Respuesta :
Answer:
1. Mass = 2070 kg
2.Total length of strip = 2556 m
3. Total length of dislocation = 7.67 X[tex]10^{14}[/tex] m
Explanation:
Given:
Aluminium coil thickness, t = 0.3 mm
= 0.3 X [tex]10^{-3}[/tex] m
Width of the coil,w = 1 m
Drum diameter, d = 15 cm
= 0.15 m
Coil outer diameter, d = 1 m
Dislocation density = [tex]10^{15}[/tex] [tex]m^{2}[/tex]
1). Area of the coil, A = [tex]\frac{\pi }{4}\times[/tex] ( [tex]d_{coil} ^{2}[/tex]-[tex]d_{drum} ^{2}[/tex])
A = [tex]\frac{\pi }{4}\times (1^{2}-0.15^{2})[/tex]
A = 0.767 [tex]m^{2}[/tex]
Volume of the coil,V = A X w
= 0.767 X 1
= 0.767 [tex]m^{3}[/tex]
We know density of aluminum at STP = 2.7 X [tex]10^{3}[/tex]
Therefore, mass of the aluminum coil is,
Mass,m = Density of aluminium X Volume
= 2.7 X [tex]10^{3}[/tex] X 0.767
= 2070 kg
Mass = 2070 kg
2). Total length of trip of coil is given by
L = Volume of coil / area of strip
= [tex]\frac{0.767}{1\times 0.3\times 10^{-3}}[/tex]
= 2556 m
Total length of strip = 2556 m
3). Total length of dislocation of the coiled strip = volume X dislocation density
= 0.767 X [tex]10^{15}[/tex]
= 7.67 X [tex]10^{14}[/tex]
Total length of dislocation = 7.67 X[tex]10^{14}[/tex] m
