course phy201 I need some understanding on the uncertainty calculations ֊DȘo
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13:57:12 Given the initial velocity, final velocity and time duration of a uniformly accelerating object, how do we reason out the corresponding acceleration and change in the position of an object?
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RESPONSE --> well if the average acceleration is the change of velocity / the elapsed time then we plug into the equation the information that we have and solve and this gives us our required information.
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13:57:49 02-05-2008 13:57:49 ** COMMON ERRONEOUS STUDENT RESPONSE: To find the average acceleration, we divide the change in veolocity by the time interval. To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. You are not given the average velocity or the change in velocity. You have to first determine the average velocity; then your strategy will work. Since acceleration is constant you can say that average velocity is the average of initial and final velocities: vAve = (v0 + vf) / 2. Change in velocity is `dv = vf - v0. Now we can do as you say: To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. **
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NOTES -------> ** COMMON ERRONEOUS STUDENT RESPONSE: To find the average acceleration, we divide the change in veolocity by the time interval. To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval. You are not given the average velocity or the change in velocity. You have to first determine the average velocity; then your strategy will work. Since acceleration is constant you can say that average velocity is the average of initial and final velocities: vAve = (v0 + vf) / 2. Change in velocity is `dv = vf - v0. Now we can do as you say: To find the change in position or displacement of the object over any time interval multiply the average velocity over that interval by the duration of the interval.
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14:08:09 In terms of the meanings of altitudes, area, slope and width, how does a velocity vs. clock time trapezoid represent change in position and acceleration?
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RESPONSE --> altitude = average velocity average velocities are found by the area under the velocity graph vs. clock time having slopes.
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14:08:16 02-05-2008 14:08:16 ** If you multiply the average altitude by the width (finding the area) of the trapezoid you are multiplying the average velocity by the time interval. This gives you the displacement during the time interval. The rise of the triangle represents the change in velocity and the run represents the time interval, so slope = rise / run represents change in velocity / time interval = acceleration. **
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assignment #001 ֊DȘo Physics I Class Notes 02-05-2008 Ǹϯ|ǹPx assignment #001 ֊DȘo Physics I Class Notes 02-05-2008
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14:26:03 What do we mean by velocity?
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RESPONSE --> velocity is how fast an object is moving and its direction
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14:26:31 02-05-2008 14:26:31 ** STUDENT RESPONSE:Velocity is the speed and direction an object is moving. INSTRUCTOR COMMENT: Good. More succinctly and precisely velocity is the rate at which position is changing. obtained by dividing change in position by change in clock time **
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NOTES -------> ** STUDENT RESPONSE:Velocity is the speed and direction an object is moving. INSTRUCTOR COMMENT: Good. More succinctly and precisely velocity is the rate at which position is changing. obtained by dividing change in position by change in clock time **
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14:28:05 How can we determine the velocity of a ball rolling down an incline?
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RESPONSE --> by determining how long it takes the ball to roll past a certain point
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14:28:16 02-05-2008 14:28:16 ** We divide displacement by the time interval to get average velocity. If the ball happens to be starting from rest and acceleration is uniform we can also conclude that the final velocity attained on the incline is double the average velocity. **
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NOTES -------> ** We divide displacement by the time interval to get average velocity. If the ball happens to be starting from rest and acceleration is uniform we can also conclude that the final velocity attained on the incline is double the average velocity. **
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14:31:57 We anticipate from our experience that a ball traveling down a greater incline, starting from rest, will experience a greater change in velocity. How can we determine whether the velocity actually changes, and whether the velocity increases in the manner that we expect?
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RESPONSE --> well we could graph the speed and the time and see what our curve looks like
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14:32:08 02-05-2008 14:32:08 ** We divide displacement by the time interval to get average velocity. We time the ball down one incline, then down the other and determine average velocity for each. We then infer that since both balls started from rest, the greater average velocity implies a greater change in velocity. **
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NOTES -------> ** We divide displacement by the time interval to get average velocity. We time the ball down one incline, then down the other and determine average velocity for each. We then infer that since both balls started from rest, the greater average velocity implies a greater change in velocity. **
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14:36:46 How could we determine the velocity of the ball at a specific point? The specific points are measured for distance and the ball is timed when it reaches these specific points. The distance is then divided by the time.
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RESPONSE --> divide distance by time
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14:44:51 How do we determine the rate at which the velocity changes? How can we understand the concept of the rate at which velocity changes?
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RESPONSE --> the average rate at which velocity changes is the change in velocity / the change in time. just like in your car and the speedometer needle
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14:45:10 02-05-2008 14:45:10 ** We find the change in velocity then divide by the change in the clock time. Any rate consists of the change in one quantity divided by the change in another. **
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NOTES -------> ** We find the change in velocity then divide by the change in the clock time. Any rate consists of the change in one quantity divided by the change in another. **
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assignment #001 ֊DȘo Physics I Vid Clips 02-05-2008
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22:49:08 Physics video clip 01: A ball rolls down a straight inclined ramp. It is the velocity the ball constant? Is the velocity increasing? Is the velocity decreasing?
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RESPONSE --> velocity is increasing
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22:49:24 ** It appears obvious, from common experience and from direct observation, that the velocity of the ball increases. A graph of position vs. clock time would be increasing, indicating that the ball is moving forward. Since the velocity increases the position increases at an increasing rate, so the graph increases at an increasing rate. **
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RESPONSE -->
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22:49:57 If the ball had a speedometer we could tell. What could we measure to determine whether the velocity of the ball is increase or decreasing?
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RESPONSE --> time it takes to pass a set distance
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22:59:18 02-05-2008 22:59:18 ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity. INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **
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NOTES -------> ** STUDENT RESPONSE: By measuring distance and time we could calculate velocity. INSTRUCTOR COMMENTS: The ball could be speeding up or slowing down--all you could get from the calculation you suggest is the average velocity. You could measure the time to travel the first half and the time to travel the second half of the ramp; if the latter is less then we would tend to confirm increasing velocity (though those are still average velocities and we wouldn't get certain proof that the velocity was always increasing). You would need at least two velocities to tell whether velocity is increasing or decreasing. So you would need two sets of distance and time measurements. **
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23:08:07 What is the shape of the velocity vs. clock time graph for the motion of the ball?
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RESPONSE --> a line sloping upward
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23:08:57 ** While on the curved end the ball will be speeding up, and the graph will therefore rise. By the time the ball gets to the level part the velocity will no longer be increasing and the graph will level off; because of friction the graph will actually decrease a bit, along a straight line. As long as the ball is on the ramp the graph will continue on this line until it reaches zero, indicating that the ball eventually stops. In the ideal frictionless situation on an infinite ramp the line just remains level forever. **
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RESPONSE --> Enter, as appropriate, an answer to the question, a critique of your answer in response to a given answer, your insights regarding the situation at this point, notes to yourself, or just an OK. k Always critique your solutions by describing any insights you had or errors you makde, and by explaining how you can make use of the insight or how you now know how to avoid certain errors. Also pose for the instructor any question or questions that you have related to the problem or series of problems.
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23:09:40 For the ball on the straight incline, we would certainly agree that the ball's velocity is increasing. Is the velocity increasing at a constant, an increasing, or a decreasing rate? What does the graph of velocity vs. clock time look like?
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RESPONSE --> increasing straight line
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23:09:58 ** It turns out that on a straight incline the velocity increases at a constant rate, so the graph is a straight line which increases from left to right. Note for future reference that a ball on a constant incline will tend to have a straight-line v vs. t graph; if the ball was on a curved ramp its velocity vs. clock time graph would not be straight, but would deviate from straightness depending on the nature of the curvature (e.g., slope decreasing at increasing rate implies v vs. t graph increasing at increasing rate).**
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RESPONSE --> half right \
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_Ca{{b䜣ؾׂs assignment #002 002. `Query 2 Physics I 02-05-2008
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23:26:33 ** Questions about velocity, average velocity, acceleration, etc. are very confusing because so many of the concepts have similar definitions. People have trouble distinguishing things like average velocity, which for uniform acceleration can be obtained in a process that adds two velocities, from average acceleration, which involves subtracting two velocities; one of these processes involves dividing by 2 and the other dividing by the time interval `dt.
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RESPONSE --> k.
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23:27:02 It is essential to keep the definitions and the meanings of the terms very clear and to work everything from definitions. It is equally important to have a good common-sense understanding of every definition so you can develop the intuition to make sense of everything you do.
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RESPONSE --> k
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23:27:15 That inevitably takes people a little time. But in the process you develop the habits you will need to succeed in the course. **
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RESPONSE --> good
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23:30:47 How is acceleration an example of a rate of change?
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RESPONSE --> because acceleration is motion advancing. It is changing position confidence assessment: 0
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23:32:28 02-05-2008 23:32:28 ** Velocity is the rate of change of position. Acceleration is rate of change of velocity--change in velocity divided by the time period. To find the acceleration from a v vs. t graph you take the rise, which represents the change in the average velocity, and divide by the run, which represents the change in time. The average rate of change of velocity with respect to clock time is the same as the acceleration **
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NOTES -------> ** Velocity is the rate of change of position. Acceleration is rate of change of velocity--change in velocity divided by the time period. To find the acceleration from a v vs. t graph you take the rise, which represents the change in the average velocity, and divide by the run, which represents the change in time. The average rate of change of velocity with respect to clock time is the same as the acceleration **
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23:34:36 If you know average acceleration and time interval what can you find?
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RESPONSE --> I can find the change in Velocity confidence assessment: 2
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23:35:36 02-05-2008 23:35:36 ** Accel = change in vel / change in clock time, so if you know accel and time interval (i.e., change in clock time) you can find change in vel = accel * change in clock time. In this case you don't know anything about how fast the object is traveling. You can only find the change in its velocity. COMMON ERROR (and response): Average acceleration is the average velocity divided by the time (for the change in the average velocity)so you would be able to find the average velocity by multiplying the average acceleration by the change in time. INSTRUCTOR RESPONSE: Acceleration is rate of change of velocity--change in velocity divided by the change in clock time. It is not average velocity / change in clock time. COUNTEREXAMPLE TO COMMON ERROR: Moving at a constant 60 mph for 3 hours, there is no change in velocity so acceleration = rate of change of velocity is zero. However average velocity / change in clock time = 60 mph / (3 hr) = 20 mile / hr^2, which is not zero. This shows that acceleration is not ave vel / change in clock time. COMMON ERROR and response: You can find displacement INSTRUCTOR RESPONSE: From average velocity and time interval you can find displacement. However from average acceleration and time interval you can find only change in velocity. Acceleration is the rate at which velocity changes so average acceleration is change in velocity/change in clock time. From this it follows that change in velocity = acceleration*change in clock time. **
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NOTES -------> ** Accel = change in vel / change in clock time, so if you know accel and time interval (i.e., change in clock time) you can find change in vel = accel * change in clock time. In this case you don't know anything about how fast the object is traveling. You can only find the change in its velocity. COMMON ERROR (and response): Average acceleration is the average velocity divided by the time (for the change in the average velocity)so you would be able to find the average velocity by multiplying the average acceleration by the change in time. INSTRUCTOR RESPONSE: Acceleration is rate of change of velocity--change in velocity divided by the change in clock time. It is not average velocity / change in clock time. COUNTEREXAMPLE TO COMMON ERROR: Moving at a constant 60 mph for 3 hours, there is no change in velocity so acceleration = rate of change of velocity is zero. However average velocity / change in clock time = 60 mph / (3 hr) = 20 mile / hr^2, which is not zero. This shows that acceleration is not ave vel / change in clock time. COMMON ERROR and response: You can find displacement INSTRUCTOR RESPONSE: From average velocity and time interval you can find displacement. However from average acceleration and time interval you can find only change in velocity. Acceleration is the rate at which velocity changes so average acceleration is change in velocity/change in clock time. From this it follows that change in velocity = acceleration*change in clock time. **
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23:37:16 Can you find velocity from average acceleration and time interval?
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RESPONSE --> yes confidence assessment: 0
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23:37:51 Can you find change in velocity from average acceleration and time interval?
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RESPONSE --> so is that a no confidence assessment: 0
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23:38:49 02-05-2008 23:38:49 **Good student response: Yes, the answer that I provided previously is wrong, I didn't consider the 'change in velocity' I only considered the velocity as being the same as the change in velocity and that was not correct. Change in velocity is average accel * `dt. CALCULUS-RELATED ANSWER WITH INSTRUCTOR NOTE(relevant mostly to University Physics students) Yes, you take the integral with respect to time INSTRUCTOR NOTE: That's essentially what you're doing if you multiply average acceleration by time interval. In calculus terms the reason you can't get actual velocity from acceleration information alone is that when you integrate acceleration you get an arbitrary integration constant. You don't have any information in those questions to evaluate c. **
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NOTES -------> i was wrong **Good student response: Yes, the answer that I provided previously is wrong, I didn't consider the 'change in velocity' I only considered the velocity as being the same as the change in velocity and that was not correct. Change in velocity is average accel * `dt. CALCULUS-RELATED ANSWER WITH INSTRUCTOR NOTE(relevant mostly to University Physics students) Yes, you take the integral with respect to time INSTRUCTOR NOTE: That's essentially what you're doing if you multiply average acceleration by time interval. In calculus terms the reason you can't get actual velocity from acceleration information alone is that when you integrate acceleration you get an arbitrary integration constant. You don't have any information in those questions to evaluate c. **
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23:42:29 Can you find average velocity from average acceleration and time interval?
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RESPONSE --> not sure because you find the average velocity is displacement / elapsed time and average acceration is change in velocity / elapsed time so if you know the ave acceraltion and the time we can solve for the change of velocity but is the change in the velocity the same as teh average velocity? confidence assessment: 0
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23:43:01 ** CORRECT STATEMENT BUT NOT AN ANSWER TO THIS QUESTION: The average acceleration would be multiplied by the time interval to find the change in the velocity INSTRUCTOR RESPONSE: Your statement is correct, but as you say you can find change in vel, which is not the same thing as ave vel. You cannot find ave vel. from just accel and time interval. There is for example nothing in accel and time interval that tells you how fast the object was going initially. The same acceleration and time interval could apply as well to an object starting from rest as to an object starting at 100 m/s; the average velocity would not be the same in both cases. So accel and time interval cannot determine average velocity. CALCULUS-RELATED ERRONEOUS ANSWER AND INSTRUCTOR CLAIRIFICATION(relevant mostly to University Physics students: Yes, you take the integral and the limits of integration at the time intervals CLARIFICATION BY INSTRUCTOR: A definite integral of acceleration with respect to t gives you only the change in v, not v itself. You need an initial condition to evaluate the integration constant in the indefinite integral. To find the average velocity you would have to integrate velocity (definite integral over the time interval) and divide by the time interval. **
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RESPONSE --> ** CORRECT STATEMENT BUT NOT AN ANSWER TO THIS QUESTION: The average acceleration would be multiplied by the time interval to find the change in the velocity INSTRUCTOR RESPONSE: Your statement is correct, but as you say you can find change in vel, which is not the same thing as ave vel. You cannot find ave vel. from just accel and time interval. There is for example nothing in accel and time interval that tells you how fast the object was going initially. The same acceleration and time interval could apply as well to an object starting from rest as to an object starting at 100 m/s; the average velocity would not be the same in both cases. So accel and time interval cannot determine average velocity. CALCULUS-RELATED ERRONEOUS ANSWER AND INSTRUCTOR CLAIRIFICATION(relevant mostly to University Physics students: Yes, you take the integral and the limits of integration at the time intervals CLARIFICATION BY INSTRUCTOR: A definite integral of acceleration with respect to t gives you only the change in v, not v itself. You need an initial condition to evaluate the integration constant in the indefinite integral. To find the average velocity you would have to integrate velocity (definite integral over the time interval) and divide by the time interval. ** self critique assessment: 2
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23:44:18 You can find only change in velocity from average acceleration and time interval. To find actual velocity you have to know at what velocity you started. Why can't you find average velocity from acceleration and time interval?
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RESPONSE --> because acceleration and time elapsed only show you the change in velocity but it does not tell you where you started or ended confidence assessment: 1
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23:44:33 02-05-2008 23:44:33 ** Average velocity is change in position/change in clock time. Average velocity has no direct relationship with acceleration. CALCULUS-RELATED ANSWER you dont know the inital velocity or the final velocity INSTRUCTOR COMMENT: . . . i.e., you can't evaluate the integration constant. **
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NOTES -------> ** Average velocity is change in position/change in clock time. Average velocity has no direct relationship with acceleration. CALCULUS-RELATED ANSWER you dont know the inital velocity or the final velocity INSTRUCTOR COMMENT: . . . i.e., you can't evaluate the integration constant. **
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23:47:35 General College Physics only: Problem #10 Summarize your solution Problem 1.10 (approx. uncertainty in area of circle given radius 2.8 * 10^4 cm).
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RESPONSE --> A=pir^2 A=pi(38000)^2 A=(14.44x10^8 A=45.3416x10^8 gives a 1% uncertainty honestly I have no idea how to solve foro the uncertainy confidence assessment: 0
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23:48:17 ** 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. **
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RESPONSE --> I still dont get teh concept of the uncertainty self critique assessment: 0
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23:49:50 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I forgot in doing the rest of the assignments that in the beginning in my reading the text and working the text problems and doing the internet problems that I still have no clue about the uncertainty i know how to use the equation if the uncertainty is given but none on how to find the uncertainty confidence assessment: 0
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