course Phy 201 Fridays and Saturdays are the days I set aside to work on Physics, but the server was down Friday evening and that is why I am late submitting some of these assignments. ѾǗ{ڑyԂassignment #003
......!!!!!!!!...................................
12:12:11 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)
......!!!!!!!!...................................
RESPONSE --> Convert all measurements to meters. Then add together. 1.80 m + 1.425 m + .534 m = 3.76 m confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:14:49
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
12:15:19 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).
......!!!!!!!!...................................
RESPONSE --> I'm in Phy 201 confidence assessment: 3
.................................................
......!!!!!!!!...................................
12:15:34 ** 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 --> ok self critique assessment: 3
.................................................
nEƤe~ۿ assignment #004 004. `Query 4 Physics I 02-03-2007
......!!!!!!!!...................................
16:59:56 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?
......!!!!!!!!...................................
RESPONSE --> vf = vO + `dv `ds = vAve * `dt confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:00:49 **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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
......!!!!!!!!...................................
17:03:29 What is the displacement `ds associated with uniform acceleration from velocity v0 to velocity vf in clock time `dt?
......!!!!!!!!...................................
RESPONSE --> `ds = `dv * `dt confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:05:00 ** 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. **
......!!!!!!!!...................................
RESPONSE --> Why is it `ds = vAve * `dt as opposed to `ds = `dv / `dt? self critique assessment: 2
.................................................
......!!!!!!!!...................................
17:21:23 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> We have vo, vf and `dt at the top of the diagram. From vo and vf, we can find vAve and `dv. From `dt amd `dv we can find `ds and aAve. confidence assessment: 1
.................................................
......!!!!!!!!...................................
17:25:08 ** 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. **
......!!!!!!!!...................................
RESPONSE --> OK this is what I had in my graph, but I didn't word it as well with stating which level everything is on. self critique assessment: 0
.................................................
......!!!!!!!!...................................
17:25:31 Query Add any surprises or insights you experienced as a result of this assignment.
......!!!!!!!!...................................
RESPONSE --> That helped me understand the levels flow diagram a little better confidence assessment: 2
.................................................
......!!!!!!!!...................................
17:25:44 ** 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. **
......!!!!!!!!...................................
RESPONSE --> ok self critique assessment: 3
.................................................
"