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  1. pmd28

    Quantum Chemistry: Theory based questions

    Awesome! Thank you very much. :)
  2. pmd28

    Quantum Chemistry: Theory based questions

    For 1, right. I think their way of drawing that imaginary horizontal line is a little weird. But your elaboration makes perfect sense. So I have another question now regarding anharmonic oscillators: It seems that when R goes to infinity this graph will become asymptotic. Is the limit of this...
  3. pmd28

    Quantum Chemistry: Theory based questions

    Hey all, I want to preface this with I wasn't sure if I should put this in the physics thread since it is quantum mechanics or in chemistry thread since this is for my Physical Chemistry course. I will gladly move the thread if the community feels it is more appropriate in the physics thread. I...
  4. pmd28

    Eletric Potential

    Nvm I solved it. Thanks :D
  5. pmd28

    Eletric Potential

    The problem states a "-q2" not a "+q2".
  6. pmd28

    Eletric Potential

    Ok, sorry I had a spark of genius. And I've been kind of going with it. I just want to know if I'm on the right track. I set the +q1 charges equal to each other and got: r=.11/(V1-V2) Then I set the -q2 charges equal to each other and got: .035V1=.0765V2 Substitute to get r=.11/(V1-.035V1)...
  7. pmd28

    Eletric Potential

    V=kq/r
  8. pmd28

    Eletric Potential

    See that's the thing, i'm still iffy on the relationshiphs. My best educated guess would be: Since V @ .0335m is equal to zero. That would mean the net electric field at that point is also 0 because E=-ΔV/Δs. Or am I going in a completely wrong direction.
  9. pmd28

    Eletric Potential

    Homework Statement A positive charge +q1 is located to the left of a negative charge −q2. On a line passing through the two charges, there are two places where the total potential is zero. The first place is between the charges and is 3.35 cm to the left of the negative charge. The second...
  10. pmd28

    Equivalent Electri Force, Find the charge.

    ok now I just got a completely wrong answer. My new quadratic equation is q22-9.59e-6+4.8e-6. I got answer of like .00348
  11. pmd28

    Equivalent Electri Force, Find the charge.

    Homework Statement Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.3 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 2.3 N. The objects are brought into contact, so...
  12. pmd28

    Diffraction from a loud speaker

    So I set that ratio equal to the ratio of vf/vi and then canceled out sqrt(\gammak/m) because it is constant in both cases therefore the ratio of Sinfθ/Siniθ is equal to sqrt(Tf)/sqrt(Ti) It was correct.
  13. pmd28

    Diffraction from a loud speaker

    Homework Statement Sound exits a diffraction horn loudspeaker through a rectangular opening like a small doorway. Such a loudspeaker is mounted outside on a pole. In winter, when the temperature is 273 K, the diffraction angle θ has a value of 11°. What is the diffraction angle for the same...
  14. pmd28

    Speed of a Wave

    There are 4 wavelengths.
  15. pmd28

    Speed of a Wave

    Homework Statement A person fishing from a pier observes that five wave crests pass by in 7.4 s and estimates the distance between two successive crests to be 4.6 m. The timing starts with the first crest and ends with the fifth. What is the speed of the wave? Homework Equations f=1/T...
  16. pmd28

    A person doing a pushup.

    OK I solved it. Thanks :)
  17. pmd28

    A person doing a pushup.

    would my F in torque be W=509?
  18. pmd28

    A person doing a pushup.

    Yea I caught on to the fact that it said per hand. And how does that play into the center of gravity forumla.
  19. pmd28

    A person doing a pushup.

    I had to draw out my own daigram. L1 is the distance from the feet to the cg, L2 is the distance from cg to the hand and X1 and X2 are just how I denoted distance in my notes, they're not given in the equations So yes johnny you are correct.
  20. pmd28

    A person doing a pushup.

    Homework Statement a person (weight W = 590 N, L1 = 0.830 m, L2 = 0.409 m) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position. Homework Equations Xcg= (W1X1 + W2X2)/(W1+W2) τ=Fl The Attempt at a...
  21. pmd28

    Mini-golf Windmill

    I knew that much, I just couldn't get θ of ω. But I slept on it and figured it out. Thanks :smile:
  22. pmd28

    Mini-golf Windmill

    Homework Statement A golf ball passing through a windmill at a miniature golf course. The windmill has 6 blades and rotates at an angular speed of 1.65 rad/s. The opening between successive blades is equal to the width of a blade. A golf ball (diameter 4.50x10-2 m) has just reached the edge...
  23. pmd28

    Mini-Golf Windmill

    Oh, pfft. The golf ball has to go through the hole without getting hit by the windmill with it's slowest possible velocity.
  24. pmd28

    Mini-Golf Windmill

    Go through the hole without getting hit by the windmill?
  25. pmd28

    Mini-Golf Windmill

    Go through the hole.
  26. pmd28

    Mini-Golf Windmill

    Homework Statement A golf ball passing through a windmill at a miniature golf course. The windmill has 6 blades and rotates at an angular speed of 1.65 rad/s. The opening between successive blades is equal to the width of a blade. A golf ball (diameter 4.50x10-2 m) has just reached the edge of...
  27. pmd28

    Car going up a hill. Find the Angle of the hill.

    I just put my nunmbers in my calculator wrong. No big deal thanks again.
  28. pmd28

    Car going up a hill. Find the Angle of the hill.

    I got a domain error... [never mind calculator error]
  29. pmd28

    Car going up a hill. Find the Angle of the hill.

    So just checking my equation: v(f+mgSinθ)=v(f-mgSinθ) + 32810.8w Now I can just plug and chug.
  30. pmd28

    Car going up a hill. Find the Angle of the hill.

    o right becasue going uphill is 44 more hp you would add it to the downhill side. and no 1hp=745watt
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