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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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Anna35 [415]

Answer:

Condition A

Heat flux is 1400 W/M^2

Condition B

Heat flux is 12800 w/m^2

Explanation:

Given that:

T_s is given as  30 degree celcius

condition A

Air temperature =  - 5 degree c

convection coefficient h = 40 w/m^2. k

heat\ flux = \frac{Q}{a}= h\Delta = 40{30 - (-5)} = 1400 w/m^2

condition A

water temperature  = 10 degree c

convection coefficient = 800 w/m^2.k

heat\ flux = \frac{Q}{A} = H(\Delta} = 800\times (30-14) = 12800w/m^2

7 0
3 years ago
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Answer:

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Explanation:

It doesnt make sense.

7 0
4 years ago
g Let the charges start infinitely far away and infinitely far apart. They are placed at (6 cm, 0) and (0, 3 cm), respectively,
irina1246 [14]

Answer:

a) V =10¹¹*(1.5q₁ + 3q₂)

b) U = 1.34*10¹¹q₁q₂

Explanation:

Given

x₁ = 6 cm

y₁ = 0 cm

x₂ = 0 cm

y₂ = 3 cm

q₁ = unknown value in Coulomb

q₂ = unknown value in Coulomb

A) V₁ = Kq₁/r₁

where   r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m

V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁

V₂ = Kq₂/r₂

where   r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m

V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂

The electric potential due to the two charges at the origin is

V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)

B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows

U = Kq₁q₂/r₁₂

where

r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m

then

U = 9*10⁹q₁q₂/(3√5/100)

⇒ U = 1.34*10¹¹q₁q₂

5 0
3 years ago
Can glasses be formed using metallic materials?​
Westkost [7]
Metallic glasses are sometimes formed when molten metal is cooled too fast for its atoms to arrange themselves in a structured, crystalline order. "The material can't decide which crystalline structure it wants to converge to, and a metallic glass emerges
5 0
3 years ago
An engineer is testing the shear strength of spot welds used on a construction site. The engineer's null hypothesis at a 5% leve
lilavasa [31]

Answer:

b) The null hypothesis should be rejected.

Explanation:

The null hypothesis is  that the mean shear strength of spot welds is at least

3.1 MPa

H0: u ≥3.1 MPa  against the claim Ha: u< 3.1 MPa

The alternate hypothesis is  that the mean shear strength of spot welds is less than 3.1 MPa.

This is one tailed test

The critical region Z(0.05) < ± 1.645

The Sample mean= x`= 3.07

The number of welds= n= 15

Standard Deviation= s= 0.069

Applying z test

z= x`-u/s/√n

z= 3.07-3.1/0.069/√15

z= -0.03/0.0178

z= -1.68

As the calculated z= -1.68  falls in the critical region Z(0.05) < ± 1.645 the null hypothesis is rejected and the alternate hypothesis is accepted that the mean shear strength of spot welds is less than 3.1 MPa

8 0
3 years ago
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