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Softa [21]
3 years ago
5

A Si sample contains 1016 cm-3 In acceptor atoms and a certain number of shallow donors, the In acceptor level is 0.16 eV above

Ev , and EF is 0.26 eV above Ev at 300 K. How many (cm-3) In atoms are un-ionized (i.e., neutral)

Engineering
2 answers:
matrenka [14]3 years ago
6 0

Answer:

Explanation: see attachment below

creativ13 [48]3 years ago
5 0

Answer:

6.5 × 10¹⁵/ cm³

Explanation:

Thinking process:

The relation N_{o} = N_{i} * \frac{E_{f}-E_{i}  }{KT}

With the expression Ef - Ei = 0.36 × 1.6 × 10⁻¹⁹

and ni = 1.5 × 10¹⁰

Temperature, T = 300 K

K = 1.38 × 10⁻²³

This generates N₀ = 1.654 × 10¹⁶ per cube

Now, there are 10¹⁶ per cubic centimeter

Hence, N_{d}  = 1.65*10^{16}  - 10^{16} \\           = 6.5 * 10^{15} per cm cube

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An asphalt concrete mixture includes 94% aggregates by weight. The specific gravity of aggregate and asphalt are 2.7 and 1.0, re
riadik2000 [5.3K]

Answer:

The correct solution is "5.74%".

Explanation:

The given values are:

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G_{agg}=2.7

Gravity of asphalt,

G_{asp}=1.0

Asphalt concrete mixture,

W_{agg}=0.94 \ W_m

We know that,

W_{asp}=W_m-W_{agg}

        =0.06 \ W_m

Now,

The theoretical specific gravity of mix,

⇒ G_t=\frac{W_{agg}+W_{asp}}{\frac{W_{agg}}{G_{agg}} +\frac{W_{asp}}{G_{asp}} }

By putting the values, we get

         =\frac{0.94 \ Wm+0.06 \ Wm}{\frac{0.94 \ Wm}{2.7} +\frac{0.06 \ Wm}{1} }

         =2.45

hence,

The percentage of voids will be:

⇒  %V = \frac{G_t-G_m}{G_t}\times 100

           = \frac{2.45-2.317}{2.45}\times 100

           = \frac{0.133}{2.317}\times 100

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A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reserv
Dahasolnce [82]

This question is incomplete, the complete question is;

A centrifugal pump is used to extract water from a reservoir at 14,000 gal/min. The pipe connecting the pump inlet to the reservoir is 12 inches in diameter and is 65 ft long.

The average friction factor in this pipe is 0.018. The pump performance curves indicate that, at this flow rate, the head rise across the pump is 320 ft, the efficiency is 81% and the required NPSH is 25 ft.

Please estimate: The required brake horsepower.

Answer:

The required brake horsepower is 1400.08

Explanation:

Given the data in the question;

Power required to drive the pump can be determined using the formula;

P = r_wQH / η₀(0.745)

given that; centrifugal pump is used to extract water from a reservoir at 14,000 gal/min.

Q = 14,000 gal/min = ( 14,000 × 0.00006309 )m³/sec = 0.883 m³/sec

the head rise across the pump is 320 ft,

H = 320 ft = ( 320 × 0.3048 )m = 97.536 m

the efficiency η₀ = 81% = 0.81

r_w = 9.81 kN/m³

so we substitute our values into the formula

P = [ 9.81 × 0.883 × 97.536 ] / 0.81(0.745)

P = 844.87926528 / 0.60345

P = 1400.08 HP

Therefore, The required brake horsepower is 1400.08

5 0
3 years ago
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