course Phy 121 Could you please email me my grade in Physics 121 as of right now? I would like to keep track of how I am doing. `щwv\}Student Name:
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16:44:17 `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 --> m1 = 6*10^24 kg m2 = 3000 kg r = 10,000 km = 10^7 m (constant) G = 6.67*10^-11 N*m^2/kg^2 F = G (m1*m2)/ r^2 = (6.67*10^-11 N*m^2/kg^2)(6*10^24 kg)(3000 kg) / (10^7m)^2 = 12006 Newtons
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16:45:03 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 understand this problem.
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16:47:43 `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 --> m1 = 6*10^24 kg m2 = 3000 kg r = 6400 km = 6.4 * 10^6 m F = G (m1 * m2) / r^2 = (6.67*10^-11 N*m^2/kg^2)(6*10^24 kg)(3000kg) / (6.4*10^6m)^2 = 29311 Newtons
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16:48:20 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 --> I understand this problem.
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16:50:27 `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 --> m = 3000 kg r = 10,000 km = 10^7 m F = 12006 N In order to find the acceleration of the satellite toward the center of the Earth, we use the force at 10,000 km from the center of the Earth and the satellite's mass in the following expression: a = F/m = 12006N/3000kg = 4.002 m/s^2
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16:51:04 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 --> Ok.
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16:57:33 `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 --> aCent = v^2/r If v = 5000 m/s and r = 10^7 meters, therefore, aCent = (5000m/s)^2 / (10^7m) = 2.5 m/s^2 This means that the acceleration of gravity of the object is greater than its centripetal acceleration.
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16:57:45 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 understand this problem.
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17:00:40 `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 --> v = 10,000 m/s r = 10^7 m aCent = v^2 / r = (10,000m/s)^2 / 10^7m = 10 m/s^2 The centripetal acceleration is greater than the acceleration that would be experienced by an object this distance from the center of the Earth. If the acceleration of gravity is 4m/s^2, then the centripetal acceleration of the object is greater.
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17:00:48 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 --> I understand.
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17:04:05 `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 --> m = 3000 kg r = 10 ^7 m a grav = 4 m/s^2 aCent = v^2/r v^2 = aCent * r v = sqrt(aCent * r) = sqrt(4m/s^2 * 10^7m) = 6325 m/s
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17:04:20 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 feel that I solved this problem accurately.
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17:06:35 `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 --> m1 = 5*10^24 kg m2 = 5 kg r = 10^7 m aCent = g` F = G (m1 * m2) / r^2 = (6.67*10^-11 N*m^2/kg^2)(6*10^24 kg)(5 kg) / (10^7 m)^2 = 20.01 N a = F/m = 20.01 N / 5 kg = 4.002 m/s^2 If we look back to question #3, we see that this is equal to the 3000 kg object's gravitational acceleration.
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17:07:01 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 --> Ok.
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17:08:18 `q008. What therefore will be the orbital velocity of the 5-kg piece of junk?
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RESPONSE --> a_cent = 4.002 m/s^2 r = 10^7 m a cent = v^2 / r v = sqrt(aCent*r) = sqrt(4.002m/s^2 * 10^7m) = 6326 m/s We see that this is nearly the same velocity which was found for the 3000 kg object.
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17:08:55 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 --> I understand this problem.
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17:16:23 `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 --> We find gravitational acceleration of an object by Newton's Second Law of Motion to be, a = Fgrav/mObject. If we know the mass of the object and the mass of the Earth we can find the gravitational force by the formula, Fgrave = G(m1*m2)/r^2. In this equation, G is the gravitational constant equal to 6.67*10^-11 N*m^2/kg^2, r is the distance from the center of the Earth, m1 is the mass of the Earth, and m2 is the mass of the object. If we combine these two equations and substitute G(m1*m2)/r^2 into the acceleration equation we will get an equation that looks like this: a = (G(mEarth*mObject)/r^2) / mObject = (G * mEarth) / r^2
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17:17:38 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 --> Ok. This equation is based on the assumptions that: G is the universal constant, Earth's mass is unchanging, and the distance (r) from the center of the Earth is unchanging as well.
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17:19:07 `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 --> F = 12006 N `ds = 10002km - 10000km = 2km = 2000 m The distance change by a factor of 1.0002 this means that the force will also change by a factor of 1.0002. Therefore, the new force is 12006N * 1.0002 = 12008 N. Now we use the change in distance and the corresponding new force to find the work done against gravity to move the satellite. `dW = F * `ds = 12008N * 2000m = 2.4*10^7 Joules
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17:20:31 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 --> I think I understand this problem.
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17:22:55 `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 --> No, the mass, distance, and acceleration of the object will affect the force required to move it. A change in any of these would change the amount of work done and KE gain for the satellite. This amount of work could result from a number of combinations of distance, mass, and force. The values do not have to be exactly what they were in this case. Since work is product of force and change in distance as well as the sum of potential and kinetic energy, there could be a variety of combinations that produce the same amount of work, 2.4*10^7 Joules.
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17:25:45 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.
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RESPONSE --> I'm not sure I explained this accurately.
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r݁R assignment #028 028. `query 28 Physics I 07-15-2006
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23:11:36 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 --> If r is the radius of Earth, rE, the gravitational acceleration is equal to the centripetal acceleration. We know that gravitational acceleration is directly proportional to the square of the radius of Earth. Therefore, if r = rE, the acceleration is directly proportional to r^2 as well. When r = rE, the gravitational acceleration is 9.8m/s^2. aGrav = 9.8m/s^2 = k*rE^2 = k*r^2 Rearrange the equation to solve for the proportional constant: k = (9.8m/s^2)/r*2 Now we plug this expression for the constant back into the original proportionality: aGrav = [(9.8m/s^2)/r*2] * rE^2 = (9.8m/s^2 * rE^2)/r^2 = 9.8m/s^2 (rE/r)^2 The equation for centripetal acceleration is aCent = v^2/r. We know that under these circumstances, aGrav = aCent. In order to solve for velocity, we simply set these two expressions equal to each other and solve for velocity. 9.8m/s^2 (rE/r)^2 = v^2/r v^2 = [9.8m/s^2 (rE/r)^2] * r v^2 = (9.8m/s^2 * rE^2) / r v = sqrt[(9.8m/s^2 * rE^2)/r]
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23:11:45 ** 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 --> I feel that my answer for this question is adequate.
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23:15:30 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 --> r = 1.74*10^6 m m = 7.35*10^22 kg aGrav = G*m/r^2 = (6.67*10^-11 N*m^2/kg^2)(7.35*10^22 kg) / (1.74*10^6 m)^2 = 1.619 m/s^2
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23:15:41 ** 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 understand this problem.
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23:19:32 Query gen phy problem 5.40 force due to planets (Mv, Mj, Ms, are .815, 318, 95.1 Me; orb radii 108, 150, 778, 1430 million km). What is the total force on Earth due to the planets, assuming perfect alignment?
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RESPONSE --> we have previously seen that the mass of Earth is about 6*10^24 kg. We can use this information to solve for the force on Earth due to each of the planets. F = G * mE * mP / (r2 - r1)^2 The radii are in km, so we must convert them in order to use them in the equation. I TRIED TO WORK A FEW OF THESE BUT I CAN TELL FROM THE NUMBERS THAT THEY ARE WRONG.
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23:20:37 ** Using F = G m1 m2 / r^2 we get Force due to Venus: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (.815 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.08 * 10^11 m)^2 = 1.1 * 10^18 N, approx. Force due to Jupiter: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (318 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 7.78 * 10^11 m)^2 = 1.9 * 10^18 N, approx. Force due to Saturn: F = 6.67 * 10^-11 N m^2 / kg^2 * (5.97 * 10^24 kg) * (95.7 * 5.97 * 10^24 kg) / (1.5 * 10^11 m - 1.43 * 10^11 m)^2 = 1.4 * 10^17 N, approx. Venus being 'inside' the Earth's orbit pulls in the direction of the Sun while Jupiter and Saturn pull in the opposite direction so the net force is -1.1 * 10^18 N + 1.9 * 10^18 N + 1.4 * 10^17 N = .9 * 10^18 N = 9 * 10^17 N, approx.. **
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RESPONSE --> I wanted to try this problem, and I used Example 5-12 on page 120 to guide me, but my answers were wrong. I know this question is above my course level, but could you explain to me the method you used to solve it?
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23:25:41 Univ. 12.50 (12.44 10th edition). 25 kg, 100 kg initially 40 m apart, deep space. Both objects have identical radii of .20 m. When 20 m apart what is the speed of each (relative to the initial common speed, we presume), and what is the velocity relative to one another? Where do they collide? Why does position of center of mass not change?
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RESPONSE --> This question is above my course level.
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23:26:38 The force would be F = (6.67 * 10^-11 * 25 * 100) / 20^2 F = 4.17 * 10^-10 a1 = 4.17 * 10^-10 / 25 a1 = 1.67 * 10^-11 m/s/s a2 = 4.17 * 10^-10 / 100 a2 = 4.17 * 10^-12 m/s/s The position of center of mass doesn't change because the two spheres are the same size. ** At separation r the force is F = G m1 m2 / r^2. For any small increment `dr of change in separation the approximate work done by the gravitational force is F `dr = G m1 m2 / r^2 * `dr. We take the sum of such contributions, between the given separations, to form an approximation to the total work done by the gravitational force. We then take the limit as `dr -> 0 and obtain the integral of G m1 m2 / r^2 with respect to r from separation r1 to separation r2. An antiderivative is - G m1 m2 / r; evaluating between the two separations we get - G m1 m2 / r1 - (-G m1 m2 / r2) = G m1 m2 ( 1/r2 - 1 / r1). This expression is evaluated at r1 = 40 m and r2 = 20 m to get the change G m1 m2 ( 1/(20 m) - 1 / (40 m) ) in KE. I get around 1.49 * 10^-9 Joules but it isn't guaranteed so you should verify that carefully. Assuming a reference frame initially at rest with respect to the masses the intial momentum is zero. If the velocities at the 20 m separation are v1 and v2 we know that m1 v1 + m2 v2 = 0, so that v2 = -(m1 / m2) * v1. The total KE, which we found above, is .5 m1 v1^2 + .5 m2 v2^2. Substituting v2 = - (m1 / m2) v1 and setting equal to the KE we can find v1; from this we easily find v2. You might get something like 4.1 * 10^-6 m/s for the velocity of the 100 kg mass; this number is again not guaranteed so verify it yourself. The position of the center of mass does not change because there is no external force acting on the 2-mass system. The center of mass is at position r with respect to m1 (take m1 to be the 25 kg object) such that m1 r - m2 (40 meters -r) = 0; substituting m1 and m2 you get 25 r - 100 (40 meters - r ) = 0. I believe you get r = 4 / 5 * 40 meters = 32 m, approx., from the 25 kg mass, which would be 8 meters from the 100 kg mass.
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RESPONSE --> I couldn't do this one. I didn't even know where to start.
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23:29:05 Query gen phy problem 5.50 24 m diam wheel, rot period 12.5 s, fractional change in apparent weight at top and at bottom. What is the fractional change in apparent weight at the top and that the bottom of the Ferris wheel?
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RESPONSE --> This question is above my course level.
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23:38:04 ** Centripetal acceleration is a = v^2 / r. For a point on the rim of the wheel, v = dist in 1 rev / time for 1 rev = `pi * 24 m / (12.5 sec) = 1.9 m/s, approx. Thus v^2 / r = (`pi * 1.9 m/s)^2 / 12 m = 3 m/s^2, approx. At the top the only accel is the centripetal, and it is acting toward the center, therefore downward. The forces acting on any mass at the top are the gravitational force and the force exerted by the wheel on the mass. At the top of the wheel the latter force is the apparent weight. Thus grav force + apparent weight = centripetal force - m * 9.8 m/s^2 + wtApparent = m * (-3 m/s^2 ) wtApparent = m (-3 m/s^2) + m ( 9.8 m/s^2) = m (6.8 m/s^2). A similar analysis at the bottom, where the centripetal force will be toward the center, therefore upward, gives us - m * 9.8 m/s^2 + wtApparent = m * (+3 m/s^2 ) wtApparent = m (+3 m/s^2) + m ( 9.8 m/s^2) = m (12.8 m/s^2). The ratio of weights is thus 12.8 / 6.8, approx. ** A more elegant solution obtains the centripetal force for this situation symbolically: Centripetal accel is v^2 / r. Since for a point on the rim we have v = `pi * diam / period = `pi * 2 * r / period, we obtain aCent = v^2 / r = [ 4 `pi^2 r^2 / period^2 ] / r = 4 `pi^2 r / period^2. For the present case r = 12 meters and period is 12.5 sec so aCent = 4 `pi^2 * 12 m / (12.5 sec)^2 = 3 m/s^2, approx. This gives the same results as before. **
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RESPONSE --> Ok. NOTE: a = v^2 / r. v = dist in 1 rev / time for 1 rev FCent = Fgrav + apparent weight
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23:38:35 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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RESPONSE --> I tried to answer any questions that I thought I was capable of answering. A few of them were just out of my league. The equation about the different planets had me stumped. What was the formula for force that was used?
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