hw 37 - homework 37 HETTIGER CHRISTOF Due 4:00 am Question...

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homework 37 – HETTIGER, CHRISTOF – Due: Apr 29 2008, 4:00 am 1 Question 1, chap 18, sect 5. part 1 of 1 10 points A block of mass 8 . 6 kg is completely im- mersed in a liquid of density 1150 kg / m 3 . The block is suspended by a thin wire which experiences a tension of 23 . 3 N. The acceleration of gravity is 9 . 8 m / s 2 . 8 . 6 kg Scale What is the volume of the submerged block? Correct answer: 0 . 00541083 m 3 (tolerance ± 1 %). Explanation: Let : m = 8 . 6 kg , block s mass ρ L = 1150 kg / m 3 , and g = 9 . 8 m / s 2 , Basic Concepts: Archimedes’ Principle: Buoyant Force = Weight of Fluid Displaced Solution: The block is in equilibrium, so summationdisplay F y = 0 = T + ρ L V g - m g , where ρ L V g is the buoyant force on the block. Solving for the volume V gives V = m g - T ρ L g = (8 . 6 kg) (9 . 8 m / s 2 ) - (23 . 3 N) (1150 kg / m 3 ) (9 . 8 m / s 2 ) = 0 . 00541083 m 3 . Question 2, chap 18, sect 5. part 1 of 1 10 points A block of volume 0 . 39 m 3 floats with a fraction 0 . 64 of its volume submerged in a liquid of density 1150 kg / m 3 , as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . V 1 liquid Find the magnitude of the buoyant force on the block. Correct answer: 2812 . 99 N (tolerance ± 1 %). Explanation: Basic Concepts: Archimedes’ Principle: Buoyant Force on Object=Weight of Fluid Displaced by Object. F Buoyant,Part 1 = Weight of displaced liquid = ρ L parenleftbigg ρ 1 ρ L parenrightbigg V 1 g = (1150 kg / m 3 ) (0 . 64) × (0 . 39 m 3 ) (9 . 8 m / s 2 ) = 2812 . 99 N , where ρ 1 ρ L = 0 . 64 . Question 3, chap 18, sect 5. part 1 of 1 10 points A sample of an unknown material weighs 305 . 7 N in air and 222 . 0 N when submerged in an alcohol solution with a density of 0 . 644 × 10 3 kg / m 3 .
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