course Phy 231 6/8 10 003. `Query 3 Question: What do the coordinates of two points on a graph of position vs. clock time tell you about the motion of the object? What can you reason out once you have these coordinates?
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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): change in position is the change in the y values; change in time is the change in the x values Self-critique rating #$&* 3 Question: Pendulums of lengths 20 cm and 25 cm are counted for one minute. The counts are respectively 69 and 61. To how many significant figures do we know the difference between these counts? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: You would know the difference to 1 significant figure because with addition and subtraction you use the number of decimal places least uncertainty: 0decimal places. The difference is 8 cycles with 1 significant figure. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* 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 units for position are meters, centimeters, kilometers. Some units for time are seconds are hours. Some units for rates of change are meters/second, centimeters/second, kilometers/hour, centimeters/hour. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ********************************************* Question: `qQuery Principles of Physics and General College Physics: Summarize your solution to Problem 1.19 (1.80 m + 142.5 cm + 5.34 * 10^5 `micro m to appropriate # of significant figures) Your solution: Convert 142.5cm to meters (divide by 100) = 1.425 m Convert 5.34*10^5 microns to meters (multiply by 10^-6 because a micron is 10^-6 of a meter) = .534 m You are going to have 2 decimal places because 1.80m has the least uncertainty. 1.80m+1.425m+.534m = 3.759m = 3.76 meters confidence rating #$&*: 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: 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). Your solution: First you find the 3 components: x: 3.1km(cos45) = 3.1*1/sqrt2 = 3.1sqrt2 /2 = 2.19 Magnitude = sqrt(0^2 +4^2) = sqrt(16) = 4 Sqrt(4^2 + 2.19^2) = sqrt(20.7961) = 4.56 4+4.56 = 8.56 km Direction = arctan(2.19 / 4) = 15.323 90-15.323 = 74.7 degrees confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The components of vectors A (2.6km in the y direction) and B (4.0km in the x direction) are known. We find the components of vector C(of length 3.1km) by using the sin and cos functions. Cx was 3.1 km * cos(45 deg) = 2.19. Adding the x component of the second vector, 4.0, we get 6.19km. Cy was 2.19 and i added the 2.6 km y displacement of the first vector to get 4.79. So Rx = 6.19 km and Ry = 4.79 km. To get vector R, i used the pythagorean theorem to get the magnitude of vector R, which was sqrt( (6.29 km)^2 + (4.79 km)^2 ) = 7.9 km. The angle is theta = arctan(Ry / Rx) = arctan(4.79 / 6.19) = 37.7 degrees. ** Self-critique (if necessary): I got the components for the vector with the 45 degrees right. You first have to make 2 vectors: I know that the component of one was 2.16. You add the x component to that (4) and get 6.19km. Then you add the y displacement to the vector (2.6) to get 4.79km. Now that you have the components, you can find the magnitude and direction. I got the magnitude and direction formulas correct. Magnitude: sqrt(2.16^2 + 6.19^2) = 7.8 km Direction = theta= arctan(4.79/6.19) = 37.7 degrees Self-critique rating #$&* 3 ********************************************* Question: A ball rolls from rest down a book, off that book and 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. How would you use your information to determine the clock time at each of the three points, 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? You know that the clock time at the top of the first book is 0 because the clock started there. To get the clock time at the end of the first book, you just use the time it takes for the ball to travel down the book, because it started at 0. To get the time at the end of the second book, you add the time it takes to go down the second book and you add it to the first time. To get the graph of the balls position, you plot the point 0,0 for the initial distance, because it is at time 0 and distance 0. Then, you plot the distance at the end of the first book and the time it took to get there (the time it takes for the ball to travel down the book). The final point would be the distance down the first book plus the distance down the second book and the two times added together. The tree points will give you a rough sketch of the graph of position versus time. This graph should be increasing at an increasing rate (concave up) because the velocity is increasing as the ball goes down the book. The graph of the velocity versus time is also going to be increasing but with a constant slope because the velocity increases at a steady rate. You have to plot the speed of the ball at the different clock times. (0,0) for the first one because the initial velocity was 0. Then, you can plot the other velocities from the other 2 points with the clock times you found before; this will make a straight line. Another thing you could do is find the equation of the position vs. time graph, take the derivative and then graph that equation, which would be the velocity graph. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ "