course PHY 121 ?M???e???????S????Student Name: assignment #026
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11:09:56 `q001. Note that this assignment contains 3 questions. . Water has a density of 1 g per cm^3. If an object is immersed in water, it experiences a buoyant force which is equal to the weight of the water it displaces. Suppose that an object of mass 400 grams and volume 300 cm^3 is suspended from a string of negligible mass and volume, and is submerged in water. If the mass is suspended in equilibrium, what will be the tension in the string?
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RESPONSE --> m=400g v=300cm^3 400g/1g cm^3 = 400cm^3 cos 400cm^3 = .76
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11:12:22 The 400 g mass will experience a downward gravitational force of .4 kg * 9.8 meters/second^2 = 3.92 Newtons. It will also experience in upward buoyant force equal to the weight of the 300 cm^3 of water it displaces. This volume of water, at 1 g per cm^3, will have a mass of 300 grams and therefore a weight of .3 kg * 9.8 meters/second^2 = 2.94 Newtons. The forces acting on the mass are therefore the downward 3.92 Newtons of gravity, the upward 2.94 Newtons of the buoyant force and the tension, which we will call T, in the string. Since the system is in equilibrium these forces must add up to 0. We thus have -3.92 Newtons + 2.94 Newtons + T = 0, which has solution T = .98 Newtons. In common sense terms, gravity pulls down with 3.92 Newtons of force and the buoyant force pushes of with 2.94 Newtons of force so to keep all forces balanced the string must pull up with a force equal to the .98 Newton difference.
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RESPONSE --> multiply the mass by acceleration of gravity and multiply volume by acceleration of gravity. The two forces add to the tension should equal zero.
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11:13:55 `q002. A solid cylinder has a cross-sectional area of 8 cm^2. If this cylinder is held with its axis vertical and is immersed in water to a depth of 12 cm, what will be the buoyant force on the cylinder?
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11:15:20 At a depth of 12 cm, the volume of the immersed portion will be 12 cm * 8 cm^2 = 96 cm^3. This portion will therefore displace 96 grams of water. The weight of this displace water will be .096 kg * 9.8 meters/second^2 = .94 Newtons. This will be the buoyant force on the cylinder.
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RESPONSE --> find the volume and multiply by the gravitational force.
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11:26:23 `q003. The solid cylinder in the preceding problem has a total length of 18 cm and a mass of 80 grams. If the cylinder is immersed as before to a depth of 12 cm then released, what will be the net force acting on it at the instant of release?
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RESPONSE --> .08kg * 9.8m/s^2 = .78N .096kg * 9.8m/s^2 = .94N 6cm * 8cm/s^2= 48cm/s^3 .048kg * 9.8cm/s^2 = .47N .94N - .78N +.47N = .63N
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11:29:04 The buoyant force on the cylinder is still .94 Newtons, directed upward. Gravity exerts a downward force of .080 kg * 9.8 meters/second^2 = .79 Newtons, approximately. The net force on the cylinder is therefore .94 N - .79 N = .15 N, directed upward. This will give its 80 gram mass and acceleration a = F / m = .15 N / .080 kg = 1.875 m/s^2. Note that as the cylinder rises less and less of its volume is submerged, so the buoyant force will decrease while the weight remains the same. Until the buoyant force has decreased to become equal and opposite to the weight, the net force will continue to be upward and the cylinder will continue to gain velocity. After this instant the cylinder will continue to rise, but the net force will be downward so that the cylinder will begin slowing down. Eventually the cylinder will come to rest and the net downward force will cause it to start descending once more. It will continue descending until the net force is again 0, at which the time it will have a downward velocity that will carry it beyond this point until it again comes to rest and the cycle will start over again.
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RESPONSE --> I did not have to do anything with the 6cm that was not submerged why?
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????????P?~? Student Name: assignment #027
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12:01:54 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 --> F = 10000N ds = 8000km
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12:02:46 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 --> 10,000N because they are equal and opposite forces.
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12:03:24 `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 --> gravity pulls the object to the planet.
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12:04:14 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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12:05:21 `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 --> The force will also be cut in half because they are equal and opposite.
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12:05:31 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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12:08:05 `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 force from the original planet was 10000N then half that would be 5000N. If the satellite is 6 times the mass and they are equal and opposite, then the force would be 5000N * 6 = 30000N
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12:08:47 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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12:10:20 `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 --> The original force of 10,000N but the obj. increases 5 times as far so the force would now be 10000N * 5 = 50,000N
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12:11:43 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 --> Divide the new distance by the old which is 5 times as far. Object will be at 1/5^2 for the new distance. 1/25 * 10000N = 400N.
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12:17:38 `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 --> F= 6.67^-11 * 100kg * (200kg/.5^2) = .000069 ???
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12:19:38 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 --> F=6.67 * 10^-11N m^2/kg^2 * 100kg *200kg/(.5m)^2 = 5.3*10^-6N
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12:20:06 `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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12:22:50 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 --> net force divided my mass use the 3rd equation of motion to find dt
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12:25:06 `q008. At what rate would the second object accelerate toward the first?
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RESPONSE --> f=m*a 5.3 *10^-6N = 200kg * a a= 5.3*10^-6N / 200kg = 2.65 * 10^-8
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12:25:32 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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