course Phy 201 ۑ]Ӝ[HZassignment #003
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10:26:32 `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 meters per second. confidence assessment: 3
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10:27:17 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 must have read the information wrong because I thought it said the units for 'ds are meters, but if it said cm then the units of vAve would be cm/s. self critique assessment: 3
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10:28:06 `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 will be measured in centimeters. confidence assessment: 3
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10:28:13 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 --> self critique assessment: 3
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10:29:36 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> Multiplying cm/sec by sec is the same as cm/sec * sec/1. Since sec is in the numerator of one fraction and the denominator of the other fraction they cancel out, leaving cm/1, which is the same as cm. confidence assessment: 3
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10:29:52 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 --> self critique assessment: 3
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10:31:31 `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. If you set up the equation only using units you can find this: 'dt = km / (km/sec) , in which case the km's will cancel out and leave 'dt being measured in seconds. confidence assessment: 3
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10:31:36 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 --> self critique assessment: 3
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10:34:55 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> When you divide km by km/sec, you are essentially dividing the fractions km/1 by km/s. When you divide fractions, you flip the second one and multiply the first by the reciprocal of the second. This gives us (km/1) * (sec/km), where the km in the numerator of one fraction and the denominator of the second fraction cancel out, giving us sec/1, which is the same thing as sec. confidence assessment: 3
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10:35:07 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 --> self critique assessment: 3
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11:02:52 `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= 6m / 3sec = 2 m/sec. 'ds is 6 meters, and 'dt is 3 seconds. We divide 'ds by 'dt to obtain the average velocity. confidence assessment: 3
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11:03:02 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 --> self critique assessment: 3
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11:04:16 `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 --> 'ds, or change in position, would be represented by s2 - s1. 'dt, or change in clock time, would be represented by t2 - t1. confidence assessment: 3
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11:04:55 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 --> Average velocity= (s2-s1) / (t2-t1) self critique assessment: 3
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11:12:00 `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 --> Position s would be on the y-axis and clock time t would be on the x-axis. The rise of the triangle represents the change in position because position is located on the y-axis, and that quantity for this specific triangle is 6 meters. The run of the triangle represents the change in clock time on the x-axis, and that quantity is 3 seconds. confidence assessment: 3
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11:12:11 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 --> self critique assessment: 3
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11:13:12 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> Slope=rise/run = 6/3 = 2. The slope represents the change in position divided by the change in time, which is the average velocity. confidence assessment: 3
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11:13:29 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> The slope has units meters per second. self critique assessment: 3
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11:35:06 `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 --> The slope of any graph of position vs. clock time will be found by dividing the change in position by the change in clock time, which is the equation used to find the average velocity of the object. A greater slope implies greater velocity because a greater slope would mean that there is a greater change in the two positions, which would correspond with a greater velocity. confidence assessment: 3
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11:35:15 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 --> self critique assessment: 3
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11:36:38 `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 --> The graph would be a curved line that gets steeper as time goes on. The graph is increasing at an increasing rate. The slope is increasing. The slope indicates that the velocity is increasing because it gets steeper and steeper, which corresponds to a greater change in position relative to the clock time. confidence assessment: 3
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11:36:49 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 --> self critique assessment: 3
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G[]̲xԩ~q assignment #003 003. `Query 3 Physics I 05-28-2008
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21:35:32 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 --> I converted all of the numbers to meters, which gave 1.80 m + 1.425 m + .0000534 m = 3.23 m. Only 3 significant figures are used because 1.80 has the smallest number of significant figures within the problem and that is 3. confidence assessment: 3
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21:35:40 ** 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 --> self critique assessment: 3
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21:35:46 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 --> confidence assessment: 3
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21:35:57 ** 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 --> self critique assessment: 3
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