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 1.915 g, is pressed into a leaf of 7.331 μm thickness, what is the area of the leaf?
(b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.100 μm, what is the length of the fiber?

a) 0.014 m2

b) 7.1 Km

Explanation:

Density = mass/volume

volume - mass/density

volume = 1.915 g/19.32 g/cm3 = 0.099 * 10^-6 m3

Volume = Area * thickness

thickness =  7.331 * 10^-6 m

Area = Volume/thickness

Area= 0.099 * 10^-6 m3/7.331 * 10^-6 m

Area=  0.014 m2

b. Volume = πr^2h

h = volume/πr^2

h = 0.099 * 10^-6 m3/3.142 * ( 2.100 * 10^-6)^2

h =  0.099 * 10^-6 m3/3.142 *4.41 * 10^-12

h = 7.1 * 10^3 m

h = 7.1 Km

The answer is below

Explanation:

Density is the ratio of mass to volume. It is given by the formula:

Density = mass / volume

a) Density of gold = 19.32 g/cm³

Mass of gold = 1.915 g

Therefore the volume (V) occupied by the gold is:

Density = mass / volume

19.32 = 1.915 / volume

volume = 1.915 / 19.32 = 0.0991 cm³

The volume of the leaf = Area * thickness

Volume of leaf = Area * (7.331 * 10⁻⁴ cm)

Volume of gold = volume of leaf

0.0991 cm³ = Area * (7.331 * 10⁻⁴ cm)

Area = 135.2 cm²

b) The volume of the cylinder = πr²h = π(2.1 * 10⁻⁴ cm)² * h

Volume of gold = volume of cylinder

0.0991 cm³ = π(2.1 * 10⁻⁴ cm)² * h

h = 0.0991 cm³ / [π(2.1 * 10⁻⁴ cm)²]

h = 340617 cm

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j