Area of a Gold Leaf? | Calculate with Density

  • Thread starter Thread starter bosburn93
  • Start date Start date
  • Tags Tags
    Area Gold
AI Thread Summary
To calculate the area of a gold leaf with a mass of 27.63g and a thickness of 1.000 µm, the volume of the gold must first be determined using its density of 19.32 g/cm^3. The volume can then be related to area through the equation V = area × thickness. Since the thickness is minimal, the gold leaf can be approximated as a short prism, allowing for the use of this formula. Additionally, for a cylindrical fiber with a radius of 2.500 µm, the length can be calculated using the same volume derived from the mass of gold. These calculations demonstrate the application of density and volume relationships in determining the area and length of gold forms.
bosburn93
Messages
3
Reaction score
0

Homework Statement



Gold, which has a density of 19.32 g/cm^3, 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, witha mass of 27.63g, is pressed into a leaf 1.000 µm thickness, what is the area of the leaf? (b) If, instead, the gold is drawn out into a cylindrical fiber of radius 2.500 µm, what is the length of the fiber?


Homework Equations



ρ=m/V and others?

The Attempt at a Solution



Not sure what to do. Got volume but how do i use that to get area? I also do not know what equations would apply to a gold leaf.
 
Physics news on Phys.org
isnt volume =area * thickness
 
Punkyc7 said:
isnt volume =area * thickness

Yes, if i assume that it is a cube or rectangular prism.
is that right?
 
In general, a prism does not have to be rectangular in cross section for the volume relation to hold, so long as all cross sections are similar. For the gold leaf, the thickness is so small, it would be reasonable to assume that the areas of the front and back of the leaf are equal, hence the leaf could be considered to be a very short prism.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top