course Phy 121
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17:37:37 `q001. Note that there are 11 questions in this assignment. vAve = `ds / `dt, which is the definition of average velocity and which fits well with our intuition about this concept. If displacement `ds is measured in meters and the time interval `dt is measured in seconds, in what units will vAve be obtained?
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RESPONSE --> vAve will be obtained in cm/s confidence assessment: 1
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17:37:54 vAve = `ds / `dt. The units of `ds are cm and the units of `dt are sec, so the units of `ds / `dt must be cm / sec. Thus vAve is in cm/s.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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17:38:16 `q002. If the definition equation vAve = `ds / `dt is to be solved for `ds we multiply both sides of the equation by `dt to obtain `ds = vAve * `dt. If vAve is measured in cm / sec and `dt in sec, then in what units must `ds be measured?
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RESPONSE --> 'ds is measured in cm because the seconds cancel out. confidence assessment: 1
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17:38:28 Since vAve is in cm/sec and `dt in sec, `ds = vAve * `dt must be in units of cm / sec * sec = cm.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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17:39:22 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> cm/sec * sec/1 = cm/1 = cm The seconds cancel out. confidence assessment: 2
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17:39:44 When we multiply cm/sec by sec we are multiplying the fractions cm / sec and sec / 1. When we multiply fractions we will multiply numerators and denominators. We obtain cm * sec / ( sec * 1). This can be rearranged as (sec / sec) * (cm / 1), which is the same as 1 * cm / 1. Since multiplication or division by 1 doesn't change a quantity, this is just equal to cm.
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RESPONSE --> My answer is mostly correct. self critique assessment: 3
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17:42:46 `q004. If the definition vAve = `ds / `dt is to be solved for `dt we multiply both sides of the equation by `dt to obtain vAve * `dt = `ds, then divide both sides by vAve to get `dt = `ds / vAve. If vAve is measured in km / sec and `ds in km, then in what units must `dt be measured?
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RESPONSE --> 'dt will be measured in seconds confidence assessment: 1
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17:43:07 Since `dt = `ds / vAve and `ds is in km and vAve in km/sec, `ds / vAve will be in km / (km / sec) = seconds.
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RESPONSE --> I think I got the correct answer. self critique assessment: 3
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17:45:11 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> If you look at the fraction that would result from the previous problem (km)/(km/sec). It is a rule of algebra that the like terms, in this case km, would cancel out. confidence assessment: 1
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17:45:59 The division is km / (km / sec). Since division by a fraction is multiplication by the reciprocal of the fraction, we have km * (sec / km). This is equivalent to multiplication of fractions (km / 1) * (sec / km). Multiplying numerators and denominators we get (km * sec) / (1 * km), which can be rearranged to give us (km / km) * (sec / 1), or 1 * sec / 1, or just sec.
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RESPONSE --> I knew there was a rule involved, but I couldn't quite remember enough about it to explain it properly. self critique assessment: 2
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17:50:12 `q006. If an object moves from position s = 4 meters to position s = 10 meters between clock times t = 2 seconds and t = 5 seconds, then at what average rate is the position of the object changing (i.e., what is the average velocity of the object) during this time interval? What is the change `ds in position, what is the change `dt in clock time, and how do we combine these quantities to obtain the average velocity?
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RESPONSE --> vAve = 'ds / 'dt = (10m-4m)/(5sec-2sec) = 6m/3sec = 2 m/s More specifically vAve = (d2-d1)/(t2-t1) Plug in the numbers as shown above. confidence assessment: 2
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17:50:33 We see that the changes in position and clock time our `ds = 10 meters - 4 meters = 6 meters and `dt = 5 seconds - 2 seconds = 3 seconds. We see also that the average velocity is vAve = `ds / `dt = 6 meters / (3 seconds) = 2 meters / second. Comment on any discrepancy between this reasoning and your reasoning.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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17:53:19 `q007. Symbolize this process: If an object moves from position s = s1 to position s = s2 between clock times t = t1 and t = t2, when what expression represents the change `ds in position and what expression represents the change `dt in the clock time?
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RESPONSE --> Change in position: vAve = 'ds/'dt = (s2-s1)/(t2-t1) Individually, change in position is s2-s1, and change in time is t2-t1. confidence assessment: 2
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17:53:32 We see that the change in position is `ds = s2 - s1, obtained as usual by subtracting the first position from the second. Similarly the change in clock time is `dt = t2 - t1. What expression therefore symbolizes the average velocity between the two clock times.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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18:01:05 `q008. On a graph of position s vs. clock time t we see that the first position s = 4 meters occurs at clock time t = 2 seconds, which corresponds to the point (2 sec, 4 meters) on the graph, while the second position s = 10 meters occurs at clock time t = 5 seconds and therefore corresponds to the point (5 sec, 10 meters). If a right triangle is drawn between these points on the graph, with the sides of the triangle parallel to the s and t axes, the rise of the triangle is the quantity represented by its vertical side and the run is the quantity represented by its horizontal side. This slope of the triangle is defined as the ratio rise / run. What is the rise of the triangle (i.e., the length of the vertical side) and what quantity does the rise represent? What is the run of the triangle and what does it represent?
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RESPONSE --> If you draw out the graph and draw out the triangle I determined the rise (or the change in the position) to be 6 and the run (or the change in time) to be 3. This results in a slope of 2. confidence assessment: 1
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18:01:35 The rise of the triangle represents the change in the position coordinate, which from the first point to the second is 10 m - 4 m = 6 m. The run of the triangle represents the change in the clock time coordinate, which is 5 s - 2 s = 3 s.
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RESPONSE --> My answer is at least partially correct. self critique assessment: 3
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18:03:53 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> The slope of the triangle is 2. It is representative of the rise (change in position) over the run (change in time). confidence assessment: 1
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18:04:17 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> I didn't include the units, but the concept still applies. self critique assessment: 2
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18:07:10 `q010. In what sense does the slope of any graph of position vs. clock time represent the velocity of the object? For example, why does a greater slope imply greater velocity?
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RESPONSE --> It isn't representative of the velocity, but it is representative of the average velocity. And, the definition of average velocity is change in position over change in time. Greater slope implies greater velocity because according to the graph, position increases as time increases. confidence assessment: 1
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18:07:26 Since the rise between two points on a graph of velocity vs. clock time represents the change in `ds position, and since the run represents the change `dt clock time, the slope represents rise / run, or change in position /change in clock time, or `ds / `dt. This is the definition of average velocity.
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RESPONSE --> I got the correct answer. self critique assessment: 3
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18:15:37 `q011. As a car rolls from rest down a hill, its velocity increases. Describe a graph of the position of the car vs. clock time. If you have not already done so, tell whether the graph is increasing at an increasing rate, increasing at a decreasing rate, decreasing at an increasing rate, decreasing at a decreasing rate, increasing at a constant rate or decreasing at a constant rate. Is the slope of your graph increasing or decreasing? How does the behavior of the slope of your graph indicate the condition of the problem, namely that the velocity is increasing?
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RESPONSE --> This graph would show a linear relationship in this shape \ I say this because the car is rolling down the hill in a downward position vs. time. The graph is decreasing at a decreasing rate. The slope is decreasing. confidence assessment: 1
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18:16:35 The graph should have been increasing, since the position of the car increases with time (the car gets further and further from its starting point). The slope of the graph should have been increasing, since it is the slope of the graph that indicates velocity. An increasing graph within increasing slope is said to be increasing at an increasing rate.
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RESPONSE --> I see where I made my mistake. I focused more on how the car was moving instead of what was happening with each respective variable. self critique assessment: 2
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????Q???????????assignment #003 003. `Query 3 Physics I 06-12-2007
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18:20:45 Query Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 `micro m to appropriate # of significant figures)
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RESPONSE --> First of all, all of the units must be the same to do a proper addition. Conversions are as follows: 142.5cm *(10^ -2m / 1cm) = 1.425m 5.34*(10^5)um *(1m / 1,000,000um) = .534m Add: 1.80m+1.425m+.534m = 3.76m confidence assessment: 2
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18:21:54 ** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore nothing smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .001 m. 5.34 * `micro m means 5.34 * 10^-6 m, or .00000534 m, accurate to within .00000001 m. When these are added you get 3.22500534 m; however the 1.80 m is not resolved beyond .01 m so the result is 3.23 m. Remaining figures are meaningless, since the 1.80 m itself could be off by as much as .01 m. **
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RESPONSE --> I didn't go into specifics of significant figures in my answer, but throughout the problem the smallest number of significant digits was 3 which was the number of significant digits in my answer. self critique assessment: 2
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18:22:23 University Physics #34: Summarize your solution to Problem 1.34 (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch).
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RESPONSE --> This doesn't apply to me; I am in principles of physics. confidence assessment: 1
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18:22:36 ** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. }Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.3 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. **
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RESPONSE --> Not applicable. self critique assessment: 3
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