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About psyclones

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  • Favorite Area of Science
    Mechanical Engineering.
  1. Cosine sequence

    Hi, I've been pulling my hair out trying to solve this. If we treat it as a series similar to proof by induction it can be simplified. Am I on the right track. But the LHS doesn't appear to factorize and reduce in any easy way! 2017-10-24.pdf
  2. Fibonacci related Proof

    Hi, Can you give me any tips of how to begin this proof using the Fibonacci sequence? Fn-1Fn+1 - Fn2 = (-1)n , where Fibonacci numbers can be calculated by Fn = Fn-1 + Fn-2 If I let n= k+1 then, but Fk Fk+2 - Fk+12 = (-1)k+1 If k = 0 , 0x1 - 12 = -1 (true) , but how do I take it further.
  3. Sandwich prob, combinations

    I've paraphrased the question: Yes it's 8 filling ingredients. I thought too that it was 1/8, but I don't believe that's the answer. The topic is combination(s) (nCr) within the text. so the sandwich can have ham only, or ham and lettuce, or ham and tomato, or ham tomato and lettuce and so on.. up to 7 other ingredient fillings combined with ham. Does that help to clarify the problem? I can't remember how to use Latex code on your site or I'd show my working using nCr notation. [ nCr(7,0) + nCr(7,1)+nCr(7,2) + ... +nCr(7,6) +nCr(7,7) ] / [ nCr(8,0) + nCr(8,1)+nCr(8,2) + ... +nCr(8,7) +nCr(8,8) ] = 2^7 / 2^8 given: nCr(n,0) + nCr(n,1)+nCr(n,2) + ... +nCr(n, n-1) +nCr(n ,n ) = 2^n
  4. Sandwich prob, combinations

    Did I complete the above post incorrectly?
  5. Sandwich prob, combinations

    I found a problem, I just like to check it. Question: From 8 total ingredients, (ham being one)- what's the probability of choosing a sandwich with ham? My solution: 27/28 = 128/256 = 1/2
  6. Deflection (structural)

    Thanks for your post. I can assure you I've used 'three moment equation(s)' (solving three equations simultaneously) to arrive at the given moments at b and c, and reaction at b (reaction b is under the support b). My question is, given the manner in which I've formulated the M(x) equation. Is it correct- given I have fixed unequal (in this case) BM's at points b and c, with a free moment due to dist load? [edit] Using the integration method to solve for angle = 0, to solve for deflection. [edit]
  7. Deflection (structural)

    Please excuse [deflection] error. Could you help me with the following. Using continuous beam theory, constructing BM diagram from points b to c, to calculate the max deflection. I only found a have a single solution, though the BM digram show two points of zero bending. I can provide the solution. your thoughts. [edit: Rb = 685 N] Please find revised attachment. attach1(revised).pdf
  8. deflection of continuous beam.

    Hi, The problem uses a continuous beam, but I'm unsure if I'm correct in solving for the max deflection between pts A and B (I've included the BM and SF diagrams btw pts A snd B, only). Deflection should be a maximum when SF = 0. I've also formulated an equation to solve for max deflection using the second area-moment, where 'x' is measured as the distance from A. The BM at B ([latex] M_{B} [/latex]) is straight forward, but the BM due to load is non-linear, I believe this needs revision, but your thoughts. [latex] \Sigma A\bar{x} = \int_{0}^{X}wx\frac{x}{2} \,dx - M_{B}\frac{x}{L}\frac{x}{2} [/latex] BM cont beam.pdf
  9. Physics - Mag/Elec fields.

    Thanks swansont, solved it.
  10. Physics - Mag/Elec fields.

    Thank-you swansont for you post. so using Lorentz force, if a negative ion is introduced the electric field also must be changed? Please find attached. With the attachment, I meant to say E and B2 need to be reversed, if I assume B1 doesn't. hwork-phys(rev).pdf
  11. Physics - Mag/Elec fields.

    Hi, Can you help with the following (please refer to attached). If I understand the problem, both magnetic fields B1 & B2 need to change direction, due to the change in current (I) by replacing the positive with a negative ion. Your thoughts, My apologies, but I can't appear to upload the attached file. I'll try again later!! Please find attached!
  12. diff-charged particles (easy)

    If they conduct, charges (electrons) are transfered. (please refer to attach). What am I doing wrong though, I still can't get a coherant answer? But how can a particle feel the charge of another particle without calculating the Force or Field btw them? and without knowing the distance... Thanks for you help. If the average charge of each pair is calculated as so; Particle R + -2Q = -1/2Q (averaged) Particle S - 1/2Q = +3/4Q (averaged) Particle T + 3/4Q = +15/8Q (averaged) Ans: +15/8Q Therefore force & fields aren’t included in this problem. hw-problem2.pdf
  13. diff-charged particles (easy)

    ok, I see what your saying, the metal sphere's don't have static charges?
  14. diff-charged particles (easy)

    Hi, Can you assist with the following (refer to attachment). I've sketched out how I think the particles will behave, given a -2Q particle interacts with each of the +ve particles (in turn). But my problem is; if the particles of opposite charge are touching. The attractive force become infinite. F = k (q(1). q(2) )/ r^2 r --> 0, F --> inf. Do the particles move? the problem states that they're fixed(??) Your thoughts. Any assistance will me much appreciated. Just add to that. To calculate the applied charges, I'd have to use a field equ, F = E/q, where E is the electric field as a result of the other particles acting on +3Q, But how do I (or would you) formulate a field equation for all particles? hw-problem.pdf
  15. Series

    Thank-you for your post, I solved it! [latex] \frac{P}{p} = \frac{1}{2}(a + d(p-1)) [/latex] [latex] (eq 1) [/latex] Use sum of Q, to let a be the subject. [latex] a = 2\frac{Q}{q} - d(q-1) [/latex] [latex] (eq 2) [/latex] Sub, eq 1 into eq 2 [latex] d = 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) [/latex] Do the same for Sum Q & R, solve for d. [latex] d = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/latex] [latex] 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) = 2 \frac{Q}{(p-r)} ( \frac{Q}{q} - \frac{R}{r} ) [/latex] Simplify, [latex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/latex]