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iwanttogotostanford

  • one year ago

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  1. iwanttogotostanford
    • one year ago
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    @zepdrix @triciaal

  2. iwanttogotostanford
    • one year ago
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    I need to know the acceleration due to gravity and the average G value for each of them given what I have

  3. iwanttogotostanford
    • one year ago
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    E Earth Moon Mars Distance (m) 2.25 3.08 4.04 4.57 4.68 3.55 Time (s) 0.680 0.793 2.20 2.26 1.58 1.38 Acceleration due to Gravity (m/s^2) 9.732 9.795 1.669 1.789 3.749 3.728 Average G value for the location (m/s^2)

  4. iwanttogotostanford
    • one year ago
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    I NEED TO KNOW THE AVERAGE G VALUE FOR THE LOCATION

  5. iwanttogotostanford
    • one year ago
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    I need the equation or formula please I am confused

  6. jdoe0001
    • one year ago
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    a picture of the chart/table will work better methinks but the acceleration could be obtained from the slope I'd think

  7. jdoe0001
    • one year ago
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    and average... is more or less rate of change, which brings us back to slope again

  8. iwanttogotostanford
    • one year ago
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    Earth Moon Mars Distance (m) 2.25 3.08 4.04 4.57 4.68 3.55 Time (s) 0.680 0.793 2.20 2.26 1.58 1.38 Acceleration due to Gravity (m/s^2) 9.732 9.795 1.669 1.789 3.749 3.728 Average G value for the location (m/s^2)

  9. jdoe0001
    • one year ago
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    then again, a picture of the table woiuld help :)

  10. iwanttogotostanford
    • one year ago
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    This is the chart but it looks a little messed up but it is in line

  11. jdoe0001
    • one year ago
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    well, surely doesn't look like that on your screen... so just screenshot it methink

  12. iwanttogotostanford
    • one year ago
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    I'm on a macbook

  13. iwanttogotostanford
    • one year ago
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    but I will try

  14. iwanttogotostanford
    • one year ago
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  15. iwanttogotostanford
    • one year ago
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    Here it is

  16. iwanttogotostanford
    • one year ago
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    it is a download did you get it

  17. anonymous
    • one year ago
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    http://prntscr.com/8q0dt3 There's a screenshot if you can't download the document for some strange reason. For future reference, press command + shift + 4 to screenshot something on a Mac.

  18. jdoe0001
    • one year ago
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    yes.... well you have two values per each location so the average "G"ravity will be the sum of both, divided by 2 say for earth will be \(\bf \cfrac{9.732+9.795}{2}\)

  19. iwanttogotostanford
    • one year ago
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    So would they be in order= 14.6295 2.5635 5.613

  20. iwanttogotostanford
    • one year ago
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    @jdoe0001

  21. iwanttogotostanford
    • one year ago
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    @jdoe0001 Please check

  22. iwanttogotostanford
    • one year ago
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    @triciaal Did i do the table right?? :-)

  23. triciaal
    • one year ago
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    no idea haven't look yet

  24. iwanttogotostanford
    • one year ago
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    @triciaal could you check please

  25. triciaal
    • one year ago
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    not sure of final for each location same process velocity = distance /time acceleration = velocity/time for Earth (3.08 - 2.25)/ (0.793-0.680) = velocity velocity/time(1) = 9.732 velocity/time(2) = 9.795 average G = velocity/(t2 - t1)

  26. triciaal
    • one year ago
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    |dw:1444528438601:dw|

  27. triciaal
    • one year ago
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    @mathmate can you help with this?

  28. triciaal
    • one year ago
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    @jim_thompson5910 ?

  29. mathmate
    • one year ago
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    @iwanttogotostanford You would need the kinematics equation for an object in free fall, \(\Delta x= (1/2)gt^2\) where g=acceleration due gravity, t=time it takes, (initial velocity = 0 for free fall) Solve for g to get \(g=2\Delta x /t^2\) That's how @jdoe001 got 9.732 and 9.795 for the first pair of data. Continue with the two other pairs for moon and mars, and you'll have a nice table.

  30. mathmate
    • one year ago
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    Average value would be, as @jdoe0001 stated, the sum divided by 2, for each pair.

  31. iwanttogotostanford
    • one year ago
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    @mathmate I know but what about for the last row of the table- i am confused

  32. iwanttogotostanford
    • one year ago
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    @triciaal

  33. iwanttogotostanford
    • one year ago
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    @jim_thompson5910 how do you figure out the last row i know the third row but did i do the last row correctly?

  34. mathmate
    • one year ago
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    Location Earth Moon Mars Acceleration due to Gravity (m/s^2) \(\color{red}{(}\)9.732 9.795\(\color{red}{)}\) \(\color{red}{(}\)1.669 1.789\(\color{red}{)}\) \(\color{red}{(}\)3.749 3.728\(\color{red}{) }\) Average acceleration 9.764 1.729 3.739 For earth, the average is (9.732+9.795)/2=9.764 For moon, the average is (1.669+1.789)/2=1.729 For mars, the average is (3.749+3.728)/2=3.739

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