course phy 231 ?vZ??????????assignment #003
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20:04:44 `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= m/seconds confidence assessment: 3
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20:05:28 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 thought the question stated meters not cm self critique assessment: 2
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20:18:48 `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 --> cm confidence assessment: 3
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20:23:19 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> vAve*dt=ds cm/sec is multiplied by seconds which would give the answer with the unit cm confidence assessment: 3
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20:23:58 `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 would be measured in seconds confidence assessment: 3
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20:25:20 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> dt=ds/vAve km is multiplied and then divided by km/sec to give the answer dt in the unit seconds confidence assessment: 3
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20:26:14 `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 --> forgot to mention the division of km/sec confidence assessment:
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?????????~????assignment #003 003. Velocity Relationships Physics I 01-30-2008
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20:28:28 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> confidence assessment: 0
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20:29:03 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 --> forgot to mention the division of km/sec the first time self critique assessment: 2
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20:33:22 `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 --> 10-4=6 meters 5-2=3 seconds 6meters/3seconds would be 2 m/sec. even if the measurements were added, the answer would be the same. confidence assessment: 3
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20:35:32 `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 --> s2-s1=s1 t2-t1=t1 position(s1)/clock time(t1) confidence assessment: 2
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20:36:41 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 --> s1/t1=1vAve self critique assessment: 3
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20:53:00 `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 --> a=1/2*b*h 5=1/2*(10)*h 2=h h=2. so this would be the lenght of the vertical side of the triangle. the rise represents seconds. b*h 10*2=20. the run of the triangle is 20. and it represents meters confidence assessment: 2
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20:54:04 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 --> was way off with this one. self critique assessment: 1
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???????????| assignment #003 003. Velocity Relationships Physics I 01-30-2008
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21:02:15 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> rise/run of the slope 6m/3sec would give you 2m/sec confidence assessment: 3
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21:10:41 `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 --> by dividing the the position of the object by time would give the vAve. confidence assessment: 3
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21:14:15 `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 --> velocity would be increasing at a decreasing rate. so from the veritcal side of the graph, it would slowly curve downards as it decreases confidence assessment: 2
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