PHYSICS MIDDLE SCHOOL

Find the elongation produced in a copper wire of length 2m and radius 5mm, when suspended by a blockof mass 500Kg. (Take Young's modulus of elasticity of copper Y = 1,2 ×109 Pa).

Answers

Answer 1
Answer:

Answer:

0.104 m

Explanation:

Stress, \sigma=\frac {F}{A}

Where F is force and A is area. Also, F=mg where m is mass and g is acceleration due to gravity

Area= \pi r^(2)

Strain=\frac {\triangle l}{l} where \triangle l is the elongation and l is the original length

E=\frac {stress}{strain}=\frac {\frac {F}{A}}{\frac {\triangle l}{l}}=\frac {Fl}{A\triangle l}

Making \triangle l the subject then

\triangle l=\frac {Fl}{AE}=(mg l)/(\pi r^(2) E)

By substituting the given values and taking g as 9.81 then

\triangle l=\frac {Fl}{AE}=(500* 9.81* 2 m)/(\pi * 0.005^(2)* 1.2* 10^(9))=0.104087333  m\approx 0.104 m

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Answers


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