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3 years ago

PLS HURRYY!!! Look at the image below

Engineering

1 answer:

Marysya12 [62]3 years ago

6 0

Answer:

1.) Bachelor's degree in aerospace engineering

2.)High school diploma

3.)Bachelor's degree in physics

Explanation:

yeah had a hard time so I can't explain

It's hard knowing what training level they are in when that's not the career your focusing on so yeah

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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

Thoughts about drinking and driving

poizon [28]

Answer: it’s very bad behavior and can be dangerous

Explanation:

8 0

3 years ago

Read 2 more answers

At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slide

Xelga [282]

Answer:

a) aA = - 13.33 mm/s²

aB = - 20 mm/s²

b) aD = - 13.33 mm/s²

c) vB = 70 mm/s

d) xB = 440 mm

Explanation:

Given

The initial speed of B is: v₀B = 150 mm/s

Distance moved by A is: xA = 240 mm

Velocity of A is: vA = 60 mm/s

Assuming:

Displacement of blocks are denoted by:

A = xA

B = xB

C = xC

D = xD

From the pic shown, the total length of the cable is:

xB + (xB - xA) + 2*(d - xA) = L

⇒ 2*xB - 3*xA = L - 2*d

where L - 2*d is constant. Differentiating the above equation with respect to time:

d(2*xB)/dt - d(3*xA)/dt = 0

⇒ 2*vB - 3*vA = 0 (i)

Substituting in equation (i)

2*(150 mm/s) - 3*vA = 0

⇒ v₀A = 100 mm/s (initial speed of A)

Then, we use the equation

vA² = v₀A² + 2*aA*xA

Substituting the values in above equation:

(60 mm/s)² = (100 mm/s)² + 2*aA*(240 mm)

⇒ aA = - 13.33 mm/s²

If 2*vB - 3*vA = 0

Differentiating the above equation with respect to time:

d(2*vB)/dt - d(3*vA)/dt = 0

⇒ 2*aB - 3*aA = 0 (ii)

Substituting in equation (ii)

2*aB - 3*(- 13.33 mm/s²) = 0

⇒ aB = - 20 mm/s²

b) From the pic shown,

xD - xA = constant

If we apply

d(xD)/dt - d(xA)/dt = 0

⇒ vD - vA = 0

then

d(vD)/dt - d(vA)/dt = 0

⇒ aD - aA = 0

⇒ aD = aA = - 13.33 mm/s²

c) We use the formula

vB = v₀B + aB*t

Substituting the values in above equation:

vB = 150 mm/s + (- 20 mm/s²)*(4 s)

⇒ vB = 70 mm/s

d) We apply the equation

xB = v₀B*t + 0.5*aB*t²

Substituting the values in above equation:

xB = (150 mm/s)*(4 s) + 0.5*(- 20 mm/s²)*(4 s)²

⇒ xB = 440 mm

4 0

3 years ago

The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to

Sergeeva-Olga [200]

Answer:

hello below is missing piece of the complete question

minimum size = 0.3 cm

answer : 0.247 N/mm2

Explanation:

Given data :

section span : 10.9 and 13.4 cm

minimum load applied evenly to the top of span : 13 N

maximum load for each member ; 4.5 N

lets take each member to be 4.2 cm

Determine the max value of P before truss fails

Taking average value of section span ≈ 12 cm

Given minimum load distributed evenly on top of section span = 13 N

we will calculate the value of by applying this formula

= $\frac{Wl^2}{12} = (0.013 * 0.0144 )/ 12$ = 1.56 * 10^-5

next we will consider section ; 4.2 cm * 0.3 cm

hence Z (section modulus ) = BD^2 / 6

= ( 0.042 * 0.003^2 ) / 6 = 6.3*10^-8

Finally the max value of P( stress ) before the truss fails

= M/Z = ( 1.56 * 10^-5 ) / ( 6.3*10^-8 )

= 0.247 N/mm2

5 0

3 years ago

In a surface grinding operation, the wheel diameter = 8.0 in, wheel width = 1.0 in, wheel speed = 6000 ft/min, work speed = 40 f

Katen [24]

Answer: the answer will be d because it is the right one to be

Explanation:

7 0

3 years ago