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Why things float -. Archimedes’ Principle. Hydrostatics Part 2.

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Archimedes’ principle.

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Archimedes’ principle.

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Related image. King Hieron of Syracuse suspects that his new gold crown is not pure. Archimedes measures the crown’s mass to be 1,200 kg in air and to be apparently 1,127 kg when immersed in water. Was the king cheated?.

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King Hieron of Syracuse suspects that his new gold crown is not pure. Archimedes measures the crown’s mass to be 1,200 kg in air and to be apparently 1,127 kg when immersed in water. Was the king cheated?.

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King Hieron of Syracuse suspects that his new gold crown is not pure. Archimedes measures the crown’s mass to be 1,200 kg in air and to be apparently 1,127 kg when immersed in water. Was the king cheated?.

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King Hieron of Syracuse suspects that his new gold crown is not pure. Archimedes measures the crown’s mass to be 1,200 kg in air and to be apparently 1,127 kg when immersed in water. Was the king cheated?.

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Try this example before continuing with the presentation. Hit the down arrow when ready to continue..

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What fraction of an iceberg’s volume is above sea level?.

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What fraction of an iceberg’s volume is above sea level?.

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A steel ball 0,15 m in diameter hangs from the end of an aluminium wire, 0,2 mm in diameter and 1,2 m long. Find the change in length of the wire when the ball is immersed in water..

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A steel ball 0,15 m in diameter hangs from the end of an aluminium wire, 0,2 mm in diameter and 1,2 m long. Find the change in length of the wire when the ball is immersed in water..

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A steel ball 0,15 m in diameter hangs from the end of an aluminium wire, 0,2 mm in diameter and 1,2 m long. Find the change in length of the wire when the ball is immersed in water..

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A steel ball 0,15 m in diameter hangs from the end of an aluminium wire, 0,2 mm in diameter and 1,2 m long. Find the change in length of the wire when the ball is immersed in water..

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Lecture Example 2. Image result for beaker. A patient has a mass of 60 kg and density of 960 kg.m -3 . She is supported by 4 rubber straps, each of length 1.2 m and cross-sectional area of 2×10 -4 m 2 , and is then lowered into water until 60% of her volume is under water. Given that Y rubber = 5×10 6 N.m -2 , Calculate the upthrust on the woman. Calculate the change in length of each strap caused by the immersion..

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Lecture Example 2. Image result for beaker. A patient has a mass of 60 kg and density of 960 kg.m -3 . She is supported by 4 rubber straps, each of length 1.2 m and cross-sectional area of 2×10 -4 m 2 , and is then lowered into water until 60% of her volume is under water. Given that Y rubber = 5×10 6 N.m -2 , Calculate the upthrust on the woman. Calculate the change in length of each strap caused by the immersion..

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Lecture Example 2. Image result for beaker. A patient has a mass of 60 kg and density of 960 kg.m -3 . She is supported by 4 rubber straps, each of length 1.2 m and cross-sectional area of 2×10 -4 m 2 , and is then lowered into water until 60% of her volume is under water. Given that Y rubber = 5×10 6 N.m -2 , Calculate the upthrust on the woman. Calculate the change in length of each strap caused by the immersion..

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Lecture Example 2. Image result for beaker. A patient has a mass of 60 kg and density of 960 kg.m -3 . She is supported by 4 rubber straps, each of length 1.2 m and cross-sectional area of 2×10 -4 m 2 , and is then lowered into water until 60% of her volume is under water. Given that Y rubber = 5×10 6 N.m -2 , Calculate the upthrust on the woman..

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Lecture Example 2. Image result for beaker. A patient has a mass of 60 kg and density of 960 kg.m -3 . She is supported by 4 rubber straps, each of length 1.2 m and cross-sectional area of 2×10 -4 m 2 , and is then lowered into water until 60% of her volume is under water. Given that Y rubber = 5×10 6 N.m -2 , Calculate the upthrust on the woman. Calculate the change in length of each strap caused by the immersion..