Answer:
0.0173 mm
Explanation:
The maximum length of eliptical flow will be given by
[tex]l=\frac {2Er}{\pi\sigma^{2}}[/tex] where l is the maximum length of elliptical flow, E is the modulus of elasticity, r is the surface energy of magnesium oxide and nickel, [tex]\sigma[/tex] is the tensile stress.
Substituting 208 Gpa for E, 1.23 for r and 97 Mpa for tensile stress then
[tex]l=\frac {2*208*10^{9}*1.23}{\pi\times (97*10^{6})^{2}}=0.0000173 m= 0.0173 mm[/tex]
