course phy201 ۑK㬬Ҧwٰassignment #003
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20:27:04 ** 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 --> ok
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20:28:13 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 required
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ꭁx۞|yw assignment #004 o[Βɕއ Physics I 06-21-2006
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20:48:26 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 --> The final velocity is vf=v0+a*'dt. The displacement is 'ds=v0*'dt+.5a*'dt^2
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20:53:57 **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 --> ok
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21:01:51 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> 'ds=v0*'dt*.5a*'dt^2
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21:04:46 ** 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 --> The first equation of uniformly accelerated motion is a=(vf-v0)/'dt
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21:05:53 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> I'm not sure how to describe this.
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21:06:20 ** 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 --> Now I understand
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21:06:45 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> the flow diagram,need more insight on that question.
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21:06:58 ** 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 --> pk
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course phy201 ۑK㬬Ҧwٰassignment #003
......!!!!!!!!...................................
20:27:04 ** 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 --> ok
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20:28:13 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 required
.................................................
ꭁx۞|yw assignment #004 o[Βɕއ Physics I 06-21-2006
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20:48:26 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 --> The final velocity is vf=v0+a*'dt. The displacement is 'ds=v0*'dt+.5a*'dt^2
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20:53:57 **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 --> ok
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21:01:51 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
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RESPONSE --> 'ds=v0*'dt*.5a*'dt^2
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21:04:46 ** 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 --> The first equation of uniformly accelerated motion is a=(vf-v0)/'dt
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21:05:53 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
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RESPONSE --> I'm not sure how to describe this.
.................................................
......!!!!!!!!...................................
21:06:20 ** 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 --> Now I understand
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21:06:45 Query Add any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> the flow diagram,need more insight on that question.
.................................................
......!!!!!!!!...................................
21:06:58 ** 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 --> pk
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