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r.maliak
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i misused the mass from the original cubes mass to the iron cubes mass so 3.9x10^-3 - 3.68x10^-3 = 2.2 x10^-4
And then just found the volume from there which gave 2.82x10^-8
They got 3.2x10^-8, by doing:
any mass divided by 7800 ✔
V × 7800 + (4.0 × 10^–6 –V) × 920 = 3.9 × 10^–3 ✔
6880 V = 3.9 × 10^–3 −3.68 × 10^–3 ✔
V = 3.2 × 10^–8 m3 ✔
The density of water is 1000 and the density of the original ice cube was 920
And then just found the volume from there which gave 2.82x10^-8
They got 3.2x10^-8, by doing:
any mass divided by 7800 ✔
V × 7800 + (4.0 × 10^–6 –V) × 920 = 3.9 × 10^–3 ✔
6880 V = 3.9 × 10^–3 −3.68 × 10^–3 ✔
V = 3.2 × 10^–8 m3 ✔
The density of water is 1000 and the density of the original ice cube was 920
Last edited by r.maliak; 1 year ago
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UnknowNAlegend
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#2
do u have the starting question
btw u cant minus them since theyre not the same density; theyre different materials
btw u cant minus them since theyre not the same density; theyre different materials
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Eimmanuel
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(Original post by r.maliak)
i misused the mass from the original cubes mass to the iron cubes mass so 3.9x10^-3 - 3.68x10^-3 = 2.2 x10^-4 .....
i misused the mass from the original cubes mass to the iron cubes mass so 3.9x10^-3 - 3.68x10^-3 = 2.2 x10^-4 .....
2.2 x10^-4 is NOT the entire mass of iron and the “entire mass” 3.68x10^-3 (=920×4.0×10^-6) is derived from the density of ice solely.
(Original post by r.maliak)
....And then just found the volume from there which gave 2.82x10^-8 ....
....And then just found the volume from there which gave 2.82x10^-8 ....
Note that
weight of water displaced = weight of the cube (ice + iron)
Note that V is 4.0 × 10-6 m3.
Last edited by Eimmanuel; 1 year ago
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