True or false Osha engineering controls are the last strategy of control an employer should use for job hazards

How many ase certifications are there for automotive technicians?

romanna [79]

Answer:

There are 50 ASE certification tests, covering almost every imaginable aspect of the automotive repair and service industry.

Explanation:

yww <33

5 0

3 years ago

Are there any danganronpa enjoyers on here :o

Mariana [72]

Indeed :D hello baylee

8 0

3 years ago

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Germanium forms a substitutional solid solution with silicon. Compute the number of germanium atoms per cubic centimeter for a g

brilliants [131]

Answer:

The number of germanium atoms per cubic centimeter for this germanium-silicon alloy is 3.16 x 10²¹ atoms/cm³.

Explanation:

Concentration of Ge ( $C_{Ge}$ ) = 15%

Concentration of Si (C $_{Si}$ ) = 85%

Density of Germanium (ρ $_{Ge}$ ) = 5.32 g/cm³

Density of Silicon (ρ $_{Si}$ ) = 2.33 g/cm³

Atomic mass of Ge (A $_{Ge}$ )= 72.64 g/mol

To calculate the number of Ge atoms per cubic centimeter for the alloy, we will use the formula:

No of Ge atoms/cm³=[Avogadro's Number* $C_{Ge}$ ]/([ $C_{Ge}$ *A $_{Ge}$ /ρ $_{Ge}$ )+(C $_{Si}$ *A $_{Ge}$ /ρ $_{Si}$ )]

= (6.02x10²³ * 15%) / [(15% * 72.64/5.32)+(72.64*85%/2.33)]

= (9.03x10²²)/(2.048+26.499)

= (9.03x10²²)/(28.547)

No of Ge atoms/cm³ = 3.16 x 10²¹ atoms/cm³

3 0

3 years ago

A counter-flow double-piped heat exchange is to heat water from 20oC to 80oC at a rate of 1.2 kg/s. The heating is to be accompl

lawyer [7]

Answer:

110 m or 11,000 cm

Explanation:

let mass flow rate for cold and hot fluid = Mc and Mh respectively
let specific heat for cold and hot fluid = Cpc and Cph respectively
let heat capacity rate for cold and hot fluid = Cc and Ch respectively

Mc = 1.2 kg/s and Mh = 2 kg/s

Cpc = 4.18 kj/kg °c and Cph = 4.31 kj/kg °c

Using effectiveness-NUT method

First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate

Cc = Mc × Cpc = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

Ch = Mh × Cph = 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= Cmin/Cmax

C = 5.016/8.62 = 0.582

.2 Second, we determine the maximum heat transfer rate, Qmax

Qmax = Cmin (Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Qmax = (5.016 kW/°c)(160 - 20) °c

Qmax = (5.016 kW/°c)(140) °c = 702.24 kW

.3 Third, we determine the actual heat transfer rate, Q

Q = Cmin (outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Qmax = (5.016 kW/°c)(60) °c = 303.66 kW

.4 Fourth, we determine Effectiveness of the heat exchanger, ε

ε = Q/Qmax

ε = 303.66 kW/702.24 kW

ε = 0.432

.5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

.6 sixth, we determine Heat Exchanger surface area, As

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

.7 Finally, we determine the length of the heat exchanger, L

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago

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

creativ13 [48]

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

5 0

3 years ago

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