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3 years ago

What year was the international union of brick layers and allied craftsman founded? *

Engineering

1 answer:

Butoxors [25]3 years ago

4 0

Answer:

October 17, 1865

Explanation:

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An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610

KatRina [158]

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + $\frac{q}{h\pi Di}$

Ts = $30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000} }$

$Ts=30+\frac{294}{0.3928} \\$

$Ts =30+748.4\\Ts = 778.4$

The surface temperature is therefore 778.4°C if the cable is bare

6 0

3 years ago

Since the application can be the prospective employers first impression, it is important to a. always answer questions truthfull

loris [4]

Answer:

Definitely D

Explanation:

You should do all of those things in an application.

7 0

4 years ago

Read 2 more answers

A one-dimensional slab without heat generation has a thickness of 20 mm with surfaces main- tained at temperatures of 275 K and

vlada-n [284]

Answer:

a) 512.5 KW/m2

b) 40.75 KW/m2

c) 2 KW/m2

Explanation:

Given data;

T_2 = 325 K

T_1 = 275 K

dx = 0.20 mm

a) for aluminium K = 205 W/m k

heat flux $= k \frac{dt}{dx}$

$= 205 \frac{325 - 275}{0.02}$

= 512.5 KW/m2

b) for AISI 316 stainless steel

k = 16.3 W/ m k

heat flux $= k \frac{dt}{dx}$

$= 16.3 \frac{325 - 275}{0.02}$

= 40.75 KW/m2

C) for Concrete

k = 0.8 W/ m k

heat flux $= k \frac{dt}{dx}$

$= 0.8 \times \frac{325 - 275}{0.02}$

= 2 KW/m2

6 0

3 years ago

Consider a single crystal oriented such that the slip direction and normal to the slip plane are at angles 42.7° and 48.3°, resp

icang [17]

Answer:

A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

5 0

3 years ago

A 1 m wide continuous footing is designed to support an axial column load of 250 kN per meter of wall length. The footing is pla

creativ13 [48]

Answer:

correct option is (A) 0.5

Explanation:

given data

axial column load = 250 kN per meter

footing placed = 0.5 m

cohesion = 25 kPa

internal friction angle = 5°

solution

we know angle of internal friction is 5° that is near to 0°

so it means the soil is almost cohesive soil.

and for a pure cohesive soil

$N_{\gamma }$ = 0

and we know formula for $N_{\gamma }$ is

$N_{\gamma }$ = (Nq - 1 ) × tan(Ф) ..................1

so here Ф is very less $N_{\gamma }$ should be nearest to zero

and its value can be 0.5

so correct option is (A) 0.5

7 0

4 years ago