course Mth 173
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Given Solution: Average velocity is defined as the average rate of change of position with respect to clock time. The average rate of change of A with respect to B is (change in A) / (change in B). Thus the average rate of change of position with respect to clock time is • ave rate = (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok Question: Why can it not be said that average velocity = position / clock time? Your solution: This can’t be said because average velocity is the change in position/the change in clock time. Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The definition of average rate involves the change in one quantity, and the change in another. Both position and clock time are measured with respect to some reference value. For example, position might be measured relative to the starting line for a race, or it might be measured relative to the entrance to the stadium. Clock time might be measure relative to the sound of the starting gun, or it might be measured relative to noon. So position / clock time might, at some point of a short race, be 500 meters / 4 hours (e.g., 500 meters from the entrance to the stadium and 4 hours past noon). The quantity (position / clock time) tells you nothing about the race. There is a big difference between (position) / (clock time) and (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok Self-critique Rating: ok ********************************************* Question: Give your solution to the following, which should be in your notes: Find the approximate uncertainty in the area of a circle given that its radius is 2.8 * 10^4 cm. STUDENT COMMENT: I understand how squaring the problem increases uncertainty and I understand the concept of a range of uncertainty but I am having trouble figuring out how the range of 2.75 * 10^4 and 2.85*10^4 were established for the initial uncertainties in radius. INSTRUCTOR RESPONSE: The key is the first sentence of the given solution: 'Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm.' You know this because you know that any number which is at least 2.75, and less than 2.85, rounds to 2.8. Ignoring the 10^4 for the moment, and concentrating only on the 2.8: Since the given number is 2.8, with only two significant figures, all you know is that when rounded to two significant figures the quantity is 2.8. So all you know is that it's between 2.75 and 2.85. Your solution: After reading the student and instructor response I understand where the 2.75 and 2.85 come from. The area uncertainty would be pi * (2.75 *10 ^4 cm)^2 and pi * (2.85 *10^4 cm)^2. Confidence Assessment: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** Radius 2.8 * 10^4 cm means that the radius is between 2.75 * 10^4 cm and 2.85 * 10^4 cm. This means that the area is between pi * (2.75 * 10^4 cm)^2 = 2.376 * 10^9 cm^2 and pi * (2.85 * 10^4 cm)^2 = 2.552 * 10^9 cm^2. The difference is .176 * 10^9 cm^2 = 1.76 * 10^8 cm^2, which is the uncertainty in the area. Note that the .1 * 10^4 cm uncertainty in radius is about 4% of the radius, which the .176 * 10^9 cm uncertainty in area is about 8% of the area. This is because the area is proportional to the squared radius. A small percent uncertainty in the radius gives very nearly double the percent uncertainty in the squared radius. ** Self-critique (if necessary): I couldn’t get a correct answer for the uncertainty in the area. I kept getting too high of numbers because I didn’t know how to plug the information into a calculator. I see how it was solved now. I just kind of forgot to multiply by the pi. Self-critique Rating: 2 ********************************************* Question: What is your own height in meters and what is your own mass in kg? Explain how you determined these? What are your uncertainty estimates for these quantities, and on what did you base these estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: My height in inches is 69 therefore; my height in meters would be approximately 1.75 meters. I found this by multiply my height in inches by 2.54 which is the amount of centimeters in an inch. I then converted centimeters to meters. To find my mass in kg I have to divide my weight in pounds, 135, by the amount of pounds in a kilogram which is 2.205 pounds. My weight in kilograms would then be 61.2 kilograms. I have read about uncertainties in the book, but I am having trouble finding them on the CD’s in the notes. After realizing this I am not too sure on how to solve the uncertainty problems. Confidence Assessment: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Presumably you know your height in feet and inches, and have an idea of your ideal weight in pounds. Presumably also, you can convert your height in feet and inches to inches. To get your height in meters, you would first convert your height in inches to cm, using the fact that 1 inch = 2.54 cm. Dividing both sides of 1 in = 2.54 cm by either 1 in or 2.54 cm tells us that 1 = 1 in / 2.54 cm or that 1 = 2.54 cm / 1 in, so any quantity can be multiplied by 1 in / (2.54 cm) or by 2.54 cm / (1 in) without changing its value. Thus if you multiply your height in inches by 2.54 cm / (1 in), you will get your height in cm. For example if your height is 69 in, your height in cm will be 69 in * 2.54 cm / (1 in) = 175 in * cm / in. in * cm / in = (in / in) * cm = 1 * cm = cm, so our calculation comes out 175 cm. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m still looking for the uncertainty notes, so if you have any hints on where I should look let me know. Self-critique Rating: 3