#$&* course PHY 201 2/19 about 10 am 003. `Query 3
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Given Solution: The coordinates a point on the graph include a position and a clock time, which tells you where the object whose motion is represented by the graph is at a given instant. If you have two points on the graph, you know the position and clock time at two instants. Given two points on a graph you can find the rise between the points and the run. On a graph of position vs. clock time, the position is on the 'vertical' axis and the clock time on the 'horizontal' axis. • The rise between two points represents the change in the 'vertical' coordinate, so in this case the rise represents the change in position. • The run between two points represents the change in the 'horizontal' coordinate, so in this case the run represents the change in clock time. The slope between two points of a graph is the 'rise' from one point to the other, divided by the 'run' between the same two points. • The slope of a position vs. clock time graph therefore represents rise / run = (change in position) / (change in clock time). • By the definition of average velocity as the average rate of change of position with respect to clock time, we see that average velocity is vAve = (change in position) / (change in clock time). • Thus the slope of the position vs. clock time graph represents the average velocity for the interval between the two graph points. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Two significant figures. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What are some possible units for position? What are some possible units for clock time? What therefore are some possible units for rate of change of position with respect to clock time? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: m, cm, km s, min, hr m/s, cm/min, km/hr confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The earths diameter is about 10,000,000m and the greatest depth ocean is about 10,000m. 10,000m/10,000,000m = 1/1000 Mt. Everst is listed in the text at 29,028 ft (actually now measure at 29,035 ft.). 29,028 ft * (1km/3281 ft.) = 29028/3281 km * (1000m/1km) = 29028000/3281 m (29028000/3281)m / 10,000,000 m = 7257/8202500 or about 9/10000 On this scale, it would be about .0009 m or 0.9 mm. I am not sure what you are asking “How might you construct a ridge of this height?” confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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):OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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: 534000 micro meters * 10^-6 = 0.534 m 142.5 cm * 10^-2 = 1.425 m 1.80 m + 1.425 m + 0.534 m = 3.759 m = 3.76 m to two significant figures to the right of the decimal point for a total of three significant figures. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ********************************************* Question: Openstax: A generation is about one-third of a lifetime. Approximately howmany generations have passed since the year 0 AD? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: A lifetime is about 70 years and 1/3 of that is 23.3 years/generation. Since 0 AD, 2013 years have passed. 2013 years * (1 generation / 23.3 years) = 86.4 generations = 86 generations confidence rating #$&*: 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:OK #$&* ********************************************* Question: Openstax: How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10^(-22) s .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a life span is 70 years: 70 years * (365 days / year) * (24 hours / day) * (60 min./hour) * (60 sec./min) = 2207520000 sec in a life span or 2.21 x 10^9 seconds. (2.21 x 10^9)s / (10^-22)s = 2.21 x 10 ^31 times longer confidence rating #$&*: 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):I got the total of seconds for the 70 year span to be much larger than yours. I think your total is for the number of seconds in one year but I thought the question asked for the entire life span. ------------------------------------------------ Self-critique Rating:3 #$&* Openstax: Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10−27 kg and the mass of a bacterium is on the order of 10−15 kg. ) Average mass of bacterium is 12.5 kg. We could find the number of hydrogen atoms in the mass of the bacterium by (10^(-27)kg) / 12.5 kg which equals 8*10^(-29). This is the number of hydrogen atoms in this amount of mass. Since the mass of an atom in the bacterium is ten times the mass of a hydrogen atom, we multiply this by 10 and get 8*10^(-28) atoms per bacterium. ********************************************* Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ 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? The time at the first point would be 0 with the given assumption. Take ‘dt for the first book and find the clock time at the end of the first book or beginning of second book. Add both ‘dts together to get the time at the bottom of the second book. To make a graph, we have the first point at (0,0) since clock time starts at the top of the first book when the ball begins to move. Plot the point ‘dt,’ds for the first book. I will say that point is (‘dt_1, ‘ds_1), with 1 representing the first or top book. The final point would be the sum of ‘ds for each book and the sum of ‘dt for each book. I will say that point is ((‘dt_1+’dt_2),(‘ds_1+’ds_2)). Three points, connect the dots. This will likely be a curve, not linear since there is acceleration and not constant velocity. For the velocity graph, we have to determine the velocity at these three points. Use v_ave = ‘ds/’dt to find these. The first point will still be at the origin (0,0). Second point (‘dt_1,(‘ds_1/’dt_1)). Third point will be (‘dt_2,((‘ds_1+’ds_2)/(‘dt_1+’dt_2))). This should be linear since acceleration is constant. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " 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? The time at the first point would be 0 with the given assumption. Take ‘dt for the first book and find the clock time at the end of the first book or beginning of second book. Add both ‘dts together to get the time at the bottom of the second book. To make a graph, we have the first point at (0,0) since clock time starts at the top of the first book when the ball begins to move. Plot the point ‘dt,’ds for the first book. I will say that point is (‘dt_1, ‘ds_1), with 1 representing the first or top book. The final point would be the sum of ‘ds for each book and the sum of ‘dt for each book. I will say that point is ((‘dt_1+’dt_2),(‘ds_1+’ds_2)). Three points, connect the dots. This will likely be a curve, not linear since there is acceleration and not constant velocity. For the velocity graph, we have to determine the velocity at these three points. Use v_ave = ‘ds/’dt to find these. The first point will still be at the origin (0,0). Second point (‘dt_1,(‘ds_1/’dt_1)). Third point will be (‘dt_2,((‘ds_1+’ds_2)/(‘dt_1+’dt_2))). This should be linear since acceleration is constant. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!