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BigorU [14]
3 years ago
13

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

ticity of the metal and oxide are, respectively, 60 GPa and 380 GPa, what is the (a) upper-bound, and (b) lower-bound modulus of elasticity values (in GPa) for a composite that has a composition of 33 vol% of oxide particles.
Engineering
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

A) Upper bound modulus of elasticity; E = 165.6 GPa

B) Lower bound modulus of elasticity; E = 83.09 GPa

Explanation:

A) Formula for upper bound modulus is given as;

E = E_m(1 - V_f) + E_f•V_f

We are given;

E_m = 60 GPa

E_f = 380 GPa

V_f = 33% = 0.33

Thus,

E = 60(1 - 0.33) + 380(0.33)

E = (60 x 0.67) + 125.4

E = 165.6 GPa

B) Formula for lower bound modulus is given as;

E = 1/[(V_f/E_f) + ((1 – V_f)/E_m)]

E = 1/[(0.33/380) + ((1 – 0.33)/60)]

E = 1/(0.0008684 + 0.01116667)

E = 1/0.01203507

E = 83.09 GPa

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Complete Question:

If both the ram air input and drain hole of the pitot system become blocked, the indicated airspeed will

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The 10-lb block is pressed against the spring so as to compress it 2 ft when it is at a point A. If the plane is smooth, determi
leva [86]

Answer:

The distance measure from the wall = 36ft

Explanation:

Given Data:

w = 10

g =32.2ft/s²

x = 2

Using the principle of work and energy,

T₁ +∑U₁-₂ = T₂

0 + 1/2kx² -wh = 1/2 w/g V²

Substituting, we have

0 + 1/2 * 100 * 2² - (10 * 3) = 1/2 * (10/32.2)V²

170 = 0.15528V²

V² = 170/0.15528

V²     = 1094.796

V = √1094.796

V = 33.09 ft/s

But tan ∅ = 3/4

∅ = tan⁻¹3/4

   = 36.87°

From uniform acceleration,

S = S₀ + ut + 1/2gt²

It can be written as

S = S₀ + Vsin∅*t + 1/2gt²

Substituting, we have

0 = 3 + 33.09 * sin 36.87 * t -(1/2 * 32.2 *t²)

19.85t - 16.1t² + 3 = 0

16.1t² - 19.85t - 3 = 0

Solving it quadratically, we obtain t = 1.36s

The distance measure from the wall is given by the formula

d = VCos∅*t

Substituting, we have

d = 33.09 * cos 36. 87 * 1.36

d = 36ft

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3 years ago
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