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2 years ago

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You are investigating surface hardening in iron using nitrogen gas. Two 5 mm thick slabs of iron are separately exposed to nitro

gen environments at 850°C and 950°C, respectively. Both situations have achieved a steady-state diffusion condition. The concentration of nitrogen at the higher pressure surface at each temperature is 4 kg/m3 . What is the concentration of nitrogen on the lower pressure surface of each slab slab given that the diffusion flux is 9.0 x 10 -8 kg/(m2 *s)?

1 answer:

Luden [163]2 years ago

4 0

Answer and Explanation:

The explanation is attached below

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Calculate the differential pressure in kPa across the hatch of a submarine 320m below the surface of the sea. Assume the atmosph

kicyunya [14]

Answer:

The pressure difference across hatch of the submarine is 3217.68 kpa.

Explanation:

Gauge pressure is the pressure above the atmospheric pressure. If we consider gauge pressure for finding pressure differential then no need to consider atmospheric pressure as they will cancel out. According to hydrostatic law, pressure varies in the z direction only.

Given:

Height of the hatch is 320 m

Surface gravity of the sea water is 1.025.

Density of water 1000 kg/m³.

Calculation:

Step1

Density of sea water is calculated as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$

Here, density of sea water is $\rho_{sw}$ , surface gravity is S.G and density of water is $\rho_{w}$ .

Substitute all the values in the above equation as follows:

$S.G=\frac{\rho_{sw}}{\rho_{w}}$

$1.025=\frac{\rho_{sw}}{1000}$

$\rho_{sw}=1025$ kg/m³.

Step2

Difference in pressure is calculated as follows:

$\bigtriangleup p=rho_{sw}gh$

$\bigtriangleup p=1025\times9.81\times320$

$\bigtriangleup p=3217680$ pa.

Or

$\bigtriangleup p=(3217680pa)(\frac{1kpa}{100pa})$

$\bigtriangleup p=3217.68$ kpa.

Thus, the pressure difference across hatch of the submarine is 3217.68 kpa.

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2 years ago

the AADT for a section of suburban freeway is 150000 veh/day. Assuming this is an urban radial facility, what range of direction

igomit [66]

Answer:

design hour volumes will be 4000 to 6000

Explanation:

given data

AADT = 150000 veh/day

solution

we get here design hour volumes that is express as

design hour volumes = AADT × k × D ..............1

here k is factor and its range is 8 to 12 % for urban

and D is directional distribution i.e traffic equal divided by the direction

so here design hour volumes will be 4000 to 6000

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3 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

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3 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

Ratling [72]

Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

where;

$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

FOS = 2.183

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2$

$\dfrac{350}{2 FOS} =88.51$

$\dfrac{350}{ FOS} =2 \times 88.51$

${ FOS} =\dfrac{350}{2 \times 88.51}$

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

7 0

3 years ago

Because assembly language is so close in nature to machine language, it is referred to as a ____________.

My name is Ann [436]

Answer:

symbolic machine code.

Explanation:

The instructions in the language are closely linked to the machine's architecture.

5 0

3 years ago