when a wire of lenth 5 m and radius 0.5 mm is stretched by a load of 49 n the elongation produsced in the wire is 0.1 cm . find energy stored per unit volume of the wire
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Expert's answer
2013-07-17T08:18:34-0400
the volume of stretched wire is length (5m) multiplied by cross-section area (pi r^2 where r is radius): V = L*S = L*pi * r^2 5.1*3.14*0.0005^2 = 4.0035 * 10^{-6} m^3
The force per unit volume then is total force (49 N) divided by total volume ( 4.0035 * 10^{-6} m^3): f = F/V = 49/(4.0035*10^{-6}) = 12.2 * 10^6 N/m^3
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