course Phy 231 I answered the questions for this assignment over a period of two different days; therefore, I copied both send files in the order they were created as pasted them below. lkپ{w⩁assignment #002
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15:50:07 `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 --> 12 m / 4 s = 3 m/ s I found my anwser as shown abover by dividing the distance traveled by the time. confidence assessment: 3
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15:53:57 `q002. How is this problem related to the concept of a rate?
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RESPONSE --> rate is one quantity compared to another as a ratio, and when I found the average speed I found the rate of travel expressed in m/s. confidence assessment: 3
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`q003. Is object position dependent on time or is time dependent on object position?
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RESPONSE --> Object position is dependent upon time, because the distance traveled varies depending upon the time interval. Also, time is always labeled on the x-axis of a graph because it is the independant variable that does not change due to a causual relationship with another variable. confidence assessment: 3
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11:08:37 `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 did not miss any concepts explained for the previous questions confidence assessment: 3
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13:50:12 `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 --> 6 m / 3 s = 2 m/s average velocity I obtained this unit by understanding that if an object moves 6 meters and during this time a stopwatch records that this took 3 seconds, then during each 2 meter interval on average one second will pass on the stopwatch. confidence assessment: 3
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13:50:29 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 --> self critique assessment: 3
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13:51:32 `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 confidence assessment: 3
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13:51:39 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 --> self critique assessment: 3
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13:53:27 `q007. How do you write the expressions `ds and `dt on your paper?
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RESPONSE --> I would write 'ds as ""change in 's', or delta symbol s"", and the same goes for 'dt. confidence assessment: 3
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13:53:49 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 --> self critique assessment: 3
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17:01:19 `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 --> 5 m/s * 10s = 50 m This problem discusses the concept of rate by giving the rate of movement for an object and for how long it moved at this rate, which allowed me to find how far the object moved during that time interval. confidence assessment: 3
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17:04:24 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 --> We obtain the resulting change in the first by canceling out the other quantities variable and multiplying the quantities given. self critique assessment:
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18:17:35 `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 ;therefore, 'ds = vAve * 'dt confidence assessment: 3
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18:17:50 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 --> self critique assessment: 3
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18:25:42 `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 --> rate is the change of one quantity in relation to the change in another quantity as expressed in a ratio. This is vAve is related to 'dt and 'ds as a ratio of the two forming a rate of objects movement in relationship to time. confidence assessment: 3
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18:25:50 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 --> self critique assessment: 3
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18:27:30 `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 = 'ds / 'dt can be solved for 'ds by muliplying both sides of the expression by 'dt. This leaves us with the result of 'ds = 'dt * vAve. confidence assessment: 3
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18:27:46 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 --> self critique assessment: 3
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18:30:36 `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 --> It makes sense to solve for the distance an object moves if you know how fast it is going and for how long it is traveling; therefore, when we solve for the change in time multiplied by the average velocity and find the change in distance that fits with our expectations. confidence assessment: 3
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18:30:52 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 --> self critique assessment: 3
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18:32:38 `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 --> To make vAve = 'ds / 'dt solve for 'dt we first multiply by 'dt leaving us with 'dt * vAve = 'ds. We then isolate 'dt by dividing by vAve which gives us the final expression as solved for 'dt as 'dt = 'ds / vAve. confidence assessment: 3
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18:33:00 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 --> self critique assessment: 3
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18:37:13 `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 --> it makes sense in terms of units, because we divided by s^-1, which is the same as muliply by the and divinding by the distance unit, which gives us seconds as the remaining for our answer which we expect to be a unit of time. For example if we travel 30 miles at a speed of 30 mph, then it makes sense that it would take us 1 hour to make that distance. confidence assessment: 3
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