course phy201 some of these questions i did not understand 䂎z}a`KBxҶ~assignment #006
......!!!!!!!!...................................
19:02:06 How do flow diagrams help us see the structure of our reasoning processes?
......!!!!!!!!...................................
RESPONSE --> The flow diagrams help us to see how something moves from one point to another. Like in a reasoning process it would be from 1 thought to another point.
.................................................
......!!!!!!!!...................................
19:02:18 ** They help us to visualize how all the variables are related. Flow diagrams can also help us to obtain formulas relating the basic kinematic quantities in terms of which we have been analyzing uniformly accelerated motion. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:03:53 How do the two most fundamental equations of uniformly accelerated motion embody the definitions of average velocity and of acceleration?
......!!!!!!!!...................................
RESPONSE --> It shows change in velocity compared to the relationship of time to equal average velocity.
.................................................
......!!!!!!!!...................................
19:04:28 ** Velocity tells us the rate at which the position changes whereas the acceleration tells us the rate at which the velocity is changing. If acceleration is uniform ave velocity is the average of initial and final velocities. The change in position is found by taking the average velocity vAve = (vf+ v0) / 2 and multiplying by the'dt to get the first fundamental equation `ds = (v0 + vf)/2 * `dt. The acceleration is accel = rate of change of velocity = change in velocity / `dt = (vf - v0) / `dt. In symbols this equation is a = (vf + v0) / `dt. Algebraic rearrangement gives us this equation in the form vf = v0 + a `dt. This form also has an obvious interpretation: a `dt is the change in velocity, which when added to the initial velocity gives us the final velocity. **
......!!!!!!!!...................................
RESPONSE --> I understand I had them backwards.
.................................................
......!!!!!!!!...................................
19:10:23 How can we interpret the third fundamental equation of uniformly accelerated motion?
......!!!!!!!!...................................
RESPONSE --> Uniform accelerated motion can be interpreted as initial velocity plus acceleration multiplied by elapsed time.
.................................................
......!!!!!!!!...................................
19:14:53 ** The third equation says that `ds = v0 `dt + .5 a `dt^2. This means that the displacement `ds arises independently from initial velocity v0 and acceleration a: v0 `dt is the displacement of an object with uniform velocity moving at velocity v0, and 1/2 a `dt^2 the distance moved from rest by a uniformly accelerating object. The two contributions are added to get the total `ds. **
......!!!!!!!!...................................
RESPONSE --> ok can ou describe this a little more?
.................................................
......!!!!!!!!...................................
19:16:35 Why can we not directly reason out the basic 'impossible situation'?
......!!!!!!!!...................................
RESPONSE --> The basic 'impossible situation' is a problem constructed that really has no answer or can be reasoned out.
.................................................
......!!!!!!!!...................................
19:16:50 ** In this situation we know v0, a and `ds. From v0 and a we cannot draw any conclusions, and the same is true for v0 and `ds and also for a and `ds. No pair of variables allows us to draw any additional information. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:17:50 What strategy will we use to reconcile the basic 'impossible situation'? WE cann write down the 2 most fundamental equations and see what we do know.
......!!!!!!!!...................................
RESPONSE --> We can write down the 2 most fundamental equations and see what we know
.................................................
......!!!!!!!!...................................
19:19:50 ** We can use the fourth equation vf^2 = v0^2 + 2 a `ds to obtain vf, then knowing the values of v0, vf, a and `ds we easily find `dt either by direct reasoning or by using one of the fundamental equations. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:21:57 What is the difference between understanding uniformly accelerated motion and analyzing it with the use of equations?
......!!!!!!!!...................................
RESPONSE --> understanding uniformly accerlerated motion is a common belief or something that you know really happens. Analyzing it with use of equations places numerical value on the accelerated motion to prove that it exists.
.................................................
......!!!!!!!!...................................
19:22:08 ** You can use equations without understanding much of anything. To use the equations you don't even need to understand things like average velocity or change in velocity. You just have to be able to identify the right numbers and plug them in, which is an important task in itself but which doesn't involve understanding of the physical concepts behind the equations. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
19:25:49 How do we extrapolate our acceleration vs. ramp slope data to obtain an estimate of the acceleration of gravity?
......!!!!!!!!...................................
RESPONSE --> Not sure what ramp slope is.
.................................................
......!!!!!!!!...................................
19:27:19 How do the unavoidable timing errors due to the uncertainty in the computer timer affect our estimate of the acceleration of gravity?
......!!!!!!!!...................................
RESPONSE --> It affects our estimate because it depends on the amount of significant figures used to round off the time
.................................................
......!!!!!!!!...................................
19:28:06 ** STUDENT ANSWER: This error causes the slope to increase at an increasing rate rather than form a linear line. INSTRUCTOR COMMENT: Good answer. A systematic error would do that. Even random, non-systematic errors affect the placement of points on the graph, and this tends to affect the slope of the straight line approximating the graph, and also to reduce the accuracy of this slope. **
......!!!!!!!!...................................
RESPONSE --> ok could you explain the acceleration and ramp slope.
.................................................
......!!!!!!!!...................................
19:29:57 How could the slight slope of the table on which the ramp rests, if not accounted for, affect our graph of acceleration vs. ramp slope but not our estimate of the acceleration of gravity?
......!!!!!!!!...................................
RESPONSE --> I you don' t allow for the slope of the table then your measurements would not be accurate. You would be figuring the measurements from the ramp being equal to zero. when there is more of a slope
.................................................
......!!!!!!!!...................................
19:31:22 How could anticipation of the instant at which a cart reaches the end of the ramp, but not of the instant at which it is released, affect our graph as well as our estimate of the acceleration of gravity?
......!!!!!!!!...................................
RESPONSE --> When you anticipate hitting the timer when the cart reaches the end of the ramp you are more likely to hit the timer before the cart reaches the end of the ramp.
.................................................
......!!!!!!!!...................................
19:31:33 ** GOOD STUDENT ANSWER: The timer is started with a slight delay due to the reaction time of the person doing the timing. This would be OK if the individual's reaction time caused the individual to stop the timer with the same delay. However, the person doing the timing often anticipates the instant when the cart reaches the end of the ramp, so that the delay is not added onto the end time as it was to the starting time. The anticipating individual often triggers the timer slightly before the cart reaches the end, compounding the error even further and also causing the graph to curve rather than remain linear. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
]pJr}\ԃS assignment #005 o[Βɕއ Physics I 06-22-2006
......!!!!!!!!...................................
19:54:37 Intro Prob 6 Intro Prob 6 How do you find final velocity and displacement given initial velocity, acceleration and time interval?
......!!!!!!!!...................................
RESPONSE --> First take a('dt) = 'dv. this gives you the change in velocity. Next take v0+'dv = vf. To find the displacement use the equation 'ds = v0'dt+.5a'dt.
.................................................
......!!!!!!!!...................................
19:55:31 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
......!!!!!!!!...................................
RESPONSE --> ok, that is what was used
.................................................
......!!!!!!!!...................................
19:55:32 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:55:32 ** To find final velocity from the given quantities initial velocity, acceleration and `dt: Multiply `dt by accel to get `dv. Then add change in velocity `dv to init vel , and you have the final velocity**
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
19:58:49 Describe the flow diagram we obtain for the situation in which we know v0, vf and `dt.
......!!!!!!!!...................................
RESPONSE --> vf-v0/'dt we take vf and subtract v0 and then divide by 'dt toget the vAve
.................................................
......!!!!!!!!...................................
19:59:46 ** The flow diagram shows us the flow of information, what we get from what, usually by combining two quantites at a time. How we get each quantity may also be included. From vf and v0 we get `dv, shown by lines from vf and v0 at the top level to `dv. From vf and v0 we also get and vAve, shown by similar lines running from v0 and vf to vAve. Then from vAve and `dt we get `ds, with the accompanying lines indicating from vAve and `dt to `ds, while from `dv and `dt we get acceleration, indicated similarly. **
......!!!!!!!!...................................
RESPONSE --> ok, worded in different way I showed the flow diagrame in the form of an equation.
.................................................
......!!!!!!!!...................................
20:09:17 Principles of Physics and General College Physics Students: Prob. 1.26: Estimate how long it would take a runner at 10 km / hr to run from New York to California. Explain your solution thoroughly.
......!!!!!!!!...................................
RESPONSE --> do not know how far it is from New York to California. But you would take the measurement converting miles to km and divide by ten. I esitmate that it would take 800 hrs. 10km/hr is approximately 6 mph.
.................................................
......!!!!!!!!...................................
20:10:49 It is about 3000 miles from coast to coast. A km is about .62 mile, so 3000 miles * 1 km / (.62 miles) = 5000 km, approximately. At 10 km / hr, the time required would be 5000 km / (10 km / hr) = 500 km / (km/hr) = 500 km * (hr / km) = 500 (km / km) * hr = 500 hr.
......!!!!!!!!...................................
RESPONSE --> I quessed that it was 5000 miles from coast to coast. I thought that is was 3000 miles but was not for sure.
.................................................
......!!!!!!!!...................................
20:13:50 All Students: Estimate the number heartbeats in a lifetime. What assumptions did you make to estimate the number of heartbeats in a human lifetime, and how did you obtain your final result?
......!!!!!!!!...................................
RESPONSE --> There would be an estimate of 3 billion heartbeats in a human lifetime. This is fiquring that the heart beats 66 time per minute and the average lifespan being 85 years.
.................................................
......!!!!!!!!...................................
20:14:47 ** Typical assumptions: At 70 heartbeats per minute, with a lifetime of 80 years, we have 70 beats / minute * 60 minutes/hour * 24 hours / day * 365 days / year * 80 years = 3 billion, approximately. **
......!!!!!!!!...................................
RESPONSE --> using my assumptions we both had the same esitmate of 3 billion.
.................................................
......!!!!!!!!...................................
20:15:30 University Physics Students Only: Problem 1.52 (i.e., Chapter 1, Problem 52): Angle between -2i+6j and 2i - 3j. What angle did you obtain between the two vectors?
......!!!!!!!!...................................
RESPONSE --> not required for me to do I'm in General Phy.
.................................................
......!!!!!!!!...................................
20:15:39 ** For the given vectors we have dot product =-2 * 2 + 6 * (-3) = -22 magnitude of first vector = sqrt( (-2)^2 + 6^2) = sqrt(40) magnitude of second vector = sqrt( 2^2 + (-3)^2 ) = sqrt(13) Since dot product = magnitude of 1 st vector * magnitude of 2d vector * cos(theta) we have cos(theta) = dot product / (magnitude of 1 st vector * magnitude of 2d vector) so that theta = arccos [ dot product / (magnitude of 1 st vector * magnitude of 2d vector) ] = arccos[ -22 / ( sqrt(40) * sqrt(13) ) ] = arccos ( -.965) = 164 degrees, approx.. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
J|Ϫ̙m淆dz assignment #006 o[Βɕއ Physics I 06-22-2006
......!!!!!!!!...................................
20:36:58 General Physics 1.42. At 1200 liters/day per family how much would level of 50 km^2 lake fall in a year if supplying town of population 40000
......!!!!!!!!...................................
RESPONSE --> I'm a little confused on this question.
.................................................
......!!!!!!!!...................................
20:41:48 ** A liter is 1/1000 of a cubic meter. It can be thought of as a cube 10 cm on a side. To fill a 1-meter cube t would take 10 rows with 10 cubes in each row to make a single layer 10 cm high, and 10 layers to fill the cube--i.e., 10 * 10 * 10 = 1000 one-liter cubes fill a 1-meter cube. A km is 1000 meters so a km^2 is (1000 m)^2 = 10^6 m^2. 1200 liters / day per family, with 40,000 people at 4 persons / family, implies 10,000 families each using 1200 * .001 m^3 = 1.2 m^3 per day. Total usage would be 10,000 families * 1.2 m^3 / day / family * 365 days / year = 4.3 * 10^6 m^3 / year. The volume of the lake corresponding to a depth change `dy is `dy * A, where A is the area of the lake. The area of the lake is 50 km^2 * 10^6 m^2 / km^2 = 5 * 10^7 m^2. `dy * A = Volume so `dy = Volume / A = 4.3 * 10^6 m^3 / (5 * 10^7 m&2) = .086 m or 8.6 cm. This estimate is based on 4 people per family. A different assumption would change this estimate. STUDENT QUESTION: 40000 people in town divided by average household of 4 = 10,000 families 10,000 families * 1200 liters /day = 12000000 liters used per day * 365 days in a year = 4380000000 liter used. Here is where I get confused changing from liters to level of 50 km^2 to subtract. INSTRUCTOR RESPONSE: If you multiply the area of the lake by the change in depth you get the volume of water used. You know the area of the lake and the volume of the water used, from which you can find the change in depth. Of course you need to do the appropriate conversions of units. Remember that a liter is the volume of a cube 10 cm on a side, so it would take 10 rows of 10 such cubes to make one layer, then 10 layers, to fill a cube 1 meter = 100 cm on a side. You should also see that a km^2 could be a square 1 km on a side, which would be 1000 meters on a side, to cover which would require 1000 rows of 1000 1-meters squares. If you end of having trouble with the units or anything else please ask some specific questions and I will try to help you clarify the situation. ANOTHER INSTRUCTOR COMMENT: The water used can be thought of as having been spread out in a thin layer on top of that lake. That thin layer forms a cylinder whose cross-section is the surface of the lake and whose altitude is the change in the water level. The volume of that cylinder is equal to the volume of the water used by the family in a year. See if you can solve the problem from this model. COMMON ERROR: Area is 50 km^2 * 1000 m^2 / km^2 INSTRUCTOR COMMENT: Two things to remember: You can't cover a 1000 m x 1000 m square with 1000 1-meter squares, which would only be enough to make 1 row of 1000 squares, not 1000 rows of 1000 squares. 1 km^2 = 1000 m * 1000 m = 1,000,000 m^2. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:42:26 univ 1.70 univ sailor 2 km east, 3.5 km SE, then unknown, ends up 5.8 km east find magnitude and direction of 3d leg, explain how diagram shows qualitative agreement
......!!!!!!!!...................................
RESPONSE --> not required
.................................................
......!!!!!!!!...................................
20:42:37 **** query univ 1.82 (1.66 10th edition) vectors 3.6 at 70 deg, 2.4 at 210 deg find scalar and vector product
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:43:07 ** GOOD STUDENT SOLUTION WITH INSTRUCTOR COMMENT: A * B = |A||B| cos(theta) = AxBx + Ay By + AzBz 3.6 * 2.4 * cos (140 deg) = -6.62 To check for consistency we can calculate the components of A and B: Ax = 3.6 * cos(70 deg) = 1.23 Ay = 3.6 * sin(70 deg) = 3.38 Bx = 2.4 * cos (210 deg) = -2.08 By = 2.4 * sin(210 deg) = -1.2 dot product = 1.23 *-2.08 + 3.38 * -1.2 = -2.5584 + -4.056 = -6.61. Close enough. ---------------------- Cross product: | A X B | = |A| |B| sin (theta) = 3.6 * 2.4 * sin(140) = 5.554. Finding the components we have (Ay * Bz - Az * By) i + (Az * Bx - Ax + Bz) j + (Ax * By - Ay * Bx) k = ((3.38*0)-(0*-1.2)] i + [(0 * -2.08) - (1.23 * 0)] j + [(1.23*-1.2)-(3.38 * -2.08)] k = 0 i + 0 j + 5.55 k, or just 5.55 k, along the positive z axis ('upward' from the plane). INSTRUCTOR COMMENT: Your conclusion is correct. The cross product must be at a right angle to the two vectors. To tell whether the vector is 'up' or 'down' you can use your result, 5.55 k, to see that it's a positive multiple of k and therefore upward. The other way is to use the right-hand rule. You place the fingers of your right hand along the first vector and position your hand so that it would 'turn' the first vector in the direction of the second. That will have your thumb pointing upward, in agreement with your calculated result, which showed the cross product as being in the upward direction. **
......!!!!!!!!...................................
RESPONSE --> Jˀzlܘā Student Name: assignment #007
.................................................
......!!!!!!!!...................................
20:46:42 `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 --> i am having troublw with the acceleration vs ramp slope graph. could you please explain how this graph works.
.................................................
......!!!!!!!!...................................
20:49:29 We can find the accelerations either using equations or direct reasoning. To directly reason the acceleration for the five-second case, we note that the average velocity in this case must be 50 cm/(5 seconds) = 10 cm/s. Since the initial velocity was 0, assuming uniform acceleration we see that the final velocity must be 20 cm/second, since 0 cm/s and 20 cm/s average out to 10 cm/s. This implies a velocity change of 20 cm/second a time interval of 5 seconds, or a uniform acceleration of 20 cm/s / (5 s) = 4 cm/s^2. The acceleration in the 3-second case could also be directly reasoned, but instead we will note that in this case we have the initial velocity v0 = 0, the time interval `dt = 3 sec, and the displacement `ds = 50 cm. We can therefore find the acceleration from the equation `ds = v0 `dt + .5 a `dt^2. Noting first that since v0 = 0 the term v0 `dt must also be 0,we see that in this case the equation reduces to `ds = .5 a `dt^2. We easily solve for the acceleration, obtaining a = 2 `ds / `dt^2. In this case we have a = 2 * (50 cm) / (3 sec)^2 = 11 cm/s^2 (rounded to nearest cm/s^2). For the 2-second case we can use the same formula, obtaining a = 2 * (50 cm) / (2 sec)^2 = 25 cm/s^2.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:50:24 `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 --> ok
.................................................
......!!!!!!!!...................................
20:50:24 `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 -->
.................................................
......!!!!!!!!...................................
20:50:37 The graph will have acceleration in cm/s^2 on the vertical axis (the traditional y-axis) and ramp slope on the horizontal axis (the traditional x-axis). The graph points will be (.01, 4 cm/s^2), (.02, 11.1 cm/s^2), (.03, 25 cm/s^2). The second point lies somewhat lower than a line connecting the first and third points, so the best possible line will probably be lower than the first and third points but higher than the second. The graph indicates that acceleration increases with increasing slope, which should be no surprise. It is not clear from the graph whether a straight line is in fact the most appropriate model for the data. If timing wasn't particularly accurate, these lines could easily be interpreted as being scattered from the actual linear behavior due to experimental errors. Or the graph could indicate acceleration vs. ramp slope behavior that is increasing at an increasing rate.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
20:53:48 `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 --> ok
.................................................
......!!!!!!!!...................................
20:54:02 A pretty good straight line goes through the points (0, -6 cm/s^2) and (.05, 42 cm/s^2). Your y coordinates might differ by a few cm/s^2 either way. For the coordinates given here, the rise is from -6 cm/s^2 to 42 cm/s^2, a rise of 48 cm/s^2. The run is from 0 to .05, a run of .05. The slope of the straight line is approximately 48 cm/s^2 / .05 = 960 cm/s^2. Note that this is pretty close to the accepted value, 980 cm/second^2, of gravity. Carefully done, this experiment will give us a very good estimate of the acceleration of gravity.
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
21:06:47 `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 the pendulum 59.0781 second to complete 100 back and forth cycles . Each cycle took .590781 seconds.
.................................................
......!!!!!!!!...................................
21:08:30 100 cycles of a pendulum of this length should require approximately 108 seconds. This would be 108 seconds per 100 cycles, or 108 sec / (100 cycles) = 1.08 sec / cycle. If you didn't count very carefully or didn't time very accurately, you might differ significantly from this result; differences of up to a couple of cycles are to be expected.
......!!!!!!!!...................................
RESPONSE --> I may have counted 1 complete cycle has 2 cycles this would explain why I was only at 59 seconds instead of 108.
.................................................
......!!!!!!!!...................................
21:17:39 `q006. You now have values for the period T and the length L, so you can use the relationship T = 2 `pi / `sqrt(g) * `sqrt(L) to find the acceleration g of gravity. Solve the equation for g and then use your values for T and L to determine the acceleration of gravity.
......!!!!!!!!...................................
RESPONSE --> .1939 m/s^2 is the acceleration of gravity
.................................................
......!!!!!!!!...................................
21:24:46 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 concluded that g=1015.391399cm/s^2
.................................................
ok understand
.................................................