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Unformatted text preview: Physics 317K General Physics I Fall 2007 Homework 5 SOLUTIONS 1. d By definition 1Pa = 1N / m 2 = 1kg · m / s 2 · m 2 = 1kg / m · s 2 2. e Δ h = 10 3 m ρ = 8 . 5 × 10 2 kg / m 3 g = 9 . 8m / s 2 p = ρgh ⇒ Δ p = p f p i = ρg ( h f h i ) = ρg Δ h Δ p = (8 . 5 × 10 2 kg / m 3 )(9 . 8m / s 2 )(10 3 m) Δ p = 8 . 3Pa 3. e The pressure at a given depth depends only on that depth. Because the distance from the surface of the fluid to the bottom of the vessel is the same for all three vessels, the pressures at the bottom are all equal. 4. a Because points 1 and 2 are at the same height, the pressures are equal. p 1 = p 2 Let F cyl be the weight of the cylinder p 1 = ρgL p 2 = F cyl A p 1 = p 2 ⇒ F cyl = ALρg 1 5. b The pressures at points 1 and 2 are equal because they are at the same height. ρ H 2 O = 1 . 0g / cm 3 h = 5cm L = 2cm p 1 = ρ oil gh p 2 = ρ H 2 O g ( h L ) p 1 = p 2 ⇒ ρ oil gh = ρ H 2 O g ( h L ) ρ oil = ρ H 2 O ( h L ) h ρ oil = (1 . 0g / cm 3 )( 3cm 5cm ) ρ oil = 0 . 60g / cm 3 6. b Because the object floats in each fluid the buoyant force provided by each fluid must be equal to the weight of the object. Let V 1 , V 2 , V 3 be the volumes displaced by the object in each of the three fluids. Then because buoyant forces are equal F A = F B = F C we have . 9 ρ V 1 g = ρ V 2 g = 1 . 1 ρ V 3 g So . 9 ρ < ρ < 1 . 1 ρ ⇒ V 3 < V 2 < V 1 7. c The blocks are identical so this means that the weight of block A equals the weight of block B. Let V A , V B be the volumes of fluid displaced by the two blocks. Then, because the blocks are floating F buoyant = ρ H 2 O V A g = m A g and F buoyant = ρ H 2 O V B g = m B g but m A g = m B g ⇒ ρ H 2 O V A g = ρ H 2 O V B g So V A = V B 2 8. b Let F s be the spring force and V the volume of the submerged object. In air F s,air = 30N F s,air mg = 0 ⇒ F s,air = mg = 30N In water, let F H 2 O be the buoyant force provided by the water F s,H 2 O = 20N F s,H 2 O + F H 2 O mg = 0 ⇒ F H 2 O = ρ H 2 O gV = 10N In the unknown fluid, let F f be the buoyant force provided by the unknown ρ = 1 2 ρ H 2 O So F f = ρgV = 1 2 ( ρ H 2 O gV ) = 1 2 F H 2 O therefore F s,f + F f mg = 0 ⇒ F s,f = mg 1 2 F H 2 O = 30N 20N = 10N 9. c r 1 = 2 . 0cm r 2 = 8 . 0cm = 4 r 1 F 2 = 1600N Then...
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 Spring '07
 KOPP
 Physics, Work

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