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# hw32 - Muraj Hamza – Homework 32 – Due 4:00 am – Inst...

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Unformatted text preview: Muraj, Hamza – Homework 32 – Due: Apr 21 2006, 4:00 am – Inst: Florin 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points A 2 . 2 cm thick bar of soap is floating on a water surface so that 1 . 46 cm of the bar is underwater. Bath oil (specific gravity 0 . 6) is poured into the water and floats on top of the water. 1 . 46cm 2 . 2cm SOAP water y oil 2 . 2cm SOAP water oil What is the depth of the oil layer when the top of the soap is just level with the upper surface of the oil? Correct answer: 1 . 85 cm. Explanation: Given : h = 2 . 2 cm , y = 1 . 46 cm , specific gravity = ρ oil ρ water = 0 . 6 . Let A be the surface area of the top or bottom of the bar. The weight of the soap bar is equal to the buoyant force when it floats in water alone: F net = B- W soap = 0 B = m fluid g ρ = m V V = A y SOAP SOAP B water W soap W soap B water + B oil in water in water and oil Before the oil is added: W soap = B water (1) = ρ water ( A y ) g . After the oil is added: W soap = B water + B oil (2) = ρ water [ A ( h- y oil )] g + ρ oil ( A y oil ) g , since y oil is the depth of the oil layer. Setting Eq. 1 equal to Eq. 2, we have ρ water ( A y ) g = ρ water [ A ( h- y oil )] g + ρ oil ( A y oil ) g ρ water y = ρ water h- ρ water y oil + ρ oil y oil ( ρ water- ρ oil ) y oil = ρ water ( h- y ) y oil = h- y 1- ρ oil ρ water (3) = (2 . 2 cm)- (1 . 46 cm) 1- . 6 = 1 . 85 cm 002 (part 1 of 2) 10 points A block of volume 0 . 48 m 3 floats with a frac- tion 0 . 61 of its volume submerged in a liquid of density 1270 kg / m 3 , as shown in the figure below. The acceleration of gravity is 9 . 8 m / s 2 . V B liquid Find the magnitude of the buoyant force on the block. Correct answer: 3644 . 19 N....
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hw32 - Muraj Hamza – Homework 32 – Due 4:00 am – Inst...

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