#$&* course Phy 201 11/19/11 around 8:43pm 003. `Query 3
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Given Solution: The coordinates a point on the graph include a position and a clock time, which tells you where the object whose motion is represented by the graph is at a given instant. If you have two points on the graph, you know the position and clock time at two instants. Given two points on a graph you can find the rise between the points and the run. On a graph of position vs. clock time, the position is on the 'vertical' axis and the clock time on the 'horizontal' axis. * The rise between two points represents the change in the 'vertical' coordinate, so in this case the rise represents the change in position. * The run between two points represents the change in the 'horizontal' coordinate, so in this case the run represents the change in clock time. The slope between two points of a graph is the 'rise' from one point to the other, divided by the 'run' between the same two points. * The slope of a position vs. clock time graph therefore represents rise / run = (change in position) / (change in clock time). * By the definition of average velocity as the average rate of change of position with respect to clock time, we see that average velocity is vAve = (change in position) / (change in clock time). * Thus the slope of the position vs. clock time graph represents the average velocity for the interval between the two graph points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 2 significant figures. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Position: cm, m, cm/sec Clock time: seconds, minutes Rate of change: cm/sec, m/min, cm/s/s These are just a few of the units that could be given confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 * 10^5 `micro m to appropriate # of significant figures) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1cm=0.01m 5.34*10^5 micro m = (5.34*10^5) *10^-6= 0.534m 142.5cm*0.01= 1.425m 1.80m + 1.425m + 0.534m = 3.76m this is because 1.80 is only accurate to 0.01m confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ok I had to look up the unit of micro m. I'll need to refresh some on my units. ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: * How far the ball rolled along each book. * The time interval the ball requires to roll from one end of each book to the other. * How fast the ball is moving at each end of each book. How would you use your information to determine the clock time at each of the three points, if we assume the clock started when the ball was released at the 'top' of the first book? How would you use your information to sketch a graph of the ball's position vs. clock time? (This question is more challenging that the others): How would you use your information to sketch a graph of the ball's speed vs. clock time, and how would this graph differ from the graph of the position? I would determine clock times at each of the points by dividing the change in position from the average velocity. `ds/aVe= `dt I would sketch the velocities on the y-axis, time x-axis. The slope of the graph would be the average velocity. The points would cover when the ball started rolling till it rolled off the end of the book. I would graph the balls acceleration at different points to get the speed vs. clock time. speed=y clock time=x. The first graph would be increasing at a near constant rate. This graph would start to decrease at some point. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 2 significant figures. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Position: cm, m, cm/sec Clock time: seconds, minutes Rate of change: cm/sec, m/min, cm/s/s These are just a few of the units that could be given confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 * 10^5 `micro m to appropriate # of significant figures) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1cm=0.01m 5.34*10^5 micro m = (5.34*10^5) *10^-6= 0.534m 142.5cm*0.01= 1.425m 1.80m + 1.425m + 0.534m = 3.76m this is because 1.80 is only accurate to 0.01m confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):Ok I had to look up the unit of micro m. I'll need to refresh some on my units. ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: * How far the ball rolled along each book. * The time interval the ball requires to roll from one end of each book to the other. * How fast the ball is moving at each end of each book. How would you use your information to determine the clock time at each of the three points, if we assume the clock started when the ball was released at the 'top' of the first book? How would you use your information to sketch a graph of the ball's position vs. clock time? (This question is more challenging that the others): How would you use your information to sketch a graph of the ball's speed vs. clock time, and how would this graph differ from the graph of the position? I would determine clock times at each of the points by dividing the change in position from the average velocity. `ds/aVe= `dt I would sketch the velocities on the y-axis, time x-axis. The slope of the graph would be the average velocity. The points would cover when the ball started rolling till it rolled off the end of the book. I would graph the balls acceleration at different points to get the speed vs. clock time. speed=y clock time=x. The first graph would be increasing at a near constant rate. This graph would start to decrease at some point. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!