Which of the following can not be used to store an electrical charge?

A simply supported wood roof beam is loaded with single point dead and roof live loads applied at midspan (PD = 400 lb, PLr = 16

Lynna [10]

Answer:mold i belive

Explanation:

3 0

3 years ago

An iron-carbon alloy initially containing 0.286 wt% C is exposed to an oxygen-rich and virtually carbon-free atmosphere at 1200°

Fantom [35]

Answer:

Explanation:

Given data:

initial construction co = 0.286 wt %

concentration at surface position cs = 0 wt %

carbon concentration cx = 0.215 wt%

time = 7 hr

$D = 7.5 \times 10^{-11} m^2/s$

for 0.225% carbon concentration following formula is used

$\frac{cx -co}{cs -co} = 1 - erf(\frac{x}{2\sqrt{DT}})$

where, erf stand for error function

$\frac{cx -co}{cs -co} = \frac{0.215 -0.286}{0 -0.286} =0.248$

$0.248 = 1 - erf(\frac{x}{2\sqrt{DT}})$

$erf(\frac{x}{2\sqrt{DT}}) = 1 - 0.248$

$erf(\frac{x}{2\sqrt{DT}}) = 0.751$

from the table erf(Z) value = 0.751 lie between (z) = 0.80 and z = 0.85 so by inteerpolation we have z = 0.815

from given table

$\frac{x}{2\sqrt{DT}} = 0.815$

$x = 2\times 0.815 \times \sqrt{7.5 \times 10^{-11}\times (7\times 3600)$

$x = 2.39\times 10^{-3} m$

x = 0.002395 mm

8 0

3 years ago

4. What are these parts commonly called?

patriot [66]

These parts are commonly called carburetor emulsion tubes. These tubes maintain the air-fuel ratio at different speeds.

The carburetor is a device of the combustion engine power supply system that mixes fuel and air in order to facilitate internal combustion.

The carburetor emulsion tubes are tubes that maintain the air-fuel ratio at different velocities.

These tubes (carburetor emulsion tubes) are small brass cylinders where the metering needle slides into them.

Learn more about carburetors here:

brainly.com/question/4237015

7 0

3 years ago

Which of the following best describes the role of engineers

Fantom [35]

Problem Solvers

Explanation:

Engineers find problems in the world, and then they find solutions for them.

8 0

3 years ago

A rectangular block of material with shear modulus G= 620 MPa is fixed to rigid plates at its top and bottom surfaces. Thelower

PIT_PIT [208]

Answer:

γ $_{xy}$ =0.01, P=248 kN

Explanation:

Given Data:

displacement = 2mm ;

height = 200mm ;

l = 400mm ;

w = 100 ;

G = 620 MPa = 620 N//mm²; 1MPa = 1N//mm²

a. Average Shear Strain:

The average shear strain can be determined by dividing the total displacement of plate by height

γ $_{xy}$ = displacement / total height

= 2/200 = 0.01

b. Force P on upper plate:

Now, as we know that force per unit area equals to stress

τ = P/A

Also, τ = Gγ $_{xy}$

By comapring both equations, we get

P/A = Gγ $_{xy}$ ------------ eq(1)

First we need to calculate total area,

A = l*w = 400 * 100= 4*10^4mm²

By putting the values in equation 1, we get

P/40000 = 620 * 0.01

P = 248000 N or 2.48 *10^5 N or 248 kN

6 0

3 years ago