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IRISSAK [1]
3 years ago
3

A continuous and aligned fiber-reinforced composite having a cross-sectional area of 1130 mm2 is subjected to an external tensil

e load. If the stresses sustained by the fiber and matrix phases are 156 MPa and 2.75 MPa, respectively, the force sustained by the fiber phase is 74,000 N and the total longitudinal strain is 1.25 x 10-3, what is the value of the modulus of elasticity of the composite material in the longitudinal direction?
Physics
1 answer:
Assoli18 [71]3 years ago
6 0

Answer:

Ec=53.7×10⁹N/m² =53.7Gpa

Explanation:

To calculate the modulus of elasticity in the longitudinal direction.  This is possible realizing Ec=σ/ε where σ=(Fm+Ff)/Ac

Ec=Sigma/E\\Ec=\frac{(Fm+Ff)/E}{Ac}\\ Ec=\frac{1802+74,000}{(1.25*10^{-3})(1130)(1/1000)^{2}  }\\ Ec=53.7*10^{9}N/m^{2}\\or\\Ec=53.7GPa

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We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

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 Q=\frac{E}{Nr^2}

 

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