course Phy 201
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14:07:24 Query Principles of Physics and 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 --> 1.80m + (142.5 cm * 100) + (5.34mm * 1000) = 19591.8
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14:07:43 ** 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. **
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RESPONSE --> Correct
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14:11:47 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 --> Not University PHysics
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14:11:53 ** 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. **
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RESPONSE --> Ok
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J ˤϕĐ assignment #002 呌騶~Q Physics I 06-08-2006
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14:12:29 NOTE PRELIMINARY TO QUERY:
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RESPONSE --> Ok
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14:12:55 ** Questions about velocity, average velocity, acceleration, etc. are very confusing because so many of the concepts have similar definitions. People have trouble distinguishing things like average velocity, which for uniform acceleration can be obtained in a process that adds two velocities, from average acceleration, which involves subtracting two velocities; one of these processes involves dividing by 2 and the other dividing by the time interval `dt.
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RESPONSE --> Ok
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14:13:01 It is essential to keep the definitions and the meanings of the terms very clear and to work everything from definitions. It is equally important to have a good common-sense understanding of every definition so you can develop the intuition to make sense of everything you do.
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RESPONSE --> Ok
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14:13:06 That inevitably takes people a little time. But in the process you develop the habits you will need to succeed in the course. **
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RESPONSE --> Ok
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ÒVvՐN⡻PNԩM| assignment #003 呌騶~Q Physics I 06-08-2006 cՏ퀸 assignment #004 呌騶~Q Physics I 06-08-2006
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14:27:57 Intro Prob 6 given init vel, accel, `dt find final vel, dist If initial velocity is v0, acceleration is a and time interval is `dt, then in symbols what are the final velocity vf and the displacement `ds?
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RESPONSE --> Vf = Dv + v0 Dt = Dv/aAve
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14:28:06 **You would use accel. and `dt to find `dv: a * `dt = `dv. Adding `dv to initial vel. vo you get final vel. Then average initial vel. and final vel. to get ave. vel.: (v0 + vf) / 2 = ave. vel. You would then multiply ave. vel. and `dt together to get the distance. For example if a = 3 m/s^2, `dt = 5 s and v0 = 3 m/s: 3 m/s^2 * 5 s = 15 m/s = `dv 15 m/s + 3 m/s = 18 m/s = fin. vel. (18 m/s + 3 m/s) / 2 = 10.5 m/s = vAve 10.5 m/s * 5 s = 52.5 m = dist. In more abbreviated form: a * `dt = `dv v0 + `dv = vf (vf + v0) /2 = vAve vAve * `dt = `ds so `ds = (vf + v0) / 2 * `dt. **
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RESPONSE --> Correct
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14:30:00 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> the time required to move through a given displacement at a given velocity is found by dividing displacesment by velocity Dt= Ds/aAve
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14:45:03 ** Since accel is uniform vAve = (v0 + vf) / 2. Thus displacement is `ds = vAve * `dt = (v0 + vf) / 2 * `dt, which is the first equation of uniformly accelerated motion. **
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RESPONSE --> If the acceleration is uniform, velocity increases from inital velocity v0 to final velocity vf, then you can caluclate the average velocity
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14:48:08 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> the flow diagram obtained is that and formaula can be arranged by algebra to give another. the change of time is v/position whole 'dt can be caluculated by change in v over position/clocktime and 'ds can can becaulcated by the change in velocoty multiplied by clock time
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14:48:34 ** The first level in the diagram would contain `dt, v0 and vf. Then v0 and vf would connect to `dv in the second level. The second level would also contain vAve, connected from vf in the first level to v0 in the first level. The third level would contain an a, connected to `dv in the second level and `dt in the first level. The third level would also contain `ds, connected to vAve in the fourth level and `dt in the first level. **
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RESPONSE --> I was not sure what this question is asking for- however I wrote out this diagram for future use
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14:49:07 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> Several insights including the flow diagram and the complexities of volumes caluclated in cubical units
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14:49:19 ** Student Response: I think I really confused what information stood for what in the Force and Pendulum Experiment. However, I enjoy doing the flow diagrams. They make you think in a different way than you are used to. INSTRUCTOR NOTE: These diagrams are valuable for most people. Not all--it depends on learning style--but most. **
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RESPONSE --> Ok
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