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  1. U

    Rotational Kinematics & Energy

    Thanks for all the wonderful help!!
  2. U

    Rotational Kinematics & Energy

    what is wrong with the formula? Only one speed is given.
  3. U

    Rotational Kinematics & Energy

    A 10.7 g CD with a radius of 6.06 cm rotates with an angular speed of 30.9 rad/s. What is its kinetic energy? What angular speed must the CD have if its kinetic energy is to be doubled? Here is my work: kinetic energy energy = 1/2 m v2 = 1/2 m ( r ω )2...
  4. U

    Potential Energy & Conservative Forces #18

    what's the conservation of energy equation i should use? thanks.
  5. U

    Potential Energy & Conservative Forces #18

    what's the equation i should use?
  6. U

    Potential Energy & Conservative Forces #21

    yeah, my math was off. Thanks, i figured it out.
  7. U

    Potential Energy & Conservative Forces #18

    f_cp = ma_cp = mv^2 / r okay, I still have the velocity missing in this equation, so what do I do?
  8. U

    Potential Energy & Conservative Forces #21

    1/2kd^2 - u_k(mgx) = 1/2mv_f^2 +mg(0) Solve for d I got d= 0.090595 incorrect answer can anyone else help me with this one?
  9. U

    Potential Energy & Conservative Forces #20

    i get 0.423 = 0.529308 + 1.7672 x 10^-4 Which gives a negative answer. Using what you wrote above I get k = 1166.9 and that answer was not correct.
  10. U

    Potential Energy & Conservative Forces #20

    .5k(0^2) = 0 so how do I get anything when I can't even solve for k?
  11. U

    Potential Energy & Conservative Forces #20

    yeah, i did, then you have to square it. I wrote my numbers up above that I was using in the equation. YOu end up with -0.105862 = 706.88(k)
  12. U

    Potential Energy & Conservative Forces #20

    i end up with a negative answer.
  13. U

    Potential Energy & Conservative Forces #18

    how do i figure out the velocity and the acceleration? I need the velocity for the centripetal force equation & I need the acceleration for the f=ma equation. thanks so much.
  14. U

    Potential Energy & Conservative Forces #20

    .5(0.348)(1.56^2) + .5(.348)(0^2) + .5k(0^2) = .5(.348)(1.56^2) + .5(.348)(v/2)^2 + .5k(1.88 x 10^-2)^2 those are all my numbers....is that right?
  15. U

    Potential Energy & Conservative Forces #21

    I get 0.0641 m when i figure it out my way and that answer is incorrect. Maybe I am doing some math wrong, but I double checked? What answer do you get if you use my above equation? Are you sure it is the right equation to use?
  16. U

    Potential Energy & Conservative Forces #20

    Two blocks, each with a mass m = 0.348 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.56 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs...
  17. U

    Potential Energy & Conservative Forces #21

    In the figure below a 1.24 kg block is held at rest against a spring with a force constant k = 700 N/m. Initially, the spring is compressed a distance d. When the block is released it slides across a surface that is frictionless, except for a section of width x = 4.85 cm that has a...
  18. U

    Potential Energy & Conservative Forces #18

    A skateboard track has the form of a circular arc with a 4.00 m radius, extending to an angle of 90.0° relative to the vertical on either side of the lowest point, as shown in the figure below. A 55.9 kg skateboarder starts from rest at the top of the circular arc. What is the normal force...
  19. U

    Work & kinetic energy problem

    thanks so much, you guys really helped me out! I knew it didn't make sense, but I couldn't figure out what I was doing wrong.
  20. U

    Work & kinetic energy problem

    The human brain consumes about 21.9 W of power under normal conditions, though more power may be required during exams. How long can one Snickers bar (280 cal per bar) power the normally functioning brain?[/B] (Note: The nutritional calorie, 1 Cal, is equivalent to 1000 calories (1000 cal) as...
  21. U

    One-Dimensional Kinematics Problem

    I have gotten hints from the teacher also. we have emailed back and forth. Before I did get help from him I tried a lot of numbers I came up with on LON CAPA, therefore, I only have one more try left, so I want to make sure I am doing it the right way. I know I am close and I have the right...
  22. U

    One-Dimensional Kinematics Problem

    is it 0.572? that is twice 0.286
  23. U

    One-Dimensional Kinematics Problem

    i tried 0.286 and it was incorrect and i only have one more try left.
  24. U

    One-Dimensional Kinematics Problem

    So I figured out the first answer. It is 0.391 m. Now I need to figure out the time. When i put it in the equation i get two answers...0.286 and 0.278...my equation is 0.391 = 2.77t - .5(9.81)t^2 What is the correct answer for the second part of the problem originally posted in the...
  25. U

    One-Dimensional Kinematics Problem

    Hey, thank you so much for that advice Hoot. I was putting in -4.06 and i tried it without the negative and it worked. THANK YOU. Can you look at my other problem I posted ??? please!
  26. U

    One-Dimensional Kinematics Problem

    anyone else have any advice on this one? the above advice didnt really help me. I need some step by step instructions or something. i have been working on this problem for too long.
  27. U

    One-Dimensional Kinematics Problem

    even if i switch the two v's i end up with the same answer just different sign, unless i am not supposed to put in 0.510 for x. I am ready to quit physics.
  28. U

    One-Dimensional Kinematics Problem

    so v^2 is 40.61, what should I put in for v_o? and what should I put in for the change in x in that equation. If i put in 0 as initial velocity and change in x as 0.510 i end up with acceleration -39.81, divide that by 9.81 and i get -4.06 which was incorrect according to lon capa.
  29. U

    One-Dimensional Kinematics Problem

    I don't know any of those formulas or anything you listed above. I just started physics so we haven't gotten to those yet. Thanks for letting me know I got the first part right. I need things in simple terms I guess.
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