course Phy 201 X`d_tassignment #026
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10:47:12 `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 --> The force of gravity on the object is 9.8m/s^2*.4kg=3.92 N Now, the other thing we know is the object has a volume of 300cm^3. Since water is 1 g to cm^2, we know that the 300cm^3 has a mass of 300 g or .3kg. We can now find the force of gravity on this as we did above: .3kg*9.8m/s^2=2.94N We can use the formula that we have been using where we add all the components (plus tension) and set them equal to zero, enabling us to solve for T. So, since gravity is pushing downward and the buoyancy is pushing upward, gravitationa force will be negative, whereas buoyancy will be positive. -3.92N + 2.94N + T =0 -.98 + T=0 T=.98N confidence assessment: 2
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10:47:34 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 --> I figured I messed up somewhere, but I feel as though I actually understand this well. self critique assessment: 3
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10:51:12 `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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RESPONSE --> I kept rereading this and I think we are supposed to use the information given to find the volume first. So, 8 cm^2 * 12 cm (depth)=96 cm^2--the volume Since water is 1 g to one cm^2, we know the mass is 96 g. Then, to find the buoyancy force, I think I just use the gravitational thing again. .096g*9.8cm^2=.94 N I guess this is the answer.. confidence assessment: 1
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10:51:17 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 --> I understand. self critique assessment: 2
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11:01:01 `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 --> Fgrav= .08 kg * 9.8 cm^2= .79 N Since gravity is pushing downward, this will be negative. Now, the buoyant force we found before to be positive .94 N. Fnet=-.79+.94=+.15N confidence assessment: 2
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11:01:43 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 didn't finish out the problem I should have filled in the Fnet and the mass to find the acceleration at the time of release. self critique assessment: 2
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