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anonymous

  • one year ago

I'm stuck on how to start this one problem. I am told the following: "A man walks at 5 km/h in the direction of a 20-km/h wind. Raindrops fall vertically at 6.6 km/h in still air. Determine the direction in which the drops appear to fall with respect to the man, measured clockwise from the direction of the wind." I know that for velocity comparisons, v_b = v_a + v_(b/a), but I don't know how to apply this logic for three different velocities in two different directions. Can someone please help me?

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  1. anonymous
    • one year ago
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    And before anyone else says it, I know this question would be better suited for the physics and engineering sections, but from my experience, nobody ever seems to check those for questions, so here I am.

  2. anonymous
    • one year ago
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    Is anybody there?

  3. anonymous
    • one year ago
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    Hello?

  4. anonymous
    • one year ago
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    @Shalante , I notice you're reading my question. Could you perchance be able to help me out?

  5. anonymous
    • one year ago
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    Nevermind, just saw your message. Will wait for further instruction.

  6. anonymous
    • one year ago
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    Again, not looking for answers, here. Just want an idea of how to approach this.

  7. triciaal
    • one year ago
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    |dw:1443676636690:dw|

  8. triciaal
    • one year ago
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    vertical angles are equal - tan theta = - tan ^-1 (6.6/20)

  9. triciaal
    • one year ago
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    do you have answer choices?

  10. anonymous
    • one year ago
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    No, just a set number of attempts.

  11. triciaal
    • one year ago
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    do you have an idea what my sketch is?

  12. anonymous
    • one year ago
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    From what I can tell, you've gor the speed of the man (6.6 km/h), and some arrow as (20 km/hr). Not sure where you got 20 from though. Also, tan^-1(6.6/20) is incorrect.

  13. anonymous
    • one year ago
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    *got

  14. triciaal
    • one year ago
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    |dw:1443677500413:dw|

  15. triciaal
    • one year ago
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    you skipped the leading minus

  16. triciaal
    • one year ago
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    it said measured clockwise

  17. anonymous
    • one year ago
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    Shalante said they figured it out. Hold please,

  18. triciaal
    • one year ago
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    @dan815 what is your take on this?

  19. anonymous
    • one year ago
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    If two objects are traveling at the same direction, then the relative velocity is the difference between those two. The wind would think the man is traveling 15km/h slower. The man would think the wind is traveling 15km/h faster. With respect to is using the man as a reference. |dw:1443677931462:dw| So relative velocity of man 15km/h Relative velocity of wind would be -15km/h \[\tan^{-1} \frac{ opposite }{ adjacent }=\tan^{-1} \frac{ 6.6 }{ 15 }=\] 23.74 degrees

  20. anonymous
    • one year ago
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    By the way, I am pretty active in the physics section, but not engineering though.

  21. anonymous
    • one year ago
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    If two objects are traveling in opposite we would add them up. Just like how cars move very fast in opposite direction.

  22. anonymous
    • one year ago
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    Where'd you get 20 from though, when you were finding the difference? That's the only thing I don't get at this point. I get choosing the man as your reference point, but why'd you subtract 5 from 20? Where did 20 km/hr come from?

  23. anonymous
    • one year ago
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    Wait, nevermind. I just realized my mistake.

  24. triciaal
    • one year ago
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    @Shalante thanks I forgot to adjust for the man's velocity @The_Silent_One the rain-drop (vertical) was 6.6 the wind was 20

  25. anonymous
    • one year ago
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    I wrote the wrong value for that variable on my physical paper, so no wonder I've been getting this wrong.

  26. anonymous
    • one year ago
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    No problem! I got the angles messed up earlier. I forgot to read that it was respect to man, so the angle is toward the x component. I did this earlier |dw:1443678723277:dw| Thats why my first answer was wrong.

  27. anonymous
    • one year ago
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    Okay, that makes sense. Thanks for helping me out! I gave you a medal for your explanation. I'm going to close this question now and get some sleep.

  28. anonymous
    • one year ago
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    Yea, no problem. Solving another difficult physics question right now.

  29. anonymous
    • one year ago
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    Thanks for the medals

  30. anonymous
    • one year ago
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    medal*

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