course Phy 210
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12:18:55 `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 --> You would divide 12/4 to get 3m/sec.
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12:19:01 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 -->
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12:19:47 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> It is related because you have to find out at what rate the object is moving by dividing.
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12:19:55 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
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12:20:24 `q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> Time is dependent on object position
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12:20:45 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 --> OK
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12:22:36 `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 had missed that the clock runs whether the object is moving or not.
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12:22:44 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
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12:23:45 `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 --> Its speed is 2 meters/sec. You would divide 6/3 to get 2.
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12:23:58 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 -->
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12:24:58 `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 --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. 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. vAVE
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12:25:32 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 --> Oh ok, i wasn't quite sure what the question was asking. So it is vAve = ds/dt
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12:26:05 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> you would draw the delta sign with the s and t beside it
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12:26:10 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 -->
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12:27:37 `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 --> It moves 50 meters. You would multiply 5x10 to get 50 which is the rate at which the object has moved.
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12:28:15 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 --> You would subtract.
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12:29:31 `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 --> this is the rate at which the object is being moved
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12:29:43 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 -->
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12:29:43 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 -->
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12:30:41 `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 --> They are all related to rate because you either divide or multiply or subtract two times first and then divide to get the rates.
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12:31:02 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 --> Oh ok, again i wasn't sure what the question was asking
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12:32:00 `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 --> You could cross mulitply and get the vAve divided by dt
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12:32:13 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
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12:33:03 `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 --> You can always switch the steps around to solve for any one of these but its still using rates.
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12:33:12 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 -->
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12:34:56 `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 --> You can multiply both sides by 'dt and get dt(vAve)=ds
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12:35:19 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 --> Ok, i left out the last step where you divide vAve
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12:36:00 `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 --> Again you can use all of these as rates by switching them around and multiplying and dividing to solve for one of them
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12:36:16 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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zaׂw͂L Student Name: assignment #002 002. Volumes
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12:55:06 `q001. There are 9 questions and 4 summary questions in this assignment. What is the volume of a rectangular solid whose dimensions are exactly 3 cm by 5 cm by 7 cm?
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RESPONSE --> 105 cm
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12:56:00 If we orient this object so that its 3 cm dimension is its 'height', then it will be 'resting' on a rectangular base whose dimension are 5 cm by 7 cm. This base can be divided into 5 rows each consisting of 7 squares, each 1 meter by 1 meter. There will therefore be 5 * 7 = 35 such squares, showing us that the area of the base is 35 m^2. Above each of these base squares the object rises to a distance of 3 meters, forming a small rectangular tower. Each such tower can be divided into 3 cubical blocks, each having dimension 1 meter by 1 meter by 1 meter. The volume of each 1-meter cube is 1 m * 1 m * 1 m = 1 m^3, also expressed as 1 cubic meter. So each small 'tower' has volume 3 m^3. The object can be divided into 35 such 'towers'. So the total volume is 35 * 3 m^3 = 105 m^3. This construction shows us why the volume of a rectangular solid is equal to the area of the base (in this example the 35 m^2 of the base) and the altitude (in this case 3 meters). The volume of any rectangular solid is therefore V = A * h, where A is the area of the base and h the altitude. This is sometimes expressed as V = L * W * h, where L and W are the length and width of the base. However the relationship V = A * h applies to a much broader class of objects than just rectangular solids, and V = A * h is a more powerful idea than V = L * W * h. Remember both, but remember also that V = A * h is the more important.
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RESPONSE --> ok
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