Recent content by amy andrews

Aha! I see. Thank you :) I was going to ask about preparation for F=ma (I know it's been asked before, but my case is slightly unusual) and that doesn't seem to fit under academic guidance...or does it?

I came on PF after about a month to ask a question in the Math and Science Learning Materials forum. However, it seems I'm not allowed to start a new topic in this forum (unlike in, say, the homework forum). Is there a reason for this-- only certain members can post, etc? Thanks! Amy

Exactly. The weight of the bike, the normal force, and the force the girl exerts. These forces are equal, if the girl pushes the bike up with constant velocity. Take a look at the attached free body diagram, write a couple of equations (with the x and y components), and the force the girl...

Think of the free body diagram for this. Apart the force the girl exerts on the bike (and I'm assuming there's no friction, since it isn't included in the problem) what are the forces acting on the bicycle?

Basic constant acceleration equations: v=v_0+at x=x_0+v_0t+(1/2)at^2 v^2=(v_0)^2+2aΔx

Use the same equations as for horizontal velocity, with the acceleration equal to g.

I agree. Well, I don't know about finding exact roots, but you can get a general idea of where the zeroes are by plugging in a few numbers. For example, with x=1 you get a positive answer, but with x=-1 you get a negative answer. So in between x=1 and x=-1 there is a root.

Well...this is what I think. Remember that the full equation for work is F*d cos theta. If theta is 90 degrees (the force is perpendicular to the distance) cos theta is 0. So, I think you shouldn't use the kinetic energy for this, it is changing perpendicular to the distance you are looking at...

Well I'm not absolutely sure on this one. But I think it is convention to put the units in parentheses...it's a sort of side note, so it should be in parentheses. It would look wrong to have it without.

Abbreviations for metric units don't end in periods, but abbreviations for everything else, except those with "per," (such as mph) do. Who knows why...

Gravitational energy looks fine. I'm pretty sure your kinetic calculation is all right too, unless they want you to find the total distance traveled (using the height and the horizontal distance). There seems to be some confusion in the third part. Remember Work=FORCE*distance, not...

I think you got your equation slightly off. Starting with F/\DeltaL=\mu0I1I2/2pi*r, I ended up with I2=(F/\DeltaL)(2pi*r/\mu0I1).

I can see how you tried this problem. However, when a block is on a ramp the situation is slightly different. Take a look at the attached diagram. This is how the forces are acting on this block. The force normal comes into play, and of course in this case, there is no force friction. So the...

If it's interesting for you, you could try writing on how thin films are used today and the principles behind them. Scientists are still creating novel thin films for solar panels, lenses etc. I also know NASA is experimenting with many optics topics, you could take a look and see if anything...

Hi everyone, I'm currently studying a bit of quantum theory and I came across this paragraph: When an electron collides with a positron, it moves with an angle theta_1 with the normal. Ditto for the positron, except that it has an angle theta_2. But the two angles must be equal. Why...