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18:56:37 Query Video experiment 3
what were the average velocities on the first and second ramp -- express as ordered pairs (1st ramp vel, 2d ramp vel).......!!!!!!!!...................................
RESPONSE --> (9.014084507, 25.09803922), (12.39669421, 23.13253012), (13.91304348, 27.23404255), (15.09433962, 28.44444444), (17.14285714, 28.13186813), (19.82300885, 31.2195122), (20, 35.55555556), (20.86956522, 35.55555556), (20.51282051, 40)
Overall averages: (16.52960151,30.48572753).................................................
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19:07:03 ** Suppose that for a certain ramp setup the first-ramp ave. vel. Is 9 cm/s and the second-ramp ave. vel. is 21 cm/s. They you would plot the point (9, 21), provided the graph specifies units of cm/s. You would plot one such point for each ramp setup. You will get several points, which will lie more or less near a 'best-fit- straight line.
The ideal slope of this line is 2, and accurate measurements will give a slope close to 2. While most students obtain results in which the second-ramp ave velocity is generally pretty close to double that first-ramp average velocity, slopes ranging from about 1.2 to 2.3 have been obtained by various students. Slopes reported by most students being a bit low (the most common results are in the range from 1.3 to 1.6). The discrepancy between the graph slope and the fact that 2d-ramp velocities are usually about double those obtained for the first ramp could indicate a systematic error in timing--something like a slight anticipation of, say, the 'click' as the ball goes from the first ramp to the second. It could also indicate a slight 'boost', due to one effect or another, at the start of the motion, although a slight delay due to friction would seem more likely. You sould understand why the second ramp should give you double the ave vel on the first, which is the main point. It is also important to understand how different sorts of errors creep in and throw results off. **......!!!!!!!!...................................
RESPONSE --> My slope is low at 1.14 and I'm pretty sure it is problems with the timing. I took three measurements for each ramp length but I could tell that my timing wasn't perfect that often. Although my results did show up as ramp 2 being nearly double ramp 1, I understand why the slope is important when it comes to graphing the information.
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19:07:23 Does the result support the hypothesis that the final velocity is double the average velocity if acceleration is uniform and if initial velocity is zero?
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RESPONSE --> Yes my result does support the hypothesis.
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19:07:41 ** 2d-ramp velocities generally run close to double the first-ramp averages, which does support the hypothesis. However due to experimental errors graph slopes are often a bit less than 2 and do not always support the hypothesis. **
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RESPONSE --> ok
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19:23:29 Query General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> I decided to change the units to all be in centimeters. (180cm + 142.5cm + .000534cm) = 322.5. In significant figures, 323cm
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19:34:17 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter).
Therefore nothing below .01 m can be distinguished. 142.5 cm is .01425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, good to within .00000001 m. Then theses are added you get 1.81425534 m; however the 1.80 m is only good to within .01 m so the result is 1.81 m. The rest of the number is meaningless, since the first number itself could be off by as much as .01 m. **......!!!!!!!!...................................
RESPONSE --> ok I put mine in cm.
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19:40:48 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> I can't find the problem in my text. 1.34 for me asks for the accuracy in the distance of a satelite that's 20,000km away...
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19:43:53 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT:
The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **......!!!!!!!!...................................
RESPONSE --> No wonder I'm so lost. I don't think I could have figured any of this out even if I could find the problem in the book.
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