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vovikov84 [41]
3 years ago
5

The mass of a single gold atom is 3.27×10-22 grams. how many gold atoms would there be in 236 milligrams of gold?

Physics
2 answers:
blsea [12.9K]3 years ago
8 0

The amount of gold atoms could be calculated by dividing the total weight of the gold with the mass of a single gold atom. Just convert the given weight to grams then divide it with 3.27x10^-22 grams. The answer would be 7.22x10^20. 

Arisa [49]3 years ago
3 0

Answer:

6.38*10²⁰ atoms of gold

Explanation:

236mg of gold = 236*10^-3g = 0.236g.

Now from the question,

1 atoms contains = 3.7*10^-22g.

X atoms will contain = 0.236g of gold.

Now we'll cross multiply and make x the subject of formula to find the amount of gold in 0.236g of gold.

X = (0.236 * 1) / 3.7*10^-22

X = 6.378*10^20 atoms

0.236g of gold contains 6.378*10^20 atoms of gold

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a) The shear stress is 0.012

b) The shear stress is 0.0082

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Explanation:

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p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

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a) The Reynold number is equal to:

Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar

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\delta=\frac{4.91*1}{Re^{0.5} }  =\frac{4.91*1}{246507^{0.5} } =0.0098 ft

The shear stress is equal to:

\tau=0.332(\frac{2.359x10^{-5}*3 }{1}  )(246507)^{0.5} =0.012

b) If the railing edge is 2 ft, the Reynold number is:

Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar

The boundary layer is equal to:

\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft

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\tau=0.332(\frac{2.359x10^{-5}*3 }{2}  )(493015.6^{0.5} )=0.0082

c) The drag coefficient is equal to:

C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019

The friction drag is equal to:

F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf

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