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3 years ago

Which of the following hazards is shared by surface mining and sub-surface mining?

Engineering

2 answers:

lubasha [3.4K]3 years ago

4 0

B I guess. since they both have potential to collapse

Alex777 [14]3 years ago

4 0

Answer: Hazardous chemicals

Explanation:

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Answer:

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3 years ago

In the truss below, the vertical reaction force at A is

Montano1993 [528]

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3 years ago

Draw a 3-D physical structure of an NMOS transistor. Label four terminals: body, drain, gate, and source. And also label silicon

Vanyuwa [196]

Answer:

Answer is attached.

Explanation:

A NMOS is a n-channel MOSFET or Metal Oxide

Semiconductor Field Effect Transistor. This type

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layers of Metal-oxide, Silicon-oxide and Silicon

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4 0

3 years ago

A jetliner flying at an altitude of 10,000 m has a Mach number of 0.5. If the jetliner has to drop down to 1000 m but still main

andreev551 [17]

Answer

Assuming

At 10000 m height temperature T = -55 C = 218 K

At 1000 m height temperature T = 0 C = 273 K

$\dfrac{V_1}{C_1} =\dfrac{V_2}{C_2} = 0.5$

R = 287 J/kg K

$C_1 = \sqrt{\gamma RT_1} = \sqrt{1.4\times 287\times 218} = 295 m/s$

$C_2 = \sqrt{\gamma RT_2} = \sqrt{1.4\times 287\times 273} = 331 m/s$

$V_2 = \dfrac{V_1}{C_1}\timesC_2$

V₂ = V₁ ×1.1222

V₁ = 0.5 × C₁ = 0.5 × 295 = 147.5 m/s

V₂ = 1.1222 × 147.5 = 165.49 m/s

so, the jetliner need to increase speed by ( V₂ -V₁ )

= 165.49 - 147.5

= 17.5 m/s

6 0

4 years ago

A diesel engine with CR= 20 has inlet at 520R, a maximum pressure of 920 psia and maximum temperature of 3200 R. With cold air p

Stella [2.4K]

Answer:

Cut-off ratio $\dfrac{V_3}{V_2}=6.15$

Cxpansion ratio $\dfrac{V_4}{V_3}=3.25$

The exhaust temperature $T_4=1997.5R$

Explanation:

Compression ratio CR(r)=20

$\dfrac{V_1}{V_2}=20$

$P_2=P_3=920 psia$

$T_1=520 R ,T_{max}=T_3,T_3=3200 R$

We know that for air γ=1.4

If we assume that in diesel engine all process is adiabatic then

$\dfrac{T_2}{T_1}=r^{\gamma -1}$

$\dfrac{T_2}{520}=20^{1.4 -1}$

$T_2=1723.28R$

$\dfrac{V_3}{V_2}=\dfrac{T_3}{T_2}$

$\dfrac{V_3}{V_2}=\dfrac{3200}{520}$

So cut-off ratio $\dfrac{V_3}{V_2}=6.15$

$\dfrac{V_1}{V_2}=\dfrac{V_4}{V_3}\times\dfrac{V_3}{V_2}$

Now putting the values in above equation

$\dfrac20=\dfrac{V_4}{V_3}\times 6.15$

$\dfrac{V_4}{V_3}=3.25$

So expansion ratio $\dfrac{V_4}{V_3}=3.25$ .

$\dfrac{T_4}{T_3}=(expansion\ ratio)^{\gamma -1}$

$\dfrac{T_3}{T_4}=(3.25)^{1.4 -1}$

$T_4=1997.5R$

So the exhaust temperature $T_4=1997.5R$

3 0

4 years ago