#$&*
course phy 231 June 29th, 4:53 pm 003. `Query 3*********************************************.............................................
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: Since ALL nonzero digits are ALWAYS significant, then 69 and 61 both have 2 significant digits; therefore the difference between 69 and 61 is 8. 8 is considered significant. 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, in, ft, miles Clock time: seconds, minutes, hours Rates of change of position: m/s, km/hr, mph confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 * 10^5 `micro m to appropriate # of significant figures) Your solution: 1.80 has 3 significant digits and is 180 cm; 142.5 has 4; 5.34 * 10^5 * 10-6 has 3 and is equal to 53.4 cm. The sum of all three = 375.90, after converting all of them to centimeters. But the answer can only have as many significant digits as the least number of significant digits after the decimal in each addend, and is limited by 180 cm (it has no digits after the decimal), therefore the answer is 376 cm. confidence rating #$&*: 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): I calculated it and left the unit in centimeters. My logic behind the significant digits was a little different than your explanation, but we arrived at the same answer. I’m pretty confident I was thinking correctly concerning significant digits. ********************************************* Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the starting point, you have a vector1 of 2.6 due North, a vector2 of 4.0 due East, then a vector3 of 3.1 km at a degree of 45 north from east. You must find the components on the third vector and the components to the North and East vectors. Therefore the components of vector3 are: x-comp = 3.1 * cos 45 degrees = 2.19 km, and the y-comp = 3.1* sin 45 degrees = 2.19 km Add these results x-comp + vector1 = 2.19 + 2.6 = 4.79 km, y-comp + vector2 = 2.19 + 4.0 = 6.19 km You now have a total North vector and a total East vector; together you have a right triangle, missing the value of the hypotenuse. Finding the hypotenuse will give you the total magnitude of the displacement of the postal delivery truck. Use Pythagorean Thm: hyp = sqrt (4.79^2 + 6.19^2) = 7.83 km is the total magnitude displacement at tan (opposite/adjacent) = tan (4.79/6.19). The angle of displacement = tan-1(.7738) = 37.73 degrees. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 ) = 7.9 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK
********************************************* 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? To find the clock time of the three points (starting, end of first book, end of second book), you would need the distance the ball had traveled at the second and third point and the velocity at those points. We are given both, how far the ball rolled along each book AND how fast the ball was moving at the end of each book. Therefore `dt = `ds/v_Ave and you could calculate the time. At the start, the time would be 0. A graph of ball’s position vs. clock time would have “position” on the y-axis and “clock time” on the x-axis. We could then start the graph with distance and time both at (0,0). The first point to plot would be the distance at the end of the first book and the time that was just calculated (time book1, distance book1). The second point that could be plotted would be (time book2, distance book2). The slope will not be constant because the ball picked up speed when it rolled down the second book. A graph of speed vs clock time would have speed on the y-axis and clock time on the x-axis. You know the speed of the ball at the end of each book and time at the end of each book that was calculated from `ds/v_Ave; therefore at least two points beyond (0,0) could be plotted. The graph again will not be constant because of the change in the rate of the velocity on the second book. confidence rating #$&*: If I am interpreting the problem correct, I believe my thinking on this is OK. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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? To find the clock time of the three points (starting, end of first book, end of second book), you would need the distance the ball had traveled at the second and third point and the velocity at those points. We are given both, how far the ball rolled along each book AND how fast the ball was moving at the end of each book. Therefore `dt = `ds/v_Ave and you could calculate the time. At the start, the time would be 0. A graph of ball’s position vs. clock time would have “position” on the y-axis and “clock time” on the x-axis. We could then start the graph with distance and time both at (0,0). The first point to plot would be the distance at the end of the first book and the time that was just calculated (time book1, distance book1). The second point that could be plotted would be (time book2, distance book2). The slope will not be constant because the ball picked up speed when it rolled down the second book. A graph of speed vs clock time would have speed on the y-axis and clock time on the x-axis. You know the speed of the ball at the end of each book and time at the end of each book that was calculated from `ds/v_Ave; therefore at least two points beyond (0,0) could be plotted. The graph again will not be constant because of the change in the rate of the velocity on the second book. confidence rating #$&*: If I am interpreting the problem correct, I believe my thinking on this is OK. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#$&* course phy 231 June 29th, 4:53 pm 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: Since ALL nonzero digits are ALWAYS significant, then 69 and 61 both have 2 significant digits; therefore the difference between 69 and 61 is 8. 8 is considered significant. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Position: cm, m, in, ft, miles Clock time: seconds, minutes, hours Rates of change of position: m/s, km/hr, mph confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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) Your solution: 1.80 has 3 significant digits and is 180 cm; 142.5 has 4; 5.34 * 10^5 * 10-6 has 3 and is equal to 53.4 cm. The sum of all three = 375.90, after converting all of them to centimeters. But the answer can only have as many significant digits as the least number of significant digits after the decimal in each addend, and is limited by 180 cm (it has no digits after the decimal), therefore the answer is 376 cm. 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): I calculated it and left the unit in centimeters. My logic behind the significant digits was a little different than your explanation, but we arrived at the same answer. I’m pretty confident I was thinking correctly concerning significant digits. ********************************************* Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: From the starting point, you have a vector1 of 2.6 due North, a vector2 of 4.0 due East, then a vector3 of 3.1 km at a degree of 45 north from east. You must find the components on the third vector and the components to the North and East vectors. Therefore the components of vector3 are: x-comp = 3.1 * cos 45 degrees = 2.19 km, and the y-comp = 3.1* sin 45 degrees = 2.19 km Add these results x-comp + vector1 = 2.19 + 2.6 = 4.79 km, y-comp + vector2 = 2.19 + 4.0 = 6.19 km You now have a total North vector and a total East vector; together you have a right triangle, missing the value of the hypotenuse. Finding the hypotenuse will give you the total magnitude of the displacement of the postal delivery truck. Use Pythagorean Thm: hyp = sqrt (4.79^2 + 6.19^2) = 7.83 km is the total magnitude displacement at tan (opposite/adjacent) = tan (4.79/6.19). The angle of displacement = tan-1(.7738) = 37.73 degrees. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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 ) = 7.9 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? To find the clock time of the three points (starting, end of first book, end of second book), you would need the distance the ball had traveled at the second and third point and the velocity at those points. We are given both, how far the ball rolled along each book AND how fast the ball was moving at the end of each book. Therefore `dt = `ds/v_Ave and you could calculate the time. At the start, the time would be 0. A graph of ball’s position vs. clock time would have “position” on the y-axis and “clock time” on the x-axis. We could then start the graph with distance and time both at (0,0). The first point to plot would be the distance at the end of the first book and the time that was just calculated (time book1, distance book1). The second point that could be plotted would be (time book2, distance book2). The slope will not be constant because the ball picked up speed when it rolled down the second book. A graph of speed vs clock time would have speed on the y-axis and clock time on the x-axis. You know the speed of the ball at the end of each book and time at the end of each book that was calculated from `ds/v_Ave; therefore at least two points beyond (0,0) could be plotted. The graph again will not be constant because of the change in the rate of the velocity on the second book. confidence rating #$&*: If I am interpreting the problem correct, I believe my thinking on this is OK. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* 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? To find the clock time of the three points (starting, end of first book, end of second book), you would need the distance the ball had traveled at the second and third point and the velocity at those points. We are given both, how far the ball rolled along each book AND how fast the ball was moving at the end of each book. Therefore `dt = `ds/v_Ave and you could calculate the time. At the start, the time would be 0. A graph of ball’s position vs. clock time would have “position” on the y-axis and “clock time” on the x-axis. We could then start the graph with distance and time both at (0,0). The first point to plot would be the distance at the end of the first book and the time that was just calculated (time book1, distance book1). The second point that could be plotted would be (time book2, distance book2). The slope will not be constant because the ball picked up speed when it rolled down the second book. A graph of speed vs clock time would have speed on the y-axis and clock time on the x-axis. You know the speed of the ball at the end of each book and time at the end of each book that was calculated from `ds/v_Ave; therefore at least two points beyond (0,0) could be plotted. The graph again will not be constant because of the change in the rate of the velocity on the second book. confidence rating #$&*: If I am interpreting the problem correct, I believe my thinking on this is OK. ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!