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anonymous

  • one year ago

I need help with this

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  1. anonymous
    • one year ago
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  2. nikato
    • one year ago
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    do you have the answer?

  3. nikato
    • one year ago
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    |dw:1433306744103:dw|

  4. nikato
    • one year ago
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    |dw:1433306925729:dw|

  5. nikato
    • one year ago
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    using this formula \[v ^{2}=v _{0}^{2}+2a(\Delta x)\]

  6. nikato
    • one year ago
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    |dw:1433307141300:dw|

  7. nikato
    • one year ago
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    |dw:1433307348337:dw| i hope this is right

  8. IrishBoy123
    • one year ago
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    simplest approach, using best equation of motion for the job, and resolving gravity down the slope as g sin(30.9) as suggested by @nikato: \(x = u t + 1/2 a t^2, u = 0, t = \sqrt{2x/a}, a = g sin (30.9), x = 208, t = 9.09s\) [error in previous attempt is in using constant velocity down the slope, which gives half the correct answer: as the snow starts at zero and accelerates to final velocity.]

  9. anonymous
    • one year ago
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    yeah I talked with my professor and I found that I actually set the problem up incorrectly, but I understand it now. I got the 9.0917 s.

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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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