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amistre64
 3 years 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.
amistre64
 3 years 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.

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amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348235064295:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i used the law of cosines to determine the magnitude of the resultant. Was that correct to do?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0Teacher 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.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i did try it, and since the hypotenuse is smaller than the "leg", it results in a nonreal value.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0his assumption is that the wind never blows things downward i spose

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0speed = distance/time speed up = 1.05 m/s speed along the angle = .97 m/s dw:1348235842574:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i followed per instructions; he noted on the lab report that the time of the outside balloon cannot be greater than the inside record

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348235955911:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Vy is the same for both

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0Im wondering if windspeed is only a horizontal measure? or does directionality not matter?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0"assume a horizontal component only"

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0there was noting stated prior in the lab to indicate that assumption tho.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the 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 :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i 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.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the velocity along the angle is relatively quicker. just like the boat crossing a stream moves quicker than the boats speedometer measuers

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the 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)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0did you really get 23°?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0well, it was the best i could do with the little plastic protractor :)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0if you were getting <11° you could use the small angle approximation

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the 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.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i really dont think i understand this experiment , what results did you get /

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{.97^2+1.05^22(.97)(1.05)~cos(23^o)}\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the wind speed itself was immaterial. The lab was designed to get us familar with using vectors and vector math.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0was it a helium ballon ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0yes, 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 :)

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0reminds me some of walter lewin 's experiments

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0what were you timing exactly ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i held the balloon on the ground, released it, and timed how long it took it to tighten the string it was attached to.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0so we were calculating the speed along a vector path

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0using 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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0so your vectors are ; \(\{ (1.05,0°) , (0.97, 23°) \} \) ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0those are more like the polar coordinants of the vectors

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

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348240487279:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the theta is in the wrong spot

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0spose to be 23 degrees west of north

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0so theta is 9023 ?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0you can find the resultant right?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0spose we just use this setup, and focus on the vectors dw:1348240769620:dw the resultant can be determined by the law of cosines

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0i thought you were ment to use pythag. ?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0he 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.

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.0do you get the same answer both ways /

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0which leads me to wonder, is windspeed only a horizontal measurement? if so, then the use of 2 vectors is a bad setup

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0no, you dont get the same answer both ways

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1348241029039:dw this would have been the resultant using pythag thrm; but my data does not present itself as a right triangle

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0there 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0how can i use the pythag thrm with the data collected? i cant.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0\[0.97^21.05^2=0.1616\] \[\sqrt{0.1616}=nonReal~\#\]

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0lol, i just noticed a typo in the original post :) assume a *vertical* component only

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0im not sure what to make of demitris reply ...

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the data wasnt "given". I went out and collected the data by performing the experiment

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the 23 im using of course is from a faulty memory.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0but not by use of pythag

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0you cant have a hypotenuse shorter than a leg ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2does wind only have horizontal component?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0experiment: i dont know :/

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the lab just wanted us to determine the resultant vector using the data we collected.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2well .. maybe the wind is blowing from dw:1348242000106:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0well, magnitude of the resultant that is

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0i agree, if the wind has a downward component, than the pythag will not be useful

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0nothing stated in the lab said to find a horizontal component ....

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2still even we consider wind, this is not just Pythagoras problem ...

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2any info given on acceleration?

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0no acceleration given. just direction and speed to define the vectors with

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0it 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.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2dw:1348242332765:dw

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the speed recorded outside needed to be quicker than the speed inside

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the 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.

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0thanks 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.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0the acceleration seems to be a bit outside the scope that the lab was intended for

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0yes, becasue I use the cosine rule, and included it in the doc that a wrote up

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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0experiment, I agree :)

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2lol ... anyway this was getting interesting. :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0we 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.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0this is just a physics 1 class, very basic stuff so far

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.2seems plausible. I've been having problems with physics quite a lot lately.

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0here in florida, temp outside is usually higher than inside :)

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0thats 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

amistre64
 3 years ago
Best ResponseYou've already chosen the best response.0thats what i had in mind at the start i prolly could have included that thought in the "results" section of the lab report.

anonymous
 3 years ago
Best 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
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