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YumYum247
 one year ago
someone kelp me please!!!!
YumYum247
 one year ago
someone kelp me please!!!!

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YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Please check my second question.....

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0please check my method and calculations with a calculator........

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you not have a calculator?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0is my method correct?????O_O

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 can you please check my work??????:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Part 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: \[\tan^{1} (\frac{ 1.2 }{ 2 }) \approx 31^{o}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The 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.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0t your part one doesn't look correct......hold up let me upload the first question....:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0However for the time i found 46.08s... but this problem is using 2 significant figures so 46s is probably sufficient

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0it's because i think the numbers that your inputting aren't so accurate....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Crap your right I was using the wrong horizontal velocity (2 km/s) :( lemme recalc

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0let me upoad the first question, may take some time....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok I got 2.8 (rounded from 2.77) km/s and 26 degrees (rounded from 25.6)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0yes 26 is what i got.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0try plug in those new numbers....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0darn i meant meters... 15.33 or just 15 m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I did YumYum i plugged in the right numbers and am getting slightly different answers in my calulations

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok for part c) its a little bit more involved but still not too bad hold on....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes YumYum i posted those corrected numbers above

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0let me upload my part c.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so basically speaking part C) is looking for you to correct for the downward drift by aiming the boat upwards

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0please check my east and north signs, are they correct? at the bottom of the page!!!

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i think it should have been (N29E), what do you think.????

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0don't worry about medals, your about to get plenty from me ^_^

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So 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

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i wrote 29degrees!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then you found the horizontal velocity to be 2.5^2=v^2+1.2^2 or v=sqrt(2.5^21.2^2)~2.19~2.2km/h

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i don't know why i wrote 2.2"KM/h"

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0it should have been 2.2Km up in the north.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0as 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 :(

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0does it all look right?!?!?!?!?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so if you take north as your main direction you would be heading at an angle 9029=61degrees east from the north or 29degrees north from the east... something like that

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So yea i agree it all looks good... so far so good

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0aight thanks!!! Please come online more often, i might need your help.....24/7 :"D Thanks!!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0since she will arrive directly at the farmers market that will be the TOTAL travel time to the farmers market using this route

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0it ok, i'll ask irishboy123....chill out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok i looked back up above and in the first possible path she ended up approx 15m to the south of the market

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The conclusion should be clear at that point which is the best route to talk (quickest)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0don't mind but i need to confirm with irishboy ^_^

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0cuz 2 is better than 1 :D

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0i do it all the time....:)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yea 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

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0take it easy man!!! ^_^ come more often, we need people like you here on OS!!! :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yea I do when I have some spare time, but i am in university currently and have my own work and research to be doing.

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Thanks for all your help!!!! you sound very very smart tho!!! :)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.0Alright Thanks for everything, i hope i see you more often ^_^ Cheers!!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thank you :D and you are very welcome I will when I can
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