The si unit of force is newton.
so, F is eqal to m*g
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:
Explanation:
Y = 5 Sin27( .2x-3t)
= 5 Sin(5.4x - 81 t )
Amplitude = 5 m
Angular frequency ω = 81
frequency = ω / 2π
= 81 / (2 x 3.14 )
=12.89
Wave length λ = 2π / k ,
k = 5.4
λ = 2π / 5.4
= 1.163 m
Phase velocity =ω / k
= 81 / 5.4
15 m / s.
The wave is travelling in + ve x - direction.
Complete question:
A college dormitory room measures 14 ft wide by 13 ft long by 6 ft high. Weight density of air is 0.07 lbs/ft3. What is the weight of air in it under normal conditions?
Answer:
the weight of the air is 76.44 lbs
Explanation:
Given;
dimension of the dormitory, = 14 ft by 13 ft by 6 ft
density of the air, = 0.07 lbs/ft³
The volume of the air in the dormitory room = 14 ft x 13 ft x 6 ft
= 1092 ft³
The weight of the air = density x volume
= 0.07 lbs/ft³ x 1092 ft³
= 76.44 lbs
Therefore, the weight of the air is 76.44 lbs
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