course Phy 121 볏JNjed~Nassignment #006
......!!!!!!!!...................................
09:32:57 `q001. Note that there are 9 questions in this assignment. Using the equation vf = v0 + a * `dt determine the acceleration of an object whose velocity increases at a uniform rate from 10 m/s to 30 m/s in 15 seconds. Begin by solving the equation for the acceleration a, then 'plug in' your initial and final velocities. Describe your work step y step.
......!!!!!!!!...................................
RESPONSE --> a=20m/s 30m/s=10m/s + a x 15s 20m/s=15a I'm confused confidence assessment: 0
.................................................
......!!!!!!!!...................................
09:34:36 `q002. It wasn't necessary to use a equation to solve this problem. How could this problem had been reasoned out without the use of an equation?
......!!!!!!!!...................................
RESPONSE --> we had the same problem on both sides of the equation, we could have just divided 20 by 15 confidence assessment: 1
.................................................
......!!!!!!!!...................................
09:40:51 We begin by solving the equation for v0. Starting with `ds = (vf + v0) / 2 * `dt, we can first multiply both sides of the equation by 2 / `dt, which gives us `ds * 2 / `dt = (vf + v0) / 2 * `dt * 2 / `dt. The right-hand side can be rearranged to give (vf + v0) * `dt / `dt * 2 / 2; since `dt / `dt = 1 and 2 / 2 = 1 the right-hand side becomes just vf + v0. The equation therefore becomes 2 * `ds / `dt = vf + v0. Adding -vf to both sides we obtain v0 = 2 * `ds / `dt - vf. We now plug in `ds = 80 meters, `dt = 10 sec and vf = 6 m/s to get v0 = 2 * 80 meters / 10 sec - 6 m/s = 160 meters / 10 sec - 6 m/s = 16 m/s - 6 m/s = 10 m/s.
......!!!!!!!!...................................
RESPONSE --> i got the answer of 10 m/s but through deductive reasoning, not the right way to get the answer self critique assessment: 3
.................................................
......!!!!!!!!...................................
09:41:11 `q004. We can reconcile the above solution with straightforward reasoning. How could the initial velocity have been reasoned out from the given information without the use of an equation? Hint: two of the quantities given in the problem can be combined to give another important quantity, which can then be combined with the third given quantity to reason out the final velocity.
......!!!!!!!!...................................
RESPONSE --> this is exactly how i got the answer confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:45:11 The unknown quantity is the initial velocity v0. To solve for v0 we start with `ds = v0 `dt + .5 a `dt^2. We first add -.5 a `dt^2 to both sides to obtain `ds - .5 a `dt^2 = v0 `dt. We then divide both sides by `dt to obtain (`ds - .5 a `dt^2) / `dt = v0. Then we substitute the given displacement `ds = 80 meters, acceleration a = -2 m/s^2 and time interval `dt = 10 seconds to obtain v0 = [ 80 meters - .5 * (-2 m/s^2) * (10 sec)^2 ] / (10 sec) = [ 80 meters - .5 * (-2 m/s^2) * 100 s^2 ] / (10 sec) = [ 80 m - (-100 m) ] / (10 sec) = 180 m / (10 s) = 18 m/s.
......!!!!!!!!...................................
RESPONSE --> my algrebra has left me this morning. i'm getting the answers but i'm not getting them by working it out right self critique assessment: 2
.................................................
......!!!!!!!!...................................
09:46:42 `q006. Check the consistency of this result by verifying, by direct reasoning rather than equations, that an object whose initial velocity is 18 m/s and which accelerates for 10 seconds at an acceleration of -2 m/s^2 does indeed experience a displacement of 80 meters.
......!!!!!!!!...................................
RESPONSE --> in second 1 it goes 18m in second 2, 16m 3,14m 4,12m 5,10, 6,8 7,6 that's already 80 m, and we still have one second to go confidence assessment: 3
.................................................
......!!!!!!!!...................................
09:54:50 `q007. Using the equation vf^2 = v0^2 + 2 a `ds determine the initial velocity of an object which attains a final velocity of 20 meters/second after accelerating uniformly at 2 meters/second^2 through a displacement of 80 meters. Begin by solving the equation for the unknown quantity and show every step.
......!!!!!!!!...................................
RESPONSE --> vf^2=v0^2 + 2a`ds vf^2 - 2a`ds = v0^2 20ms^2 - 2(8ms^2*80m) = v0^2 20m/s^2 -16m/s^2 * 160m = v0^2 4m/s^2 * 160m i know the final answer should be 12m/s^2 confidence assessment: 2
.................................................
......!!!!!!!!...................................
09:56:56 `q009. Describe in commonsense terms the motion of the object in this example if its initial velocity is indeed -8.9 m/s. Assume that the object starts at the crossroads between two roads running North and South, and East and West, respectively, and that the object ends up 80 meters North of the crossroads. In what direction does it start out, what happens to its speed, and how does it end up where it does?
......!!!!!!!!...................................
RESPONSE --> it starts out going south, its speed reverses and accelerates and it goes north confidence assessment: 2
.................................................
|aHxآ㤟 assignment #007 007. Acceleration of Gravity Physics I 06-18-2007
......!!!!!!!!...................................
18:57:18 `q001. We obtain an estimate of the acceleration of gravity by determining the slope of an acceleration vs. ramp slope graph for an object gliding down an incline. Sample data for an object gliding down a 50-cm incline indicate that the object glides down the incline in 5 seconds when the raised end of the incline is .5 cm higher than the lower end; the time required from rest is 3 seconds when the raised end is 1 cm higher than the lower end; and the time from rest is 2 seconds when the raised end is 1.5 cm higher than the lower end. What is the acceleration for each trial?
......!!!!!!!!...................................
RESPONSE --> 1. 50cm in 5sec 2. 50cm in 3s 3. 50cm in 2s 1. a=10cm/s 2. a= 16.7cm/s 3. a=25cm/s confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:58:14 `q002. What are the ramp slopes associated with these accelerations?
......!!!!!!!!...................................
RESPONSE --> the ramp slopes associate with the slope of the accelerations if we were to graph the three trials confidence assessment: 2
.................................................
......!!!!!!!!...................................
18:59:47 `q003. Sketch a reasonably accurate graph of acceleration vs. ramp slope and give a good description and interpretation of the graph. Be sure to include in your description how the graph points seem to lie with respect to the straight line that comes as close as possible, on the average, to the three points.
......!!!!!!!!...................................
RESPONSE --> the graph slope is steep and increasing. the graph for trial 3 is increasing at an ever increasing rate, moreso than graph for trial 1 confidence assessment: 2
.................................................
......!!!!!!!!...................................
19:02:55 `q004. Carefully done experiments show that for small slopes (up to a slope of about .1) the graph appears to be linear or very nearly so. This agrees with theoretical predictions of what should happen. Sketch a vertical line at x = .05. Then extend the straight line you sketched previously until it intersects the y axis and until it reaches past the vertical line at x = .05. What are the coordinates of the points where this line intersects the y-axis, and where it intersects the x =.05 line? What are the rise and the run between these points, and what therefore is the slope of your straight line?
......!!!!!!!!...................................
RESPONSE --> the point it intersects the y axis is the 50cm the slope of this straight line appears to be 33.3 confidence assessment: 1
.................................................
......!!!!!!!!...................................
19:06:57 `q005. The most accurate way to measure the acceleration of gravity is to use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) for the period of a pendulum. Use your washer pendulum and time 100 complete back-and-forth cycles of a pendulum of length 30 cm. Be sure to count carefully and don't let the pendulum swing out to a position more than 10 degrees from vertical. How long did it take, and how long did each cycle therefore last?
......!!!!!!!!...................................
RESPONSE --> it took about 84 seconds 100 cycles in 84s 1.19cycles per second each cycle took .84 seconds confidence assessment: 3
.................................................
......!!!!!!!!...................................
19:08:42 Solving T = 2 `pi / `sqrt(g) * `sqrt(L) for g, we can first multiply both sides by `sqrt(g) to obtain T * `sqrt(g) = 2 `pi `sqrt(L). Then dividing both sides by T we obtain `sqrt(g) = 2 `pi `sqrt(L) / T. Squaring both sides we finally obtain {}g = 4 `pi^2 L / T^2. Plugging in the values given here, L = 30 cm and T = 1.08 sec, we obtain g = 4 `pi^2 * 30 cm / (1.08 sec)^2 = 1030 cm/s^2. You should check these calculations for accuracy, since they were mentally approximated.
......!!!!!!!!...................................
RESPONSE --> i had to look ahead on this one. it was tough for me. self critique assessment: 3
.................................................
"