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2 years ago

Which option shows the most valuable metallic properties

Engineering

1 answer:

Rina8888 [55]2 years ago

6 0

Malleable and ductile

non metals like plastic also have other properties but can't be malleable and ductile so they r most valuable metallic properties

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How should you move your board through the planer? (Pick two choices.)

iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0

3 years ago

Refer to the following distribution of commissions:Monthly Commissions Class Frequencies$600 up to $800 3800 up to 1,000 71,000

erica [24]

Answer:

21.81 %

Explanation:

given,

$600 up to $800 3

800 up to 1,000 7

$1,000 up to $1,200 11

$1,200 up to $1,400 12

$1,400 up to $1,600 40

$1,600 up to $1,800 24

$1,800 up to $2,000 9

$2,000 up to $2,200 4

Total = 3 + 7 + 11 + 12 + 40 + 24 + 9 + 4 = 110

the frequency of $1,600 to $1800

= $\dfrac{24}{110}$

=21.81 %

7 0

3 years ago

The outer surface of a spacecraft in space has an emissivity of 0.6 and an absorptivity of 0.2 for solar radiation. If solar rad

kondaur [170]

Answer:

T surface = 3.9°C

Explanation:

given data

emissivity 0.6

absorptivity = 0.2

solar radiation is incident rate = 1200 W/m²

solution

we get here surface temperature by equality of emitted and absorbed heat rate that is

Q (absorbed) = Q (heat ) .................1

α Qinc = $\epsilon * \sigma *A*T^4(surface)$

T surface = $\sqrt[4]{\frac{\alpha Qinc}{\epsilon *\sigma * A} }$ ..........................2

put here value and we get

T surface = $\sqrt[4]{\frac{0.2*1000}{0.6*5.67**10^{-8}} }$

T surface = 276.9 K

T surface = 3.9°C

8 0

3 years ago

Air is drawn from the atmosphere into a turbomachine. At the exit, conditions are 500 kPa (gage) and 130oC. The exit speed is 10

finlep [7]

Answer:

P=- 88.41 KW

Negative sign indicates that power is given to the system.

Explanation:

Given that

P₂=500 KPa

T₂=130°C

V₂=100 m/s

mass flow rate ,m= 0.8 kg/s

Lets take inlet condition for air

T₁=25°C

P₁=100 KPa

V₁=0 m/s

We know that

Heat capacity for air Cp=1.005 KJ/kg.k

We know that for air change in enthalpy only depends only on temperature

Now from first law for open system

$h_1+\dfrac{V_1^2}{2000}=h_2+\dfrac{V_2^2}{2000}+W$

$1.005\times 298+\dfrac{0^2}{2000}=1.005\times 403+\dfrac{100^2}{2000}+W$

$W=1.005\times 298-1.005\times 403-\dfrac{100^2}{2000}$

W=-110.52 KJ/kg

Shaft power P = m .W

P = -110.52 x 0.8

P=- 88.41 KW

Negative sign indicates that power is given to the system.

3 0

3 years ago

Ammonia gas is diffusing at a constant rate through a layer of stagnant air 1 mm thick. Conditions are such that the gas contain

fiasKO [112]

Answer:

The solution to this question is 5.153×10⁻⁴(kmol)/(m²·s)

That is the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

Explanation:

The diffusion through a stagnant layer is given by

$N_{A} = \frac{D_{AB} }{RT} \frac{P_{T} }{z_{2} - z_{1} } ln(\frac{P_{T} -P_{A2} }{P_{T} -P_{A1} })$

Where

$D_{AB}$ = Diffusion coefficient or diffusivity

z = Thickness in layer of transfer

R = universal gas constant

$P_{A1}$ = Pressure at first boundary

$P_{A2}$ = Pressure at the destination boundary

T = System temperature

$P_{T}$ = System pressure

Where $P_{T}$ = 101.3 kPa $P_{A2} =0$ , $P_{A1} =y_{A}$ , $P_{T} =$ 0.5×101.3 = 50.65 kPa

Δz = z₂ - z₁ = 1 mm = 1 × 10⁻³ m

R = $\frac{kJ}{(kmol)(K)} ,$ T = 298 K and $D_{AB}$ = 1.18 $\frac{cm^{2} }{s}$ = 1.8×10⁻⁵ $\frac{m^{2} }{s}$

$N_{A} = \frac{1.8*10^{-5} }{8.314*295} *\frac{101.3}{1*10^{-3} }* ln(\frac{101.3-0}{101.3-50.65})$ = 5.153×10⁻⁴ $\frac{kmol}{m^{2}s }$

Hence the rate of diffusion of ammonia through the layer is

5.153×10⁻⁴(kmol)/(m²·s)

5 0

3 years ago