course Phy201 ȣҧOԅvCassignment #020
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10:20:48 Explain how we get the components of the resultant of two vectors from the components of the original vectors.
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RESPONSE --> ok confidence assessment:
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10:21:06 ** If we add the x components of the original two vectors we get the x component of the resultant. If we add the y components of the original two vectors we get the y component of the resultant. **
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RESPONSE --> ok self critique assessment:
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10:22:20 Explain how we get the components of a vector from its angle and magnitude.
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RESPONSE --> ok confidence assessment:
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10:22:28 ** To get the y component we multiply the magnitude by the sine of the angle of the vector (as measured counterclockwise from the positive x axis). To get the x component we multiply the magnitude by the cosine of the angle of the vector (as measured counterclockwise from the positive x axis). **
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RESPONSE --> ook self critique assessment:
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}kSLC棤 assignment #021 ÝzlvzزmwW Physics I 10-26-2008 ۂ׃ӳ assignment #021 ÝzlvzزmwW Physics I 10-26-2008
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22:04:49 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 --> sqrt((2MG)/pACd
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22:05:24 ** 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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22:06:17 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 --> hills, dropping and tossing up
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22:06:25 ** 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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}y튛쭂咕 assignment #021 ÝzlvzزmwW Physics I 10-26-2008 念ˎ\σԑlבy assignment #020 020. Forces (inclines, friction) Physics II 10-26-2008
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10:16:47 `q001. Note that this assignment contains 3 questions. . A 5 kg block rests on a tabletop. A string runs horizontally from the block over a pulley of negligible mass and with negligible friction at the edge of the table. There is a 2 kg block hanging from the string. If there is no friction between the block in the tabletop, what should be the acceleration of the system after its release?
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RESPONSE --> 9.8m/s^2 confidence assessment:
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10:17:02 Gravity exerts a force of 5 kg * 9.8 meters/second = 49 Newtons on the block, but presumably the tabletop is strong enough to support the block and so exerts exactly enough force, 49 Newtons upward, to support the block. The total of this supporting force and the gravitational force is zero. The gravitational force of 2 kg * 9.8 meters/second = 19.6 Newtons is not balanced by any force acting on the two mass system, so we have a system of total mass 7 kg subject to a net force of 19.6 Newtons. The acceleration of this system will therefore be 19.6 Newtons/(7 kg) = 2.8 meters/second ^ 2.
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RESPONSE --> ok self critique assessment:
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10:18:45 `q002. Answer the same question as that of the previous problem, except this time take into account friction between the block in the tabletop, which exerts a force opposed to motion which is .10 times as great as the force between the tabletop and the block. Assume that the system slides in the direction in which it is accelerated by gravity.
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RESPONSE --> 3.08m/s^2 confidence assessment:
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10:19:03 Again the weight of the object is exactly balance by the upward force of the table on the block. This force has a magnitude of 49 Newtons. Thus friction exerts a force of .10 * 49 Newtons = 4.9 Newtons. This force will act in the direction opposite that of the motion of the system. It will therefore be opposed to the 19.6 Newton force exerted by gravity on the 2 kg object. The net force on the system is therefore 19.6 Newtons -4.9 Newtons = 14.7 Newtons. The system will therefore accelerate at rate a = 14.7 Newtons/(7 kg) = 2.1 meters/second ^ 2.
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RESPONSE --> ok self critique assessment:
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]]zЧxh assignment #021 021. projectiles 2 Physics II 10-26-2008
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21:45:46 `q001. Note that this assignment contains 3 questions. . A projectile has an initial velocity of 12 meters/second, entirely in the horizontal direction. After falling to a level floor three meters lower than the initial position, what will be the magnitude and direction of the projectile's velocity vector?
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RESPONSE --> 4.74m at 230.5 degrees confidence assessment:
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21:46:22 To answer this question we must first determine the horizontal and vertical velocities of the projectile at the instant it first encounters the floor. The horizontal velocity will remain at 12 meters/second. The vertical velocity will be the velocity attained by a falling object which is released from rest and allowed to fall three meters under the influence of gravity. Thus the vertical motion will be characterized by initial velocity v0 = 0, displacement `ds = 3 meters and acceleration a = 9.8 meters/second ^ 2. The fourth equation of motion, vf^2 = v0^2 + 2 a `ds, yields final vel in y direction: vf = +-`sqrt( 0^2 + 2 * 9.8 meters/second ^ 2 * 3 meters) = +-7.7 meters/second. Since we took the acceleration to be in the positive direction the final velocity will be + 7.7 meters/second. This final velocity is in the downward direction. On a standard x-y coordinate system, this velocity will be directed along the negative y axis and the final velocity will have y coordinate -7.7 m/s and x coordinate 12 meters/second. The magnitude of the final velocity is therefore `sqrt((12 meters/second) ^ 2 + (-7.7 meters/second) ^ 2 ) = 14.2 meters/second, approximately. The direction of the final velocity will therefore be arctan ( (-7.7 meters/second) / (12 meters/second) ) = -35 degrees, very approximately, as measured in the counterclockwise direction from the positive x axis. The direction of the projectile at this instant is therefore 35 degrees below horizontal. This angle is more commonly expressed as 360 degrees + (-35 degrees) = 325 degrees.
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RESPONSE --> ok self critique assessment:
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21:48:23 `q002. A projectile is given an initial velocity of 20 meters/second at an angle of 30 degrees above horizontal, at an altitude of 12 meters above a level surface. How long does it take for the projectile to reach the level surface?
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RESPONSE --> 4.3s confidence assessment:
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21:48:37 To determine the time required to reach the level surface we need only analyze the vertical motion of the projectile. The acceleration in the vertical direction will be 9.8 meters/second ^ 2 in the downward direction, and the displacement will be 12 meters in the downward direction. Taking the initial velocity to be upward into the right, we situate our x-y coordinate system with the y direction vertically upward and the x direction toward the right. Thus the initial velocity in the vertical direction will be equal to the y component of the initial velocity, which is v0y = 20 meters/second * sine (30 degrees) = 10 meters/second. Characterizing the vertical motion by v0 = 10 meters/second, `ds = -12 meters (`ds is downward while the initial velocity is upward, so a positive initial velocity implies a negative displacement), and a = -9.8 meters/second ^ 2, we see that we can find the time `dt required to reach the level surface using either the third equation of motion `ds = v0 `dt + .5 a `dt^2, or we can use the fourth equation vf^2 = v0^2 + 2 a `ds to find vf after which we can easily find `dt. To avoid having to solve a quadratic in `dt we choose to start with the fourth equation. We obtain vf = +-`sqrt ( (10 meters/second) ^ 2 + 2 * (-9.8 meters/second ^ 2) * (-12 meters) ) = +-18.3 meters/second, approximately. Since we know that the final velocity will be in the downward direction, we choose vf = -18.3 meters/second. We can now find the average velocity in the y direction. Averaging the initial 10 meters/second with the final -18.3 meters/second, we see that the average vertical velocity is -4.2 meters/second. Thus the time required for the -12 meters displacement is `dt = `ds / vAve = -12 meters/(-4.2 meters/second) = 2.7 seconds.
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RESPONSE --> ok self critique assessment:
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21:51:45 `q003. What will be the horizontal distance traveled by the projectile in the preceding exercise, from the initial instant to the instant the projectile strikes the flat surface.
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RESPONSE --> 155.7m confidence assessment:
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21:51:55 The horizontal velocity of the projectile will not change so if we can find this horizontal velocity, knowing that the projectile travels for 2.7 seconds we can easily find the horizontal range. The horizontal velocity of the projectile is simply the x component of the velocity: horizontal velocity = 20 meters/second * cosine (30 degrees) = 17.3 meters/second. Moving at this rate for 2.7 seconds the projectile travels distance 17.3 meters/second * 2.7 seconds = 46 meters, approximately.
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RESPONSE --> ok self critique assessment:
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㌧d assignment #022 022. Motion in force field Physics II 10-26-2008
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22:11:50 `q001. Note that this assignment contains 2 questions, which relate to a force-field experiment which is done using a computer simulation, and could for example represent the force on a spacecraft, where uphill and downhill are not relevant concepts. . An object with a mass of 4 kg is traveling in the x direction at 10 meters/second when it enters a region where it experiences a constant net force of 5 Newtons directed at 210 degrees, as measured in the counterclockwise direction from the positive x axis. How long will take before the velocity in the x direction decreases to 0? What will be the y velocity of the object at this instant?
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RESPONSE --> 8s 30.3 confidence assessment:
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22:12:03 A constant net force of 5 Newtons on a 4 kg object will result in an acceleration of 5 Newtons/(4 kg) = 1.25 meters/second ^ 2. If the force is directed at 210 degrees then the acceleration will also be directed at 210 degrees, so that the acceleration has x component 1.25 meters/second ^ 2 * cosine (210 degrees) = -1.08 meters/second ^ 2, and a y component of 1.25 meters/second ^ 2 * sine (210 degrees) = -.63 meters/second ^ 2. We analyze the x motion first. The initial velocity in the x direction is given as 10 meters/second, we just found that the acceleration in the x direction is -1.08 meters/second ^ 2, and since we are trying to find the time required for the object to come to rest the final velocity will be zero. We easily see that the change in the next velocity is -10 meters/second. At a rate of negative -1.08 meters/second ^ 2, the time required for the -10 meters/second change in velocity is `dt = -10 meters/second / (-1.08 meters/second ^ 2) = 9.2 seconds. We next analyze the y motion. The initial velocity in the y direction is zero, since the object was initially moving solely in the x direction. The acceleration in the y direction is -.63 meters/second ^ 2. Therefore during the time interval `dt = 9.2 seconds, the y velocity changed by (-.63 meters/second ^ t) * (9.2 seconds) = -6 meters/second, approximately. Thus the y velocity changes from zero to -6 meters/second during the 9.2 seconds required for the x velocity to reach zero.
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RESPONSE --> ok self critique assessment:
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22:14:03 `q002. Suppose that the mass in the preceding problem encounters a region in which the force was identical to that of the problem, but that this region extended for only 30 meters in the x direction (assume that there is the limit to the extent of the field in the y direction). What will be the magnitude and direction of the velocity of the mass as it exits this region?
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RESPONSE --> 340 degrees 35.4m/s^2 confidence assessment:
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22:14:18 As we have seen in the preceding problem the object will have an acceleration of -1.08 meters/second ^ 2 in the x direction. Its initial x velocity is 10 meters/second and it will travel 30 meters in the x direction before exiting the region. Thus we have v0, a and `ds, so that you to the third or fourth equation of uniform accelerated motion will give us information. The fourth equation tells us that vf = +-`sqrt( (10 meters/second) ^ 2 + 2 * (-1.08 meters/second ^ 2) * (30 meters) ) = +-6 meters/second. Since we must exit the region in the positive x direction, we choose vf = + 6 meters/second. It follows that the average x velocity is the average of the initial 10 meters/second and the final 6 meters/second, or eight meters/second. Thus the time required to pass-through the region is 30 meters/(8 meters/second) = 3.75 seconds. During this time the y velocity is changing at -.63 meters/second ^ 2. Thus the change in the y velocity is (-.63 meters/second ^ 2) * (3.75 seconds) = -2.4 meters/second, approximately. Since the initial y velocity was zero, the y velocity upon exiting the region will be -2.4 meters/second. Thus when exiting the region the object has velocity components +6 meters/second in the x direction and -2.4 meters/second in the y direction. Its velocity therefore has magnitude `sqrt ( (6 meters/second) ^ 2 + (-2.4 meters/second) ^ 2) = 6.4 meters/second. The direction of velocity will be arctan ( (-2.4 meters/second) / (6 meters/second) ) = -22 degrees, approximately. Thus the object exits at 6.4 meters/second at an angle of 22 degrees below the positive x axis, or at angle -22 degrees + 360 degrees = 338 degrees.
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RESPONSE --> ok self critique assessment:
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