course phy 201 ںFL\|xNnassignment #000
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14:55:43 The Query program normally asks you questions about assigned problems and class notes, in question-answer-self-critique format. Since Assignments 0 and 1 consist mostly of lab-related activities, most of the questions on these queries will be related to your labs and will be in open-ended in form, without given solutions, and will not require self-critique. The purpose of this Query is to gauge your understanding of some basic ideas about motion and timing, and some procedures to be used throughout the course in analyzing our observations. Answer these questions to the best of your ability. If you encounter difficulties, the instructor's response to this first Query will be designed to help you clarify anything you don't understand. {}{}Respond by stating the purpose of this first Query, as you currently understand it.
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RESPONSE --> ok
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14:56:36 If, as in the object-down-an-incline experiment, you know the distance an object rolls down an incline and the time required, explain how you will use this information to find the object 's average speed on the incline.
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RESPONSE --> distance divided by time = speed confidence assessment: 3
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15:01:57 If an object travels 40 centimeters down an incline in 5 seconds then what is its average velocity on the incline? Explain how your answer is connected to your experience.
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RESPONSE --> 40/5=8centimeters per second this is connected to the experiment because we had to find the speed at which the oblect rolled down our book. confidence assessment: 3
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15:06:06 If the same object requires 3 second to reach the halfway point, what is its average velocity on the first half of the incline and what is its average velocity on the second half?
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RESPONSE --> first half 20/3=6.67 cm per sec. average speed second half 20/2=10 cm per sec. average speed confidence assessment: 3
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15:07:40 Using the same type of setup you used for the first object-down-an-incline lab, if the computer timer indicates that on five trials the times of an object down an incline are 2.42 sec, 2.56 sec, 2.38 sec, 2.47 sec and 2.31 sec, then to what extent do you think the discrepancies could be explained by each of thefollowing: {}{}a. The lack of precision of the TIMER program{}{}b. The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse){}{}c. Actual differences in the time required for the object to travel the same distance.{}{}d. Differences in positioningthe object prior to release.{}{}e. Human uncertainty in observing exactly when the object reached the end of the incline.
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RESPONSE --> B and E confidence assessment: 3
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15:16:14 How much uncertainty do you think each of the following would actually contribute to the uncertainty in timing a number of trials for the object-down-an-incline lab? {}{}a. The lack of precision of the TIMER program{}{}b. The uncertain precision of human triggering (uncertainty associated bLine$(lineCount) =with an actual human finger on a computer mouse){}{}c. Actual differences in the time required for the object to travel the same distance.{}{}d. Differences in positioning the object prior to release.{}{}e. Human uncertainty in observing exactly when the object reached the end of the incline.
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RESPONSE --> a. not much uncertainty b.human interpretation is the biggest factor in uncertainty c.there will be no difference in the ""actual"" time it takes. d.unless you place a backboard up against the top of the incline it is hard to place it in the exact same spot every time. e. this has a lot to do with the time differences confidence assessment: 3
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15:21:02 What, if anything, could you do about the uncertainty due to each of the following? Address each specifically. {}{}a. The lack of precision of the TIMER program{}{}b. The uncertain precision of human triggering (uncertainty associated with an actual human finger on a computer mouse){}{}c. Actualdifferences in the time required for the object to travel the same distance.{}{}d. Differences in positioning the object prior to release.{}{}e. Human uncertainty in observing exactly when the object reached the end of the incline.
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RESPONSE --> a not really anything b.keep doing it over and over and try to get better c.nothing, it doesn't change d. put a backborad at the top of the incline or make a mark on the incline e. there is really nothing we can do about human judgement but practicing and repetition would help. confidence assessment: 3
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wfͺI^٪`}T┒ assignment #002 002. Velocity Physics I 01-27-2007
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14:05:37 `q001. Note that there are 14 questions in this assignment. If an object moves 12 meters in 4 seconds, then at what average rate is the object moving? Explain how you obtained your result in terms of commonsense images.
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RESPONSE --> 3 meters per sec. 12m/4sec=3m/sec confidence assessment: 3
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14:05:58 Moving 12 meters in 4 seconds, we move an average of 3 meters every second. We can imagine dividing up the 12 meters into four equal parts, one for each second. Each part will have 3 meters, corresponding to the distance moved in 1 second, on the average.
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RESPONSE --> ok self critique assessment: 2
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14:06:28 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> it asks us to find the rate of speed confidence assessment: 2
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14:06:40 A rate is obtained by dividing the change in a quantity by the change in another quantity on which is dependent. In this case we divided the change in position by the time during which that change occurred.
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RESPONSE --> ok self critique assessment: 2
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14:07:15 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> time dependent on object position confidence assessment: 2
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14:07:36 Object position is dependent on time--the clock runs whether the object is moving or not so time is independent of position. Clock time is pretty much independent of anything else.
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RESPONSE --> oh yeah i get it self critique assessment: 2
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14:08:24 `q004. So the rate here is the average rate at which position is changing with respect to clock time. Explain what concepts, if any, you missed in your explanations.
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RESPONSE --> I didnt misunderstand any concepts confidence assessment: 3
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14:08:37 You should always self-critique your work in this manner. Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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RESPONSE --> ok self critique assessment: 3
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14:11:26 `q005. If an object is displaced -6 meters in three seconds, then what is the average speed of the object what is its average velocity? Explain how you obtained your result in terms of commonsense images and ideas.
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RESPONSE --> 6m/3sec=2m/sec split up the six meters into three 2m sections, one for each second. confidence assessment: 3
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14:12:10 Speed is the average rate at which distance changes, and distance cannot be negative. Therefore speed cannot be negative. Velocity is the average rate at which position changes, and position changes can be positive or negative.
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RESPONSE --> ok self critique assessment: 2
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14:13:12 `q006. If `ds stands for the change in the position of an object and `dt for the time interval during which the position changes, then what expression stands for the average velocity vAve of the object during this time interval?
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RESPONSE --> 'ds/'dt=vAve confidence assessment: 3
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14:13:21 Average velocity is rate of change of position. Change in position is `ds and change in clock tim is `dt, so vAve = `ds / `dt.
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RESPONSE --> ok self critique assessment: 2
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14:14:05 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> 'ds and 'dt, i don't know confidence assessment: 1
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14:15:09 You use the Greek capital Delta symbol Delta. `d is often used here because the symbol for Delta is not interpreted correctly by some Internet browsers. You should get in the habit of thinking and writing Delta when you see `d. You may use either `d or Delta when submitting work and answering questions.
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RESPONSE --> ok i remember, I forgot about the whole delta deal self critique assessment: 2
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14:16:12 `q008. If an object changes position at an average rate of 5 meters/second for 10 seconds, then how far does it move? How is this problem related to the concept of a rate?
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RESPONSE --> 50 meters 10*5=50 we are given a rate and a time and asked how far it is moved. confidence assessment: 3
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14:17:31 In this problem you are given the rate at which position changes with respect to time, and you are given the time interval during which to calculate the change in position. Given the rate at which one quantity changes with respect to another, and the change in the second quantity, how do we obtain the resulting change in the first?
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RESPONSE --> ok self critique assessment: 2
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14:18:50 `q009. If vAve stands for the rate at which the position of the object changes (also called velocity) and `dt for the time interval during which the change in position is to be calculated, then how to we write the expression for the change `ds in the position?
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RESPONSE --> vAve*'ds='dt confidence assessment: 3
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14:19:06 To find the change in a quantity we multiply the rate by the time interval during which the change occurs. We therefore obtain the change in position by multiplying the velocity by the time interval: `ds = vAve * `dt. The units of this calculation pretty much tell us what to do: Just as when we multiply pay rate by time (dollar / hr * hours of work) or automobile velocity by the time interval (miles / hour * hour), when we multiply vAve, in cm / sec or meters / sec or whatever, by `dt in seconds, we get displacement in cm or meters, or whatever, depending on the units of distance used.
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RESPONSE --> ok self critique assessment: 2
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14:22:47 `q010. Explain how the quantities average velocity vAve, time interval `dt and displacement `ds are related by the definition of a rate, and how this relationship can be used to solve the current problem problem.
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RESPONSE --> vAve ='ds/'dt is a way to find the average rate of something moving. confidence assessment: 3
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14:23:00 vAve is the average rate at which position changes. The change in position is the displacement `ds, the change in clock time is `dt, so vAve = `ds / `dt.
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RESPONSE --> ok self critique assessment: 2
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14:24:00 `q011. The basic rate relationship vAve = `ds / `dt expresses the definition of average velocity vAve as the rate at which position s changes with respect to clock time t. What algebraic steps do we use to solve this equation for `ds, and what is our result?
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RESPONSE --> vAve=s/t average velocity is our result confidence assessment: 3
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14:24:10 To solve vAve = `ds / `dt for `ds, we multiply both sides by `dt. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds . Switching sides we have `ds = vAve * `dt.
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RESPONSE --> ok self critique assessment: 2
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14:25:02 `q012. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> the questions deal with all that stuff confidence assessment: 0
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14:25:17 Our most direct intuition about velocity probably comes from watching an automobile speedometer. We know that if we multiply our average velocity in mph by the duration `dt of the time interval during which we travel, we get the distance traveled in miles. From this we easily extend the idea. Whenever we multiply our average velocity by the duration of the time interval, we expect to obtain the displacement, or change in position, during that time interval.
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RESPONSE --> ok self critique assessment: 2
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14:26:03 `q013. What algebraic steps do we use to solve the equation vAve = `ds / `dt for `dt, and what is our result?
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RESPONSE --> multiply vAve and 'dt result is 'ds confidence assessment: 3
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14:26:53 To solve vAve = `ds / `dt for `dt, we must get `dt out of the denominator. Thus we first multiply both sides by the denominator `dt. Then we can see where we are and takes the appropriate next that. The steps: vAve = `ds / `dt. Multiply both sides by `dt: vAve * `dt = `ds / `dt * `dt Since `dt / `dt = 1 vAve * `dt = `ds. We can now divide both sides by vAve to get `dt = `ds / vAve.
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RESPONSE --> oops I solved for d's instead of 'dt self critique assessment: 2
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14:27:13 `q014. How is this result related to our intuition about the meanings of the terms average velocity, displacement and clock time?
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RESPONSE --> the same as before confidence assessment: 0
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14:27:35 If we want to know how long it will take to make a trip at a certain speed, we know to divide the distance in miles by the speed in mph. If we divide the number of miles we need to travel by the number of miles we travel in hour, we get the number of hours required. We extend this to the general concept of dividing the displacement by the velocity to get the duration of the time interval.
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RESPONSE --> ok self critique assessment: 2
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