex.z-dn.net/?f=u%3D1.85%2B2.05x%2B0.656y)
![v=0.754-2.18x-2.05y](https://tex.z-dn.net/?f=v%3D0.754-2.18x-2.05y)
The x-component and y-component of the acceleration field can be found using the following equations:
![a_x=\frac{du}{dt}+u\frac{du}{dx}+v\frac{du}{dy}](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7Bdu%7D%7Bdt%7D%2Bu%5Cfrac%7Bdu%7D%7Bdx%7D%2Bv%5Cfrac%7Bdu%7D%7Bdy%7D)
![a_y=\frac{dv}{dt}+u\frac{dv}{dx}+v\frac{dv}{dy}](https://tex.z-dn.net/?f=a_y%3D%5Cfrac%7Bdv%7D%7Bdt%7D%2Bu%5Cfrac%7Bdv%7D%7Bdx%7D%2Bv%5Cfrac%7Bdv%7D%7Bdy%7D)
The derivatives in this problem are:
![\frac{du}{dt}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdt%7D%3D0)
![\frac{dv}{dt}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdt%7D%3D0)
![\frac{du}{dx}=2.05](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdx%7D%3D2.05)
![\frac{du}{dy}=0.656](https://tex.z-dn.net/?f=%5Cfrac%7Bdu%7D%7Bdy%7D%3D0.656)
![\frac{dv}{dx}=-2.18](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdx%7D%3D-2.18)
![\frac{dv}{dy}=-2.05](https://tex.z-dn.net/?f=%5Cfrac%7Bdv%7D%7Bdy%7D%3D-2.05)
Substituting, we find:
![a_x=0+(1.85+2.05x+0.656y)(2.05)+(0.754-2.18x-2.05y)(0.656)=\\a_x=4.287+2.772x](https://tex.z-dn.net/?f=a_x%3D0%2B%281.85%2B2.05x%2B0.656y%29%282.05%29%2B%280.754-2.18x-2.05y%29%280.656%29%3D%5C%5Ca_x%3D4.287%2B2.772x)
And
![a_y=0+(1.85+2.05x+0.656y)(-2.18)+(0.754-2.18x-2.05y)(-2.05)=\\a_y=-5.579+2.772y](https://tex.z-dn.net/?f=a_y%3D0%2B%281.85%2B2.05x%2B0.656y%29%28-2.18%29%2B%280.754-2.18x-2.05y%29%28-2.05%29%3D%5C%5Ca_y%3D-5.579%2B2.772y)
2)
In this part of the problem, we want to find the acceleration at the point
(x,y) = (-1,5)
So we have
x = -1
y = 5
First of all, we substitute these values of x and y into the expression for the components of the acceleration field:
![a_x=4.287+2.772x\\a_y=-5.579+2.772y](https://tex.z-dn.net/?f=a_x%3D4.287%2B2.772x%5C%5Ca_y%3D-5.579%2B2.772y)
And so we find:
![a_x=4.287+2.772(-1)=1.515\\a_y=-5.579+2.772(5)=8.281](https://tex.z-dn.net/?f=a_x%3D4.287%2B2.772%28-1%29%3D1.515%5C%5Ca_y%3D-5.579%2B2.772%285%29%3D8.281)
And finally, we find the magnitude of the acceleration simply by applying Pythagorean's theorem:
![a=\sqrt{a_x^2+a_y^2}=\sqrt{1.515^2+8.281^2}=8.418](https://tex.z-dn.net/?f=a%3D%5Csqrt%7Ba_x%5E2%2Ba_y%5E2%7D%3D%5Csqrt%7B1.515%5E2%2B8.281%5E2%7D%3D8.418)