course phy201 ֒ٯӰFꈆassignment #021
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05:09:51 Explain how to obtain the final speed and direction of motion of a projectile which starts with known velocity in the horizontal direction and falls a known vertical distance, using the analysis of vertical and horizontal motion and vectors.
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RESPONSE --> Use the equation y=y0+v0t+.5at^2 where a=gravity and y=the known vertical distance to calculate the time the projectile takes to fall. Then, multiply this by g to get the final vertical velocity. The pythagorean theorem gives the 2D velocity based on v(x) and v(y). The final direction may be calculated as the arctan of (y/x).
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05:10:12 ** The horizontal velocity is unchanging so the horizontal component is always equal to the known initial horizontal velocity. The vertical velocity starts at 0, with acceleration thru a known distance at 9.8 m/s^2 downward. The final vertical velocity is easily found using the fourth equation of motion. We therefore know the x (horizontal) and y (vertical) components of the velocity. Using the Pythagorean Theorem and arctan (vy / vx) we find the speed and direction of the motion. **
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RESPONSE --> OK
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05:24:39 Give at least three examples of vector quantities for which we might wish to find the components from magnitude and direction. Explain the meaning of the magnitude and the direction of each, and explain the meaning of the vector components.
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RESPONSE --> A ship's sailing route may be broken down into north-south and east-west movement which would correspond to the vertical and horizontal components, respectively, oriented on the surface of the ocean. An helicopter's flight path may be broken down into up-down and forward-reverse movement which would correspond to the vertical and horizontal components, respectively oriented in two dimensions as an observer would see if there were no side to side flight. A whale's swim path may be broken down into up-down and forward-reverse movement which would correspond to the vertical and horizontal components, respectively, oriented in two dimensions as an observer would see if there were no side to side swimming.
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05:25:11 ** GOOD STUDENT RESPONSE: Examples might include: A force acting on an object causing it to move in an angular direction. A ball falling to the ground with a certain velocity and angle. A two car collision; velocity and momentum are both vector quantities and both important for analyzing the collision.. The magnitude and directiohn of the relsultant is the velocity and direction of travel. The vector components are the horizontal and vertical components that would produce the same effect as the resultant.
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RESPONSE --> OK
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oxjEۭ assignment #022 ֒ٯӰFꈆ Physics I 05-13-2007 ԋ쪒p|ġkI assignment #001 001. Only assignment: prelim asst
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04:37:00 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> yes
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04:37:37 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> I have been able to locate the SEND file.
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04:38:08 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> My info stays in the SEND file.
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04:38:24 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> previously done
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04:38:47 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> I should backup frequently
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04:39:19 Once more, locate the SEND file in your c:\vhmthphy folder, and open the file. Copy its contents to the clipboard (this is a common operation, but in case you don't know how, just use CTRL-A to highlight the contents of the file and CTRL-C to copy the contents to the clipboard). Then return to the form that instructed you to run this program, and paste the contents into the indicated box (just right-click in the box and select Paste). You may now click on the Quit button, or simply close the program.
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RESPONSE --> ԋ쪒p|ġkI assignment #001 001. Only assignment: prelim asst
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04:37:00 `q001. Part 1 includes six activities. If you have completed an activity, just enter the answer 'completed'. This question is appearing in the Question box. The box to the right is the Answer box, where you will type in your answers to the questions posed here. To use this program you read a question, then enter your answer in the Answer box and click on Enter Answer. In your answers give what is requested, but don't go into excruciating detail. Try to give just enough that the instructor can tell that you understand an item. After entering an answer click on Next Question/Answer above the Question box. Do you understand these instructions?
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RESPONSE --> yes
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04:37:37 This program has created the folder c:\vhmthphy on your hard drive. Browse to that folder and locate the file whose name begins with SEND. The name of this file will also include your name, as you gave it to the program, and the file will show as a Text file. Never tamper with a SEND file in any way. It contains internal codes as if these codes are tampered with you won't get credit for the assignment. However you are welcome to copy this file to another location and view it, make changes, etc. Just be sure that when requested to do so you send the instructor the original, tamper-free file. State in the Answer box whether or not you have been able to locate the SEND file. Don't send the SEND file yet. Note that more questions/instructions remain in the q_a_prelim.
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RESPONSE --> I have been able to locate the SEND file.
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04:38:08 `q002. Note that every time you click on Enter Answer the program writes your response to your SEND file. Even if the program disappears all the information you have entered with the Enter Answer button will remain in that file. This program never 'unwrites' anything. Even if this program crashes your information will still be there in the SEND file. Explain this in your own words.
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RESPONSE --> My info stays in the SEND file.
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04:38:24 Any time you do not receive a reply from the instructor by the end of the following day, you should resubmit your work using the Resubmit Form at http://www.vhcc.edu/dsmith/genInfo/. You have already seen that page, but take another look at that page and be sure you see the Submit Work form, the Resubmit Form and a number of other forms that will be explained later. Enter a sentence or two describing the related links you see at that location.
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RESPONSE --> previously done
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04:38:47 `q003. If you are working on a VHCC computer, it is probably set up in such a way as to return to its original configuration when it is rebooted. To avoid losing information it is suggested that you back up your work frequently, either by emailing yourself a copy or by using a key drive or other device. This is a good idea on any computer. Please indicate your understanding of this suggestion.
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RESPONSE --> I should backup frequently
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ّh}[J Student Name: assignment #024
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05:31:31 `q001. Note that this assignment contains 4 questions. . Note that this assignment contains 4 questions. When an object moves a constant speed around a circle a force is necessary to keep changing its direction of motion. This is because any change in the direction of motion entails a change in the velocity of the object. This is because velocity is a vector quantity, and if the direction of a vector changes, then the vector and hence the velocity has changed. The acceleration of an object moving with constant speed v around a circle of radius r has magnitude v^2 / r, and the acceleration is directed toward the center of the circle. This results from a force directed toward the center of the circle. Such a force is called a centripetal (meaning toward the center) force, and the acceleration is called a centripetal acceleration. If a 12 kg mass travels at three meters/second around a circle of radius five meters, what centripetal force is required?
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RESPONSE --> 21.6N b/c ((3^2)/5)*12=21.6
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05:31:54 The centripetal acceleration of the object is v^2 / r = (3 meters/second) ^ 2/(5 meters) = 1.8 meters/second ^ 2. The centripetal force, by Newton's Second Law, must therefore be Fcent = 12 kg * 1.8 meters/second ^ 2.
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RESPONSE --> OK
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05:33:22 `q002. How fast must a 50 g mass at the end of a string of length 70 cm be spun in a circular path in order to break the string, which has a breaking strength of 25 Newtons?
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RESPONSE --> 4.18 m/s b/c sq.rt.(.7*25)=4.18
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05:35:05 The centripetal acceleration as speed v will be v^2 / r, where r = 70 cm = .7 meters. The centripetal force will therefore be m v^2 / r, where m is the 50 g = .05 kg mass. If F stands for the 25 Newton breaking force, then we have m v^2 / r = F, which we solve for v to obtain v = `sqrt(F * r / m). Substituting the given values we obtain v = `sqrt( 25 N * .7 meters / (.05 kg) ) = `sqrt( 25 kg m/s^2 * .7 m / (.05 kg) ) = `sqrt(350 m^2 / s^2) = 18.7 m/s.
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RESPONSE --> I just forgot to include the mass in the calculation.
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05:41:34 `q003. What is the maximum number of times per second the mass in the preceding problem can travel around its circular path before the string breaks?
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RESPONSE --> 4.25 times b/c circumference=2*pi*.7=4.40m. 18.7/4.40=4.25
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05:41:38 The maximum possible speed of the mass was found in the preceding problem to be 18.7 meters/second. The path of the mass is a circle of radius 70 cm = .7 meters. The distance traveled along this path in a single revolution is 2 `pi r = 2 `pi * .7 meters = 4.4 meters, approximately. At 18.7 meters/second, the mass will travel around the circle 18.7/4.4 = 4.25 times every second.
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RESPONSE --> ok
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05:42:25 `q004. Explain in terms of basic intuition why a force is required to keep a mass traveling any circular path.
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RESPONSE --> Because the inertia of the mass would otherwise compel it to keep travelling a straight path.
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05:42:36 We simply can't change the direction of motion of a massive object without giving it some sort of a push. Without such a force an object in motion will remain in motion along a straight line and with no change in speed. If your car coasts in a circular path, friction between the tires and the road surface pushes the car toward the center of the circle, allowing it to maintain its circular path. If you try to go too fast, friction won't be strong enough to keep you in the circular path and you will skid out of the circle. In order to maintain a circular orbit around the Earth, a satellite requires the force of gravity to keep pulling it toward the center of the circle. The satellite must travel at a speed v such that v^2 / r is equal to the acceleration provided by Earth's gravitational field.
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RESPONSE --> ok
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̷ Student Name: assignment #025
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06:02:35 `q001. Note that this assignment contains 5 questions. . A pendulum consists of a 150 g mass suspended from a light string. Another light string is attached to the mass, which is then pulled back from its equilibrium position by that string until the first string makes an angle of 15 degrees with vertical. The second string remains horizontal. Let the x axis be horizontal and the y axis vertical. Assume that the mass is pulled in the positive x direction. If T stands for the tension in the pendulum string, then in terms of the variable T what are the x and y components of the tension?
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RESPONSE --> x=Tsin15deg y=Tcos15deg
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06:13:15 The pendulum string makes an angle of 15 degrees with vertical. Since we have assumed that the pendulum is pulled in the positive x direction, the direction of the tension in the string will be upward and to the left at an angle of 15 degrees with vertical. The tension force will therefore be directed at 90 degrees + 15 degrees = 105 degrees as measured counterclockwise from the positive x axis. The tension will therefore have x component T cos(105 degrees) and y component T sin(105 degrees).
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RESPONSE --> I see that I should have added 90 to 15 instead of using 15degress for my calculations. I also needed to reverse my orientation with sin and cos
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06:18:09 `q002. Continuing the preceding problem, we see that we have a vertical force of T sin(105 deg) from the tension. What other vertical force(s) act on the mass? What is the magnitude and direction of each of these forces? What therefore must be the magnitude of T sin(105 deg).
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RESPONSE --> Gravity also acts as a vertical force on the mass with a magnitude and direction exactly opposite to that of the vertical component of the tension being applied to the mass.
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06:26:01 The only other vertical force acting on the mass will be the gravitational force, which is .150 kg * 9.8 meters/second ^ 2 = 1.47 Newtons. The direction of this force is vertically downward. Since the mass is in equilibrium, i.e., not accelerating, the net force in the y direction must be zero. Thus T sin(105 deg) - 1.47 Newtons = 0 and T sin(105 deg) = 1.47 Newtons.
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RESPONSE --> I forgot that mass had already been given earlier in this problem set. I understand the calculation F=mg and that this must equal the vertical tension force.
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06:44:26 `q003. Continuing the preceding two problems, what therefore must be the tension T, and how much tension is there in the horizontal string which is holding the pendulum back?
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RESPONSE --> T=1.47/sin105=1.52N horizontal component=1.47cos105=.380N
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06:44:45 If T sin(105 deg) = 1.47 Newtons then T = 1.47 Newtons / (sin(15 deg)) = 1.47 Newtons/.97 = 1.52 Newtons. Thus the horizontal component of the tension will be T cos(105 deg) = 1.52 Newtons * cos(105 deg) = 1.52 Newtons * (-.26) = -.39 Newtons, approximately. Since the mass is in equilibrium, the net force in the x direction must be zero. The only forces acting in the x direction are the x component of the tension, to which we just found to be -.39 Newtons, and the tension in the second string, which for the moment will call T2. Thus T2 + (-.39 N) = 0 and T2 = .39 N. That is, the tension in the second string is .39 Newtons. STUDENT COMMENT: I'm really confused now. If we started out with a .15 kg mass that is equal to 1.47 Newtons. How did we create more weight to get 1.52 Newtons? Is the horizontal string not helping support the weight or is it puling on the weight adding more force? INSTRUCTOR RESPONSE: A horizontal force has no vertical component and cannot help to support an object against a vertical force. The vertical component of the tension is what supports the weight, so the tension has to be a bit greater than the weight. The tension in the string is resisting the downward weight vector as well as the horizontal pull, so by the Pythagorean Theorem it must be greater than either.
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RESPONSE --> ok
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