#$&* course PHY 241 9/23 10 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): ------------------------------------------------ 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: The difference of 69 and 61 is 8. Since 20 cm and 1 minute only have one significant figure, the answer should have one significant figure. 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: meters, feet, miles, kilometers, centimeters, inches Clock time: seconds, hours, minutes Velocity: meters/minute, feet/second, kilometers/hour confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The diameter of the earth is 12 742 km. The greatest ocean depth is 10 924 meters. That would make it 10 924 / 12 742 000 (about 0.0857%) of the Earth’s diameter. The greatest mountain height is 8850 meters, which is 8850 / 12 742 000 (about 0.0695%) of Earth’s diameter. On a globe 1 meter in diameter, the mountain would be 0.0695 meters (6.95 cm) high. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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.
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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: The truck travels 2.6 km in the y-direction, 4.0 km in the x-direction, and 3.1 km at an angle of 45 degrees counterclockwise with respect to the x-axis. The total displacement in the y-direction is 2.6 + 3.1 * sin(45 degrees), which is about 4.79 km. The total displacement in the x-direction is 4.0 + 3.1 * cos(45 degrees), which is about 6.19 km. Now that I have the displacement in the x and y directions, I can use Pythagorean Theorem to find the magnitude of the displacement and trigonometry to find the angle. sqrt(4.79^2 + 6.19^2) = 7.8 km tan^-1 (4.79 / 6.19) = 38 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): ------------------------------------------------ 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: I already know the time interval of all three books, but to find the total change in clock time, I need the time it took for the ball to fall from one book to another. One possibility is, since I already know the final velocity for each book and the ball will not be accelerating in the horizontal direction while falling, to measure how far it travelled in the horizontal direction before landing on the next book and divide that by the horizontal velocity (`dt = `ds / v). Another possibility, if I knew the vertical distance between the books, would be to use the equation v_f^2 = v_0^2 + 2a `ds, v_0 being 0, a being gravity, and ‘ds being the vertical distance between the books. After finding v_f, I could then find the average velocity and get the change in time. Assuming position is measured in the horizontal direction, the graph of the ball’s position vs. clock time would start at 0 and slowly curve upward until the ball reached the end of the first book. The graph would then become a straight line with the same slope as the curve the instant the ball left the book. When the ball landed on the second book, the line would become a curve again until the ball once again fell off the book and the line would become straight again. For a velocity vs. clock time graph, assuming constant acceleration, the graph would start at 0 and form an increasing straight line until the ball left the first book, at which point the line would instantly become horizontal. When the ball landed on the second book, the line would start increasing again, probably with the same slope, until the ball left the second book and the line would become level once more. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " 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: I already know the time interval of all three books, but to find the total change in clock time, I need the time it took for the ball to fall from one book to another. One possibility is, since I already know the final velocity for each book and the ball will not be accelerating in the horizontal direction while falling, to measure how far it travelled in the horizontal direction before landing on the next book and divide that by the horizontal velocity (`dt = `ds / v). Another possibility, if I knew the vertical distance between the books, would be to use the equation v_f^2 = v_0^2 + 2a `ds, v_0 being 0, a being gravity, and ‘ds being the vertical distance between the books. After finding v_f, I could then find the average velocity and get the change in time. Assuming position is measured in the horizontal direction, the graph of the ball’s position vs. clock time would start at 0 and slowly curve upward until the ball reached the end of the first book. The graph would then become a straight line with the same slope as the curve the instant the ball left the book. When the ball landed on the second book, the line would become a curve again until the ball once again fell off the book and the line would become straight again. For a velocity vs. clock time graph, assuming constant acceleration, the graph would start at 0 and form an increasing straight line until the ball left the first book, at which point the line would instantly become horizontal. When the ball landed on the second book, the line would start increasing again, probably with the same slope, until the ball left the second book and the line would become level once more. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!