<span> Transuranium are chemical elements with atomic numbers greater than 92 that can be found in nature having stable like hydrogen, and very long life.<span>The exceptions elements are 43,61,85,87, but not only in minor branches of the uranium and thorium decay chains.
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Answer:
required diameter is 7.65 mm
Explanation:
Given the data in the question;
F = 6660 N
l₀ = 380 mm = 0.38 m
E = 110 GPa = 110 × 10⁹ N/m²
Δl = 0.50 mm = 0.0005 m
So, lets assume the deformation is elastic;
d₀ = √( [4l₀F] / [πEΔl] )
we substitute
d₀ = √( [4 × 0.38 × 6660] / [π × (110 × 10⁹) × 0.0005]] )
d₀ = √( 10123.2 / 172787595.947 )
d₀ = √( 5.85875 × 10⁻⁵ )
d₀ = 0.007654 m
d₀ = ( 0.007654 × 1000 )mm
d₀ = 7.65 mm
Therefore, required diameter is 7.65 mm
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I'm thinking its answer choice D. But it might be a good idea to wait and see if someone agrees