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calculusxy

  • one year ago

MEDAL!!!!!!!!!!!!! Problem stated below.

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  1. calculusxy
    • one year ago
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    I did an experiment today relating to torques. The materials that I had were a meter stick, masking tape, and weights. We had to use the ruler and place the two weights on each side, to try to make it balance to each other. Also, we had to note how far the tape (holding the weight) was from the center (or 50cm). Here is my first data: Mass 1: 200g Distance from 50cm: 19cm Mass 2: 100g Distance from 50cm: 36 cm Then we had to multiply the mass times the distance for each and were asked to look for a pattern. Here are my products: Mass 1: 200g x 19 cm = 3800 Mass 2: 100g x 36 cm = 3600 I don't notice a pattern and need help in finding one.

  2. calculusxy
    • one year ago
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    @Nnesha

  3. calculusxy
    • one year ago
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    @ikram002p

  4. DanJS
    • one year ago
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    looks pretty close

  5. anonymous
    • one year ago
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    okay so you say your lab involved torque right

  6. calculusxy
    • one year ago
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    Yes

  7. DanJS
    • one year ago
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    The real world will have small errors, not ideal... you did it correct

  8. calculusxy
    • one year ago
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    @DanJS So can my conclusion be like since they are both near of each other, it means that they are almost balanced to each other.

  9. DanJS
    • one year ago
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    Are you discovering torque moments for the first time here, or do you know already

  10. DanJS
    • one year ago
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    torque = (force perpendicular to the line of action) *(distance)

  11. calculusxy
    • one year ago
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    Well it is the second to last day of class, and my teacher didn't want to explain torques to us fully and just let us do the experiment. So we needed to find out pattern with the products.

  12. DanJS
    • one year ago
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    The rotational moment about a point (here 50cm mark) = distance from that point, times the perpendicular force

  13. anonymous
    • one year ago
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    here this is what i used to study torque last year https://www.youtube.com/watch?v=sAMNoRBBjzg i hope it helps :)

  14. DanJS
    • one year ago
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    so if you have a force on an angle, you need to break into components first..

  15. calculusxy
    • one year ago
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    I just need to know if the statement that I wrote earlier would be fine for my conclusion or not.

  16. DanJS
    • one year ago
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    hah, if your teacher is cramming it in this late, probably wont be much of it on the final

  17. calculusxy
    • one year ago
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    I already took the final. It's just an extra thing.

  18. DanJS
    • one year ago
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    oh, is that the only single trial point you did? did you try different distances?

  19. calculusxy
    • one year ago
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    yes until the weights were balanced (almost)

  20. DanJS
    • one year ago
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    I would say something like, The distance multiplied by the force seems almost the same... Need to try further experiments with different masses and distances to test if this is true

  21. DanJS
    • one year ago
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    one test isn't enough

  22. calculusxy
    • one year ago
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    I did another trial, but with three different weights ( two on one side and one on the other). Mass 1 : 100g Distance from 50cm: 28cm Mass 2: 200g Distance from 50cm : 20cm Mass 3: 300g( this weight was kept on the other side) Distance from 50cm: 20cm

  23. calculusxy
    • one year ago
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    @DanJS

  24. DanJS
    • one year ago
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    ok, apply the same reasoning as before... If a torque is clockwise it is negative, and if it is counter clockwise about a point, it is positive... The sum of all the torques should add to zero if the thing is balanced and not moving

  25. DanJS
    • one year ago
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    so 100*28 + 200*20 ideally would be 300*20

  26. calculusxy
    • one year ago
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    Okay I get it thanks. I have some other questions, will you be able to help me on them?

  27. calculusxy
    • one year ago
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    This is an example of a velocity a. 45 miles per hour b. 34 meters per second c. 23 centimeters per year northeast d. 12 days per mile north e. 0.1 milliseconds per kilometer f. all of the above are examples of velocity

  28. calculusxy
    • one year ago
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    It's C right?

  29. DanJS
    • one year ago
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    back

  30. DanJS
    • one year ago
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    yes, a velocity is a vector quantity

  31. DanJS
    • one year ago
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    has a magnitude and a direction for distance/time

  32. calculusxy
    • one year ago
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    second one austin is out for a run. he starts at 98th street and 5th ave and runs 1km north to 110th street in 10 minutes. then, he turns left and runs 0.8km in 9 minutes. then, he turns left and runs down central park west. this takes him 11 min to run down the remaining 1.0km down to 98th street. what is the average velocity?

  33. calculusxy
    • one year ago
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    @DanJS

  34. DanJS
    • one year ago
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    sorry, here

  35. DanJS
    • one year ago
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    First need to draw all those vectors

  36. DanJS
    • one year ago
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    |dw:1438729327608:dw|

  37. DanJS
    • one year ago
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    The average velocity will be the resultant of those 3 vectors, the sum of them

  38. DanJS
    • one year ago
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    |dw:1438729501422:dw|

  39. DanJS
    • one year ago
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    The average velocity is the total displacement/ total time Displacement is the start point to the end point, independent of the path

  40. calculusxy
    • one year ago
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    I got the velocity of 33.33m/s

  41. calculusxy
    • one year ago
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    I am wrong, right?

  42. DanJS
    • one year ago
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    since they made a square, the magnitude or R is the same as the top piece

  43. calculusxy
    • one year ago
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    0.8km/9min = V_avg

  44. DanJS
    • one year ago
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    |dw:1438729828843:dw|

  45. DanJS
    • one year ago
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    His net displacement is 0.8 km straight west, the total time is (10 + 9 + 11) min

  46. DanJS
    • one year ago
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    Average Velocity = Total Disp[lacement / Total Time

  47. calculusxy
    • one year ago
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    0.8 / 1800 sec

  48. calculusxy
    • one year ago
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    0.0004

  49. DanJS
    • one year ago
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    I would just leave it in the units the prob already is in , km/min, since it didn't ask to change it

  50. calculusxy
    • one year ago
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    Our teacher likes seconds to be used that's why.

  51. DanJS
    • one year ago
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    0.8km / 30 min

  52. DanJS
    • one year ago
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    oh, then convert to whatever units you need, but that is the prob

  53. DanJS
    • one year ago
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    about 0.027 km/min

  54. DanJS
    • one year ago
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    west

  55. calculusxy
    • one year ago
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    THANK YOU!!!

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