5.2 Harder Questions 07 October 2011 07:46
· Collins p.106 Q.1-3. (Table of densities below) [cid:image001.png@01CC84CD.0F0A7F10] [cid:image002.jpg@01CC84CD.0F0A7F10]
[cid:image003.jpg@01CC84CD.0F0A7F10] ANSWERS Q1 Wood in Oil = Float
Wood in Mercury = Float
Plastic in Oil = Sink
Steel in Mercury = Float
Silver in Air = Sink
Gold in Mercury (this experiment must be done rapidly as the gold will dissolve quickly) = Sink
Helium balloon = Float Q2 Dimensions: 0.200 x 0.090 x 0.065 (0.065 = 3.d.p.) Mass of brick => m = W/g = 22.2/10 = 2.2kg Volume of Brick => V = 0.200 x 0.090 x 0.065 = 0.00126 Density of Brick => p = mass / volume = 2.2/0.00126 = 1746.031..... = 1746 kg/m^3 Q3 (1.93kg = 1930g) volume of crown = mass / density => 1930 / 19.3 = 100cm3 new water level = 800 + 100 = 900cm3 (if pure gold) IF not pure gold, the water level will be different because density affects the volume displaced (v = m/p) 
· Collins p.106 Q.1-3. (Table of densities below) [cid:image001.png@01CC84CD.0F0A7F10] [cid:image002.jpg@01CC84CD.0F0A7F10]
[cid:image003.jpg@01CC84CD.0F0A7F10] ANSWERS Q1 Wood in Oil = Float
Wood in Mercury = Float
Plastic in Oil = Sink
Steel in Mercury = Float
Silver in Air = Sink
Gold in Mercury (this experiment must be done rapidly as the gold will dissolve quickly) = Sink
Helium balloon = Float Q2 Dimensions: 0.200 x 0.090 x 0.065 (0.065 = 3.d.p.) Mass of brick => m = W/g = 22.2/10 = 2.2kg Volume of Brick => V = 0.200 x 0.090 x 0.065 = 0.00126 Density of Brick => p = mass / volume = 2.2/0.00126 = 1746.031..... = 1746 kg/m^3 Q3 (1.93kg = 1930g) volume of crown = mass / density => 1930 / 19.3 = 100cm3 new water level = 800 + 100 = 900cm3 (if pure gold) IF not pure gold, the water level will be different because density affects the volume displaced (v = m/p)
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