Homework 35 - homework 35 BAUTISTA ALDO Due 4:00 am Version...

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homework 35 – BAUTISTA, ALDO – Due: Apr 26 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 1. 10 points. A 2 . 1 cm thick bar of soap is floating on a water surface so that 1 . 75 cm of the bar is underwater. Bath oil (specific gravity 0 . 8) is poured into the water and floats on top of the water. 1 . 75 cm 2 . 1 cm SOAP water y oil 2 . 1 cm 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 . 75 cm (tolerance ± 1 %). Explanation: Given : h = 2 . 1 cm , y = 1 . 75 cm , specific gravity = ρ oil ρ water = 0 . 8 . 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 . 1 cm) - (1 . 75 cm) 1 - 0 . 8 = 1 . 75 cm Question 2 Part 1 of 1. 10 points. A frog in a hemispherical pod finds that he just floats without sinking in a fluid with a density of 1 . 36 g / cm 3 . If the pod has a radius of 6 . 54 cm and negligible mass, what is the mass of the frog? Correct answer: 0 . 796766 kg (tolerance ± 1 %).
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homework 35 – BAUTISTA, ALDO – Due: Apr 26 2006, 4:00 am 2 Explanation: Given : ρ fluid = 1 . 36 g / cm 3 100 cm m 3 · 1 kg 1000 g = 1360 kg / m 3 and r = 6 . 54 cm = 0 . 0654 m . Since the frog floats, the buoyant force equals the weight of the frog, so the weight of the displaced fluid equals the weight of the frog: ( ρ fluid V ) g = m frog g m frog = ρ fluid V = ρ fluid 1 2 4 3 π r 3 = 2 3 ρ fluid π r 3 = 2 3 ( 1360 kg / m 3 ) π (0 . 0654 m) 3 = 0 . 796766 kg . Question 3 Part 1 of 1. 10 points. A jet of water squirts horizontally from a hole near the bottom of the tank. The hole is 1 . 2 m above the floor and the stream of water extend a distance of 0 . 6 m before hitting the floor, as shown below. h 1 . 2 m 0 . 6 m If the hole has a diameter of 3 . 54 mm and the top of the tank is open, what is the height of the water in the tank? Correct answer: 0 . 075 m (tolerance ± 1 %). Explanation: Let : Δ x = 0 . 60 m and Δ y = - 1 . 2 m Basic Concepts: P 1 + 1 2 ρ v 2 1 + ρ g h 1 = P 2 + 1 2 ρ v 2 2 + ρ g h 2 For the motion of the water after leaving the tank, Horizontally: Δ x = v x Δ t since a x = 0 m/s 2 . Vertically: Δ y = - g t ) 2 2 since v y,i = 0 m/s. Solution: From the vertical equation Δ t = s - y g , and from the horizontal equation v x = Δ x Δ t = Δ x r - 2 Δ y g = Δ x r g - 2 Δ y .
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