course Phy 121 I˵wwassignment #027
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15:31:17 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 is 6400 km we would assume that g=k/r^2 that means k=401,408,000 km m/s^2 then we use the formula g=k/distance^2 i hope that makes some sense to you, it does to me right now
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15:33:06 ** 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 --> so a distance of 500km above the surface would mean a force of F=(500/6400)^2 * 9.8m/s^2 which would be F=.0598 approx.
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15:33:39 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 --> we take the inverse of the double and square it so it would be 1/2^2 which would be 1/4
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15:33:49 ** 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 --> got it
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15:37:56 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 --> the Force will equal mass * G for a 50kg mass the force at the surface would be 490 N at 300 km above the surface we would take the total 'ds from the center and divide that by the radius we would use the formula F=G/('ds/r)^2 so in the 300 km case we would get a force of 9.37 m/s^2 which would yield a F of 468.5 N at that altitude
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15:39:04 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 --> this one i have got. i didn't know how to approximate it though.
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15:47:49 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 --> i can't imagine if they were shot parrallel to the earth's surface and gained in velocity enough to escape the gravitational field of the earth, that they would ever go down in the vertical direction. it seems their velocity wouldhave to be too great
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15:50:44 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> the one that falls parabolically at first. when it reaches near the surface of the earth, it's velocity will be great enough to ""slingshot"" it back up and around the earth and its velocity will be steady, matching the gravitational effect of the earth, keeping it in a circle. but what do i know?
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15:51:49 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 --> i would think that in order for the orbit to be exactly parrallel, the orbit would have to be circular
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15:57:49 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 --> i though we would have to know the mass to figure this but i guess not there are v^2/r or 459375 N applied to the jet the jet traveled 37.68 km, assuming the arc made half a circle after that,i'm lost
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15:58:43 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 --> whew again, i understand now i think
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I˵ww assignment #027 gzܞ Physics I 07-19-2007
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15:31:17 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 is 6400 km we would assume that g=k/r^2 that means k=401,408,000 km m/s^2 then we use the formula g=k/distance^2 i hope that makes some sense to you, it does to me right now
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15:33:06 ** 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 --> so a distance of 500km above the surface would mean a force of F=(500/6400)^2 * 9.8m/s^2 which would be F=.0598 approx.
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15:33:39 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 --> we take the inverse of the double and square it so it would be 1/2^2 which would be 1/4
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15:33:49 ** 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 --> got it
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15:37:56 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 --> the Force will equal mass * G for a 50kg mass the force at the surface would be 490 N at 300 km above the surface we would take the total 'ds from the center and divide that by the radius we would use the formula F=G/('ds/r)^2 so in the 300 km case we would get a force of 9.37 m/s^2 which would yield a F of 468.5 N at that altitude
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15:39:04 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 --> this one i have got. i didn't know how to approximate it though.
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15:47:49 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 --> i can't imagine if they were shot parrallel to the earth's surface and gained in velocity enough to escape the gravitational field of the earth, that they would ever go down in the vertical direction. it seems their velocity wouldhave to be too great
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15:50:44 How many of the velocities in the preceding question would result in a perfectly circular orbit about the Earth?
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RESPONSE --> the one that falls parabolically at first. when it reaches near the surface of the earth, it's velocity will be great enough to ""slingshot"" it back up and around the earth and its velocity will be steady, matching the gravitational effect of the earth, keeping it in a circle. but what do i know?
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15:51:49 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 --> i would think that in order for the orbit to be exactly parrallel, the orbit would have to be circular
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15:57:49 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 --> i though we would have to know the mass to figure this but i guess not there are v^2/r or 459375 N applied to the jet the jet traveled 37.68 km, assuming the arc made half a circle after that,i'm lost
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15:58:43 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 --> whew again, i understand now i think
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mВGy assignment #027 027. Newton's Law of Universal Gravitation Physics II 07-19-2007
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12:42:15 `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 --> i thought it had to be equal and opposite force, therefore i say it exerts -10,000 N on the planet confidence assessment: 2
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12:42:46 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 put negative newtons, is that right? self critique assessment: 3
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12:45:08 `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 --> are we to assume that the smaller object will always accelerate towards the larger one? is it because its mass is greater, therefore the gravitational force is greater? confidence assessment: 1
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12:45:57 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 --> it's a more complex concept than just saying the smaller one always accelerates more, but i get it. self critique assessment: 3
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12:47:23 `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 --> it would be cut in half to 5000 N or would it be inversely proportional? that would mean it would be 1/2^2 or 1/4 making it only 2500 N confidence assessment: 2
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12:47:40 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, it was just half of the original force self critique assessment: 3
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12:48:30 `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 --> there would be 6 times the force, or 60,000 N confidence assessment: 3
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12:49:25 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 just didn't read the question correctly. self critique assessment: 3
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12:50:12 `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 --> 40,000 is 5 times as far away so the force would be 1/25 as much it would be 400 N confidence assessment: 3
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12:50:18 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 --> got it self critique assessment: 3
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12:54:47 `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 --> it would equal G * 100kg * 200kg/ G*20,000/r^2 133,400 * 10^-11/.25 533,600 * 10^-11 .00000533600 is what i get. something tells me that is wrong confidence assessment: 1
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12:55:29 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 --> no, i was actually right. i just did the whole decimal out instead of leaving it in scientific notation self critique assessment: 3
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12:58:38 `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 --> since one object is half the mass of the other, i would think the smaller would accelerate at 1/4 the gravitational acceleration, so my wild guess here is 2.45 m/s^2 but i know that is too fast. confidence assessment: 0
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13:03:42 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 --> oh, i could have gotten that if i had simply applied the formula you just gave us. self critique assessment: 3
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13:04:26 `q008. At what rate would the second object accelerate toward the first?
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RESPONSE --> i thought it was 5.3 * 10^-1 m/s^2 confidence assessment: 2
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13:04:49 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 --> same force only half the mass, i got it now. self critique assessment: 3
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zoyd溴\Ϫ{Gpq assignment #028 028. Orbital Dynamics Physics II 07-19-2007
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16:17:33 `q001. Note that this assignment contains 11 questions. The planet Earth has a mass of approximately 6 * 10^24 kg. What force would therefore be experienced by a 3000 kg satellite as it orbits at a distance of 10,000 km from the center of the planet?
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RESPONSE --> F=G*m1*m2/r^2 9.8 * 18e27/1e8 9.8*18e19 Newtons confidence assessment: 3
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16:18:40 The force would be F = G m1 m2 / r^2, with m1 and m2 the masses of the planet and the satellite and r the distance of the satellite from the center of the planet. Thus we have F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (10,000,000 meters) ^ 2 = 12,000 Newtons.
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RESPONSE --> i would have gotten this if i had used the correct G instead of 9.8 self critique assessment: 3
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16:20:44 `q002. What force would the same satellite experience at the surface of the Earth, about 6400 km from the center.
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RESPONSE --> wouldn't that just be normal 9.8 * 3000 kg since it's at the surface? confidence assessment: 1
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16:21:42 The force would be F = G m1 m2 / r^2, with m1 and m2 the masses of the planet and the satellite and r the distance of the satellite from the center of the planet. Thus we have F = 6.67 * 10^-11 N m^2 / kg^2 * (6 * 10^24 kg) * 3000 kg / (6,400,000 meters) ^ 2 = 29,000 Newtons. Note that this is within roundoff error of the F = m g = 3000 kg * 9.8 m/s^2 = 29400 N force calculated from the gravitational acceleration experienced at the surface of the Earth.
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RESPONSE --> it was indeed. self critique assessment: 3
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16:24:45 `q003. What would be the acceleration toward the center of the Earth of the satellite in the previous two questions at the distance 10,000 km from the center of the Earth? We may safely assume that no force except gravity acts on the satellite.
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RESPONSE --> cent force accleration is = v^2/r if the initial velocity is 0 the Acent from 10,000 km is 442.7 m/s^2 confidence assessment: 1
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16:26:23 The force at the 10,000 km distance was previously calculated to be 12,000 Newtons, the mass of the satellite being 3000 kg. Since the only force acting on the satellite is that of gravity, the 12,000 Newtons is the net force and the acceleration of the satellite is therefore a = 12,000 N / 3000 kg = 4 m/s^2.
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RESPONSE --> i used Newtons' law and got a final v of 442 or so. i also got that it would take about 45 seconds to land. that doesn't sound like enough time. self critique assessment: 2
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16:27:57 `q004. The centripetal acceleration of an object moving in a circle of radius r at velocity v is aCent = v^2 / r. What would be the centripetal acceleration of an object moving at 5000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how this this compare to the 4 m/s^2 acceleration net would be experienced by an object at this distance from the center of the Earth?
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RESPONSE --> the Acent would be 2500 m/s^2 a lot more than 4 m/s^2 confidence assessment: 2
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16:28:21 The centripetal acceleration of the given object would be aCent = (5000 m/s)^2 / (10,000,000 m) = 2.5 m/s^2. This is less than the acceleration of gravity at that distance.
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RESPONSE --> i just forgot to convert km to meters. self critique assessment: 3
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16:29:33 `q005. What would be the centripetal acceleration of an object moving at 10,000 m/s in a circular orbit at the distance of 10,000 km from the center of a planet, and how does this compare to the 4 m/s^2 acceleration that would be experienced by an object at this distance from the center of the Earth?
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RESPONSE --> that would be 10 m/s^2 the acceleration is a lot less now being so far out. confidence assessment: 3
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16:30:03 The centripetal acceleration of this object would be aCent = v^2 / r = (10,000 m/s)^2 / (10,000,000 m) = 10 m/s^2, which is greater than the acceleration of gravity at that distance.
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RESPONSE --> interesting. so it can orbit at that distance then, right? self critique assessment: 2
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16:32:50 `q006. An object will move in a circular orbit about a planet without the expenditure of significant energy provided that the object is well outside the atmosphere of the planet, and provided its centripetal acceleration matches the acceleration of gravity at the position of the object in its orbit. For the satellite of the preceding examples, orbiting at 10,000 km from the center of the Earth, we have seen that the acceleration of gravity at that distance is approximately 4 m/s^2. What must be the velocity of the satellite so that this acceleration from gravity matches its centripetal acceleration?
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RESPONSE --> well, i know it's somewhere between 5000 m/s and 10,000 m/s from what we've seen previously i'm not sure how to get it exactly though. i messed around with some trial and errors and didn't get it. confidence assessment: 1
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16:33:30 The velocity must be such that aCent = v^2 / r matches the 4 m/s^2. Solving aCent = v^2 / r for v we obtain v = `sqrt( aCent * r ), so if aCent is 4 m/s^2, v = `sqrt( 4 m/s^2 * 10,000,000 m ) = `sqrt( 40,000,000 m) = 6.3 * 10^3 m/s.
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RESPONSE --> i knew it had to be between those two figures. very interesting. self critique assessment: 2
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16:57:31 `q007. The orbital velocity of a satellite in a circular orbit is that velocity for which the centripetal acceleration of the satellite is equal to its gravitational acceleration. The satellite in the previous series of examples had a mass of 3000 kg and orbited at a distance of 10,000 km from the center of the Earth. What would be the acceleration due to Earth's gravity of a 5-kg hunk of space junk at this orbital distance?
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RESPONSE --> the force on it would be about 20 Newtons from 10,000 km confidence assessment: 3
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16:57:51 The force of gravity on the junk hunk is easily found from Newton's Law of Universal Gravitation. Using F = G m1 m2 / r^2 we see that the force of gravity must be Fgrav = (6.67 * 10^-11 kg) * (6 * 10^24 kg) * (5 kg) / (10,000,000 m)^2 = 20 Newtons, approx.. Its acceleration due to gravity is thus a = Fgrav / m = 20 Newtons / 5 kg = 4 m/s^2. We note that this is the same gravitational acceleration experienced by the 3000 kg mass, and conjecture that any mass will experience the same gravitational acceleration at this distance from the center of the planet.
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RESPONSE --> i got that using the same formula as before, you know, when i used it incorrectly self critique assessment: 3
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17:00:20 `q008. What therefore will be the orbital velocity of the 5-kg piece of junk?
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RESPONSE --> the acceleration would be 20N/5kg= 4m/s^2 the orbital velocity must equal that so 4 m/s^2=v^2/10,000,000 v=6324.5 m/s^2 confidence assessment: 2
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17:00:32 Orbital velocity is calculated from distance and gravitational acceleration by solving a = v^2 / r for v, where a is the centripetal acceleration, which is the same as the gravitational acceleration. We get v = `sqrt( a * r), just as before, and v = `sqrt( 4 m/s^2 * 10,000,000 m) = 6.3 * 10^3 m/s, the same velocity as for the 3000 kg satellite.
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RESPONSE --> alright, i got it. self critique assessment: 3
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17:02:21 `q009. Is it true that the gravitational acceleration of any object at a distance of 10,000,000 meters from the center of the Earth must be the same as for the 3000-kg satellite and the 5-kg hunk of space junk? (Hint: We have to find the acceleration for any mass, so we're probably going to have to let the mass of the object be represented by symbol. Use mObject as a symbol for the mass of the object. While dealing in symbols, you might as well leave G and r in symbols and let mEarth stand for the mass of the Earth. Find an expression for the force, then using this expression and Newton's Second Law find an expression for the acceleration of the object).
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RESPONSE --> i believe that it is according to Mr. Newton. the rate is the same on the surface so it must be the same at 10,000,000 meters as we saw by these 2 objects. confidence assessment: 3
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17:02:44 We know that the gravitational force on the object is Fgrav = G * mEarth * mObject / r^2, where G is the universal gravitational constant, r the distance from the center of the Earth, mEarth the mass of the Earth and mObject the mass of the object. The acceleration of the object is a = Fgrav / mObject, by Newton's Second Law. Substituting the expression G * mEarth * mObject / r^2 for Fgrav we see that a = [ G * mEarth * mObject / r^2 ] / mObject = G * mEarth / r^2. We note that this expression depends only upon the following: G, which we take to be univerally constant, the effectively unchanging quantity mEarth and the distance r separating the center of the Earth from the center of mass of the object. Thus for all objects at a distance of 10,000 km from the center of the Earth the acceleration due to the gravitational force must be the same.
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RESPONSE --> that's what i thought self critique assessment: 3
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17:06:03 `q010. How much work would have to be done against gravity to move the 3000 kg satellite from a circular orbit at a distance of 10,000 km to a circular orbit at a distance of 10,002 km?
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RESPONSE --> force at 10,000,000 meters was 12,000 Newtons force at 10,000,002 meters is 1.2e11/10,000,002 doesn't seem like much of a difference confidence assessment: 0
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17:07:19 As found previously the object experiences a force of approximately 12,000 N at a distance of 10,000 km. At a distance of 10,002 km, the force of gravity will be slightly less than at 10,000 km, but only by about 5 Newtons or .0004 of the force. That is, over the 2 km distance the force of gravity doesn't change by very much. Therefore to move 2 km = 2000 m further from the center of the planet would require the application of a force very close to 12,000 N in the direction away from the center. The work done by this force is therefore `dW = 12,000 Newtons * 2000 m = 24,000,000 Joules.STUDENT QUESTION: I understand this mathmatically, I'm not sure I understand practically. How do you gain KE if one object was intially stationary? It would seem that the first object would lose and the second object would gain what was lost but not more than what was lost... INSTRUCTOR RESPONSE: As stated it isn't possible for total KE to increase unless there is some other source of energy involved. For example if there is a coiled spring on one object it could uncoil on collision and add extra KE. Momentum conservation does not say anything about energy. Momentum and energy are two completely independent quantities. **
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RESPONSE --> it sure wasn't much of a diff. in force but it was a lot of joules self critique assessment: 2
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17:07:50 `q011. Does it therefore follow that the work done to move a 3000 kg satellite from the distance of 10,000 km to a distance of 10,002 km from the center of the Earth must be 24,000,000 Joules?
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RESPONSE --> i would think so confidence assessment: 1
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17:08:22 It might seem so, but this is not the case. The net force does work, but when the radius of the orbit changes the velocity and hence the kinetic energy of the satellite also changes. The work done by the net force is equal to the sum of the changes in the KE and the gravitational PE of the satellite. The change in gravitational PE is the 24,000,000 J we just calculated, and if there is no KE change this will be equal to the work done by the net force. However if KE increases the net force must do more than 24,000,000 J of work, and if KE decreases the net force must do less than 24,000,000 J of work. In this case, as we move further away the KE decreases so the net force must do less than 24,000,000 J of work. (See also Conservation of Energy in Orbit under Q&A)
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RESPONSE --> ok, these assignments are getting tough. i'm determined to get these concepts though. self critique assessment: 1
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pz弃{Fwzh assignment #028 gzܞ Physics I 07-20-2007
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16:07:52 Query class notes #26 Explain how we use proportionality along with the radius rE of the Earth to determine the gravitational acceleration at distance r from the center of the Earth to obtain an expression for the gravitational acceleration at this distance. Explain how we use this expression and the fact that centripetal forces is equal to v^2 / r to obtain the velocity of a satellite in circular orbit.
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RESPONSE --> the force is always equal to G/(distance from rE/r)^2 we know that if it is orbiting, its orbital velocity must equal its gravitational acceleration from the altitude it is at, away from the rE
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16:09:03 ** The proportionality is accel = k r^2. When r = rE, accel = 9.8 m/s^2 so 9.8 m/s^2 = k * rE^2. Thus k = 9.8 m/s^2 / rE^2, and the proportionality can now be written accel = [ 9.8 m/s^2 / (rE)^2 ] * r^2. Rearranging this gives us accel = 9.8 m/s^2 ( r / rE ) ^2. **
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RESPONSE --> yes, that is the formula i typed in first.
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16:12:47 Principles of Physics and Gen Phy problem 5.30 accel of gravity on Moon where radius is 1.74 * 10^6 m and mass 7.35 * 10^22 kg.
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RESPONSE --> i'm trying to understand which formula to put to work here.
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16:16:21 ** The acceleration due to gravity on the Moon is found using the equation g' = G (Mass of Moon)/ radius of moon ^2 g' = (6.67 x 10^-11 N*m^2/kg^2)(7.35 X 10^22 kg) / (1.74 X 10^6 m) = 1.619 m/s^2 **
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RESPONSE --> i see we take the G, multiply it by the moon's mass, and divide that by the square of the radius of the moon. my trouble is understanding Newton's law of universal gravitation, i've got my book open to that now and i'm reading. these problems are eluding me so far.
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~i]FWwr assignment #029 029. Radian measure of angle; angular position, angular velocity Physics II 07-23-2007
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15:36:44 `q001. Note that this assignment contains 15 questions. If an object moves a distance along the arc of a circle equal to the radius of the circle, it is said to move through one radian of angle. If a circle has a radius of 40 meters, then how far would you have to walk along the arc of the circle to move through one radian of angle? How far would you have to walk to move through 3 radians?
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RESPONSE --> wouldn't that be 40 m = one radian therefore 3 radians would be 120 m? confidence assessment: 1
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15:36:53 Since 1 radian of angle corresponds to the distance along the arc which is equal to the radius, if the radius of the circle is 40 meters then a 1 radian angle would correspond to a distance of 40 meters along the arc. An angle of 3 radians would correspond to a distance of 3 * 40 meters = 120 meters along the arc. Each radian corresponds to a distance of 40 meters along the arc.
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RESPONSE --> easy enough self critique assessment: 3
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15:48:47 `q002. On a circle of radius 40 meters, how far would you have to walk to go all the way around the circle, and through how many radians of angle would you therefore travel? Through how many radians would you travel if you walked halfway around the circle? Through how many radians would you travel if you walked a quarter of the way around the circle?
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RESPONSE --> the circumf would be 251 meters so that's how far you'd walk to get around the circle that is 6.28 radians i guess halfway would be 3.14 radians, pi. one fourth would be 1.57 radians confidence assessment: 3
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15:49:12 The circumference of a circle is the product of `pi and its diameter, or in terms of the radius r, which is half the diameter, C = 2 `pi r. The circumference of this circle is therefore 2 `pi * 40 meters = 80 `pi meters. This distance can be left in this form, which is exact, or if appropriate this distance can be approximated as 80 * 3.14 meters = 251 meters (approx). The exact distance 2 `pi * 40 meters is 2 `pi times the radius of the circle, so it corresponds to 2 `pi radians of arc. Half the arc of the circle would correspond to a distance of half the circumference, or to 1/2 ( 80 `pi meters) = 40 `pi meters. This is `pi times the radius so corresponds to `pi radians of angle. A quarter of an arc would correspond to half the preceding angle, or `pi/2 radians.
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RESPONSE --> so far, so good self critique assessment: 3
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15:52:20 `q003. On a circle of radius 6 meters, what distance along the arc would correspond to 3 radians? What distance would correspond to `pi / 6 radians? What distance would correspond to 4 `pi / 3 radians?
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RESPONSE --> 3 radians would be just less than halfway 'pi/6 radians would be just over 3 meters 4'pi/3 radians would be 25.12 meters along the arc of 37 meters confidence assessment: 2
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15:55:18 3 radians along the arc would correspond to an arc distance of 3 times the radius, or 3 * 6 meters, or 18 meters. `pi / 6 radians would correspond to `pi / 6 times the radius, or `pi / 6 * 6 meters = `pi meters. 4 `pi / 3 radians would correspond to 4 `pi / 3 * 6 meters = 8 `pi meters.
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RESPONSE --> ok, answer in terms of 'pi self critique assessment: 2
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15:55:45 `q004. If you were traveling around a circle of radius 50 meters, and if you traveled through 4 radians in 8 seconds, then how fast would you have to be moving?
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RESPONSE --> 200 m in 8 sec is 25 m/s confidence assessment: 3
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15:55:51 If you travel 4 radians along the arc you half traveled an arc distance of 4 times the radius, or 4 * 50 meters = 200 meters. If you traveled this distance in 8 seconds your average speed would be 200 meters / (8 seconds) = 25 m/s.
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RESPONSE --> got it self critique assessment: 3
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15:56:11 `q005. Traveling at 3 radians / second around a circle of radius 20 meters, how fast would you have to be moving?
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RESPONSE --> you'd be going 60m/s confidence assessment: 3
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15:56:17 3 radians along the arc is a distance of 3 times the radius, or 3 * 20 meters = 60 meters. Moving at 3 radians/second, then, the speed along the arc must be 3 * 20 meters / sec = 60 meters /sec.
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RESPONSE --> got it self critique assessment: 3
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15:57:18 `q006. If you know how many radians an object travels along the arc of a circle, and if you know the radius of the circle, how do you find the distance traveled along the arc? Explain the entire reasoning process.
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RESPONSE --> one radian = the length of the radius confidence assessment: 2
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15:57:38 The distance traveled along the arc of circle is 1 radius for every radian. Therefore we multiply the number of radians by the radius of the circle to get the arc distance.
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RESPONSE --> got it self critique assessment: 3
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15:57:57 `q007. If you know the distance an object travels along the arc of a circle, and if you know the radius of the circle, how do you find the corresponding number of radians?
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RESPONSE --> divide the distance traveled by the radius confidence assessment: 3
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15:58:08 An arc distance which is equal to the radius corresponds to a radian. Therefore if we divide the arc distance by the radius we obtain the number of radians.
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RESPONSE --> got it self critique assessment: 3
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15:58:54 `q008. If you know the time required for an object to travel a given number of radians along the arc of a circle of known radius, then how do you find the average speed of the object?
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RESPONSE --> multiply the radius by the number of radians traveled and divide by the time it took to do it confidence assessment: 3
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15:59:01 If you know the number of radians you can multiply the number of radians by the radius to get the distance traveled along the arc. Dividing this distance traveled along the arc by the time required gives the average speed of the object traveling along the arc.
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RESPONSE --> got it self critique assessment: 3
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16:00:02 `q009. If you know the speed of an object along the arc of a circle and you know the radius of the circle, how do you find the angular speed of the object in radians/second?
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RESPONSE --> divide the number of meters per sec by the radius (in meters) to get radians/second confidence assessment: 3
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16:01:35 `q010. We usually let `d`theta stand for the anglular displacement in radians between two points on the arc of the circle. We usually let `omega stand for the angular velocity in radians / second. We let `ds stand for the distance traveled along the arc of a circle, and we let r stand for the radius of the circle. If we know the radius r and the arc distance `ds, what is the anglular displacement `d`theta, in radians?
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RESPONSE --> it is 'ds/r = 'd'theta (in radians) confidence assessment: 2
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16:02:39 `q011. If we know the radius r of a circle and the angular velocity `omega, how do we find the velocity v of the object as it moves around the arc of the circle?
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RESPONSE --> we multiply the 'omega by the radius to get the velocity confidence assessment: 3
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16:05:03 `q012. We can change an angle in degrees to radians, or vice versa, by recalling that a complete circle consists of 360 degrees or 2 `pi radians. A half-circle is 180 degrees or `pi radians, so 180 degrees = `pi radians. How many radians does it take to make 30 degrees, how many to make 45 degrees, and how many to make 60 degrees?
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RESPONSE --> 360 degrees= 2`pi radians 30 degrees = 1/6 `pi radians 45 degrees = 1/4 `pi radians 60 degrees = 1/3 `pi radians confidence assessment: 2
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16:06:42 `q013. Since 180 deg = `pi rad, we can convert an angle from degrees to radians or vice versa if we multiply the angle by either `pi rad / (180 deg) or by 180 deg / (`pi rad). Use this idea to formally convert 30 deg, 45 deg and 60 deg to radians.
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RESPONSE --> 30deg = .52 radians 45deg = .78 radians 60deg = 1.04 radians confidence assessment: 2
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16:06:57 To convert 30 degrees to radians, we multiply by the rad / deg conversion factor, obtaining 30 deg * ( `pi rad / 180 deg) = (30 deg / (180 deg) ) * `pi rad = 1/6 * `pi rad = pi/6 rad. To convert 45 degrees to radians we use the same strategy: {}45 deg * (`pi rad / 180 deg) = ( 45 deg / ( 180 deg) ) * `pi rad = 1/4 * `pi rad = `pi/4 rad. To convert 60 degrees: 60 deg * (`pi rad / 180 deg) = ( 60 deg / ( 180 deg) ) * `pi rad = 1/3 * `pi rad = `pi/3 rad.
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RESPONSE --> got it. self critique assessment: 3
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16:07:52 `q014. Convert 50 deg and 78 deg to radians.
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RESPONSE --> 50 deg = .28 `pi radians 78 deg = .43 'pi radians confidence assessment: 3
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16:08:19 50 deg * (`pi rad / 180 deg) = ( 50 deg / ( 180 deg) ) * `pi rad = 5/18 * `pi rad = (5 `pi/ 18) rad. 78 deg * (`pi rad / 180 deg) = ( 78 deg / ( 180 deg) ) * `pi rad = 78/180 * `pi rad = (13 `pi/ 30) rad.
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RESPONSE --> i didn't do it exactly like that but i would have gotten the same outcome self critique assessment: 3
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16:09:44 `q015. Convert (14 `pi / 9) rad to degrees.
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RESPONSE --> (14`pi / 9)radians 1.55 * 180 = 279 degrees confidence assessment: 2
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16:09:50 Since the angle is given radians, we need to multiply by deg / rad to get the angle in degrees. (14 `pi / 9) rad * ( 180 deg / (`pi rad)) = ( 14 `pi / 9 ) * (180 / `pi ) deg = ( 14 * 180 / 9) * (`pi / `pi) deg = 14 * 20 deg = 280 deg.
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RESPONSE --> got it self critique assessment: 3
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