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3 years ago

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Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, ass

uming the following: (1) an average fiber diameter of 10  10-3 mm, (2) an average fiber length of 1 mm, (3) a fiber fracture strength of 2.5 GPa, (4) a fiber-matrix bond strength of 10 MPa, (5) a matrix stress at fiber failure of 10.0 MPa, and (6) a matrix tensile strength of 75 MPa.

1 answer:

zmey [24]3 years ago

8 0

Answer:

632.5 MPa

Explanation:

$\sigma_{m}$ = Matrix stress at fiber failure = 10 MPa

$V_f$ = Volume fraction of fiber = 0.25

$\sigma_f$ = Fiber fracture strength = 2.5 GPa

The longitudinal strength of a composite is given by

$\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa$

The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa

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Answer:

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$\frac{1}{2}*m*v^2_{i} = \frac{1}{2}*m*(0.516v_{i})^2 + \frac{1}{2}*M*V^2 + 0.413*\frac{1}{2}*m*v^2_{i}\\\\0.320744*m*v^2_{i} = M*V^2\\\\M = \frac{0.320744*m*v^2_{i} }{V^2} ....... Eq 2$

Substitute Eq 2 into Eq 1

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3 years ago

A 12.0kg microwave oven is pushed 14.0m up the sloping surface of a loading ramp inclined at an angle 37 degrees above the horiz

Semenov [28]

Answer:

Answered

Explanation:

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c) increase in the internal energy

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0 + 0 + (W_F - W_f ) = U_2 + K_2

1341.71 J - 329 J - 991 J = K_2

K_2 = 21.71 J

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a = 7.55m/s^2

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viva [34]

Answer:

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$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

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Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

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g100num [7]

Force required to accelerate 10 kg object to 5.9 m/s/s ?

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3 years ago