psyclones

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!

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

ok, I see what your saying, the metal sphere's don't have static charges?

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

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]

Hi All, I found this problem, The sum of p, q, r terms of an Arithmetic Progression, are P, Q, R respectively: show that [latex] \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 [/latex] My thoughts on how to start the problem is; if [latex] S_{n} = \frac{a}{2} (n + (n-1)d ) [/latex] then the sum of say 'p' terms, would be [latex] P = S_{p} = \frac{a}{2} (p + (p-1)d ) [/latex] Therefore; [latex] Q = S_{q} = \frac{a}{2} (q + (q-1)d ) [/latex] [latex] R = S_{r} = \frac{a}{2} (r + (r-1)d ) [/latex] If I used the following series, to simplify the above P, Q & R series; [latex] S_{n} = 1 + 2 + 3 ... + n, [/latex] then [latex] S_{n} = \frac{1}{2}n(n+1) [/latex] But how to form the above equation, which contains all the terms (i.e, p, q, r, P, Q & R)?

Thankyou both for your posts, I agree, we could probably put aside the reaction forces at A, I only included them for completeness. I after some revision, I realised the direction and position of the buoyancy force was incorrect, I think I found a way around this (refer to attached). Could we employ a sum of moments of area method to calculate the centroid distance? (Ignoring effects due to Fwa). If this (sum of moments of area) isn't valid, then we have to use sum of moments (including force due to water), which will introduce another unknown, the CoB centroid.

Hi, This problem has been driving me mad! Can anyone simplify the physics of this problem? Because I can't solve due to to many unknowns. Question ref: Engineering Materials, Benham, Crawford & Armstrong. Please refer to attachments.

Solved it! [latex]2 ln{(x+4)}= x-1[/latex] Substitute, u = x + 4 [latex]\ln u = \frac{u - 5}{2}[/latex] [latex]u = e^{\frac{u-5}{2}}[/latex] [latex]ue^{-\frac{u}{2}}=e^{-\frac{5}{2}}[/latex] [latex]-\frac{u}{2}e^{-\frac{u}{2}}=-\frac{e^{-\frac{5}{2}}}{2}[/latex] [latex]Y=Xe^{X} \Rightarrow X=W(Y)[/latex] (Lambert's W function) [latex]W(-\frac{1}{2e^{\frac{5}{2}}})=-\frac{u}{2}[/latex] [latex]W_{0}(-\frac{1}{2e^{\frac{5}{2}}})=-0.043[/latex] [latex]W_{-1}(-\frac{1}{2e^{\frac{5}{2}}})=-4.752[/latex] Substitute x for u [latex]x=-2W(-\frac{1}{2e^{\frac{5}{2}}})-4[/latex] x = -3.914, 5.504

Interesting way to simplify the problem, but the domain for inverse of sin(nx); is [-1 to 1].

Hi, I had a little problem which I'd like to have answered! How to solve for x, given the following; 2Ln(x+4)+1=x ? the solutions are x= -3.91.. & 5.50.. Wolfram-alpha solves the problem using 'Lamberts W-function', which makes sense because using the inverse property of logs with exponents will not simplify this problem at all. But by using Using Lambert's Transcendental Equation, with the correct substitution of 'x' for another variable say 'u', I think it can be solved. Ln u = a u^b {b = 1} solution: u = exp [ -W(-a*b)/b ] , where W( ) is Lambert's W-function. Your thoughts? {http://mathworld.wolfram.com/LambertW-Function.html } {http://mathworld.wolfram.com/LambertsTranscendentalEquation.html }