Question

A metal wire of length 1.2 m and cross-sectional area 2.0 x 10 raised to power-7 m 2 is stretched by a force of 50 N. assuming the force constant of the metal is 6000 Nm-1. Calculate the tensile stress

Answer 1

Answer:

L=1m

$\alpha$

=2mm²=(2/1000²)=2(10^-6)m

y=4x10¹¹N/m²

∆l=2mm=2/10.00=0.002m

$F = y \alpha l \div l$

=(4x10¹¹x2x10^-6x2x10^-3)÷1m

=16x10¹¹-⁶-³

=16x10¹¹-⁹

=16x10²

=1600N

where a=cross sectional area=2x10^-6m

C=2mm= 2x10^-3m

L=1m

hope it helps…