Recent content by GreyNoise

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    How can I comment on or post a question about an Insight Article?

    Where is "here" fresh_42? I tried from the page where this thread appears (is that what you meant?). It took me to the article, but I can't find a link to comments from there.
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    How can I comment on or post a question about an Insight Article?

    Does anyone know how to leave comments for an Insight Article? I wanted to comment and ask a question, but I can't find a link (not obvious to me anyway). At the bottom it reads to login to leave a comment but just tells me that I'm already logged in when I click on it, and I cannot find any...
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    Find v_1 for projectile using start position, final position and angle

    A big thnx to everyone for your help on this problem. As all of you took the time to help out, I thought to share my final solution. Shout out to kuruman for the insight article; it was the crow bar I needed to pry out an answer. given: ##\Delta y## = 25.0 m, ##\Delta x## = 60.0 m, ##m## = 15.0...
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    Find v_1 for projectile using start position, final position and angle

    I have attempted the problem a few times, and have posted my work below. I thought to check my work using ## v_x t = v_1 \times \cos\theta \times t = \Delta x## to see if I get back expected results, and I did for the horizontal motion (the substitutions are from the numbered steps below): ##...
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    Derive the Venturi Meter eqn from the Bernoulli eqn

    Thnx so much for the response TNsy. Pointing out my contradiction between lines (1) and (2) was the big aha moment for me, and including the ##\rho gH## term makes the physical sense clear now.
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    Derive the Venturi Meter eqn from the Bernoulli eqn

    Advanced apologies for this format; I am posting my question as an the image b/c the Latex is being very buggy with me, and I lost a kind of lengthy post to it. Can anyone show me what I am doing wrong? I have attached a pdf version for easier reading if need be.
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    B You normally go through B and C when passing from A to D

    I couldn't resist the parting comment that your original question brings to my mind, what are called in physics, the equations of continuity; they arose originally in fluid mechanics but the concept shows itself in electrodynamics and energy as well. I lifted a paragraph from a Wikipedia...
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    Integrate wave function squared - M. Chester text

    Thnx Charles. I am confident of my own answer then.
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    Integrate wave function squared - M. Chester text

    Homework Statement given: A wire loop with a circumference of L has a bead that moves freely around it. The momentum state function for the bead is ## \psi(x) = \sqrt{\frac{2}{L}} \sin \left (\frac{4\pi}{L}x \right ) ## find: The probability of finding the bead between ## \textstyle...
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    Find Fourier coefficients - M. Chester text

    Sorry for delayed response; thnx are in order; I was away on vacation. Thank you Dick and Ray. The posts got me back on top of the curve (my Fourier forays are ten years ago now), so I committed them to my notes. I might be back for more regarding the M. Chester text I am reading. Thnx again.
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    Find Fourier coefficients - M. Chester text

    Thnx much for that Dick! That was the fix I needed. Now it brings to my mind a question about the form of the exponential argument. The author wrote $$ C_k = \frac{1}{L} \int_0^L e^{-ikx} f(x) dx $$ and I used the form literally, but now I am concerned that I missed a basic Fourier series...
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    Find Fourier coefficients - M. Chester text

    Homework Statement I am self studying an introductory quantum physics text by Marvin Chester Primer of Quantum Mechanics. I am stumped at a problem (1.10) on page 11. We are given f(x) = \sqrt{ \frac{8}{3L} } cos^2 \left ( \frac {\pi}{L} x \right ) and asked to find its Fourier...
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    Prove arccos(-x)+arccos(x) = pi

    Yes, I did get the sinking feeling I was begging the question after I posted So how about this? given:\hspace{10mm}\cos^{-1}(-x) = \alpha and \cos^{-1}(x) = \beta show:\hspace{10mm}\alpha = \pi - \beta from given \cos\alpha = -x \cos\beta = x\hspace{5mm} so \cos\alpha = -cos\beta \cos\alpha...
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