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weqwewe [10]
3 years ago
12

The critical resolved shear stress for iron is 27 MPa (4000 psi). Determine the maximum possible yield strength for a single cry

stal of Fe pulled in tension.

Engineering
1 answer:
jasenka [17]3 years ago
7 0

Answer:

Answer is mentioned below.

Explanation:

Answer: Calculation of maximum yield strength for a single crystal of Fe pulled in tension is shown in the image attached .

value of max. Yield strength= 54MPa

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(30 pts) A simply supported beam with a span L=20 ft and cross sectional dimensions: b=14 in; h=20 in; d=17.5 in. is reinforced
Nat2105 [25]

Answer:

Zx = 176In³

Explanation:

See attached image file

5 0
3 years ago
Draw a flowchart to represent the logic of a program that allows the user to enter values for the current year and the user’s bi
Digiron [165]

Answer:

Please, see the attachment.

Explanation:

First, we have to create two input boxes that allows the user to write the current year in one of them and his/her birth year in the another one. Also, we have to create a label that will show the result of the desired variable. We can write a message "Your age is:" and it will be attached to the result.

For the algorithm, let's call the variables as follows:

CY = Current Year

BY = Birth Year

X = Age of user

When the user inserts the current year and his/her birth year, the program will do the following operation:

X = CY - BY; this operation will give us the age of the user

After this the user will see something like "Your age is:" X.

3 0
3 years ago
Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of
Alexxandr [17]

Answer:

P_m_i_n = 442KPA

Explanation:

We are given:

m = 1.06Kg

T_H = 1.2T_L

T = 22kj

Therefore we need to find coefficient performance or the cycle

COP_R = \frac {1}{(T_R/T_l) -1}

= \frac {1 }{1.2-1}

= 5

For the amount of heat absorbed:

Q_l = COP_R Wm

= 5 × 22 = 110KJ

For the amount of heat rejected:

Q_H = Q_L + W_m

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= = \frac{132}{1.06}

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

T_L = \frac{T_H}{1.2};

T_L = \frac{342.5}{1.2}

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at T_L = 12.4°c we have 442KPa

6 0
3 years ago
Jodie bought some shirts for 6$ each marge brought some shirts for 8$ each
Alex_Xolod [135]

Answer:

you need more details but if you have to find the difference, its $2.00

Explanation:

8-6=2

3 0
3 years ago
Consider the velocity boundary layer profile for flow over u flat plate to be of the form u = C_1 + C_2 y. Applying appropriate
ra1l [238]

Answer:

The  result in terms of the local Reynolds number ⇒ Re = [μ_∞ · x] / v

Explanation:

See below my full workings so you can compare the results with those obtained from the exact solution.

4 0
3 years ago
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