Recent content by bobey

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    Measuring Alkyl Chain Cross-Sectional Area for Gibbs Adsorption

    1. Surfactant molecule is made from water-loving head and grease-loving tail (Figure 1). My question: How do we measure the cross-sectional area of the alkyl chain of surfactant? Do we measure it vertically (refer to GREEN DOUBLE ARROWS of Figure 1) or horizontally (refer to RED DOUBLE ARROWS of...
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    Contradiction of Gibbs adsoprtion?

    1. Surfactant molecule is made from water-loving head and grease-loving tail (Figure 1). http://tinypic.com/r/2j2gdah/9 (Figure 1) My question: How do we measure the cross-sectional area of the alkyl chain of surfactant? Do we measure it vertically (refer to GREEN DOUBLE ARROWS of Figure 1) or...
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    Looking for the ENGLISH translation of the JAPANESE manuscript

    Dear forumers, I'm looking for the translation in ENGLISH for the following article : On the Dispersion Mechanism in Laminar Flow through Tubes Kunio Arai1), Shozaburo Saito1), Siro Maeda1) 1) Department of Chemical Engineering, Tohoku University Chemical engineering Vol. 31 (1967) No. 1 P...
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    MHB Problems with propagation of error

    Problems with propagation of error for multiple variables please help by telling me whether my approach to solve the problems are right or wrong. please refer to the ATTACHMENT for the questions and my approaches... your help is highly appreciated! question 1 question 2
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    Problems with propagation of error for multiple variables

    please help by telling me whether my approach to solve the problems are right or wrong. please refer to the ATTACHMENT for the questions and my approaches... your help is highly appreciated!
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    MHB Problems with propagation of error

    I have problem regarding the propagation of error since the equations involving mixtures of multiplication, division, addition, subtraction, and powers. Please help me to clarify whether my attempts are right or wrong.
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    How Does Entropy Relate to the Arrangement Factor in Statistical Mechanics?

    based on this question : http://i825.photobucket.com/albums/zz175/bobey/blablabal.jpg" i tried to answer the question as follow : My answer for the first part of the question : http://i825.photobucket.com/albums/zz175/bobey/black.jpg" My answer for the second part of the...
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    Properties of cross and dot products

    why you think to use this fact : u\cdot v= |u||v|cos(\theta) and |u\times v|= |u||v| sin(\theta) in your arguments? I CAN'T SEE IT! this is my new argument :: yes v = w. u.v = v.u and u x v = u x w implies that u.(v-w)=0 and u x (v-w) = 0 implies that u perpendicular with (v-w) and u parallel...
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    Understanding Equations in Space: Lines & Planes

    Homework Statement describe the set of points in space whose coordinate satisfy the given equation or pair of equations: (i) z=2y (ii) 3x=4y, z=1 if (i) or (ii) represents a line in space, give a unit vector that is parallel to the line. If (i) or (ii) represents a plane, give a unit...
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    Properties of cross and dot products

    Homework Statement let u be a nonzero vector in space and let v and w be any two vectors in space. if u.v = u.w and u x v = u x w, can you conclude that v=w? give reason for your answer. Homework Equations The Attempt at a Solution v is not necessary equal to w since u is nonzero...
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    Distance from the line to the plane (which one is correct?)

    distance from the line to the plane (which one is correct?) Homework Statement find the distance from the line x=2+t, y=1+t, z= -(1/2)-(1/2)t to the plane x+2y+6z =10Homework Equations The Attempt at a Solution my first attempt :: by setting t = 0, we get a point on the line, say P...
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    Second order differential equation

    how to do that... still blurr...
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    Second order differential equation

    show that y1(x) = e^(2+i)x and y2(x) = e^(2-i)x, i=sqrt(-1) are two linearly independent functions hence obtain a second order linear differential equation with constant coefficients each that y1(x) and y2(x) are its two fundamental solutions. my attempt : for the first part, I use the...
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    Solve (cos x)y''-y'+y=0 Using Reduction Method

    another way of my attempt is (cos x)y''-y'+y = 0 y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0 The roots of the characteristic equation are the solutions to this problem. \lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0 If the roots don't present in nice form, like in the case of...
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    Solve (cos x)y''-y'+y=0 Using Reduction Method

    find the general solution of (cos x)y''-y'+y = 0 this i so the question, so in my opinion, the method of reduction is the best method to solve the question. is there any other way that much more easier than this?
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