Answer:
Option C is correct.
Modulus of elasticity of the composite perpendicular to the fibers = (12 × 10⁶) psi
Explanation:
For combination of materials, the properties (especially physical properties) of the resulting composite is a sum of the fractional contribution of each material thay makes up the composite.
In this composite,
The fibres = 20 vol%
Aluminium = 80 vol%
Modulus of elasticity of the composite
= [0.2 × E(fibres)] + [0.8 × E(Al)]
Modulus of elasticity of the fibers = E(fibres) = (55 × 10⁶) psi. =
Modulus of elasticity of aluminum = E(Al) = (10 × 10⁶) psi.
But modulus of elasticity of the composite perpendicular to the fibers is given in the expression.
[1 ÷ E(perpendicular)]
= [0.2 ÷ E(fibres)] + [0.8 ÷ E(Al)]
[1 ÷ E(perpendicular)]
= [0.2 ÷ (55 × 10⁶)] + [0.8 ÷ (10 × 10⁶)]
= (3.636 × 10⁻⁹) + (8.00 × 10⁻⁸)
= (8.3636 × 10⁻⁸)
E(perpendicular) = 1 ÷ (8.3636 × 10⁻⁸)
= 11,961,722.5 psi = (11.96 × 10⁶) psi
= (12 × 10⁶) psi
Hope this Helps!!!
Answer:
A because it is the most affective way to do it
Explanation:
Answer: The density is greater at point C
Explanation: At that point
The convention current slowly cools off
Answer:
19 elements
Explanation:
1 carbon + 12 hydrogen + 6 oxygen = 19
Mohs scale of mineral hardness is a qualitative scale that characterizes the scratch resistance of various minerals with the ablity of harder material to scratch softer material
