course Phy121 002. `ph1 query 2
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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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Because if it’s the average, it must be the change in position divided by the change in clock time. This would be represented with a delta. 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The uncertainty would be +-0.05 because we know that anything 2.75 or higher would round up to 2.8 and anything lower than 2.85 would round down to 2.8. Area= pi r^2= 2.8*10^9 pi Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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 need to calculate the values for the uncertainties before I calculate the area. Then I need to calculate the difference in the two to find the uncertainty. Self-critique Rating: 3 ********************************************* 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 67. My weight in pounds is 147. Next, I need to convert these into meters and kg. There are about 39.37 inches in one meter: 67 inches * (1meter/ 39.37 inches)= 1.70 meters There are about 2.20 lbs in 1 kg: 147 lbs* (1kg/ 2.20 lbs)= 66.82 kg I believe the uncertainties for both are +-0.1. This is because the meter and kg conversions I used were rounded. They actually have more decimal places. Confidence Assessment: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT SOLUTION 5 feet times 12 inches in a feet plus six inches = 66 inches. 66inches * 2.54 cm/inch = 168.64 cm. 168.64 cm * .01m/cm = 1.6764 meters. INSTRUCTOR COMMENT: Good, but note that 66 inches indicates any height between 65.5 and 66.5 inches, with a resulting uncertainty of about .7%. 168.64 implies an uncertainty of about .007%. It's not possible to increase precision by converting units. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK Self-critique Rating: OK "