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BloomLocke367

  • one year ago

more homework check xp

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  1. BloomLocke367
    • one year ago
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  2. BloomLocke367
    • one year ago
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    @Shalante

  3. anonymous
    • one year ago
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    Hard to see. >.<

  4. BloomLocke367
    • one year ago
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    sorryyyyyy

  5. anonymous
    • one year ago
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    It requires real effort to look at one line.

  6. BloomLocke367
    • one year ago
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    I can see it clearly o.o I'm sorryyyyy

  7. anonymous
    • one year ago
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    I do not want to check your work, but just do the problem. (Hard to see your work as well)

  8. BloomLocke367
    • one year ago
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    wait... my work is hard to see.. but I can see the problems XD If you get something different I'll just ask about it lol

  9. anonymous
    • one year ago
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    Only checked the first 2. Lemme see the rest? And do you want me check all the problems?

  10. BloomLocke367
    • one year ago
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    I would like it.. but you don't have to lol... I think I'm doing a better job than last time with this stuff

  11. anonymous
    • one year ago
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    Glad to hear that.

  12. anonymous
    • one year ago
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    Holy sheet. Another worksheet.

  13. BloomLocke367
    • one year ago
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    yea.. 19 problems for one night... plus a worksheet from last week that I haven't continued on yet lol (that's the one you helped with last time lol)

  14. anonymous
    • one year ago
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    1. is correct.

  15. BloomLocke367
    • one year ago
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    yayyy

  16. anonymous
    • one year ago
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    2 and 3 is correct

  17. BloomLocke367
    • one year ago
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    woooo I am getting better XD

  18. BloomLocke367
    • one year ago
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    you deserve like 50 gazillion medals btw XD

  19. anonymous
    • one year ago
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    Lol, no need.

  20. BloomLocke367
    • one year ago
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    you're awesome omg

  21. anonymous
    • one year ago
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    4 and 5 are correct.

  22. BloomLocke367
    • one year ago
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    woooo

  23. anonymous
    • one year ago
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    Only to you. :p

  24. BloomLocke367
    • one year ago
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    hehe thanks xp I don't see why tho

  25. anonymous
    • one year ago
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    6 is correct. 100% on first worksheet. Nice!!!

  26. anonymous
    • one year ago
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    Hey, I forgot to check sf. 1 and 2 are correct but not the right # of sf.

  27. BloomLocke367
    • one year ago
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    wooooooo

  28. BloomLocke367
    • one year ago
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    and he doesn't really care about sig figs right now but thanks c: I try getting them anyways for practice. I'll look and fix it

  29. BloomLocke367
    • one year ago
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    1... 50m/s? that drops it to 1 sig fig.. 2...4.0m/s^2?

  30. anonymous
    • one year ago
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    I'll check the sf at the end. For 3. You can use only one formula to find the answer. \[x _{f}=x _{i}+v _{_{i}}t+\frac{ 1 }{ 2 }at^2\]

  31. BloomLocke367
    • one year ago
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    I haven't seen that formula before O.O

  32. BloomLocke367
    • one year ago
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    We only have 4

  33. BloomLocke367
    • one year ago
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    nevermind.. yes I have XD

  34. BloomLocke367
    • one year ago
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    lol

  35. BloomLocke367
    • one year ago
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    but I still got the right answer, right?

  36. anonymous
    • one year ago
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    Same goes for 5. 4 is correct.

  37. anonymous
    • one year ago
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    Wait nvm you would still need 2. Your way is faster for 5.

  38. BloomLocke367
    • one year ago
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    haha thanks xp

  39. BloomLocke367
    • one year ago
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    did I get 3 right?

  40. anonymous
    • one year ago
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    You're not stuck on 6 right? You are just taking a break?

  41. anonymous
    • one year ago
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    Yes.

  42. anonymous
    • one year ago
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    Plug in the values to the equation I gave you and it would still be the same with by just using 1 formula.

  43. BloomLocke367
    • one year ago
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    I'm on 8 right now lol

  44. anonymous
    • one year ago
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    Nice!

  45. BloomLocke367
    • one year ago
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    I got 1.5m for 6 and -0.13m/s^2

  46. BloomLocke367
    • one year ago
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    for 7

  47. BloomLocke367
    • one year ago
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    well I got the final velocity to be 9.6m/s then I used the formula \(v^2=v_0^2+2a(x-x_0)\)

  48. anonymous
    • one year ago
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    9.6^2=92.16m 92.16m=0+2(3.2m/s^2)(d) 92.16m/6.4m/s^2=14.4m or 14m when rounded to sf.

  49. anonymous
    • one year ago
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    You can tell your answer is wrong by reading the problem and thinking like this :

  50. anonymous
    • one year ago
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    Hmm, if a car accelerate for 3.2m/s^2 for 3 sec straight, wouldn't the distance be more than like around 5m?

  51. anonymous
    • one year ago
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    It speeds up by 3.2m/s^2 every second

  52. BloomLocke367
    • one year ago
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    OMG I SEE MY MISTAKE

  53. BloomLocke367
    • one year ago
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    I didn't square 9.6 lol

  54. BloomLocke367
    • one year ago
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    I dunno if I got 8 correct... i got 25m/s after I rounded

  55. anonymous
    • one year ago
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    lol

  56. anonymous
    • one year ago
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    7 is correct.

  57. anonymous
    • one year ago
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    Foe 8, I got the same results.

  58. BloomLocke367
    • one year ago
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    oki yay

  59. BloomLocke367
    • one year ago
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    I'm starting on 9 now

  60. BloomLocke367
    • one year ago
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    I got no for 9.. because the final velocity is 5m/s

  61. BloomLocke367
    • one year ago
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    is that right?

  62. BloomLocke367
    • one year ago
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    OMG http://prntscr.com/8e69uy

  63. anonymous
    • one year ago
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    LMAO

  64. anonymous
    • one year ago
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    For 9, I got no because it would be at 25m/s 0.0

  65. BloomLocke367
    • one year ago
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    that's what I meant lol

  66. BloomLocke367
    • one year ago
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    sorry XD I have issues typing the right numbers for some reason XD

  67. anonymous
    • one year ago
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    Lol, no wonder.

  68. BloomLocke367
    • one year ago
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    what would I do for 10?

  69. anonymous
    • one year ago
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    Hint: one formula.

  70. BloomLocke367
    • one year ago
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    hang on.. I may be able to figure it out

  71. BloomLocke367
    • one year ago
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    the formula you showed me right..

  72. BloomLocke367
    • one year ago
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    ?

  73. anonymous
    • one year ago
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    yes

  74. BloomLocke367
    • one year ago
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    gimme a sec.. everyone is talking

  75. BloomLocke367
    • one year ago
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    you can't add t and t^2, can you?

  76. BloomLocke367
    • one year ago
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    OMG I KNOW WHAT TO DO.. I CAN USE THE QUADRATIC FORMULA

  77. BloomLocke367
    • one year ago
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    wait.. I got both negative answers..but it's time.. @Shalante can you walk me through this?

  78. BloomLocke367
    • one year ago
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    wait... nevermind xp I graphed it and got 4.. is it 4 seconds?

  79. anonymous
    • one year ago
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    You would need quadratic, which is a pain. So the one formula I told you to use was bad. vf^2=v^i+2ax then vf=vi+at is better.

  80. anonymous
    • one year ago
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    If you are given everything except time in the formula, then it would be a good idea to use that 1 formula.

  81. BloomLocke367
    • one year ago
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    ok.. sorry I'm getting stressed

  82. BloomLocke367
    • one year ago
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    so which one?

  83. BloomLocke367
    • one year ago
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    ohhh do the two separate ones like I have been?

  84. anonymous
    • one year ago
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    You got 4 right. Did you do it using 1 formula or 2?

  85. BloomLocke367
    • one year ago
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    I graphed the quadratic from the first equation you had me use xp

  86. anonymous
    • one year ago
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    Either way works but the one formula is annoying since it has a quadratic.

  87. anonymous
    • one year ago
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    Dont do that next time. I messed up also because I hinted you to do it in 1 formula instead of 2.

  88. anonymous
    • one year ago
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    vf^2=v^i+2ax then vf=vi+at

  89. BloomLocke367
    • one year ago
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    bc I tried the quadratic formula but messed up then I just graphed it XD

  90. BloomLocke367
    • one year ago
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    OMG I FOUND MY MISTAKE AND GOT IT WITH THE QUADRATIC FORMULA :D

  91. anonymous
    • one year ago
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    Ok next question.

  92. BloomLocke367
    • one year ago
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    ok.. idk what to do here

  93. anonymous
    • one year ago
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    Are you focusing better right now? Because this question is similar to the ones you done.

  94. BloomLocke367
    • one year ago
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    no I'm not tbh

  95. BloomLocke367
    • one year ago
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    I GOT IT

  96. BloomLocke367
    • one year ago
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    i got 3 seconds

  97. BloomLocke367
    • one year ago
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    make that 3.0

  98. anonymous
    • one year ago
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    Yea I got 3.0 seconds too.

  99. BloomLocke367
    • one year ago
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    okay.. time for the next one.. thank you so much by the way XD

  100. BloomLocke367
    • one year ago
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    for 12.. is 0 the initial velocity and 25 the final?

  101. anonymous
    • one year ago
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    Yes.

  102. BloomLocke367
    • one year ago
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    okay thank you

  103. anonymous
    • one year ago
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    No problem!

  104. BloomLocke367
    • one year ago
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    oops I forgot x.x

  105. BloomLocke367
    • one year ago
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    I got 2083.3m....

  106. BloomLocke367
    • one year ago
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    idk if that's right

  107. anonymous
    • one year ago
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    Yes, but sig figs makes it 2.1 x 10^3m

  108. anonymous
    • one year ago
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    That means the minimum distance is 2083.3m in order for the train to reach a velocity of 25 m/s. It makes since because the train is accelerating at 0.15 m/s^2

  109. BloomLocke367
    • one year ago
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    okay :)

  110. BloomLocke367
    • one year ago
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    LAST ONE.

  111. BloomLocke367
    • one year ago
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    then I can move onto precal XD

  112. BloomLocke367
    • one year ago
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    omg can I just use a ratio for the last one?

  113. BloomLocke367
    • one year ago
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    \[\frac{ 53 }{ 120}=\frac{ 85 }{ x }\]

  114. BloomLocke367
    • one year ago
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    can I do that?

  115. anonymous
    • one year ago
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    Yea, wait after, because I am helping someone on a difficult chem question.

  116. BloomLocke367
    • one year ago
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    ok

  117. anonymous
    • one year ago
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    No because it does not proportional in that kind of ratio.

  118. BloomLocke367
    • one year ago
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    what do you mean?

  119. anonymous
    • one year ago
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    speed 1/distance 1 is not proportional to speed 2/distance 2 Fraction is called a ratio. Meant to write it is not proportional* in the previous post.

  120. BloomLocke367
    • one year ago
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    oh

  121. BloomLocke367
    • one year ago
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    nevermind then

  122. BloomLocke367
    • one year ago
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    hmmm

  123. BloomLocke367
    • one year ago
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    what do I do?

  124. anonymous
    • one year ago
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    THINK!!!!

  125. BloomLocke367
    • one year ago
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    OH

  126. BloomLocke367
    • one year ago
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    wait.. idk

  127. BloomLocke367
    • one year ago
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    are you sure it's not proportional?

  128. BloomLocke367
    • one year ago
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    why isn't it proportional?

  129. anonymous
    • one year ago
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    Ok, a drag racer can reach a speed of 53 m/s can also mean that when it starts from rest the drag racer can reach a speed of 53 m/s So initial velocity is 0 m/s. Final velocity would be 53 m/s. Distance would be 120m Solve for acceleration using the formula. (Not telling you which formula) But what about 85 m/s. It needs to start from rest in order for the drag racer to move right? Initial velocity is zero, final velocity is 85, acceleration is the one you solved earlier, and distance is what you need to know for the answer.

  130. BloomLocke367
    • one year ago
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    um.. don't you need time to find acceleration?

  131. anonymous
    • one year ago
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    We would use the same acceleration since it is the same drag racer and it told us he had a max acceleration (his highest is 53m/s at 120m)

  132. BloomLocke367
    • one year ago
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    \(\large a=\frac{v-v_0}{t}\)

  133. BloomLocke367
    • one year ago
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    right? and we don't have time... wait.. yes we do... v=d/t

  134. BloomLocke367
    • one year ago
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    hang on

  135. BloomLocke367
    • one year ago
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    I got 2.26 for the time

  136. anonymous
    • one year ago
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    Why would use that formula when they dont give you the 2 units (time and acceleration) How are you going to solve the unknown variable?

  137. BloomLocke367
    • one year ago
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    I think I got this.. I gotta go..

  138. BloomLocke367
    • one year ago
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    ttyl

  139. anonymous
    • one year ago
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    Hint: what formula requires distance, initial and final velocity? (Forgot already!?)

  140. anonymous
    • one year ago
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    KK

  141. Jhannybean
    • one year ago
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    Have you made a flashcard with all the kinematic equations on it? If not, I suggest you do, and then either post it on your wall, look at it every day and then see how many you have memorized. Carry the flashcard with you to school as well, and review it every chance you get. The faster you can recall the formulas and how they are applied, the quicker you will finish your test!

  142. BloomLocke367
    • one year ago
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    V^2=V_0^2+2a(x-x_0) lol

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