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3 years ago

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2.14 (a) Using series/parallel resistance reductions, find the equivalent resistance between terminals A and B in the circuit of

Figure P2.14. (b) Repeat, but with C and D shorted together.

1 answer:

olasank [31]3 years ago

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Using data in Appendix A ,calculate the number of atoms in 1 tonne of iron

maksim [4K]

<h2>Answer:</h2>

$1.0783*10^{28}(atoms).$

<h2>Explanation:</h2>

<em>Since I don't have access to "Appendix A", I'll solve the problem using data from the periodic table.</em>

<em />

<h3>1. Determine the molar mass of iron.</h3>

<em>According to the periodic table, the molar mass of iron is:</em>

55.845g/mole.

<h3>2. Convert 1 tonne to grams.</h3>

$1(tonne)*1000=1000kg\\1000kg*1000=1000000g=10^6g$

<h3>3. Apply rule of 3.</h3>

$55.845g$ ----------- $1 mole$

$10^6g$ ----------- $x$

$x=\frac{10^6*1}{55.845}=17906.7061(moles)$

<h3>4. Determine the amount of atoms.</h3>

<em>Considering that there are, approximately, </em> $6.022*10^{23}$ atoms in a mole of any element, apply another rule of 3.

1 mole --------------------- $6.022*10^{23}(atoms)$

$17906.7061(moles)$ --------------------- x

$x=\frac{17906.7061*6.022*10^{23}}{1}=1.0783*10^{28}$ .

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2 years ago

Looking for new information is one reason you should do research for a speech.

ValentinkaMS [17]

The correct answer is False

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3 years ago

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Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 1

Elis [28]

Answer:

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W" $_{cv$ / [ (h₁ - h₂ ) - $\frac{V_2^2}{2}$ ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW × $\frac{1\frac{k^3}{s} }{1kW}$ ] / [ (932.93 - 503.02 )k³/kg - $\frac{100^2\frac{m^2}{s^2} }{2}$ | $\frac{ln}{kg\frac{m}{s^2} }$ || $\frac{1kJ}{10^3N-m}$ | ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

so

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[ $(\frac{8.314}{28.97}\frac{k^3}{kg.K})$ × $500 K$ × $(7.54 kg/s)$ ] / [(1 bar)(100 m/s )]} | $\frac{1 bar}{10N/m^2}$ || $10^3N.m/1k^3$

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²

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3 years ago

I'll give a free brainliest

spayn [35]

All I need is one more.
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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

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3 years ago