Recent content by Rats_N_Cats

  1. R

    Two bodies attracting via gravitation

    Doing as suggested, I end up with the values for v1 and v2. Then the velocity of approach should be v1 + v2. I get for that v_1 + v_2 = \sqrt{\frac{2G(m_1 + m_2)}{r}}. Is that correct?
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    Two bodies attracting via gravitation

    [FONT="Times New Roman"][SIZE="3"]Homework Statement Say, there are two bodies, mass m1 and m2, initially at rest at infinite separation. They start accelerating towards each other because of gravity. Express the relative velocity of approach as a function of the distance between the two...
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    Undergrad How can we prove 0.9999... is equal to 1?

    [FONT="Times New Roman"][SIZE="3"]Is this proof acceptable? 0.\={9} = 0.999... = 0.9 + 0.09 + 0.009 + ... = \sum_{k=1}^{\infty} \frac{9}{10^k} = \frac{9/10}{1-1/10} = 1 Using the infinite geometric series formula, of course.
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    Integrating Complexity: Solving \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}

    [FONT="TIMES NEW ROMAN"][SIZE="3"]Will anyone please confirm if my answer is correct or not? This problem's been bugging me for quite some time.
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    Undergrad Unraveling the Mystery of 4π in Coulomb's Law

    \mbox{hmmm...got that} Thanks, born2bwire! :smile: Your explanation was good.
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    Undergrad Unraveling the Mystery of 4π in Coulomb's Law

    [FONT="Times New Roman"][SIZE="3"]So both CGS and SI have the 4π, only in different places? But what is its necessity? Wouldn't it be possible to define the electromagnetic units such that the factor of 4π is eliminated?
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    Integrating Complexity: Solving \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}

    [FONT="Times New Roman"][SIZE="3"]I did that, letting (1+xn)-1/n as first function, and I ended up with : \frac{x^{1-n}}{1-n}\,(1+x^n)^{-1/n} + \frac{x}{1-n} + \frac{x^{n+1}}{1-n^2} + C Is that correct?
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    Undergrad Unraveling the Mystery of 4π in Coulomb's Law

    [FONT="Times New Roman"][SIZE="3"]And what's the "spherical geometry of the field created by a point charge"? Could anyone elaborate on that?
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    Integrating Complexity: Solving \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}

    [FONT="Times New Roman"][SIZE="3"]I tried letting xn = t, but that ended up with \frac{1}{n} \int \frac{1}{\sqrt[n]{t^2+t}}\:\mathrm{d}t , [FONT="Times New Roman"][SIZE="3"]And I don't see how to do it. Then I tried letting xn+1 = tn, and got something similarly unsolvable. Can anyone tell...
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    Undergrad Unraveling the Mystery of 4π in Coulomb's Law

    [FONT="Times New Roman"][SIZE="3"]I've read that link you provided...I understand the difference between statcoulomb and coulomb, that they're not dimensionally equivalent. however this comes from the εo, which has a dimension. It doesn't say why the 4π enters the picture
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    Integrating Complexity: Solving \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}

    You've left the dx at the bottom :wink: but how will it become (1+x^n)^{n}? Bringing it to the top will change the sign of the exponent, right?
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    Integrating Complexity: Solving \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x}

    Homework Statement An Integral : \int \frac{1}{x^n(1+x^n)^{1/n}} \;\mathrm{d}x} Homework Equations The Standard integrals. The Attempt at a Solution I'm aware that integrals like this become very easy after a clever substitution...but maybe I'm not that clever...
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    Undergrad Unraveling the Mystery of 4π in Coulomb's Law

    Why the 4[FONT="Garamond"][SIZE="4"]π in Coulomb's Law, SI version? The CGS version does well without it...:confused: \mbox{thanks in advance!}
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    Solving Difficult Integral Homework: \int_0^1\frac{ln(1+x)}{1+x^{2}} dx

    Okay, I got that. Thanks. I knew of the identity, but didn't know I'd have to use it after making a substitution.
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    Solving Difficult Integral Homework: \int_0^1\frac{ln(1+x)}{1+x^{2}} dx

    Homework Statement I'm stuck with this definite integral : \int_0^1\frac{ln(1+x)}{1+x^{2}} dx Homework Equations The various "standard integrals". The Attempt at a Solution I just don't know where to start, or how to do it. I tried various substitutions but none of them worked; I...