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JulijaS [17]
1 year ago
7

For 11.20 kg of a magnesium-lead alloy of composition 30 wt% Pb-70 wt% Mg, is it possible, at equilibrium, to have α and Mg2Pb p

hases having respective masses of 7.39 kg and 3.81 kg? If so, what will be the approximate temperature of the alloy? If such an alloy is not possible, explain why.
Engineering
1 answer:
stealth61 [152]1 year ago
3 0

Answer:

Maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration. Therefore it is not possible.

Explanation:

let's first determine the mass fraction of the phases as follows

W_{a}=\frac{m_{a} }{m_{a}+m_{p_{g2}p_{b}  }  }

Given that masses are m_{a}=7.39kg and m_{p_{g2}p_{b}  }=3.81kg

Thus, W_{a} =\frac{7.39}{7.39+3.81}=0.659

W_{M_{g2Pb} } =1-W_{a}=1-0.659=0.341

Now determine the concentration

W_{a}=\frac{C_{0}-C_{a}  }{C_{b}-C_{a}  }

At 81% PB and 70% Pb only point at which M_{g2Pb} exists, we can solve

0.659=\frac{70-81}{C_{b}-81 }

C_{b}=64.3%

Refer text book page no. 326, figure 9.20, from figure we can say maximum concentration for Pb is only 41%. we get 64.3% which exceeds the maximum concentration.

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mars1129 [50]

Answer:

Control mechanisms

Explanation:

Organizational chart of any company will give details of different aspects of the company such as the major sub-units of the organization with the names of team leaders for different sub-units, it can also give you the span of control and the division of work within the company. However, the chart can't show you control mechanisms of different departments.

4 0
2 years ago
A cylindrical drum (2 ft. dia ,3 ft height) is filled with a fluid whose density is 40 lb/ft^3. Determine (a. the total volume o
Ksivusya [100]

Answer:

a)V=9.42\ ft^3

b)Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)v=0.025\ ft^3/lb

d)w=1276 \ lb/ft.s^2

Explanation:

Given that

d= 2 ft

r= 1 ft

h= 3 ft

Density

\rho = 40\ lb/ft^3

a)

We know that volume V given as

V=\pi r^2 h

V=\pi \times 1^2\times 3

V=9.42\ ft^3

b)

Mass = Density x volume

mass =40\times 9.42\ lb

mass= 376.8 lb

We know that

1 lb = 0.031 slug

So 376.8 lb= 11.71 slug

Mass in lb = 376.8 lb

Mass in slug = 11.71 slug

c)

we know that specific volume(v) is the inverse of density.

v=\dfrac{1}{\rho}\ ft^3/lb

v=\dfrac{1}{40}\ ft^3/lb

v=0.025\ ft^3/lb

d)

Specific weight(w) is the product of density and the gravity(g).

w= ρ X g

w = 40 x 31.9

w=1276 \ lb/ft.s^2

8 0
2 years ago
An electric field is expressed in rectangular coordinates by E = 6x2ax + 6y ay +4az V/m.Find:a) VMN if point M and N are specifi
Fittoniya [83]

Answer:

a.) -147V

b.) -120V

c.) 51V

Explanation:

a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).

b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.

c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.

Honestly, these things take practice to get used to. It's really hard to explain this.

3 0
2 years ago
The legend that Benjamin Franklin flew a kite as a storm approached is only a legend—he was neither stupid nor suicidal. Suppose
Delicious77 [7]

Answer: 0.93 mA

Explanation:

In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm’s law, we need to calculate the resistance of the water layer.

We can express the resistance as follows:

R = ρ.L/A

In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:

A= π(r22 – r12) = π( (0.0025)2-(0.002)2 ) m2 = 7.07 . 10-6 m2

Replacing by the values, we get R as follows:

R = 1.4 1010 Ω

Applying Ohm’s Law, and solving for the current I:

I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA

7 0
2 years ago
A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the
morpeh [17]

The magnitude of applied stress in the direction of 101 is 12.25 MPA and in the direction of 011, it is not defined.                                          

<u>Explanation</u>:        

<u>Given</u>:

tensile stress is applied parallel to the [100] direction

Shear stress is 0.5 MPA.

<u>To calculate</u>:

The magnitude of applied stress in the direction of [101] and [011].

<u>Formula</u>:

zcr=σ cosФ cosλ

<u>Solution</u>:

For in the direction of 101

cosλ = (1)(1)+(0)(0)+(0)(1)/√(1)(2)

cos λ = 1/√2

The magnitude of stress in the direction of 101 is 12.25 MPA

In the direction of 011

We have an angle between 100 and 011

cosλ = (1)(0)+(0)(-1)+(0)(1)/√(1)(2)

cosλ  = 0

Therefore the magnitude of stress to cause a slip in the direction of 011 is not defined.                                                                                                                                                                      

                                                                                   

                                                                                                                                                   

                                                                                                                                                             

5 0
2 years ago
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