#$&* course Phy 231 2/18 4 003. `Query 3
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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): Rise between two points represents the change in position of an object. ------------------------------------------------ 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: 'dcounts = 69 - 61 = 8.0 counts The difference would have two significant figures since both counts are accurate to two significant figures.
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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): I went into the textbook and internet to pull out my figures. I should have used estimation to make the numbers easier. Diameter of the Earth being 12742 km is about 13,000 km. and the mountain about 10,000 meters, my ration would be 1 / 1000 If I estimated for the trench, which is about 11 000 meters deep, I can say that the ratio is 11 000 / 13 000 000 = 11 / 13000 or 1/10000 approximately. ------------------------------------------------ Self-critique Rating: ********************************************* 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 = 3.76 m confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: Openstax: A generation is about one-third of a lifetime. Approximately howmany generations have passed since the year 0 AD? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 generartaion = 1/3 * lifetime = (1/3) * 70 years = 23.333333 years Genereations since 0 AD = 2014 / 23.333333 = approximately 86.3 generations have passed since 0 AD. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A lifetime is about 70 years. 1/3 of that is about 23 years. About 2000 years have passed since 0 AD, so there have been about 2000 years / (23 years / generation) = 85 generations in that time &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: Openstax: How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10^(-22) s .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1 life time = 70 years *(365 days / 1 year) *( 24 hours / 1 day) * (3600 secs / 1 hour) 1 life time = 2.21 * 10^9 seconds = approx. 2.21 billion seconds # times longer = human life / nucleus life = (2.21 * 10^9 sec) / (1 * 10^-22 sec) = (2.21 * 10^31) times longer confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Assuming a 70-year human lifetime: A years is 365 days * 24 hours / day * 60 minutes / hour * 60 seconds / minute = 3 000 000 seconds. The number of seconds could be calculated to a greater number of significant figures, but this would be pointless since the 10^(-22) second is only an order-of-magnitude calculation, which could easily be off by a factor of 2 or 3. Dividing 3 000 000 seconds by the 10^-22 second lifetime of the nucleus we get 1 human lifetime = 3 000 000 seconds / (10^-22 seconds / nuclear lifetime) = 3 * 10^28 nuclear lifetimes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* Openstax: Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10−27 kg and the mass of a bacterium is on the order of 10−15 kg. ) ********************************************* 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: Magnitude R = { (4 + 3.1 cos 45 km)^2 i + (3.1 sin 45 km )^2 j }^.5 = 6.6 km Angle from origin = tan^-1 (2.192 / 6.192) = 19.5 degrees north of east. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ok ------------------------------------------------ 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: How far each book = 30 cm Time to reach end first book = 5 seconds = t1 Time to reach end second book = 5 seconds + 2.5 seconds = 7.5 seconds= t2 V1 = velocity at end of first book = 30cm / 5 seconds = 6 cm/s v2 = velocity at end second book = 30 cm / 2.5 seconds = 12 cm/s clock times, t0 = 0 seconds, t1 = 5 seconds, t2 = 7.5 seconds a graph of position vs time, I would have time on my x axis, and position on my y axis. My first point would be (0,0), and it would increase at a steady rate to (5,30). So it would take 5 seconds to reach 30 cm. After this point the slope increases, so it covers the same 30 cm distance in a shorter amount of time, in this case 2.5 seconds. so the next point would be at (7.5, 60) The rise would be the same but the run would be 1/2. A velocity vs time graph would start at (0,0) increase at steady rate to (6,5), with x =cm/s, and y = time in seconds, and then increase and move to the point (12,7.5) with an increasing slope. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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: How far each book = 30 cm Time to reach end first book = 5 seconds = t1 Time to reach end second book = 5 seconds + 2.5 seconds = 7.5 seconds= t2 V1 = velocity at end of first book = 30cm / 5 seconds = 6 cm/s v2 = velocity at end second book = 30 cm / 2.5 seconds = 12 cm/s clock times, t0 = 0 seconds, t1 = 5 seconds, t2 = 7.5 seconds a graph of position vs time, I would have time on my x axis, and position on my y axis. My first point would be (0,0), and it would increase at a steady rate to (5,30). So it would take 5 seconds to reach 30 cm. After this point the slope increases, so it covers the same 30 cm distance in a shorter amount of time, in this case 2.5 seconds. so the next point would be at (7.5, 60) The rise would be the same but the run would be 1/2. A velocity vs time graph would start at (0,0) increase at steady rate to (6,5), with x =cm/s, and y = time in seconds, and then increase and move to the point (12,7.5) with an increasing slope. confidence rating #$&*: 2" ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!