Recent content by MarkusNaslund19

Thanks for your responses. I understand.

Thanks for the π²³ ∞ ° → ~ µ ρ σ τ ω ∑ … √ ∫ ≤ ≥ ± ∃ · θ φ ψ Ω α β γ δ ∂ ∆ ∇ ε λ Λ Γ ô, Tim!

Homework Statement Hey, I was studying this problem and solution (by Rudy Arthur): http://www.feynmanlectures.info/solutions/maximum_angle_deflection_sol_1.pdf What I wasn't sure was why this solution only works for m<M. At which point did we restrict ourselves to m<M. How about when...

Thanks.

Hi, very simple question: What is the parity of the anti neutral kaon?

I hated Hecht with passion.

q1 and q2 are the charges of the spheres, which is +1.60 micro C. r is the distance between the charges, which is double the original spacing between the spheres. Keep in mind that the two k's you have written in the equations above are not the same! Anyways, you have the force F exerted onto...

You need to write r_a and r_b in terms of x. Then you can solve G01's equation for x.

Yes, sorry. I read the question wrong!

I don't think your kinetic energy is right. You should have three terms since the kinetic energy ought to be 1/2*m*v.v, where v velocity of the bead.

Cursed, The weight of the mass is completely irrelevant for this question. The mass is on some horizontal surface, and by Newton's Third Law, it is canceled out by a normal force. The unknown here is what force the person exerts. We'll call this Fext. Only the horizontal component of this...

For the hockey question, do use trig. Break the velocities for both players into vertical and horizontal components. For the player going in the x-direction: p_x = mv For the other player: p_x = -mvcos45 and p_y = mvsin45 Also, remember that momentum is conserved for both the horizontal...

Assuming that the acceleration you have in your Acceleration vs Time graph is the net acceleration, just multiply the acceleration by the mass of whatever it is you are studying to get a Net Force vs Time graph. F=ma ;).

Cursed, in your solution you subracted a vertical force, the weight of the mass, by the friction, a horizontal. This is invalid. The weight is in the vertical direction, and the force of friction always points in the direction opposite of the motion, which is in this case horizontal. To subract...

Let's say a proton is confined to some diameter, d. What would be the uncertainty in the position? In some questions they use delx as (1/2)d as the uncertainty in position. In other cases they use the entire d as the uncertainty. This is extremely confusing...