Costume

Denisa is preparing for a goldsmith's costume carnival. During the preparations, she thought she would let her hair wipe instead - she would apply a 5 μm thick layer of gold to each hair. How much gold would Denisa need? Assume that all hundred thousand Denise's hair has a length of 0.5 m a diameter 60 μm and a cylinder shape. The density of gold is 19,000 kg·m-3.

Result

m =  969.967 g

Solution:

h=19 r2=60/1000/10/2=31000=0.003 r1=r2+5/1000/10=0.003+5/1000/10=72000=0.0035 V=50⋅ (π⋅ r12−π⋅ r22)=50⋅ (3.1416⋅ 0.00352−3.1416⋅ 0.0032)≐0.0005 m1=h⋅ V=19⋅ 0.0005≐0.0097 m=100000⋅ m1=100000⋅ 0.0097≐969.9667≐969.967 gh=19 \ \\ r_{2}=60/1000/10/2=\dfrac{ 3 }{ 1000 }=0.003 \ \\ r_{1}=r_{2} +5/1000/10=0.003 +5/1000/10=\dfrac{ 7 }{ 2000 }=0.0035 \ \\ V=50 \cdot \ (\pi \cdot \ r_{1}^2-\pi \cdot \ r_{2}^2)=50 \cdot \ (3.1416 \cdot \ 0.0035^2-3.1416 \cdot \ 0.003^2) \doteq 0.0005 \ \\ m_{1}=h \cdot \ V=19 \cdot \ 0.0005 \doteq 0.0097 \ \\ m=100000 \cdot \ m_{1}=100000 \cdot \ 0.0097 \doteq 969.9667 \doteq 969.967 \ \text{g}

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