#$&* course Phy 231 003. `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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: We know the difference to one significant figure, because we are counting in whole-number cycles and the difference is reported in a single-digit whole number, 8. 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: Position should be measured in units of length, such as inches, centimeters, meters, or miles. Clock time should be measured in units of time, such as seconds, minutes, or hours. So, units for rate of change of position with respect to clock time, or (change in position)/(change in clock time), could include inches/second, meters/minute, or miles/hour. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The deepest part of the ocean is just over 10km deep. If the Earth's diameter is estimated to be about 13,000km, then the deepest ocean would be a little more than 1/1300th of the diameter of the earth. The tallest mountain is about 9km, or a little less than 1/1300th of the diameter of the earth. A 1-km globe would be 1/13,000th the size of the earth, so 1-meter globe would have a diameter 1/13,000,000th the size of the actual earth. 10,000m/13,000,000= 1/1300m, or 1/13 of a millimeter. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ???So you're rounding that much-- estimating 10,000 km instead of 13,000km? It's hard for me to imagine that the mountain would be a fraction of a mm high on the 1-m globe. Are the topographic globes we're used to THAT exaggerated? Or have I made an error here????
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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): ********************************************* Question: Openstax: A generation is about one-third of a lifetime. Approximately howmany generations have passed since the year 0 AD? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If a lifetime can be estimated at 75 years, then a generation is about 25 years. In just over 2,000 years, that means we've had about 2,000/25 generations, or 8,000 generations. Probably more, if we look at the fact that a lifetime was much shorter in the past. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: A lifetime is about 70 years. 1/3 of that is about 23 years. About 2000 years have passed since 0 AD, so there have been about 2000 years / (23 years / generation) = 85 generations in that time &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* ********************************************* 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: Such a nucleus lives [10^(-22)]s. A person living 70 years gets (70 years)(365 days/year)(24 hours/day)(3600 sec/hour)= 2.2*10^9 seconds [2.2*10^9]/[10^(-22)]= 2.2*10^31 times longer confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Assuming a 70-year human lifetime: A years is 365 days * 24 hours / day * 60 minutes / hour * 60 seconds / minute = 3 000 000 seconds. The number of seconds could be calculated to a greater number of significant figures, but this would be pointless since the 10^(-22) second is only an order-of-magnitude calculation, which could easily be off by a factor of 2 or 3. Dividing 3 000 000 seconds by the 10^-22 second lifetime of the nucleus we get 1 human lifetime = 3 000 000 seconds/year * 70 years / (10^-22 seconds / nuclear lifetime) = 2 * 10^31 nuclear lifetimes. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Rounding gets us to 30,000,000 sec/year, right? Somehow I got to essentially the same answer, even though we're off by a magnitude of 10 at the seconds-per-year stage. ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: 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. ) To understand what this means, you might think it through something like this: 10^-27, written 0.000 000 000 000 000 000 000 000 001, is smaller 10^-15 (which you should write out and think about). Ten times the mass of a hydrogen atom is 10 * 10^-27 = 10^-26. 1000 times 10^-26 is 10^-23; 1000 times 10^-23 is 10^-20; 1000 times 10^-20 is 10^-17, and you still have to multiply this by 100 to get 10^-15. So the number of atoms is about 1000 * 1000 * 1000 * 100 = 100 000 000 000 (that's 100 billion). Of course also want to calculate the result without thinking much about what it means. To do so you see how many of the smaller quantity it takes to make up the larger. In other words you'll divide the smaller quantity into the larger. The result is as follows: number of atoms in bacterium = mass of bacterium / mass of atom = 10^-15 kg / (10^-26 kg) = 10^-11 kg / kg = 10^-11. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Total mass of the bacterium = 10^-15kg Total mass of one atom in the bacterium = (10^-27kg)(10) = (10^-26kg) Total mass of the bacterium = number of atoms * mass of one atom 10^-15kg = number of atoms * 10^-26kg (10^-15kg)/(10^-26kg)= number of atoms Whenever I deal with negative exponents, I try if possible to restructure and find the reciprocal, which allows me to do 1 over the positive version of the same exponent. Here, that gives me (10^26)/(10^15) = number of atoms 10^11 = number of atoms. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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: If the clock starts at the moment the ball is released from the top of the first book, then clock time at that initial position is 0. We would find the clock time at the transition between the two books by adding the time interval for the first book to 0. We'd find the clock time at the bottom of the second book by adding the two given time intervals together. To sketch a graph of position vs. clock time, we would plot the three points listed above-- plotted as coordinates of time on the x-axis and position on the y-axis in the form (time, position)-- and connect them with two line segments, the slope of each segment representing average velocity over the span of each book (the second will be steeper than the first, as the position changes more per second after the ball moves onto the second book). To sketch a graph of speed vs. clock time, we would plot the three points listed above (using 0 for the velocity of the first point and the two given velocities for the other two points, as well as the clock times we have determined). Again, we would see the graph increase, but then increase more rapidly after the second point. Overall, speed would increase less than position increased over the same time span. In fact, I believe that the graph of position vs. clock time would actually be a curve. Speed increases from 0 to some speed after 1 second, and then increase more with each successive second. So every second sees the ball able to move further in that second than it was able to in the previous second. So my understanding is actually that the graph of position vs. clock time is more challenging. Velocity is increasing at a uniform rate on each book, which means the lines are straight on each book, with one change in slope at the point the ball moves from one book to the other. But position, on the other hand, is increasing at an increasing rate. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: #$&* ********************************************* Question: For University Physics students: Summarize your solution to Problem 1.31 (10th edition 1.34) (4 km on line then 3.1 km after 45 deg turn by components, verify by scaled sketch). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We'll call the initial due-north segment Vector A, the due-east segment Vector B, and the 45-deg-angle segment Vector C. Asubx=0 Asuby=2.6km Bsubx=4.0km Bsuby=0 Csubx=(cos45)(3.1km)=2.19km Csuby=(sin45)(3.1km)=2.19km Total_x=0+4.0km+2.19km=6.19km Total_y=2.6km+0+2.19km=4.79km Total_magnitude=sqrt[(6.19)^2+(4.79)^2]=sqrt[61.26]=7.83km Magnitude is 7.83km tan(theta)=4.79km/6.19km tan(theta)=.77 theta=atan(.77) theta=37.6deg Direction is +37.6 degrees. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): I'm not sure how to address the second part of this question, concerning the vector-addition diagram. Is the idea to simply draw the three vectors close enough to scale that we can estimate the horizontal and vertical components of each leg of the journey and use the results to get a visual estimate of total horizontal and vertical displacement? Also, I must be making rounding errors, because I'm off by .1 here and there. Is there a rule of thumb for when to round? I generally try to carry through exact calculations until the very end whenever possible, and then round just the final step. ------------------------------------------------ Self-critique Rating: