Respuesta :
Answer:
1) Mass of the grain is [tex]2.9632\times 10^{-9} kg[/tex].
2) 0.08901 kg of sand would have surface area equal the surface area of the cube.
Explanation:
1) Radius of the grain,r = 64.8 µm =[tex]6.48\times 10^{-5}m [/tex]
Volume of the sphere =[tex]\frac{4}{3}\pi r^3[/tex]
Volume of the grain of a sand:
[tex]V=\frac{4}{3}\times \pi r^3=\frac{4}{3}\times 3.14\times (6.48\times 10^{-5} m)^3[/tex]
[tex]V=1.1397\times 10^{-12} m^3[/tex]
Density of a grain of sand = [tex]d=2600 kg/m^3[/tex]
Mass of a grain of a sand = M
[tex]d=2600 kg/m^3=\frac{M}{1.1397\times 10^{-12} m^3}[/tex]
[tex]M=2.9632\times 10^{-9} kg[/tex]
Mass of the grain is [tex]2.9632\times 10^{-9} kg[/tex].
2) Surface are of sphere: [tex]4\pi r^2[/tex]
Surface area of a grain:
[tex]A=4\times 3.14\times (6.48\times 10^{-5}m)^2[/tex]
[tex]A=5.2766\times 10^{-8} m^2[/tex]
Length of the cube = a = 0.514 m
Total surface area of cube ,A'= [tex]6a^2[/tex]
[tex]A'=6\times (0.514 m)^2=1.5851 m^2[/tex]
let the number grains with area equal to total surface area of cube be x.
[tex]A'=A\times x[/tex]
[tex]x=\frac{1.5851 m^2}{5.2766\times 10^{-8} m^2}=3.003\times 10^7 [/tex]
Volume of x number of grains :V'
[tex]V'=V\times x[/tex]
[tex]V= 1.1397\times 10^{-12} m^3\times 3.003\times 10^7[/tex]
[tex]V'=3.4236\times 10^{-5} m^3[/tex]
Mass of [tex]3.4236\times 10^{-5} m^3[/tex] of sandL:
=[tex]3.4236\times 10^{-5} m^3\times 2600 kg/m^3=0.08901 kg[/tex]
0.08901 kg of sand would have surface area equal the surface area of the cube.
