Recent content by Apphysicist

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?

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.

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...

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...)...

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

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) =...

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...

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...

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...

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))...

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...

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.

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...

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

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.