## dddan Group Title 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? 2 years ago 2 years ago

1. .Sam.

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

thanks can you explain how you get from v=basexhightxlength to v=areaxheight?

3. .Sam.

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4. dddan

got it thanks