#$&* course Phy 201 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
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Given Solution: Average velocity is defined as the average rate of change of position with respect to clock time. The average rate of change of A with respect to B is (change in A) / (change in B). Thus the average rate of change of position with respect to clock time is • ave rate = (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: Why can it not be said that average velocity = position / clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Dividing one specific position by the time elapsed doesn’t actually tell you anything towards velocity. Except where that one position is at one specific time. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The definition of average rate involves the change in one quantity, and the change in another. Both position and clock time are measured with respect to some reference value. For example, position might be measured relative to the starting line for a race, or it might be measured relative to the entrance to the stadium. Clock time might be measure relative to the sound of the starting gun, or it might be measured relative to noon. So position / clock time might, at some point of a short race, be 500 meters / 4 hours (e.g., 500 meters from the entrance to the stadium and 4 hours past noon). The quantity (position / clock time) tells you nothing about the race. There is a big difference between (position) / (clock time) and (change in position) / (change in clock time). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: Explain in your own words the process of fitting a straight line to a graph of y vs. x data, and briefly discuss the nature of the uncertainties encountered in the process. For example, you might address the question of how two different people, given the same graph, might obtain different results for the slope and the vertical intercept. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The variable that is being determined “in respect of” is the one that should go on the x-axis. People could easily confuse which variable goes on which axis, and so they would have very different slopes and y-intercepts. Though neither way is wrong, axis’ should be labeled to make clear what the relationship between the two variables is. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: (Principles of Physics and General College Physics students) What is the range of speeds of a car whose speedometer has an uncertainty of 5%, if it reads 90 km / hour? What is the range of speeds in miles / hour? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: At 90km/hour, 90 * .05 = 4.5 So you would say that it could be + or - 4.5km/h. When the speedometer says 90km/h, you could be going anywhere from 85.5km/h to 94.5km/h. 90km/h = about 56mph 56 * .05 = 2.8 When the speedometer reads 56mph, you could be going anywhere from 53.2mph to 58.8mph. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ #$&*
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Given Solution: 5% of 90 km / hour is .05 * 90 km / hour = 4.5 km / hour. So the actual speed of the car might be as low as 90 km / hour - 4.5 km / hour = 85.5 km / hour, or as great as 90 km / hour + 4.5 km / hour = 94.5 km / hour. To convert 90 km / hour to miles / hour we use the fact, which you should always know, that 1 inch = 2.54 centimeters. This is easy to remember, and it is sufficient to convert between SI units and British non-metric units. Using this fact, we know that 90 km = 90 000 meters, and since 1 meter = 100 centimeter this can be written as 90 000 * (100 cm) = 9 000 000 cm, or 9 * 10^6 cm. Now since 1 inch = 2.54 cm, it follows that 1 cm = (1 / 2.54) inches so that 9 000 000 cm = 9 000 000 * (1/2.54) inches, or roughly 3 600 000 inches (it is left to you to provide the accurate result; as you will see results in given solutions are understood to often be very approximate, intended as guidelines rather than accurate solutions). In scientific notation, the calculation would be 9 * 10^6 * (1/2.54) inches = 3.6 * 10^6 inches. Since there are 12 inches in a foot, an inch is 1/12 foot so our result is now 3 600 000 *(1/12 foot) = 300 000 feet (3.6 * 10^6 * (1/12 foot) = 3 * 10^5 feet). Since there are 5280 feet in a mile, a foot is 1/5280 mile so our result is 300 000 * (1/5280 mile) = 58 miles, again very approximately. So 90 km is very roughly 58 miles (remember this is a rough approximation; you should have found the accurate result). Now 90 km / hour means 90 km in an hour, and since 90 km is roughly 58 miles our 90 km/hour is about 58 miles / hour. A more formal way of doing the calculation uses 'conversion factors' rather than common sense. Common sense can be misleading, and a formal calculation can provide a good check to a commonsense solution: We need to go from km to miles. We use the facts that 1 km = 1000 meters, 1 meter = 100 cm, 1 cm = 1 / 2.54 inches, 1 inch = 1/12 foot and 1 foot = 1 / 5280 mile to get the conversion factors (1000 m / km), (100 cm / m), (1/2.54 in / cm), (1/12 foot / in) and (1/5280 mile / ft) and string together our calculation: 90 km / hr * (1000 m / km) * (100 cm / m) * (1/2.54 in / cm) * (1/12 foot / in) * (1/5280 mile / ft) = 58 mi / hr (again not totally accurate). Note how the km divides out in the first multiplication, the m in the second, the cm in the third, the inches in the fourth, the feet in the fifth, leaving us with miles / hour. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: (Principles of Physics students are invited but not required to submit a solution) Give your solution to the following: Find theapproximateuncertainty inthearea ofacircle giventhat itsradiusis2.8 * 10^4 cm. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Radius = 2.8 * 10^4 = 28,000cm confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ #$&* ********************************************* Question: What is your own height in meters and what is your own mass in kg(if you feel this question is too personal then estimate these quantities for someone you know)? Explainhow you determinedthese. What are your uncertainty estimates for these quantities, and on what did you base these estimates? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am about 5’ 9” so in meters that is about 1.75m I weigh about 160lb so in kg that is about 72.57kg For my height I think that’s pretty solid My weight really goes anywhere from 150-160lb, being an uncertainty of: 155 being in the middle, + or - 5 pounds, = 3% uncertainty. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ #$&*
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Given Solution: Presumably you know your height in feet and inches, andhave an idea ofyouridealweight 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. STUDENT SOLUTION AND QUESTIONS My height in meters is - 5’5” = 65inches* 2.54cm/1in = 165cm*1m/100cm = 1.7m. My weight is 140lbs* 1kg/2.2lbs = 63.6kg. Since 5’5” could be anything between 5’4.5 and 5’5.5, the uncertainty in height is ???? The uncertainty in weight, since 140 can be between 139.5 and 140.5, is ?????? INSTRUCTOR RESPONSE Your height would be 5' 5"" +- .5""; this is the same as 65"" +- .5"". .5"" / 65"" = .008, approximately, or .8%. So the uncertainty in your height is +-0.5"", which is +-0.8%. Similarly you report a weight of 140 lb +- .5 lb. .5 lb is .5 lb / (140 lb) = .004, or 0.4%. So the uncertainty is +-0.5 lb, or +- 0.4%. STUDENT QUESTION I am a little confused. In the example from another student her height was 66 inches and you said that her height could be between 65.5 and 66.5 inches. but if you take the difference of those two number you get 1, so why do you divide by .5 when the difference is 1 INSTRUCTOR RESPONSE If you regard 66 inches as being a correct roundoff of the height, then the height is between 65.5 inches and 66.5 inches. This makes the height 66 inches, plus or minus .5 inches. This is written as 66 in +- .5 in and the percent uncertainty would be .5 / 66 = .007, about .7%. If you regard 66 inches having been measured only accurately enough to ensure that the height is between 65 inches and 67 inches, then your result would be 66 in +- 1 in and the percent uncertainty would be 1 / 66 = .015 or about 1.5%. STUDENT QUESTION If a doctor were to say his inch marker measured to the nearest 1/4 inch, would that be the uncertainty? Meaning, would I only have to multiply that by .0254 to find the uncertainty in meters, dividing that by my height to find the percent uncertainty? INSTRUCTOR RESPONSE That's pretty much the case, though you do have to be a little bit careful about how the rounding and the uncertainty articulate. For example I'm 72 inches tall. That comes out to 182.88 cm. It wouldn't make a lot of sense to say that I'm 182.88 cm tall, +- .64 cm. A number like 182.88 has a ridiculously high number of significant figures. It wouldn't quite be correct to just round up and say that I'm 183 cm tall +- .64 cm. We might be able to say that I'm 183 cm tall, +- .76 cm, but that .76 cm again implies more precision than is present. We would probably end up saying that I""m 183 cm tall, +- 1 cm. Better to overestimate the uncertainty than to underestimate it. As far as the percent uncertainty goes, we wouldn't need to convert the units at all. In my case we would just divide 1/4 in. by 72 in., getting about .034 or 3.4%. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: A ball rolls from rest down a book, off that book and onto another book, where it picks up additional speed before rolling off the end of that book. Suppose you know all the following information: • How far the ball rolled along each book. • The time interval the ball requires to roll from one end of each book to the other. • How fast the ball is moving at each end of each book. How would you use your information to calculate the ball's average velocity on each book? How would you use your information to calculate how quickly the ball's speed was changing on each book? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I would take the distance traveled on each book, and divide it by the time it took to roll that distance, giving me the average velocity on that book. To find how quickly the ball accelerates, I think I would use the speed at each end of the book, and divide that by the time it took to travel from each end of the book, giving me seconds^2. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ #$&* "