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We used vector to find wind speed.
A balloon is timed as it rises inside, assume a *vertical component only
The balloon is taken outside, and timed. The angle it makes is also recorded.
Using the vectors, calculate the wind speed.
 one year ago
 one year ago
We used vector to find wind speed. A balloon is timed as it rises inside, assume a *vertical component only The balloon is taken outside, and timed. The angle it makes is also recorded. Using the vectors, calculate the wind speed.
 one year ago
 one year ago

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amistre64Best ResponseYou've already chosen the best response.0
dw:1348235064295:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i used the law of cosines to determine the magnitude of the resultant. Was that correct to do?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
Teacher wanted us to use pythag thrm to determine the wind speed. But the data collected gave a sqrt(#) which would not allow the pythag thrm to be used. I decided to take the more general Law of Cosines.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i did try it, and since the hypotenuse is smaller than the "leg", it results in a nonreal value.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
his assumption is that the wind never blows things downward i spose
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
speed = distance/time speed up = 1.05 m/s speed along the angle = .97 m/s dw:1348235842574:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i followed per instructions; he noted on the lab report that the time of the outside balloon cannot be greater than the inside record
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
dw:1348235955911:dw
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
Vy is the same for both
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
Im wondering if windspeed is only a horizontal measure? or does directionality not matter?
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
"assume a horizontal component only"
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
there was noting stated prior in the lab to indicate that assumption tho.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the data recorded showed that it rose slower outside, and the note i got on the lab report afterwards was that the data cannot show that. But he also says that we cannot manipulate the data ... i get the feeling that im getting contradictory advice from him :/
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i wasnt. but if we only assume the wind blows things from a horizontal, or upwards ... then the data would show it to rise faster. Rise is prolly a bad term for the outside part.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the velocity along the angle is relatively quicker. just like the boat crossing a stream moves quicker than the boats speedometer measuers
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the lab was designed to get you working with vectors. The speed of the wind could have been calculated from the outside data alone if we were to assume a horizontal wind speed only. .97*sin(23)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
did you really get 23°?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
well, it was the best i could do with the little plastic protractor :)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
if you were getting <11° you could use the small angle approximation
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the teacher suggested the pythag thrm as a means of overcoming the inherent inaccuracies in using the protractors. There was no way to tell if it was "level" and if the person catching the balloon to read the thing hadnt moved it.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i really dont think i understand this experiment , what results did you get /
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[\sqrt{.97^2+1.05^22(.97)(1.05)~cos(23^o)}\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the wind speed itself was immaterial. The lab was designed to get us familar with using vectors and vector math.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
was it a helium ballon ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
yes, we used 3 balloons filled with helium. those shiney things you get at a party supply place. I think one of them had Dora the Explorer on it :)
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
reminds me some of walter lewin 's experiments
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
what were you timing exactly ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i held the balloon on the ground, released it, and timed how long it took it to tighten the string it was attached to.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
so we were calculating the speed along a vector path
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
using both vectors (inside and outside) we would have been able to calculate the speed (magnitude) of the resultant vector  which in turn would have been the wind speed
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
so your vectors are ; \(\{ (1.05,0°) , (0.97, 23°) \} \) ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
those are more like the polar coordinants of the vectors
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
so converting to cartesian coordinates \[\{\quad(0,1.05)\quad,\quad\left(0.97 \cos(23°), 0.97\sin (23°)\right)\quad\}\]
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
dw:1348240487279:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the theta is in the wrong spot
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
spose to be 23 degrees west of north
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
so theta is 9023 ?
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
you can find the resultant right?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
spose we just use this setup, and focus on the vectors dw:1348240769620:dw the resultant can be determined by the law of cosines
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
i thought you were ment to use pythag. ?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
he said we could use whatever method we wanted to find the resultant, and when I used the law of cosines, he said I was wrong for using it.
 one year ago

UnkleRhaukusBest ResponseYou've already chosen the best response.0
do you get the same answer both ways /
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
which leads me to wonder, is windspeed only a horizontal measurement? if so, then the use of 2 vectors is a bad setup
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
no, you dont get the same answer both ways
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
dw:1348241029039:dw this would have been the resultant using pythag thrm; but my data does not present itself as a right triangle
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
there was no "answer" to compare to. So the answer itself is irrelevant. It was the process that I used to find it that he said was wrong. and I dont know why
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
how can i use the pythag thrm with the data collected? i cant.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
\[0.97^21.05^2=0.1616\] \[\sqrt{0.1616}=nonReal~\#\]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
lol, i just noticed a typo in the original post :) assume a *vertical* component only
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
im not sure what to make of demitris reply ...
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the data wasnt "given". I went out and collected the data by performing the experiment
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the 23 im using of course is from a faulty memory.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
but not by use of pythag
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
you cant have a hypotenuse shorter than a leg ...
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
does wind only have horizontal component?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
experiment: i dont know :/
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the lab just wanted us to determine the resultant vector using the data we collected.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
well .. maybe the wind is blowing from dw:1348242000106:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
well, magnitude of the resultant that is
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
i agree, if the wind has a downward component, than the pythag will not be useful
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
nothing stated in the lab said to find a horizontal component ....
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
still even we consider wind, this is not just Pythagoras problem ...
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
is it @amistre64 ?
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
any info given on acceleration?
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
no acceleration given. just direction and speed to define the vectors with
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
it was noted when i got the lab report handed back to me, that the time outside needed to be quicker than the time inside. But my data did not reflect that.
 one year ago

Algebraic!Best ResponseYou've already chosen the best response.0
of course it does...
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1348242332765:dw
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the speed recorded outside needed to be quicker than the speed inside
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the pythag thrm was suggests as a means to circumvent the inaccuracy of degree measurement. But that only works out if the data collected does not have the wind pushing down on the balloon outside.
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
dw:1348242525949:dw we can calculate the acceleration on balloon due to wind. \[{d^2x \over dt^2} = k {dx \over dt} \]
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
thanks for reading it, but my results were calculated already using the cosine rule. I am wondering if it was appropriate to use it in the first place.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
the acceleration seems to be a bit outside the scope that the lab was intended for
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
yes, becasue I use the cosine rule, and included it in the doc that a wrote up
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
not sure ... if this is mathematics dw:1348243032602:dw you can simply get that component using cosine rule.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
experiment, I agree :)
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
lol ... anyway this was getting interesting. :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
we got a new printer in the office, and this ine allows scans to flash drives :) this is the instruction for the lab, just in case you can spot a place that I may have missed.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
this is just a physics 1 class, very basic stuff so far
 one year ago

experimentXBest ResponseYou've already chosen the best response.2
seems plausible. I've been having problems with physics quite a lot lately.
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
here in florida, temp outside is usually higher than inside :)
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
thats a good notice on the outside only part. the balloons were already a part of the lab in class; i had no control on prepping them. the inside part would have been an instruction from the teacher, since its not explicitly stated anywhere in the handout. My only idea as to why we would do the inside part would be to better establish the vertical part of the experiment, but it seems to have added more trouble than it was worth :) Thanks
 one year ago

amistre64Best ResponseYou've already chosen the best response.0
thats what i had in mind at the start i prolly could have included that thought in the "results" section of the lab report.
 one year ago

hipsterBest ResponseYou've already chosen the best response.0
@demitris, please enable private messaging so i may message you about a post that you'd tagged me in a few minutes ago
 one year ago
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