Physics, asked by Veeresh4567, 3 months ago

a 20kg load is suspended by a wire of cross section 0.4 mm² The stress produced in Nm-2 is​

Answers

Answered by BrainlyTwinklingstar
9

Given :

Mass = 20 kg

Area of cross section = 0.4 mm² = 0.4 × 10¯⁶m²

To Find :

The stress produced

Solution :

Stress is the external force per unit area of the body,

 \bf Stress =  \dfrac{Froce }{Area}

by substituting the values in the formula,

 \dashrightarrow \sf Stress =  \dfrac{Froce }{Area}

 \dashrightarrow \sf Stress =  \dfrac{mg }{0.4 \times  {10}^{ - 6} }

 \dashrightarrow \sf Stress =  \dfrac{20 \times 10 }{0.4 \times  {10}^{ - 6} }

 \dashrightarrow \sf Stress =  \dfrac{200 }{0.4 \times  {10}^{ - 6} }

 \dashrightarrow \sf Stress =  50 \times  {10}^{7}

 \dashrightarrow \sf Stress =  5 \times  {10}^{8} \:  Nm^{-2}

Thus, the stress produced by the body is 5 × 10 N/.

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