The elastic constant of the wire is about 470 N/m
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Further explanation
Let's recall Elastic Constant formula as follows:
\(\boxed{k = \frac{E A}{L}}\)
where:
k = elastic constant ( N/m )
E = Young's Modulus ( Pa )
A = cross-sectional area ( m² )
L = initial length ( m )
Let us now tackle the problem!
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Translation:
A wire, with cross sectional area A = 5 mm² and length L = 32 cm, is stretched by a force. Assuming that Young's modulus of wire E = 3 × 10⁷ N/m², calculate the elastic constant k of the wire !
Given:
cross sectional area = A = 5 mm² = 5 × 10⁻⁶ m²
length of wire = L = 32 cm = 0.32 m
Young's modulus of wire = E = 3 × 10⁷ N/m²
Asked:
elastic constant of wire = k = ?
Solution:
\(k = \frac{E A}{L}\)
\(k = \frac{ (3 \times 10^7) \times (5 \times 10^{-6})}{0.32}\)
\(k = 468.75 \texttt{ N/m}\)
\(k \approx 470 \texttt{ N/m}\)
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Learn more
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
- Young Modulus : https://brainly.com/question/9202964
- Simple Harmonic Motion : https://brainly.com/question/12069840
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Answer details
Grade: High School
Subject: Physics
Chapter: Elasticity
