Recent content by Apphysicist

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    Direction of the change of momentum

    Useful equation: F(net)=dp/dt Given: Initial VELOCITY (initial MOMENTUM) is in the direction of g^vector. A net FORCE is applied in the direction of the a^vector. Ask yourself: did you think about it appropriately at first? Can you truly add those vectors quantities?
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    Graphing Distance and Time to find Acceleration

    I have no idea why you would take ln of anything with this. You're graphing d=mt^2...m=0.5a. So you would want to graph d v. t^2 (d is y-axis, t^2 is x-axis) to give a straight line and take its slope.
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    Vectors homework problem

    Certainly you can think of (1,1,1) as the start of your vector, then (3,4,5) as the end of your vector...thus subtraction yields a vector that passes through both points, just with a certain magnitude. The magnitude of a vector is given as sqrt(x^2+y^2+z^2)...you can find that you won't get a...
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    Error Propagation

    Since they have STDEV as a separate row, I suspect they want the absolute uncertainty for the mean: x(avg) = sum(x)/N...so if you add up all the uncertainties in quadrature, and then scale it by N, you have uncert(avg)=(1/N)*sqrt(uncert(x1)^2+uncert(x2)^2+uncert(x3)^2...)...
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    Solving trig problems

    Why don't you just Arccos both sides from the beginning? (I assume you capitalized Cos to mean the principal cosine)
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    Question about power series

    I would find the power series for 1/(x^2+1) then multiply each term by x, then look at it and write the summation notation. f(x)= f(0)+f '(0)*x +f ''(0)*x^2/2 + f '''(0)*x^3/6 +... f(x) = (x^2+1)^-1 f '(x) = - 2x*(x^2+1)^-2 f'' (x) = -2x*-2*2x*(x^2+1)^-3 + (x^2+1)^-2*-2 f(0) =...
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    Spring DE- Find position at any time t.

    Without knowing the relation of dynes to newtons off the top of my head, let me write out your DE using variables (I assume this is not a simpler case of horizontal motion, rather, by saying "above" and "directed down" you are dealing with a vertical spring system so the force of gravity...
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    Series Convergence

    As far as I can tell, (a) refers to f(x) as a series (thus a sum), but since f(x) is not itself a summation, it is more a sequence, a function for discrete values of n (if you plotted f(x) v. n for a fixed value of x, it'd look like a discontinuous function...a bunch of points). When x=0...
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    How can the force NOT be zero here?

    Just a bit further down the same page, the point is that the velocity selector allows a charged particle with a certain velocity to shoot straight down the finely tuned system between the plates and in the magnetic field unimpeded. Presumably, if it veers off the middle line just a bit, the...
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    Wavefunction normalization help!

    Not knowing anything about wavefunctions, just on the information you gave me, I think you multiplied incorrectly (left out the A): (A-Ae^(ikx))*(A-Ae^(-ikx)) = 2A^2-A^2*(e^(ikx)-e^(-ikx)) = 1 From there, I'm seeing a chance to divide by 2A^2, so you have something like 1-sinh(u) = (1/(2A^2))...
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    Series Convergence

    Right, so if you rewrite that so it looks a little nicer: You have the limit AS N GOES TO INFINITY of the absolute value of x*(n+2)/(3n+3). That's the key here: you're taking the limit for n, seeing what the "final" term looks like compared to the one before it (remember to converge, the...
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    Apostol's Theorem 2.5, sine cosine example

    They used the trig identity to replace sin(x)^2 with (1-cos(2x))/2...then pulled the 1/2 out of the integral and integrated each term. Integral of 1 = x. They did a u-sub on cos(2x) with u=2x, du=2dx, thus it gets another 1/2 out front, and integral of sin = cos.
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    Force mass and acceleration

    Right. It's much easier to think about F=ma really being: \sum_{\text{all}}\bold{F}\,=\,m \bold{a} So it's a sum of all your vector forces (the bold meaning it's a vector...so for your problem, the +/- indicates the direction is all, since there are no components, it's just in one...
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    Force mass and acceleration

    No, it would be changed to (T-mg)=ma. Do you see why?
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    Force mass and acceleration

    Indeed! F=ma is your net force = m* net acceleration. If you apply your maximum tension upward, gravity still acts downward (sum the forces for your net force, don't forget that gravity gets a minus-sign because it acts in the opposite direction of your upward tension). Then use F=ma.
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