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456_PartUniversity Physics Solution

# 456_PartUniversity Physics Solution - 14-20 Chapter 14...

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14-20 Chapter 14 14.75. (a) I DENTIFY : Apply Newton°s 2nd law to the crown. The buoyancy force is given by Archimedes° principle. The target variable is the ratio c w / ρ ρ (c crown, = w water). = S ET U P : The free-body diagram for the crown is given in Figure 14.75. E XECUTE : y y F ma = 0 T B w + = T fw = w c , B V g ρ = where w density ρ = of water, c volume V = of crown Figure 14.75 Then w c 0. fw V g w ρ + = w c (1 ) f w V g ρ = Use c c , w V g ρ = where c density ρ = of crown. c c w c (1 ) f V g V g ρ ρ = c w 1 , 1 f ρ ρ = as was to be shown. 0 f gives c w / 1 ρ ρ = and 0. T = These values are consistent. If the density of the crown equals the density of the water, the crown just floats, fully submerged, and the tension should be zero. When 1, f c w ρ ρ >> and . T w = If c w ρ ρ >> then B is negligible relative to the weight w of the crown and T should equal w . (b) ±apparent weight² equals T in rope when the crown is immersed in water. , T fw = so need to compute f . 3 3 c 19.3 10 kg/m ; ρ = × 3 3 w 1.00 10 kg/m ρ = × c w 1 1 f ρ ρ = gives 3 3 3 3 19.3 10 kg/m 1 1.00 10 kg/m 1 f × = × 19.3 1/(1 ) f = and 0.9482 f = Then (0.9482)(12.9 N) 12.2 N. T fw = = = (c) Now the density of the crown is very nearly the density of lead; 3 3 c 11.3 10 kg/m . ρ = × c w 1 1 f ρ ρ = gives 3 3 3 3 11.3 10 kg/m 1 1.00 10 kg/m 1 f × = × 11.3 1/(1 ) f = and 0.9115 f = Then (0.9115)(12.9 N) 11.8 N. T fw = = = E VALUATE : In part (c) the average density of the crown is less than in part (b), so the volume is greater. B is greater and T is less. These measurements can be used to determine if the crown is solid gold, without damaging the crown. 14.76. I DENTIFY : Problem 14.75 says object fluid 1 1 f ρ ρ = , where the apparent weight of the object when it is totally immersed in the fluid is fw . S ET U P : For the object in water, water water / f w w = and for the object in the unknown fluid, fluid fluid / f w w = . E XECUTE : (a) steel fluid fluid , ρ w ρ w w = steel fluid water ρ w ρ w w = . Dividing the second of these by the first gives fluid fluid water water . ρ w w ρ w w = (b) When fluid w is greater than water, w the term on the right in the above expression is less than one, indicating that the fluids is less dense than water, and this is consistent with the buoyant force when suspended in liquid being less than that when suspended in water. If the density of the fluid is the same as that of water fluid w = water w , as expected. Similarly, if fluid w is less than water w , the term on the right in the above expression is greater than one, indicating that the fluid is denser than water.

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Fluid Mechanics 14-21 (c) Writing the result of part (a) as fluid fluid water water 1 , 1 ρ f ρ f = and solving for fluid , f ( ) ( ) ( ) fluid fluid water water 1 1 1 1.220 0.128 0.844 84.4%. ρ f f ρ = = = = E VALUATE : Formic acid has density greater than the density of water. When the object is immersed in formic acid the buoyant force is greater and the apparent weight is less than when the object is immersed in water.
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456_PartUniversity Physics Solution - 14-20 Chapter 14...

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