#$&* course phy201 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: 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: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: 2 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: meters, miles, inches, feet. minutes, seconds, hours, days. meters/minute, miles/second, inches/hour, feet/day. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Marianna Trench: 10,994m. Earth Diameter: 12,742km. Mount Everest: 8,848m. Marianna Trench: 11000/12800000=.0008 or 1/1250 Mount Everest: 8000/12800000=.0006 or 3/5000 Trench on globe. 1*.0008= .0008m or .8mm deep Everest on globe. 1*.0006= .0006m or .6mm high. Paper is .15mm thick approx. Fold paper in half twice, viola. confidence rating #$&*: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:3 ********************************************* 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: 1,800,000micro m + 1,425,000micro m + 534,000micro m=3,759,000micro m 376cm or 3.76m confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ********************************************* Question: Openstax: A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: With average lifespan being approx. 75 years each generation would be 25 years. 2014/25=80.56=81 generations. confidence rating #$&*: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):I was unsure about the life expectancy, so i looked online. Online says 77, I rounded down to make the problem easier. ------------------------------------------------ Self-critique Rating:3 #$&* Question: Openstax: How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10^(-22) s .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used 75 years as the life expectancy on the previous question. I’ll use 70 as you did for the given answer. 70yrs*365days*24hrs*60min*60sec=2,207,520,000seconds 2,000,000,000/10^(-22)=2e31 confidence rating #$&*: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):ok, i used e instead of the *10^. ------------------------------------------------ Self-critique Rating:3 " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!