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dddan

  • 3 years ago

a 1.40 g sample of gold having a density of 19.32 g/cm^3 is hammered into a very thin sheet. if the area of the sheet is 35.0ft^2, what is the average thickness of the sheet expressed in centimeters?

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  1. .Sam.
    • 3 years ago
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    Using back the formula, \[\rho= \frac{m}{V}\] \[V= \frac{m}{\rho}\] \[V= \frac{1.40 g }{19.32 g/cm^3}\] \[V= 0.072cm^3\] ------------------------------------------ Changing 35.0ft^2, \[35.0ft^2=\frac{929cm^2}{1ft^2}=32515cm^2\] ------------------------------------------- We assume that the metal sheet is either cubed or cuboid, \[V=Length \times Base \times Height\] \[V=Aree \times Height\] \[Height=\frac{Volume}{Area}\] \[Height=\frac{0.072cm^3}{32515cm^2}\] \[Height=2.21 \times 10^{-6}cm\]

  2. dddan
    • 3 years ago
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    thanks can you explain how you get from v=basexhightxlength to v=areaxheight?

  3. .Sam.
    • 3 years ago
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    |dw:1348503704260:dw|

  4. dddan
    • 3 years ago
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    got it thanks

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