What are trouble code charts?

Air is compressed slowly in a piston-cylinder assembly from an initial state where P1 = 1.4 bar, V1= 4.25 m^3, to a final state

lord [1]

Answer:

W=-940.36 KJ

Explanation:

Given that

$P_1=1\ bar,V_1=4.25 {m^3}$

$P_2=6.8\ bar$

Process follows pv=constant

So this is the isothermal process and work in isothermal process given as

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

Now by putting the values (1.4 bar =140 KPa)

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=140\times 4.25 \ln \dfrac{1.4}{6.8}$

W=-940.36 KJ

Negative sign indicates that this is a compression process and work will given to the system.

7 0

4 years ago

Iron has a BCC crystal structure, an atomic radius of 0.124 nm, and an atomic weight of 55.85 g/mol. Compute and compare its the

Andre45 [30]

Answer:

Explanation:

The density of the unit cell of a material, Iron in this case, has to be approximately equal with its experimental value of 7.87 g/cm³.

The density d = m/v, so what we need to do is calculate the volume of the unit cell and its mass and perform the calculation.

For a BCC crystal structure the length of the side of the cube is given by:

a = 4r/√3

where a is the atomic radius of Iron

first we will convert this radius to cm since we want the density in g/cm³:

0.124 nm x 1 x 10⁻⁷ cm / nm = 1.24 x 10⁻⁸ cm

a = 4 x 1.24 x 10⁻⁸ cm /√3 = 2.86 x 10⁻⁸ cm

the volume of the cubic cell is:

v = a³ = ( 2.86 x 10⁻⁸ cm )³ =2.35 x 10⁻²³ cm³

The mass of iron in the body centered cubic cell is obtained from the mass of the atoms in it:

BCC = 2 atoms / unit cell ( 1/8 from the 8 corners + 1 in the center)

m = 2 atoms/unit cell x 1 mol/ 6.022 x 10²³ atoms x 55.85 g/mol

= 1.85 x 10⁻²² g

Therefore,

d = m/v = 1.85 x 10⁻²² g / 2.35 x 10⁻²³cm³ = 7.88 g/cm³

An excelent agreement which confirms that the density of the BCC unit cell agrees with the experimental value.

4 0

4 years ago

A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal

Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X $_{L}$ , which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX $_{L}$

X $_{L}$ = $\frac{V}{I}$ ................ 2

Given that; V = 240V, f = 50Hz, $\pi$ = $\frac{22}{7}$ , I = 15A, so that;

From equation 2,

X $_{L}$ = $\frac{240}{15}$

= 16Ω

To determine the inductance of the inductor,

X $_{L}$ = 2 $\pi$ fL

L = $\frac{X_{L} }{2 \pi f}$

= $\frac{16}{2*\frac{22}{7}*50 }$

= 0.05091

The inductance of the inductor is 0.051H.

4 0

4 years ago

A powerful contribution of the 1st and 2nd laws is to determine the design feasibility of a process/device/system. Please determ

Sladkaya [172]

Answer:

Detailed step wise solution is given below:

6 0

3 years ago

Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo

deff fn [24]

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

8 0

3 years ago

Read 2 more answers