Recent content by MarkusNaslund19

  1. M

    Maximum angle of deflection

    Thanks for your responses. I understand.
  2. M

    Maximum angle of deflection

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

    Maximum angle of deflection

    Homework Statement Hey, I was studying this problem and solution (by Rudy Arthur): 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...
  4. M

    Parity of the anti neutral kaon

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

    Continuum mechanics problem: Stresses

    Homework Statement Mechanics of Deformable Media (Bhatia and Singh), 5.2: Consider a long rod of elastically isotropic material of L standing vertically in a vacuum in equilibrium under the gravitational field of the earth, then: (i) What are the boundary conditions for \sigma_{ij} on the...
  6. M

    Physics problem - Spring Constant

    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...
  7. M

    Location where charge is cancelled out

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

    Please help - mechanics of a bead on a hoop

    Yes, sorry. I read the question wrong!
  9. M

    Please help - mechanics of a bead on a hoop

    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.
  10. M

    Shopping Cart Problem

    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...
  11. M

    Angles totally mess me up

    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...
  12. M

    How to get a Fnet vs. time graph from a velocity vs time graph.

    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 ;).
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    Shopping Cart Problem

    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...