course Phy 201 €ÏžÛyÂßlÁè—©¢Ë³¤²ŠÀéÇÀassignment #026
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08:21:14 `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 --> We wil have a gravitational force of .4kg*9.8m/s^2=3.92N. Since the the volume of the cylinder is 300cm^3 and 1gram =1cm^3 then we have 1gr*300cm^3=300gr. we now find the buoyant force of .3kg*9.8m/s^2=2.94N. Since it is at equilibrium then we must not have any force. With this we wil make grav. force negitve since it is pulling down and the buoyant force positive since it is pushing up. We have -3.92N+2.94N+T=0, -.98N+T=0,T=.98N confidence assessment: 2
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08:23:26 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 --> ok self critique assessment: 3
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08:32:18 `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 --> First we take gravitational force which is .08kg*9.8m/s^2=.78N. The buoyant force from the last quesstion was .94N. We can now find the Fnet=-.78N+94N=.16N. confidence assessment: 3
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08:33:11 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 --> ok self critique assessment: 3