#$&* course PHY 201 02/07/13 at 9:55 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): I did not explain what the graph tells me about the motion as clear as this, but i understand its meaning. ------------------------------------------------ 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: 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: Units for position may include centimeters, millimeters, meters, inches, kilometers, feet, etc… Units for clock time may include seconds, minutes, hours, etc… Possible units for rate of change of position with respect to clock time include cm/s, m/s, mm/s, km/s, ft/s, cm/h, miles/h, etc... confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Not sure what is being asked in this question and the given solution doesn't help clear it up. I guess we are doing approximations. The greatest ocean depth is approximately 11000 m. Greatest mountain height on earth is approximately 9000 m 1/11000 m =9.09E-5 m and 1/9000 m = 1.11E-4 m confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The greatest mountain height is a bit less than 10 000 meters. The diameter of the Earth is a bit less than 13 000 kilometers. Using the round figures 10 000 meters and 10 000 kilometers, we estimate that the ratio is 10 000 meters / (10 000 kilometers). We could express 10 000 kilometers in meters, or 10 000 meters in kilometers, to actually calculate the ratio. Or we can just see that the ratio reduces to meters / kilometers. Since a kilometer is 1000 meters, the ratio is 1 / 1000. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `qQuery Principles of Physics and General College Physics: Summarize your solution to the following: Find the sum 1.80 m + 142.5 cm + 5.34 * 10^5 `micro m to the appropriate number of significant figures. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.80 m has 3 significant figures, 1.425 m has four significant figures, 0.534 m has three significant figures. When doing addition and subtraction, the result should have the same amount of significant figures as the number with the least amount of significant numbers. 1.80 m + 1.425 m + 0.534 m = 3.76 m confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ********************************************* Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If I know how far the ball rolled along each book, then I would know 'ds. I would know the time interval the ball required to roll from one end of each book to the other, 'dt = t2 - t1. I am assuming that by knowing how fast the ball is moving at each end of each book I would know the velocity at the end of that book. vAve = 'ds / 'dt = (s2 - s1) / (t2 - t1) => t1 = t2 - ('ds / vAve). I would create a graph of the balls position vs clock time and take the slope to calculate the average velocity. I would then create a graph of velocity vs clock time. I can think take the slope of that graph to determine acceleration. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: • How far the ball rolled along each book. • The time interval the ball requires to roll from one end of each book to the other. • How fast the ball is moving at each end of each book. How would you use your information to determine the clock time at each of the three points, if we assume the clock started when the ball was released at the 'top' of the first book? How would you use your information to sketch a graph of the ball's position vs. clock time? (This question is more challenging that the others): How would you use your information to sketch a graph of the ball's speed vs. clock time, and how would this graph differ from the graph of the position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If I know how far the ball rolled along each book, then I would know 'ds. I would know the time interval the ball required to roll from one end of each book to the other, 'dt = t2 - t1. I am assuming that by knowing how fast the ball is moving at each end of each book I would know the velocity at the end of that book. vAve = 'ds / 'dt = (s2 - s1) / (t2 - t1) => t1 = t2 - ('ds / vAve). I would create a graph of the balls position vs clock time and take the slope to calculate the average velocity. I would then create a graph of velocity vs clock time. I can think take the slope of that graph to determine acceleration. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If I know how far the ball rolled along each book, then I would know 'ds. I would know the time interval the ball required to roll from one end of each book to the other, 'dt = t2 - t1. I am assuming that by knowing how fast the ball is moving at each end of each book I would know the velocity at the end of that book. vAve = 'ds / 'dt = (s2 - s1) / (t2 - t1) => t1 = t2 - ('ds / vAve). I would create a graph of the balls position vs clock time and take the slope to calculate the average velocity. I would then create a graph of velocity vs clock time. I can think take the slope of that graph to determine acceleration. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!