course Phy 201
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12:11:56 `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 as meters/sec
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12:12:29 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 read in this question that the units were meters. However the same concept is accurate.
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12:13:22 `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 obtained as centimeters
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12:13:30 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 --> Correct
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12:14:59 `q003. Explain the algebra of multiplying the unit cm / sec by the unit sec.
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RESPONSE --> cm/sec = cm/sec * sec sec* cm/sec = cm
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12:15:07 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 --> Correct
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12:16:02 `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 --> If vAve is measured in km / sec and `ds in km, then 'dt would be meaasure in seconds, as it is the information needed regarding clock time.
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12:16:15 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 --> Correct
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12:42:58 `q005. Explain the algebra of dividing the unit km / sec into the unit km.
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RESPONSE --> km / (km / sec) = seconds. km / (km / sec) * (km/sec) = sec /(km/sec) km = sec /(km/sec)
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12:43:02 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 --> Correct
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12:50:42 `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 --> Averate rate of the position changing: averge rate = change in position/time interval vAve = 'ds/'dt average rate = 6m/3 seconds = 2 m/sec Change in 'ds position: 'ds = vAve 'dt 'ds= 2m/sec * 3 se = 6 m Change in 'dt clocktime: 'dt = 'ds/vAve 'dt = 3 sec/ 6 m/sec = 2 m Combine Quantities to find Ave: These formulas are connceted in such a way that finding the value of three quantities can be found from the values of the other two.
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12:50:54 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 --> Correct
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12:52:28 `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 the equal to the product of the average vcelocity and time interval so 'ds = vAve 'dt and the expression tha trepresents the change 'dt in the clock time is 'dt = 'ds/vAve
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12:52:34 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 --> correct
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12:55:49 `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 --> The rise of the triangle is 6 m and the run is 3 seconds
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12:55:57 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 --> Correct
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12:56:19 `q009. What is the slope of this triangle and what does it represent?
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RESPONSE --> the slope of this triangle is 6m/3 sec = 2 m/sec
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12:56:25 The slope of this graph is 6 meters / 3 seconds = 2 meters / second.
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RESPONSE --> correct
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12:58:32 `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 gives the position of vs.clock time as rise is on the x-axis with position and clock time (run) is on the y-axis. With rise over run you can caluclate the differences of coordinates to come up with the velocity. A greater slope implies a greater velocity because there is more position changing in the moving object over a shorter amount of clock time
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12:58:39 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 --> correct
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13:00:14 `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 velocity increases as the car moves a faster pace over a shorter amount of time. This graph would be decreasing at an increasing rate and the slope would be increasing as the rise and run becoem larger numbers.
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13:00:33 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 --> Correct
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ѣʉ̶m Student Name: assignment #003 003. Misc: Surface Area, Pythagorean Theorem, Density
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14:18:43 `q001. There are 10 questions and 5 summary questions in this assignment. What is surface area of a rectangular solid whose dimensions are 3 meters by 4 meters by 6 meters?
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RESPONSE --> A = l * w * h A = 3m * 6m * 4m A = 72 m^2
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14:21:01 A rectangular solid has six faces (top, bottom, front, back, left side, right side if you're facing it). The pairs top and bottom, right and left sides, and front-back have identical areas. This solid therefore has two faces with each of the following dimensions: 3 m by 4 m, 3 m by 6 m and 4 m by 6 m, areas 12 m^2, 18 m^2 and 24 m^2. Total area is 2 * 12 m^2 + 2 * 18 m^2 + 2 * 24 m^2 = 108 m^2.
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RESPONSE --> I did the caluclations for a simple rectangle, however with more dimensions this probelm must be altered as we now need additional sides of the rectangle to be calulated in.
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