Asked by COFFEE
[Given]
Fiber Linear Density = 1 denier = 1 g/9000m
Fiber Density = 1.14 g/cm^3
[Find..]
Fiber surface area in cm^2/g
Assume that the fiber strand is a uniform cylinder
[Answer]
Surface area = 3,150 cm^2/g
....... how do i get to the answer? my professor gives us these notes:
Pi*D*L
Pi*D is the diameter of the top of the cylinder when it is cut vertically in half. and L is the height.
((Pi*D^2)/(4)) * L = volume of the cylinder
((Pi*D^2)/(4)) * L * Density = weight
..please help!
Fiber Linear Density = 1 denier = 1 g/9000m
Fiber Density = 1.14 g/cm^3
[Find..]
Fiber surface area in cm^2/g
Assume that the fiber strand is a uniform cylinder
[Answer]
Surface area = 3,150 cm^2/g
....... how do i get to the answer? my professor gives us these notes:
Pi*D*L
Pi*D is the diameter of the top of the cylinder when it is cut vertically in half. and L is the height.
((Pi*D^2)/(4)) * L = volume of the cylinder
((Pi*D^2)/(4)) * L * Density = weight
..please help!
Answers
Answered by
bobpursley
If it is a cylinder of length L
volume= PI r^2 *L
Surface area= PI*2(r)L
mass=masslineardensiy*L
density= mass/volume=1.14g/cm^3
fiberarea/g= fiberarea/mass=
PI*2*r*L/massdensity*V=
PI*2r*L/(1.14g/cm^2 * PI*r^2*L)
2cm^2/(1.14g*r)= 2/1.14r g/cm^2
Now, r
linear density=1g/9000m
= mass/L= volumedensity*volume/L
1g/9000m=1.14g/cm^3*PIr^2/L
r^2=1cm^3/(1.14*9000m*PI)
=1/(1.14*900000*PI) cm^2
t hen find the sqrt root of that.
final answer= 2/(1.14r) and it will give you your answer.
volume= PI r^2 *L
Surface area= PI*2(r)L
mass=masslineardensiy*L
density= mass/volume=1.14g/cm^3
fiberarea/g= fiberarea/mass=
PI*2*r*L/massdensity*V=
PI*2r*L/(1.14g/cm^2 * PI*r^2*L)
2cm^2/(1.14g*r)= 2/1.14r g/cm^2
Now, r
linear density=1g/9000m
= mass/L= volumedensity*volume/L
1g/9000m=1.14g/cm^3*PIr^2/L
r^2=1cm^3/(1.14*9000m*PI)
=1/(1.14*900000*PI) cm^2
t hen find the sqrt root of that.
final answer= 2/(1.14r) and it will give you your answer.
Answered by
goh han liang
how to do surface area
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