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3 years ago

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Air,in a piston cylinder assembly, is initially at 300 K and 200 kPa.It is then heated at constant pressure to 600 K. Determine

the change in internal energy of air per unit mass, using:
a) The ideal gas table for air.b) Show the process on a (p-v) diagram.

2 answers:

Mandarinka [93]3 years ago

5 0

Answer:

Explanation: Please see the attached picture for answer.

Dvinal [7]3 years ago

5 0

Answer:Check the attached

Explanation:

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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

What do you own that might not be manufactured?

horrorfan [7]

Answer:

A pet

Explanation:

Latin time I checked animals aren't made by people? I honestly don't know if this helps but I'm technically not wrong.

8 0

3 years ago

Consider a 1.80-m-tall man standing vertically in water and completely submerged in a pool. Determine the difference between the

defon

Answer:

17.658 kPa

Explanation:

The hydrostatic pressure of a fluid is the weight of a column of that fluid divided by the base of that column.

$P = \frac{weight}{base}$

Also, the weight of a column is its volume multiplied by it's density and the acceleration of gravity:

$weight = \delta * v * g$

Meanwhile, the volume of a column is the area of the base multiplied by the height:

$V = base * h$

Replacing:

$P = \frac{\delta * base * h * g}{base}$

The base cancels out, so:

$P = \delta * h * g$

The pressure depends only on the height of the fluid column, the density of the fluid and the gravity.

If you have two point at different heights (or depths in the case of objects submerged in water) each point will have its own column of fluid exerting pressure on it. Since the density of the fluid and the acceleration of gravity are the same for both points (in the case of hydrostatics density is about constant for all points, it is not the case in the atmosphere), we can write:

$\Delta P = \rho * g * (h1 - h2)$

We do not know at what depth the man of this problem is, but it doesn't matter, because we know the difference in height of the two points of interes (h1 - h2) = 1.8 m. So:

$\Delta P = 9.81 \frac{m}{s^{2} } * 1000 \frac{kg}{m^3} * 1.8 m = 17658 Pa = 17.658 kPa$

4 0

3 years ago

Write a new ARMv8 assembly file called "lab04b.S" which is called by your main function. It should have the following specificat

Len [333]

Answer:

my_mul:

.globl my_mul

my_mul:

//Multiply X0 and X1

// Does not handle negative X1!

// Note : This is an in efficient way to multipy!

SUB SP, SP, 16 //make room for X19 on the stack

STUR X19, [SP, 0] //push X19

ADD X19, X1, XZR //set X19 equal to X1

ADD X9 , XZR , XZR //set X9 to 0

mult_loop:

CBZ X19, mult_eol

ADD X9, X9, X0

SUB X19, X19, 1

B mult_loop

mult_eol:

LDUR X19, [SP, 0]

ADD X0, X9, XZR // Move X9 to X0 to return

ADD SP, SP, 16 // reset the stack

BR X30

Explanation:

6 0

4 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

<h3>Flow rate in pipe B is = 0.3094 $\frac{m^{3} }{s}$ </h3>

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago