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igomit [66]
3 years ago
10

It describes the physical and social elements common to this work. Note that common contexts are listed toward the top, and less

common contexts are listed toward the bottom. According to O*NET, what are common work contexts for Construction Carpenters? Check all that apply. Spend time sitting wear common protective or safety equipment face-to-face discussions exposed to hazardous equipment spend time using hands to handle, control, or feel objects, tools, or controls public speaking
Engineering
2 answers:
QveST [7]3 years ago
7 0

Answer:

BCDE

Explanation:

just look at the link, it tells you.

Dafna1 [17]3 years ago
3 0

Answer:

bcde!!

Explanation:

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Select the correct answer.
MAVERICK [17]

Answer:crane and engine I guess

Explanation:

8 0
3 years ago
A solid circular rod that is 600 mm long and 20 mm in diameter is subjected to an axial force of P = 50 kN The elongation of the
Kay [80]

Answer:

a) V = 0.354

b)  G = 25.34 GPA

Explanation:

Solution:

We first determine Modulus of Elasticity and Modulus of rigidity

Elongation of rod ΔL = 1.4 mm

Normal stress, δ = P/A

Where P = Force acting on the cross-section

A = Area of the cross-section

Using Area, A = π/4 · d²

= π/4 · (0.0020)²  = 3.14 × 10⁻⁴m²

δ = 50/3.14 × 10⁻⁴    = 159.155 MPA

E(long) = Δl/l  = 1.4/600 = 2.33 × 10⁻³mm/mm

Modulus of Elasticity Е = δ/ε

= 159.155 × 10⁶/2.33 × 10⁻³    = 68.306 GPA

Also final diameter d(f) = 19.9837 mm

Initial diameter d(i) = 20 mm

Poisson said that V = Е(elasticity)/Е(long)

= -  <u>( 19.9837 - 20 /20)</u>

        2.33 × 10⁻³                  

= 0.354,

∴ v = 0.354

Also G = Е/2. (1+V)

=  68.306 × 10⁹/ 2.(1+ 0.354)

= 25.34 GPA

⇒ G = 25.34 GPA

5 0
3 years ago
Multiply. Write the answer in simplest form. 1 3/10×1/8
kicyunya [14]

9514 1404 393

Answer:

  13/80

Explanation:

The product is ...

  (1 3/10)×(1/8) = (13/10)×(1/8) = (13×1)/(10×8) = 13/80

4 0
2 years ago
It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
denis23 [38]

The exit temperature is 586.18K and  compressor input power is 14973.53kW

Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

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6 0
2 years ago
Which type of boot authentication is more secure?
MA_775_DIABLO [31]

The type of boot authentication that is more secure is Unified Extensible Firmware Interface

Unified Extensible Firmware Interface  help to provide a computer booting that is more secured.

Unified Extensible Firmware Interface is a computer software program that work hand in hand with an operating system,  it main function is to stop a computer system from boot with an operating system that is not secured.

For a  computer system to boot successfully it means that the Operating system support the  Unified Extensible Firmware Interface because it secured.

Inconclusion The type of boot authentication that is more secure is Unified Extensible Firmware Interface

Learn more here :

brainly.com/question/24750986

7 0
2 years ago
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