#$&* course phy121 sept 20, 2:25pm Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
.............................................
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: Since the units of 69 and 61 do not have a decimal point at the end, we cannot be sure if the last digit is certain, so I would assume that it would be to one significant digit. Therefore, 69-61=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: Some possible units for position are mm, cm, m, km. Some possible units for clock time are milliseconds, seconds, minutes, hours. Some possible units for rate of change of position respect to clock time are mm/s, cm/s, m/s, km/s, km/h. Those are more common than some of the other units we could use. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: What fraction of the Earth's diameter is the greatest ocean depth? What fraction of the Earth's diameter is the greatest mountain height (relative to sea level)? On a large globe 1 meter in diameter, how high would the mountain be, on the scale of the globe? How might you construct a ridge of this height? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I am not sure if we can use google to find these values, but I would assume that the diameter of the earth would be in units of kilometers and the ocean depths to be in meters. Since there are 1000m in 1km, I would divide the ocean depth by 1000 to have the same units and find the fraction to correspond earth’s diameter to ocean depth, where I would write it as depth/diameter. I would do the same thing for the earths diameter to the height of the greatest mountain. Convert both to kilometers and find the fraction. I would write it as height/diameter. confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Now that I know the height of the mountain and the diameter of the earth, my solution in numbers would look like: 10,000m= 10km 10km/10,000km 1/1000km. I was not sure that we could look it up, but by checking with my description without values, I can see that my description was correct. ------------------------------------------------ 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: 1micron= 0.001mm= 0.0001cm= 0.000001m (5.34*10^5 micro m) * (0.000001m/1micro m)= 0.534m 142.5cm/100= 1.425m 1.80m + 1.425m + 0.534m= 3.759m Since 1.80m has 3 significant digits, the final answer would be 3.76m confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** 1.80 m has three significant figures (leading zeros don't count, neither to trailing zeros unless there is a decimal point; however zeros which are listed after the decimal point are significant; that's the only way we have of distinguishing, say, 1.80 meter (read to the nearest .01 m, i.e., nearest cm) and 1.000 meter (read to the nearest millimeter). Therefore no measurement smaller than .01 m can be distinguished. 142.5 cm is 1.425 m, good to within .00001 m. 5.34 * `micro m means 5.34 * 10^-6 m, so 5.34 * 10^5 micro m means (5.34 * 10^5) * 10^-6 meters = 5.34 + 10^-1 meter, or .534 meter, accurate to within .001 m. Then theses are added you get 3.759 m; however the 1.80 m is only good to within .01 m so the result is 3.76 m. ** Self-critique (if necessary): OK ********************************************* Question: Openstax: A generation is about one-third of a lifetime. Approximately howmany generations have passed since the year 0 AD? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **Principle of physics student** confidence rating #$&*:: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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: #$&* ********************************************* Question: Openstax: How many times longer than the mean life of an extremely unstable atomic nucleus is the lifetime of a human? (Hint: The lifetime of an unstable atomic nucleus is on the order of 10^(-22) s .) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **Principle of physics student** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): Self-critique Rating: #$&* ********************************************* Question: Openstax: Calculate the approximate number of atoms in a bacterium. Assume that the average mass of an atom in the bacterium is ten times the mass of a hydrogen atom. (Hint: The mass of a hydrogen atom is on the order of 10−27 kg and the mass of a bacterium is on the order of 10−15 kg. ) To understand what this means, you might think it through something like this: 10^-27, written 0.000 000 000 000 000 000 000 000 001, is smaller 10^-15 (which you should write out and think about). Ten times the mass of a hydrogen atom is 10 * 10^-27 = 10^-26. 1000 times 10^-26 is 10^-23; 1000 times 10^-23 is 10^-20; 1000 times 10^-20 is 10^-17, and you still have to multiply this by 100 to get 10^-15. So the number of atoms is about 1000 * 1000 * 1000 * 100 = 100 000 000 000 (that's 100 billion). Of course also want to calculate the result without thinking much about what it means. To do so you see how many of the smaller quantity it takes to make up the larger. In other words you'll divide the smaller quantity into the larger. The result is as follows: number of atoms in bacterium = mass of bacterium / mass of atom = 10^-15 kg / (10^-26 kg) = 10^-11 kg / kg = 10^-11. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **Principle of physics student** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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: If we know all the information, to determine clock time at each of the 3 points, we know that the ball started from rest at 0s. To find the time at the end of the first book, which is the beginning of the second book, we know how fast the ball is moving at the end of the book, so using the formula vAve= ‘ds/’dt, we would re arrange it to ‘dt= ‘ds/vAve, because we know how far the ball travelled along each book and we know the speed at the end of each book. This would find the time it took to travel from rest to the end of the first book. For the second book, we would do the same formula, but we have to find the subtract the final speed by the initial speed of the second book to find the change of rate of speed, then input into the formula. You would sketch a graph of position vs clock time with the position on the Y-axis and the clock time on the X-axis. As the time progresses, the position of the ball would be going further from the initial position. We would see an increasing at an increasing rate graph. If sketching a graph of ball speed vs clock time, with the speed on the Y-axis and the clock time on the X-axis. This graph would be showing that as the time increases, the speed of the ball also increases because the ball is going down a ramp and the further it goes down the ramp the quicker the ball will go. Personally, I think the graphs would look the same. confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-critique Rating: 2 I wouldn’t know how they would differ besides the units used on the graph. #$&* ********************************************* 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: **Principle of physics student** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `a** THE FOLLOWING CORRECT SOLUTION WAS GIVEN BY A STUDENT: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.9 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #$&* " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!