YumYum247
  • YumYum247
someone kelp me please!!!!
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
YumYum247
  • YumYum247
Please check my second question.....
YumYum247
  • YumYum247
YumYum247
  • YumYum247
please check my method and calculations with a calculator........

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anonymous
  • anonymous
Do you not have a calculator?
YumYum247
  • YumYum247
is my method correct?????O_O
YumYum247
  • YumYum247
@IrishBoy123 can you please check my work??????:)
anonymous
  • anonymous
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: \[\tan^{-1} (\frac{ 1.2 }{ 2 }) \approx 31^{o}\]
anonymous
  • anonymous
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.
YumYum247
  • YumYum247
t your part one doesn't look correct......hold up let me upload the first question....:)
anonymous
  • anonymous
However for the time i found 46.08s... but this problem is using 2 significant figures so 46s is probably sufficient
YumYum247
  • YumYum247
it's because i think the numbers that your inputting aren't so accurate....
anonymous
  • anonymous
Crap your right I was using the wrong horizontal velocity (2 km/s) :( lemme recalc
YumYum247
  • YumYum247
let me upoad the first question, may take some time....
anonymous
  • anonymous
Ok I got 2.8 (rounded from 2.77) km/s and 26 degrees (rounded from 25.6)
YumYum247
  • YumYum247
yes 26 is what i got.
anonymous
  • anonymous
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)
YumYum247
  • YumYum247
try plug in those new numbers....
anonymous
  • anonymous
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)
anonymous
  • anonymous
darn i meant meters... 15.33 or just 15 m
anonymous
  • anonymous
I did YumYum i plugged in the right numbers and am getting slightly different answers in my calulations
YumYum247
  • YumYum247
anonymous
  • anonymous
Ok for part c) its a little bit more involved but still not too bad hold on....
anonymous
  • anonymous
Yes YumYum i posted those corrected numbers above
YumYum247
  • YumYum247
let me upload my part c.....
anonymous
  • anonymous
so basically speaking part C) is looking for you to correct for the downward drift by aiming the boat upwards
YumYum247
  • YumYum247
YumYum247
  • YumYum247
please check my east and north signs, are they correct? at the bottom of the page!!!
YumYum247
  • YumYum247
i think it should have been (N29E), what do you think.????
YumYum247
  • YumYum247
don't worry about medals, your about to get plenty from me ^_^
anonymous
  • anonymous
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
YumYum247
  • YumYum247
i wrote 29degrees!!!
anonymous
  • anonymous
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
YumYum247
  • YumYum247
i don't know why i wrote 2.2"KM/h"
YumYum247
  • YumYum247
it should have been 2.2Km up in the north.....
anonymous
  • anonymous
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 :(
YumYum247
  • YumYum247
does it all look right?!?!?!?!?
anonymous
  • anonymous
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
anonymous
  • anonymous
So yea i agree it all looks good... so far so good
YumYum247
  • YumYum247
aight thanks!!! Please come online more often, i might need your help.....24/7 :"D Thanks!!!!
anonymous
  • anonymous
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
anonymous
  • anonymous
since she will arrive directly at the farmers market that will be the TOTAL travel time to the farmers market using this route
YumYum247
  • YumYum247
it ok, i'll ask irishboy123....chill out
anonymous
  • anonymous
ok i looked back up above and in the first possible path she ended up approx 15m to the south of the market
anonymous
  • anonymous
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
anonymous
  • anonymous
The conclusion should be clear at that point which is the best route to talk (quickest)
anonymous
  • anonymous
take*
YumYum247
  • YumYum247
Thanks!!
anonymous
  • anonymous
Your welcome :D
YumYum247
  • YumYum247
don't mind but i need to confirm with irishboy ^_^
YumYum247
  • YumYum247
cuz 2 is better than 1 :D
YumYum247
  • YumYum247
i do it all the time....:)
anonymous
  • anonymous
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
YumYum247
  • YumYum247
take it easy man!!! ^_^ come more often, we need people like you here on OS!!! :)
anonymous
  • anonymous
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.
YumYum247
  • YumYum247
Thanks for all your help!!!! you sound very very smart tho!!! :)
YumYum247
  • YumYum247
understood!!!
YumYum247
  • YumYum247
Alright Thanks for everything, i hope i see you more often ^_^ Cheers!!!
anonymous
  • anonymous
Thank you :D and you are very welcome I will when I can
YumYum247
  • YumYum247
Yes!!!!! :D

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