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3 years ago

What three training organizations are important for Union Masons?

Engineering

1 answer:

kvasek [131]3 years ago

8 0

Answer:

The International Masonry Institute (IMI) and the International Masonry Training and Education Foundation (IMTEF) both work to provide job opportunities and training for BAC members and contractors.

Explanation:

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Stainless steel ball bearings (rho = 8085 kg/m3 and cp = 0.480 kJ/kg·°C) having a diameter of 1.2 cm are to be quenched in water

sveticcg [70]

Answer:

$\dot Q = -2.341\,kJ$

Explanation:

The rate of heat transfer from the balls to the air is:

$\dot Q = \left[(800\,\frac{1}{min} )\cdot (\frac{1\,min}{60\,s} )\cdot(8085\,\frac{kg}{m^{3}})\cdot (\frac{4}{3}\pi )\cdot (6\times 10^{-3}\,m)^{3}\right]\cdot (0.480\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (850\,^{\textdegree}C-900\,^{\textdegree}C)$ $\dot Q = -2.341\,kJ$

3 0

3 years ago

Read 2 more answers

A heat engine does 210 J of work per cycle while exhausting 440 J of waste heat. Part A What is the engine's thermal efficiency?

Oxana [17]

Answer:

The engine's thermal efficiency is 0.32

Explanation:

Thermal efficiency = work done ÷ quantity of heat supplied

Work done = 210 J

Quantity of heat supplied = work done + waste heat = 210 + 440 = 650 J

Thermal efficiency = 210 ÷ 650 = 0.32

6 0

3 years ago

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

a) Determine the yielding factor of safety

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

b) Determine the overload factor of safety

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

C) Determine the factor of safety based on joint separation

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

D) Determine the fatigue factor of safety using the Goodman criterion.

nf = 0.849

attached below is the detailed solution .

4 0

3 years ago

In a cellular phone system, a mobile phone must be paged to receive a phone call. However, paging attempts don’t always succeed

Alina [70]

Answer:

The correct response will be "0.992". The further explanation to the following question is given below.

Explanation:

The probability that paging would be beneficial becomes 0.8

Effective paging at the very first attempted is 0.8

On the second attempt the success probability will be:

⇒ $0.2\times 0.8$

⇒ $0.16$

On the third attempt the success probability will be:

⇒ $0.2\times 0.2\times 0.8$

⇒ $0.032$

So that the success probability will be:

⇒ $0.8 + 0.16 + 0.032$

⇒ $0.992$

7 0

3 years ago

You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-

Katen [24]

Answer:

The required wall thickness is $1.506 \times 10^{-3}$ m

Explanation:

Given:

Fluid density $\rho = 1200$ $\frac{kg}{m^{3} }$

Diameter of tank $d = 8$ m

Length of tank $l = 4$ m

F.S = 4

For A-36 steel yield stress $\sigma = 250$ MPa,

Allowable stress $\sigma _{allow} = \frac{\sigma}{F.S}$

$\sigma _{allow} = \frac{250}{4} = 62.5$ MPa

Pressure force is given by,

$P = \rho gh$

$P = 1200 \times 9.8 \times 4$

$P = 47088$ Pa

Now for a vertical pipe,

$\sigma _{allow} = \frac{Pd}{4t}$

Where $t =$ required thickness

$t = \frac{Pd}{4 \sigma _{allow} }$

$t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }$

$t = 1.506 \times 10^{-3}$ m

Therefore, the required wall thickness is $1.506 \times 10^{-3}$ m

8 0

3 years ago