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2 years ago

7

Technician A says that weld-through primer can be removed from the immediate weld area to improve weld quality. Technician B say

s that weld-through primer can be removed from adjacent areas to the weld site to improve penetration. Who is right

1 answer:

lisabon 2012 [21]2 years ago

4 0

Answer:

83737373777473737373738388383838

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A subway car leaves station A; it gains speed at the rate of 4 ft/s^2 for of until it has reached and then at the rate 6 then th

salantis [7]

Answer:

See attachment

1512 ft

Explanation:

Since the acceleration is either constant or zero, the a−t curve is made of horizontal straight-line segments. The values of t2 and a4 are determined as follows:

Acceleration - Time

0 < t < 6: Change in v = area under a–t curve

V_6 - 0 = (6 s)(4 ft/s2 ) = 24 ft/s

6 < t < t2: Since the velocity increases from 24 to 48 ft/s,

Change in v = area under a–t curve

48 - 24 = (t2 - 6) * 6

t2 = 10 s

t2 < t < 34: Since the velocity is constant, the acceleration is zero.

34 < t < 40: Change in v = area under a–t curve

0 - 42 = 6*a4

a4 = - 8 ft / s^2

The acceleration being negative, the corresponding area is below the t axis; this area represents a decrease in velocity.

Velocity - Time

Since the acceleration is either constant or zero, the v−t curve is made of straight-line segments connecting the points determined above.

Change in x = area under v−t curve

0 < t < 6: x6 - 0 = 0.5*6*24 = 72 ft

6 < t < 10: x10 - x6 = 0.5*4*(24 + 48) = 144 ft

10 < t < 34: x34 - x10 = 48*24 = 1152 ft

34 < t < 40: x40 - x34 = 0.5*6*48 = 144 ft

Adding the changes in x, we obtain the distance from A to B:

d = x40 - 0 = 1512 ft

8 0

2 years ago

A steady tensile load of 5.00kN is applied to a square bar, 12mm on a side and having a length of 1.65m. compute the stress in t

Shtirlitz [24]

Answer:

The stress in the bar is 34.72 MPa.

The design factor (DF) for each case is:

A) DF=0.17

B) DF=0.09

C) DF=0.125

D) DF=0.12

E) DF=0.039

F) DF=1.26

G) DF=5.5

Explanation:

The design factor is the relation between design stress and failure stress. In the case of ductile materials like metals, the failure stress considered is the yield stress. In the case of plastics or ceramics, the failure stress considered is the breaking stress (ultimate stress). If the design factor is less than 1, the structure or bar will endure the applied stress. By the opposite side, when the DF is higher than 1, the structure will collapse or the bar will break.

we will calculate the design stress in this case:

$\displaystyle \sigma_{dis}=\frac{T_l}{Sup}=\frac{5.00KN}{(12\cdot10^{-3}m)^2}=34.72MPa$

The design factor for metals is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{y}}$

The design factor for plastic and ceramics is:

$DF=\displaystyle \frac{\sigma_{dis}}{\sigma_{f}}=\frac{\sigma_{dis}}{\sigma_{u}}$

We now need to know the yield stress or the ultimate stress for each material. We use the AISI and ASTM charts for steels, materials charts for non-ferrous materials and plastics safety charts for the plastic materials.

For these cases:

A) The yield stress of AISI 120 hot-rolled steel (actually is AISI 1020) is 205 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{205MPa}=0.17$

B) The yield stress of AISI 8650 OQT 1000 steel is 385 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{385MPa}=0.09$

C) The yield stress of ductile iron A536-84 (60-40-18) is 40Kpsi, this is 275.8 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{275.8MPa}=0.125$

D) The yield stress of aluminum allot 6061-T6 is 290 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{290MPa}=0.12$

E) The yield stress of titanium alloy Ti-6Al-4V annealed (certified by manufacturers) is 880 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{880MPa}=0.039$

F) The ultimate stress of rigid PVC plastic (certified by PVC Pipe Association) is 4Kpsi or 27.58 MPa, therefore:

$DF=\displaystyle\frac{34.72MPa}{27.58 MPa}=1.26$

In this case, the bar will break.

F) You have to consider that phenolic plastics are used as matrix in composite materials and seldom are used alone with no reinforcement. In this question is not explained if this material is reinforced or not, therefore I will use the ultimate stress of most pure phenolic plastics, in this case, 6.31 MPa:

$DF=\displaystyle\frac{34.72MPa}{6.31 MPa}=5.5$

This material will break.

3 0

2 years ago

Determine the design angle ϕ (0∘≤ϕ≤90 ∘) between struts AB and AC so that the 400 lb horizontal force has a component of 600 lb

Svetllana [295]

Answer:

design angle ∅ = 4.9968 ≈ 5⁰

Explanation:

First calculate the force Fac :

Fac = $\sqrt{400^2 + 650^2 - 2(400)(650)cos30}$

= $\sqrt{160000 + 422500 - 80210}$

= 708.72 Ib

using the sine law to determine the design angle

$\frac{sin}{400} = \frac{sin 30}{Fac}$

hence ∅ = $sin^{-1} (\frac{sin 30 *400}{708.72} )$

= $sin^{-1} 0.0871$ = 4.9968 ≈ 5⁰

4 0

2 years ago

How does miep build suspense in this passage

siniylev [52]

Then I walked into Mr. and Mrs. Frank's bedroom. On the floor, amidst the chaos of papers and books, my eye lit on the little red-orange checkered, cloth-bound diary that Anne had received from her father on her thirteenth birthday. She loves Anne and respects her work.

I hope this helps you.

3 0

2 years ago

Read 2 more answers

Is an isothermal process necessarily internally reversible? Explain your answer with an example

torisob [31]

Answer:

please give me brainlist and follow

Explanation:

Example of an irreverseble isothermal process is mixing of two fluids on the same temperature - it requires a lot of energy to unmix Jack and coke. ... Example of an reversible process with changing temperature is isentropic expansion.

5 0

1 year ago