ws20 - water V disp g = m iron g = iron V iron g This means...

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Worksheet 20: Fluid Statics Questions 1. At the bottom of the sea, you are bearing the weight of the water (and air) above you, while at the top of the mountain, you are only carrying the weight of the air above you. 2. If pressure were lower at one end of the tub, then water would flow from the high pressure side of the tub to the low pressure side of the tub; this means we are not in static equilibrium. Hence water at the same vertical height is under the same pressure. 3. Both buckets weigh the same amount. Looking at the bucket with the block of wood: the amount of water that was lost by putting in the block is equal to the weight of the block (we know this because the block is floating, so F B = m block g = m disp. water g ). Hence, there is no net change in weight. 4. The water level in the lock would lower. To see this, consider the amount of water displaced before and after. Before the iron is tossed in, the weight amount of water displaced equals the weight of the iron ( F B = ρ water V disp g = m iron g = ρ iron V iron
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Unformatted text preview: water V disp g = m iron g = iron V iron g ). This means that the water displaced is V disp = iron water V iron . After the iron is thrown overboard, the volume of water displaced is equal to the volume of the iron (because the iron is completely submerged). Since iron is more dense than water, iron water V iron > V iron we see more water was displaced when the iron was on the boat; hence the water level was higher then also. Problems 1. A hollow spherical iron shell Foats almost completely submerged in water. The inner diameter is: d inner = 2 p 1- water iron P 1 / 3 R = 56 . 1 cm 2. A helium balloon lifts a person. The radius is: r = b 3( m + M ) 4 ( air- He ) B 1 / 3 = 2 . 8 m 3. A cubical block of wood Foats in water. 1 (a) wood = water ( 1-. 08 . 20 ) = 600 kg/m 3 (b) F = water L 2 yg where y is the distance from equilibrium. (c) T = 2 r wood L water g = 0 . 695 s 2...
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