How do you want me to do my test? Is it alright if i have my parents to proctor my test here. that way I can use summit work from for my answers. I promise if you let me do it here I will practice one day. Then on the next day I will get on there one time and I will not use my book or nothing. One reason why I would like to be able to do it here is because the computers are always full at the librar. But if you still want me to do it at the library how do you want me to do it?
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22:37:11 q001. Note that this assignment contains 8 questions. Masses attract each other. The forces of attraction are equal and opposite: The force exerted by one small concentrated mass on another is equal in magnitude but in the opposite direction from the force exerted on it by the other. Greater masses exert greater attractions on one another. If two such objects remain separated by the same distance while one object increases to 10 times its original mass while the other remains the same, there will be 10 times the original force. If both objects increase to 10 times their original masses, there will be 100 times the original force. The force of attraction is inversely proportional to the square of the distance between the objects. That means that if the objects move twice as far apart, the force becomes 1 / 2^2 = 1/4 as great; if they move 10 times as far apart, the force becomes 1 / 10^2 = 1/100 as great. The same statements hold for spherical objects which have mass distributions which are symmetric about their centers, provided we regard the distance between the objects as the distance between their centers. Suppose a planet exerts a force of 10,000 Newtons on a certain object (perhaps a satellite) when that object is 8000 kilometers from the center of the planet. How much force does the satellite exert on the planet?
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RESPONSE --> To answer this question we do not need to solve an equation. We know that the forces of the planet and the object are equal and in the opposite direction. If the planet exerts 10,000 Newtons on the planet, then the planet exerts 10,000 Newtons on the planet.
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22:37:21 The gravitational forces exerted by the planet and the object are equal and opposite, and are both forces of attraction, so that the object must be exerting a force of 10,000 Newtons on the planet. The object is pulled toward the planet, and the planet is pulled toward the object.
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RESPONSE --> I understand. NOTE: The force exerted by one mass is equal and opposite to the force exerted ON it by the other mass. If the mass changes by a certain factor, the force of the object also changes by that factor. The force of attration between two objects is inversely proportional to the distance squared betweent the two objects.
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22:39:40 `q002. If the object and the planet are both being pulled by the same force, why is it that the object accelerates toward the planet rather than the planet accelerating toward the object?
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RESPONSE --> The gravitational force which pulls the object towards the planet is stronger than the force which pulls the planet to the object. This is due to the fact that the planet has a much greater mass than the object. Force of the object is directly related to the mass by the equation Fnet = m*a.
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22:40:01 Presumably the planet is much more massive than the object. Since the acceleration of any object is equal to the net force acting on it divided by its mass, the planet with its much greater mass will experience much less acceleration. The minuscule acceleration of the planet toward a small satellite will not be noticed by the inhabitants of the planet.
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RESPONSE --> I understand this problem.
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22:43:10 `q003. If the mass of the object in the preceding exercise is suddenly cut in half, as say by a satellite burning fuel, while the distance remains at 8000 km, then what will be the gravitational force exerted on it by the planet?
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RESPONSE --> We can find this change in force by knowing that since the mass changed by 1/2 the force would also change by 1/2. This agrees with the assumption that if the mass changes by a certain factor, the force of the object also changes by that factor. Therefore, 1/2 mass causes 1/2 Fgrav and Fgrav = 1/2(10,000N) = 5,000 Newtons.
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22:43:43 Halving the mass of the object, while implicitly keeping the mass of the planet and the distance of the object the same, will halve the force of mutual attraction from 10,000 Newtons to 5,000 Newtons.
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RESPONSE --> Ok.
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22:46:16 `q004. How much force would be experienced by a satellite with 6 times the mass of this object at 8000 km from the center of a planet with half the mass of the original planet?
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RESPONSE --> If the distance remains the same and mass is increased by six times, the force will be directly affected because F = m*a. Therefore, the gravitational force will increase by 6 times its original amount as well. F = 6(10,000N) = 60,000 Newtons Also we see that the planet is half the mass of the original planet. Therefore, if the mass has decreased by half, the force will also decrease by half. 1/2(60,000N) = 30,000 Newtons
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22:48:13 The distance is the same as in the previous examples, so increasing the mass by a factor of 6 would to result in 6 times the force, provided everything else remained the same; but halving the mass of the planet would result in halving this force so the resulting force would be only 1/2 * 6 = 3 times is great as the original, or 3 * 10,000 N = 30,000 N.
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RESPONSE --> I understand this problem. Change in mass is directly related to change in the force.
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22:49:53 `q005. How much force would be experienced by the original object at a distance of 40,000 km from the center of the original planet?
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RESPONSE --> `ds = 40,000km This distance is 5 times the original distance which was 8,000 kilometers. Since the force is inversely related to the square of the distance between the objects, we can say that the factor of change was 1/(5^2) or 0.04. Now we multiply that factor by the original force to find the new force that results from this change in distance. F = 0.04 * 10,000N = 400 Newtons.
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22:51:11 The object is 40,000 km / (8000 km) = 5 times as far from the planet as originally. Since the force is proportional to the inverse of the square of the distance, the object will at this new distance experience a force of 1 / 5^2 = 1/25 times the original, or 1/25 * 10,000 N = 400 N.
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RESPONSE --> Ok. Force is proportional to the inverse of the distance squared.
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22:56:26 `q006. The relationship between the force of attraction and the masses and separation can be expressed by a proportionality. If the masses of two small, uniformly spherical objects are m1 and m2, and if the distance between these masses is r, then the force of attraction between the two objects is given by F = G * m1 * m2 / r^2. G is a constant of proportionality equal to 6.67 * 10^-11 N m^2 / kg^2. Find the force of attraction between a 100 kg uniform lead sphere and a 200 kg uniform lead sphere separated by a center-to-center distance of .5 meter.
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RESPONSE --> I use the formula that is given to solve for the force of attraction between these two objects. F = G * m1 * m2/r^2 m1 = 100kg m2 = 200kg r = 0.5m G = 6.67 x 10^-11N*m^2/kg^2 F = (6.67 x 10^-11N*m^2/kg^2) (100kg) (200kg/0.5^2) = 5.336 x 10^-6 Newtons
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22:56:38 We are given the two masses m1 = 100 kg and m2 = 200 kg and the separation r = .5 meter between their centers. We can use the relationship F = G * m1 * m2 / r^2 directly by simply substituting the masses and the separation. We find that the force is F = 6.67 * 10^-11 N m^2 / kg^2 * 100 kg * 200 kg / (.5 m)^2 = 5.3 * 10^-6 Newton. Note that the m^2 unit in G will be divided by the square of the m unit in the denominator, and that the kg^2 in the denominator of G will be multiplied by the kg^2 we get from multiplying the two masses, so that the m^2 and the kg^2 units disappear from our calculation.
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RESPONSE --> I understand this problem.
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22:57:48 `q007. If these two objects were somehow suspended so that the net force on them was just their mutual gravitational attraction, at what rate would the first object accelerate toward the second, and if both objects were originally are rest approximately how long would it take it to move the first centimeter?
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RESPONSE --> We know the acceleration of this attraction is equal to the force of attraction divided by the mass. F = 5.336*10^-6N / 100kg = 5.336 x 10^-8 m/s^2 Now since we know the acceleration and the distance, 1cm = 0.01m, between objects we can find the time interval using the formula, `ds = (vo)(`dt) + 1/2(a)(`dt)^2. We know that the initial velocity is 0 since the objects were both originally at rest. So the formula now reads: `ds = 1/2(a)(`dt^2) `dt^2 = (2*`ds)/a = (2*0.01m)/(5.336*10^-8 m/s^2) = 374812.6 s^2 `dt = 612 seconds = approx 10 minutes, 2 seconds.
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22:58:16 A mass of 100 kg subject a net force of 5.3 * 10^-6 N will have acceleration of a = 5.3 * 10^-6 N / (100 kg) = 5.3 * 10^-8 m/s^2. At this rate to move from rest (v0 = 0) thru the displacement of one centimeter (`ds = .01 m) would require time `dt such that `ds = v0 `dt + .5 a `dt^2; since v0 = 0 this relationship is just `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt( 2 * .01 m / (5.3 * 10^-8 m/s^2) ) = `sqrt( 3.8 * 10^5 m / (m/s^2) ) = 6.2 * 10^2 sec, or about 10 minutes. Of course the time would be a bit shorter than this because the object, while moving somewhat closer (and while the other object in turn moved closer to the center of gravity of the system), would experience a slightly increasing force and therefore a slightly increasing acceleration.
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RESPONSE --> I used one of the formulas of uniformly accelerated motion to solve this problem.
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23:01:29 `q008. At what rate would the second object accelerate toward the first?
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RESPONSE --> vo = 0 m/s m2 = 200 kg F = 5.336*10^-6 N The acceleration of the second object can be found by dividing the force of attraction by the second object's mass. Though we know the initial velocity is zero, it is not an important factor in solving this problem. a = F/m = 5.336*10^-6N / 200kg = 2.67*10^-8 m/s^2
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23:04:20 The second object, with its 200 kg mass, would also a subject to a net force of 5.3 * 10^-6 N and would therefore experience and acceleration of a = 5.3 * 10^-6 N / (200 kg) = 2.7 * 10^-8 m/s^2. This is half the rate at which the first object changes its velocity; this is due to the equal and opposite nature of the forces and to the fact that the second object has twice the mass of the first.
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RESPONSE --> I understand. NOTE: The acceleration for the second object is half the rate of the first object's acceleration. This is because the two objects forces are equal and opposite and the second object's mass is two times the first object. The first object's rate of change in velocity would therefore be twice as fast as the second's rate of change in velocity.
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V???}??g?????v??assignment #027 027. `query 27 Physics I 07-14-2006
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23:07:47 Query intro probs set 7, 1-7 Knowing the 9.8 m/s^2 gravitational field strength of the Earth's field at the surface of the Earth, and knowing the radius of the Earth, how do we find the gravitational field strength at a given distance 'above' the surface of the Earth?
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RESPONSE --> The radius of Earth is represented by the variable R. The given distance from the surface is represented by the variable, r1. The acceleration of gravity is represented by the variable, g. The field strength at this distance i represented by the equation, (R/r1)^2 * g.
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23:07:57 ** You have an inverse square force. Square the ratio of Earth radius to orbital radius and multiply by 9.8 m/s^2: Field strength=(Re/r)^2*9.8m/s^2 **
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RESPONSE --> Ok.
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23:09:13 If we double our distance from the center of the Earth, what happens to the gravitational field strength we experience?
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RESPONSE --> If we double the distance, r1, from the center of the Earth the equation will then read: Field strength = (R/2r1)^2 * g = 1/4(R/r1^2)(g) This shows that the gravitational field strength is 1/4 the original gravitational field strength when the distance is doubled.
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23:09:53 ** We have an inverse square force so if r2 = 2 * r1 the ratio of the gravitational field will be g2 / g1 = (1 / r2^2) / (1 / r1^2) = r1^2 / r2^2 = (r1 / r2)^2 = (r1 / (2 * r1))^2 = r1^2 / 4 r1^2 = 1/4. In a nutshell double the radius gives us 1 / 2^2 = 1/4 the gravitational field. **
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RESPONSE --> I understand. I tried a somewhat different approach.
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23:11:07 How do we approximate the energy required to move a given mass from the surface of the Earth to a given height 'above' the Earth, where the field strength at the given height differ significantly from that at the surface?
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RESPONSE --> We know that the energy required to move this mass can be represented by `dw = KE = Fave*`ds. The height form the surface is represented by `ds. The average force, Fave, is the force between the initial force on the surface and the final velocity at the farthest point from the surface.
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23:11:24 STUDENT SOLUTION AND INSTRUCTOR RESPONSE: mass*[(Re + distance)/Re]^2=force Force*distance=KE INSTRUCTOR RESPONSE: The first approximation would be to average the force at the surface and the force at the maximum altitude, then multiply by the distance. The result would give you the work necessary to 'raise' the object against a conservative force, which would be equal to the change in PE. ADDENDUM FOR UNIVERSITY PHYSICS STUDENTS ONLY:The exact work is obtained by integrating the force with respect to position. You can integrate either G M m / r^2 or g * (RE / r)^2 from r = RE to rMax. **
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RESPONSE --> I don't really understand the method that the student in the solution used. Is that method accurate?
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23:14:03 Describe the paths of various particles 'shot' parallel to the surface of the Earth from the top of a very high tower, starting with a very small velocity and gradually increasing to a velocity sufficient to completely escape the gravitational field of the Earth.
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RESPONSE --> If the object is shot at a low velocity, it will experience a motion like rolling of the edge in the horizontal direction and then falling in a parabolic path to the ground. If the object is shot at a slightly greater velocity, it won't travel a long distance before it starts to fall but it will go further than the slower object. If the object is shot at a very high velocity, it will move almost horizontally and its path curves slightly toward the center of the Earth. A circular orbit can be achieved if we shoot the object at just the right velocity. This object will only stop when something is in its path.
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23:16:34 GOOD STUDENT ANSWER: Each particle sets out to follow an orbit around the center of mass of the earth. But for particles shot at slower speeds, this path is interupted by the surface of the eath and simply stops there. The faster it is shot, the further x distance becomes before the particle lands. However, if it given a great enough velocity, it will fall around the curviture of the earth. If is shot even faster than that, it will follow an eliptical oribit with varying speeds and distances from center of earth. GOOD STUDENT ANSWER: With a very low velocity the projectile will not travlel as far. It will fall to earth in a nearly parabolic fashion since it gains vertical velocity as it travels horizontally at a steady pace. If the projectile is fired at a very strong velocity it will leave the earths vacinity but will still be pulled by the forces acting on it from the earths center. This will cause it to go only so far at which point it has slowed down considerabley, since it has lost most of its kinetic energy. It turns and begins to gain energy as it approaches the earths area, using the potential energy it gained on the trip out. (Causing it to speed up). The path that this projectile will take will be eliptical, and it will continue to loop around the earth. If the projectile is fired at the correct velocity to form a circular orbit, it will also fall at a parabolic fashion, although the earth's surface will also be descending at the same rate so that the object will appear to be 'not falling'. It is falling but at the same rate the earth is 'falling' under it. It will circle the earth until something causes it to stop. INSTRUCTOR RESPONSE: The path of the projectile will always be an ellipse with the center of the Earth at one focus. For low velocities and low altitude this path is very nearly parabolic before being interrupted by the surface of the Earth. One of these ellipses is a perfect circle and gives us the circular orbit we use frequently in this section. **
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RESPONSE --> I understand. Note: The path of the projectile will be elliptical with the center of the Earth at one focus. For low velocities and low altitudes, the path is parabolic in its fall to Earth.
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23:18:32 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> A circular orbit can be achieved by a certain velocity that has a centripetal acceleration which equals the gravitational acceleration.
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23:18:49 ** For a given distance from the center of the Earth, there is only one velocity for which centripetal acceleration is equal to gravitational acceleration, so there is only one possible velocity for a circular orbit of given orbital radius. The orbital radius is determined by the height of the 'tower', so for a given tower there is only one velocity which will achieve a circular orbit. **
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RESPONSE --> This makes sense.
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23:20:31 Is it necessary in order to achieve a circular orbit to start the object out in a direction parallel to the surface of the Earth?
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RESPONSE --> If the object starts out in a direction that is not parallel to the Earth's surface, the object will more than likely not have a circular orbit. It will be more likely to start falling toward the Earth's surface. It will not have a constant distance from the Earth's center.
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23:21:23 ** If you have just one 'shot' then you must start out parallel to the surface of the Earth. The reason is that any circle about the center must be perpendicular at every point to a radial line--a line drawn from the center to the circle. Any radial line will intercept the surface of the Earth and must be perpendicular to it, and the circular orbit must also be perpendicular to this line. Therefore the orbit and the surface are perpendicular to the same line and are therefore parallel. **
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RESPONSE --> Ok. NOTE: Any circle about the center must be perpendicular at every point to a radial line. The circular orbit must be perpendicular to this line.
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23:22:10 Principles of Physics and General College Physics Problem 5.2: A jet traveling at 525 m/s moves in an arc of radius 6.00 km. What is the acceleration of the jet?
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RESPONSE --> v = 1890km/h = 525 m/s r = 6 km = 6000 m We can find the centripetal acceleration in units of gravitational acceleration, g, by the formula: a cent = v^2/r * g a cent = [(525m/s)^2 / 6000m] 9.8m/s^2 = 4.59 g's
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23:23:09 The jet will have centripetal acceleration a_cent = v^2 / r, where v is its speed and r the radius of the circle on which it is traveling. In this case we have v = 525 m/s and r = 6.00 km = 6000 meters. The centripetal acceleration is therefore a_cent = v^2 / r = (525 m/s)^2 / (6000 m) = 45 m/s^2, approx.. One 'g' is 9.8 m/s^2, so this is about (45 m/s^2) / (9.8 m/s^2) = 4.6 'g's'.
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RESPONSE --> I understand this problem.
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23:23:32 Univ. Why is it that the center of mass doesn't move?
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RESPONSE --> The reason an object does not move is due to the fact that there is a net force of zero, and therefore there is no force being applied to the object to make it accelerate.
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23:23:49 ** There is no net force on the system as a whole so its center of mass can't accelerate. From the frame of reference of the system, then, the center of mass remains stationary. **
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RESPONSE --> I understand this concept.
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