#$&* course PHY 241 4:30 am 24 Feb. Trying to catch up. Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
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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:3 ********************************************* 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: Just one since we count by integers. 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: For position we could have meters feet inches centimeters among many other. For time we could use seconds, minutes, hours, days, weeks, years, decades, centuries, millennia, and many more. For rate of change of position with respect to clock better known as velocity we could use cm/s, mph, ft/min and others. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The diameter of the Earth is 12,762.2 km. The deepest part of the ocean is Challenger Deep at 10942 meters (10.942 km) below sea level. 10.942 km/12,762.2km=1/1166.3 Ratio The tallest mountain relative to sea level is Mount Everest at 8850 Meters 8.850 km. 8.850km/12762.2km=1/1442.1 ratio. On a 1 meter globe we would build a ridge at 1m * 1/1442.1=0.0006 m or 0.6 mm. Not very substantial. confidence rating #$&*: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: 3 ********************************************* 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 + 1.425 m + 0.534 m 3.759m Significant figures make it 3.76m confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ********************************************* Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I came up with an answer of 7.9 km displacement from the original start point using the formulas to calculate the components of each vector. I then added the x components together and then the y components together. This gave me x component of 6.19 and y component of 4.79. Then I converted back into vector form. The scaled sketch agreed pretty closely to my answer. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): My answer is lacking compared to the solution. My answers although correct arent explained well enough. Also I didnt even think to calculate the angle. ------------------------------------------------ Self-critique Rating: 1 ********************************************* Question: A ball rolls from rest down a book, off that book and smoothly 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. The acceleration on each book is uniform. How would you use your information to determine the clock time at each of the three points (top of first book, top of second which is identical to the bottom of the first, bottom of second book), 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 the clock started at the time the ball started rolling the simplest thing to do would be to add the time intervals to the initial clock time at the start. With this we could say that at time 0 the ball was at position 0. And at time = 0+first time interval it is at whatever the distance to the second book is. Doing the same with the second book we can connect the dots and this gives us our position vs. clock time graph. For a speed vs. clock time graph we would need to find the initial and final velocities for the time intervals. We could then plot these and again connect the dots. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: A ball rolls from rest down a book, off that book and smoothly onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: How far the ball rolled along each book. The time interval the ball requires to roll from one end of each book to the other. How fast the ball is moving at each end of each book. The acceleration on each book is uniform. How would you use your information to determine the clock time at each of the three points (top of first book, top of second which is identical to the bottom of the first, bottom of second book), if we assume the clock started when the ball was released at the 'top' of the first book? How would you use your information to sketch a graph of the ball's position vs. clock time? (This question is more challenging that the others): How would you use your information to sketch a graph of the ball's speed vs. clock time, and how would this graph differ from the graph of the position? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If the clock started at the time the ball started rolling the simplest thing to do would be to add the time intervals to the initial clock time at the start. With this we could say that at time 0 the ball was at position 0. And at time = 0+first time interval it is at whatever the distance to the second book is. Doing the same with the second book we can connect the dots and this gives us our position vs. clock time graph. For a speed vs. clock time graph we would need to find the initial and final velocities for the time intervals. We could then plot these and again connect the dots. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&! ********************************************* Question: A ball rolls from rest down a book, off that book and smoothly 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. The acceleration on each book is uniform. How would you use your information to determine the clock time at each of the three points (top of first book, top of second which is identical to the bottom of the first, bottom of second book), 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 the clock started at the time the ball started rolling the simplest thing to do would be to add the time intervals to the initial clock time at the start. With this we could say that at time 0 the ball was at position 0. And at time = 0+first time interval it is at whatever the distance to the second book is. Doing the same with the second book we can connect the dots and this gives us our position vs. clock time graph. For a speed vs. clock time graph we would need to find the initial and final velocities for the time intervals. We could then plot these and again connect the dots. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!