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Lina20 [59]
3 years ago
10

Is the housekeeping satisfactory? ​

Engineering
2 answers:
inessss [21]3 years ago
6 0

Answer:

yes

Explanation:

Nat2105 [25]3 years ago
3 0

Answer:

yess

Explanation:

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5. A non-cold-worked brass specimen of average grain size 0.01 mm has a yield strength of 150 MPa. Estimate the yield strength o
Papessa [141]

Answer:

97.17 MPa

Explanation:

Given:-

- The nominal strength of the grain, σ0  = 25 MPa

- The average grain size of the brass specimen, d* = 0.01 m

- The yield strength of the non-cold worked specimen, σy = 150 MPa

- Conditions of cold-working: T = 500°C , t = 1000 s

Find:-

Estimate the yield strength of this alloy after cold - working process

Solution:-

- The nominal strength of the grain is a function of yield strength of the material, grain yield factor ( Ky ) and the grain size.

- the following relation is used to determine the grain strength:

                             σ0  = σy  - ( Ky / √( d ) )

- We will use the above relation to determine the grain yield factor ( Ky ) for the alloy as follows. Note: here we will use the average value of grain size:

                            Ky = ( σy  - σy )*√( d* )

                            Ky = ( 150 - 25 ) * √0.01

                            Ky = 12.5 MPa - √mm

- Now we will use the cold working conditions of T = 500 C and time of the process is t = 1000 s. We will look up the elongated size of the grain after the cold-working process in lieu with its yield factor ( Ky ). Use figure 7.25.

- The cold-worked grain size with the given conditions can be read off from the figure 7.25. The new size comes out to be d = 0.03 mm.

- We will again use the nominal grain strength relation expressed initially. And compute for the new yield strength of the cold-worked alloy.

                            σ0  = σy  - ( Ky / √( d ) )

                            σy = σ0 + ( Ky / √( d ) )

                            σy = 25MPa + ( 12.5 / √( 0.03 mm ) )

                            σy = 97.17 MPa

- We see that the yield strength of the alloy decreases after cold-working process. This happens because the cold working process leaves with inter-granular strain ( dislocation of planes ) in the material structure which results from the increase in grain size.

6 0
3 years ago
A projectile is launched horizontally 1m above the ground. If it lands 300m away from the initial launch position, find: a)-the
PIT_PIT [208]

Answer:

(a): The launch velocity is Vx= 666.66 m/s.

(b): The angle wich the projectile contacts the ground is α= 0.38°

Explanation:

h= 1m

g= 9.8 m/s²

h= g*t²/2

t= 0.45 s

Vy= g*t

Vy= 4.42 m/s

d=Vx* t

Vx= 666.66 m/s (a)

α= tg⁻¹ ( Vy/Vx)

α= 0.38° (b)

4 0
3 years ago
Sedimentary rock forms at or near the
Natalija [7]
I believe it’s A bit I. could be wrong
5 0
3 years ago
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Si la demanda por un bien final es elástica, entonces la demanda de trabajo correspondiente también es elástica
shusha [124]

Answer:

You can speak English to help you

4 0
3 years ago
Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by
vladimir2022 [97]

Answer:

The answer for the question :

"Develop a chase plan that matches the forecast and compute the total cost of your plan. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Omit the "$" sign in your response.)"

is explained in the attachment.

Explanation:

4 0
3 years ago
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