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polet [3.4K]
4 years ago
5

The number of vacancies in some hypothetical metal increases by a factor of 3 when the temperature is increased from 1020 ˚C to

1290 ˚C. Calculate the energy for vacancy formation (in J/mol) assuming that the density of the metal remains the same over this temperature range.
Engineering
1 answer:
weeeeeb [17]4 years ago
6 0

Answer:

first step here is to substitute the 3 of your two equations into the second;

3 Ne^(-Q_v/k(1293)) = Ne^(-Q_v/k(1566))

Since 'N' is a constant, we can remove it from both sides.

We also want to combine our two Q_v values, so we can solve for Q_v, so we should put them both on the same side:

3 = e^(-Q_v/k(1293)) / e^(-Q_v/k(1566))

3 = e^(-Q_v/k(1293) + Q_v/k(1566) ) (index laws)

ln (3) = -Q_v/k(1293) + Q_v/k(1566) (log laws)

ln (3) = -0.13Q_v / k(1566) (addition of fractions)

Q_v = ln (3)* k * 1566 / -0.13 (rearranging the equation)

Now, as long as you know Boltzmann's constant it's just a matter of substituting it for k and plugging everything into a calculator.

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Tech A says that serviceable wheel bearings can be repacked by removing the dust cap, filling it with grease, and reinstalling i
rewona [7]

Serviceable wheel bearings can be repacked by removing the dust cap and the cotter pin must be replaced with a new one every time it is removed. Therefore, both Tech A and B are correct.

Sealed bearing assemblies are typically prefilled with lubricant. They also make a more reliable and consistent installation process due to the fact that every into comes preset at the proper clearance.

Furthermore, the serviceable wheel bearings can be repacked by removing the dust cap, filling it with grease, and reinstalling it. Also, cotter pin must be replaced with a new one every time it is removed.

Therefore, both tech A and B are correct.

Read related link on:

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8 0
2 years ago
____accelerates the rusting process A lubricant B improper storage CMoisture DOil​
Dmitriy789 [7]

Answer:

i would go with-B

Explanation:

because  Exposure to the elements has resulted in the formation of the blue-green patina

6 0
4 years ago
Read 2 more answers
A closed, rigid, 0.45 m^3 tank is filled with 12 kg of water. The initial pressure is p1 = 20 bar. The water is cooled until the
liberstina [14]

Answer:

initial quality = 0.3690

heat transfer = 979.63 kJ/kg

Explanation:

Given data:

volume of tank 0.45^3

weight of water 12 kg

Initial pressure 20 bar

final pressure 4 bar

Specific volume v = \frac {0.45}{12} = 0.0375 m^3/kg

At Pressure = 20 bar, from saturated water table

v_f = 0.01177 m^/kg

v_g = 0.099587 m^3/kg

x = \frac{v -v_f}{v_g -v_f} = \frac{0.0375 - 0.001177}{0.099587 - 0.001177}

inital quality is x =0.3690

Heat transfer is calculated as

u_1 = h_f + x(h_g - h_f) = v_f + x( h_{fg})

from saturated water table, for pressure 20 bar ,

h_f = 908.79 kJ/kg, h_{fg} = 1890.7 kJ/kg

     =908.79 + 0.0357(1890.7)

      = 979.63 kJ/kg

7 0
3 years ago
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
Nina [5.8K]

dutile is the correct answer

6 0
3 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
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