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YumYum247

  • one year ago

someone kelp me please!!!!

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  1. YumYum247
    • one year ago
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    Please check my second question.....

  2. YumYum247
    • one year ago
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  3. YumYum247
    • one year ago
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    please check my method and calculations with a calculator........

  4. anonymous
    • one year ago
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    Do you not have a calculator?

  5. YumYum247
    • one year ago
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    is my method correct?????O_O

  6. YumYum247
    • one year ago
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    @IrishBoy123 can you please check my work??????:)

  7. anonymous
    • one year ago
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    Part A) Her velocity relative to the short is equal: \[v=\sqrt{(1.2\frac{km}{h})^{2}+(2\frac{km}{h})^{2}} \approx 2.3\frac{km}{h}\] Which by my calculation should give you an angle of: \[\tan^{-1} (\frac{ 1.2 }{ 2 }) \approx 31^{o}\]

  8. anonymous
    • one year ago
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    The method you choose for part b seems correct... using the known velocity across the river coupled with the width of the river to find the travel time.... then using that travel time to calculate the vertical distance downstream the boat has gone.

  9. YumYum247
    • one year ago
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    t your part one doesn't look correct......hold up let me upload the first question....:)

  10. anonymous
    • one year ago
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    However for the time i found 46.08s... but this problem is using 2 significant figures so 46s is probably sufficient

  11. YumYum247
    • one year ago
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    it's because i think the numbers that your inputting aren't so accurate....

  12. anonymous
    • one year ago
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    Crap your right I was using the wrong horizontal velocity (2 km/s) :( lemme recalc

  13. YumYum247
    • one year ago
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    let me upoad the first question, may take some time....

  14. anonymous
    • one year ago
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    Ok I got 2.8 (rounded from 2.77) km/s and 26 degrees (rounded from 25.6)

  15. YumYum247
    • one year ago
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    yes 26 is what i got.

  16. anonymous
    • one year ago
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    so for part b like i was saying the idea is correct, but i got different numbers (see above) that I did use the correct velocity to calculate (2.5 not 2)

  17. YumYum247
    • one year ago
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    try plug in those new numbers....

  18. anonymous
    • one year ago
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    so in 46 seconds the boat travelled downstream a distance d=vt=(1.2*1000/3600) m/s * 46s = 15.33s or roughly 15 seconds (again 2 sig figs.... I dont know if your teacher is a stickler for that)

  19. anonymous
    • one year ago
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    darn i meant meters... 15.33 or just 15 m

  20. anonymous
    • one year ago
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    I did YumYum i plugged in the right numbers and am getting slightly different answers in my calulations

  21. YumYum247
    • one year ago
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  22. anonymous
    • one year ago
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    Ok for part c) its a little bit more involved but still not too bad hold on....

  23. anonymous
    • one year ago
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    Yes YumYum i posted those corrected numbers above

  24. YumYum247
    • one year ago
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    let me upload my part c.....

  25. anonymous
    • one year ago
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    so basically speaking part C) is looking for you to correct for the downward drift by aiming the boat upwards

  26. YumYum247
    • one year ago
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  27. YumYum247
    • one year ago
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    please check my east and north signs, are they correct? at the bottom of the page!!!

  28. YumYum247
    • one year ago
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    i think it should have been (N29E), what do you think.????

  29. YumYum247
    • one year ago
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    don't worry about medals, your about to get plenty from me ^_^

  30. anonymous
    • one year ago
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    So you correctly note the hypoteneuse is of magnitude 2.5 km/s because that is max speed of her boat... which gives you an angle of sin^-1(1.2/2.5) ~28.68~29degrees

  31. YumYum247
    • one year ago
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    i wrote 29degrees!!!

  32. anonymous
    • one year ago
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    then you found the horizontal velocity to be 2.5^2=v^2+1.2^2 or v=sqrt(2.5^2-1.2^2)~2.19~2.2km/h

  33. YumYum247
    • one year ago
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    i don't know why i wrote 2.2"KM/h"

  34. YumYum247
    • one year ago
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    it should have been 2.2Km up in the north.....

  35. anonymous
    • one year ago
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    as for the formatting of the direction.... that I am a little unsure about since it has been a while... but your angle is correct so i have to leave that to you unfortunately :(

  36. YumYum247
    • one year ago
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    does it all look right?!?!?!?!?

  37. anonymous
    • one year ago
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    so if you take north as your main direction you would be heading at an angle 90-29=61degrees east from the north or 29degrees north from the east... something like that

  38. anonymous
    • one year ago
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    So yea i agree it all looks good... so far so good

  39. YumYum247
    • one year ago
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    aight thanks!!! Please come online more often, i might need your help.....24/7 :"D Thanks!!!!

  40. anonymous
    • one year ago
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    now part d is confusingly worded in my opinion but if I understand it correctly than it is easy... just reproduce what you did in part b using your new horizontal velocity to calculate the time to travel across the river here

  41. anonymous
    • one year ago
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    since she will arrive directly at the farmers market that will be the TOTAL travel time to the farmers market using this route

  42. YumYum247
    • one year ago
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    it ok, i'll ask irishboy123....chill out

  43. anonymous
    • one year ago
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    ok i looked back up above and in the first possible path she ended up approx 15m to the south of the market

  44. anonymous
    • one year ago
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    so use her walking velocity to calculate the time it will take her to walk that distance north to the farmers market..... add it to the time it took her to cross the river... and that will give you the TOTAL time of the trip using this route

  45. anonymous
    • one year ago
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    The conclusion should be clear at that point which is the best route to talk (quickest)

  46. anonymous
    • one year ago
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    take*

  47. YumYum247
    • one year ago
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    Thanks!!

  48. anonymous
    • one year ago
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    Your welcome :D

  49. YumYum247
    • one year ago
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    don't mind but i need to confirm with irishboy ^_^

  50. YumYum247
    • one year ago
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    cuz 2 is better than 1 :D

  51. YumYum247
    • one year ago
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    i do it all the time....:)

  52. anonymous
    • one year ago
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    yea no probs i have seen irish boy he is good.... i couldve sworn i saw you two working earlier but when i saw this post again there was no content which is why i jumped in

  53. YumYum247
    • one year ago
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    take it easy man!!! ^_^ come more often, we need people like you here on OS!!! :)

  54. anonymous
    • one year ago
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    Yea I do when I have some spare time, but i am in university currently and have my own work and research to be doing.

  55. YumYum247
    • one year ago
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    Thanks for all your help!!!! you sound very very smart tho!!! :)

  56. YumYum247
    • one year ago
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    understood!!!

  57. YumYum247
    • one year ago
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    Alright Thanks for everything, i hope i see you more often ^_^ Cheers!!!

  58. anonymous
    • one year ago
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    Thank you :D and you are very welcome I will when I can

  59. YumYum247
    • one year ago
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    Yes!!!!! :D

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