Recent content by Kate_12

  1. K

    Kepler problem in parabolic coordinates

    I solve (1). But to solve (2), What should be the suitable separation constants? I am so confused... E=2/(m*(a+b)) * (a*(dWa/da)^2+b*(dWb/db)^2-k)+l^2/(2mab) where l(constant) is pc since c is cyclic. What should I do to solve the problem?
  2. K

    Iron block dropped into a container of water sitting on a scale

    Then, is there any general form of solution to ##\ddot y+Cy=D##?
  3. K

    Iron block dropped into a container of water sitting on a scale

    mg-##\rho##*A*h*g=m*##\ddot h## so integral 0 to h (dh^2/(m-##\rho##*A*h))= integral 0 to t (g/m)dt^2 I think I can solve this equation and there would be log term of h as a function of time. I really appreciate for your help. Have a nice day!
  4. K

    Iron block dropped into a container of water sitting on a scale

    So with drag force, the scale change would be mg(the weight of the block) regardless of time. Without drag force, the net force on the block would be phro*V*g (V: the submerged volume of the block) 1) during submerging, the change of the scale would be phro*A*h*g(h: the submerged height of the...
  5. K

    Iron block dropped into a container of water sitting on a scale

    The net force on the block is zero because it drops at constant speed. so the water+container system exerts m(the mass of the block)*g on the block. That means, the change in scale would be the weight of the block. Is it right?
  6. K

    Iron block dropped into a container of water sitting on a scale

    Before the block enters the water, there is no change in scale (just the weight of container filled with water) When the block is fully immersed, the change in scale would be the weight of the block, because the buoyancy force is just internal force.
  7. K

    Iron block dropped into a container of water sitting on a scale

    Does the scale change during the block sinks? I am so confused... Is there any difference on change of scale with the applying of drag force?
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