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HW_14_solutions

# HW_14_solutions - HW 14 Spring 2006 Physics 112 13.90 The...

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HW 14 Spring 2006 – Physics 112 13.90: The moment of inertia about the pivot is , ) 3 2 ( ) 3 1 ( 2 2 2 ML ML = and the center of gravity when balanced is a distance ) 2 (2 L d = below the pivot (see Problem 8.95). From Eq. (13.39), the frequency is f = 1 T = 1 2 " 3 g 4 2 L = 1 4 " 3 g 2 L # 13.54: At resonance, Eq. (13.46) reduces to . 2 b) . a) . 1 3 d max 1 A b F A A ! = Note that the resonance frequency is independent of the value of b (see Fig. (13.27)). 14.10: a) The pressure used to find the area is the gauge pressure, and so the total area is ! = " " 2 3 3 cm 805 ) Pa 10 205 ( ) N 10 5 . 16 ( b) With the extra weight, repeating the above calculation gives 2 cm 1250 . 14.11: a) " gh = (1.03 # 10 3 kg m 3 )(9.80 m s 2 )(250 m) = 2.52 # 10 6 Pa. b) The pressure difference is the gauge pressure, and the net force due to the water and the air is (2.52 " 10 6 Pa)( # (0.15 m) 2 ) = 1.78 " 10 5 N. 14.21: The buoyant force must be equal to the total weight; " water V g = " ice V g + mg , so V = m " water # " ice = 45.0 kg 1000 kg m 3 # 920 kg m 3 = 0.563 m 3 , or 3 m 56 . 0 to two figures. 14.54: The difference between the densities must provide the “lift” of 5800 N (see Problem 14.59). The average density of the gases in the balloon is then " ave = 1.23 kg m 3 # (5800N) (9.80 m s 2 )(2200m 3 ) = 0.96 kg m 3 .

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14.60: a) Archimedes’ principle states . so , A M L Mg gLA ! ! = = b) The buoyant force is , ) ( F Mg x L gA + = + ! and using the result of part (a) and solving for x gives x = F " gA . c) The “spring constant,” that is, the proportionality between the displacement x and the applied force F , is k = " gA , and the period of oscillation is T = 2 " M k = 2 " M # gA .
#1: Mutiny on the Canal We can reformulate the question is this way: does the scrap metal displace more water when it is thrown into the water, or when it is placed in the boat, thus making the boat sink lower into the water? The easy part first: when the metal is thrown into the water it displaces a volume of water equal to the volume of the metal. However, when the scrap metal is placed on board a different amount of water is displaced: Archimedes’ principle states that the water exerts a force on the boat equal to the weight of the water displaced by the boat. The boat is not accelerating in the vertical

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