I need help with physics as questions?
1) The weight in air of a piece of aluminum is 4.70N, and its weight in water density 1000kgm-3 is 2.95N. Calculate: a) The volume of water displaced by aluminum b) the density of the aluminum sample 2) An ice cube is used to cool a drink. The ice bucket has a section cross fleet 2.0cm x 4.0cm and 0.2cm in the drink above the surface. Calculate the total thickness of the cube. 3) A piece of jewelry is made of an alloy of gold and densities 19.3gcm silver-3 and 10.5gcm-3, respectively. The piece of jewelry in the air weighs 9.90N and 9.35N in water density 1000kgm-3. Assuming no change in volume when gold and silver combine to produce the alloy, calculate the percentage by weight of gold in the piece of jewelry. Thanks, better get 5 stars:)
1 a) The difference in weight (1.75 N) is the buoyancy of water displaced. 1.75 = 1000 V (9.8), so the displaced water volume (= volume of aluminum) is 1.75/9800 = 1.80×10 ^ -4 m ^ 3. 1b) Mass of the piece of aluminum = 4.70/9.8 = 0.48 kg, so density of aluminum = 0.48/1.80×10 ^ -4 = 2667 kg / m ^ 3. 2) The mass of the cube ice is 931 (.02 *. 04 * T) = 0.745 T, where 931 is the density of ice and T is its thickness. The water body that moves (needed to support the ice cube) is 1000 (.02 *. 04 * (T-.002)) to 0.745 T = 1000 (.02 *. 04 * (T-.002)) or 0.745T = 0.8T – .0016 and = .029 M or 2 , 9 cm. T 3) The mass of jewelry is 9.9/9.8 = 1.01 kg = 1010 g The buoyancy of water is 9.90-9.35 = 0.55 N so that the volume of the piece is 0.55 / (9.8 * 1000) Cm = 5.61×10 ^ -5 m ^ 3 or 56.1 ^ 3 The density Part is rho = [rho (Au) * V (Au) + rho (Ag) * V (Ag)] / [V (Au) + V (Ag)] Since V (TOT) = 56.1 cm ^ 3 = V (Au) + V (Ag) and rho = 1010/56.1 = 18.0 g / cm ^ 3 then 18.0 = [19.3 * V (Au) + 10.5 * (56.1 - V (Au))] / 56.1 or 1009, 8 = 8.8 * V (Au) + 589.0 and V (Au) = 47.8 cm ^ 3 Thus, weight (Au) = 47.8 * 19.3 = 922.5 g and the% of Au is 922.5/1010 = 0.914 or 91.4% of UA
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