Find the area and length of a gold leaf

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MachineInTheStone
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Homework Statement



Gold, which has a density of 19.32 g/cm3, is the most ductile metal and can be pressed into a thin leaf or drawn out into a long fiber. (a) If a sample of gold with a mass of 3.872 g, is pressed into a leaf of 5.372 μm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.300 μm, what is the length of the fiber?

Homework Equations


d = m/v
v = pi * r^2 * L
v = l*w*h

The Attempt at a Solution


d = m/v[/B]
[Part A]
1) 19.32 g / 1 cm^3 = 3.872g / v
v = 4.98967 cm^3

2) 5.372 micrometer * 1 cm / 10,000 micrometer
= 0.0005372 cm

3) 4.98967 cm^3 / 0.0005372 cm
= 9288 cm^2

4) 9288 cm^2 * 1 m^2 / 10,000 cm^2
= 0.9288 m^2

[part B]
1) 2.300 μm * 1 cm / 10,000 μm
= 0.00023 cm

2) v = pi * r^2 * L
Since v = 4.98967 cm^3 ...
4.98967 cm^3 = pi * (0.00023 cm)^2 * L
4.98967 cm^3 = 0.0000001662 cm^2 * L
L = 30022100 cm

3) 30022100 cm * 1 m / 0.01 cm
= 3,002,210,000 m

Are part A and B correct?
B seems totally wrong!
 
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MachineInTheStone said:
L = 30022100 cm

3) 30022100 cm * 1 m / 0.01 cm
= 3,002,210,000 m
How come the number is bigger in m than in cm!?
 
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ah.
L = 30022100 cm

3) 30,022,100 cm * 1 m / 100 cm
= 300,221 m

Is part A and B correct now?
 
MachineInTheStone said:
1) 19.32 g / 1 cm^3 = 3.872g / v
v = 4.98967 cm^3
No. 1/v = 4.98967 cm-3
 
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