In an RL parallel circuit, VT = 240 V, R = 330 Ω, and XL = 420 Ω. What is the Apparent Power (VA)?

Write a Nested While Loop that will increment the '*' from 1 to 10.

andrey2020 [161]

Answer:

The program is as follows:

i = 1

while(i<11):

j = 1

while(j<=i):

print('*', end = '')

j += 1

i += 1

print()

Explanation:

Initialize i to 1

i = 1

The outer loop is repeated as long as i is less than 11

while(i<11):

Initialize j to 1

j = 1

The inner loop is repeated as long as j is less than or equal i

while(j<=i):

This prints a *

print('*', end = '')

This increments j and ends the inner loop

j += 1

This increments i

i += 1

This prints a blank and ends the inner loop

print()

8 0

3 years ago

One cylinder in the diesel engine of a truck has an initial volume of 650 cm3 . Air is admitted to the cylinder at 35 ∘C and a p

kupik [55]

Answer:

1) the final temperature is T2 = 876.76°C

2) the final volume is V2 = 24.14 cm³

Explanation:

We can model the gas behaviour as an ideal gas, then

P*V=n*R*T

since the gas is rapidly compressed and the thermal conductivity of a gas is low a we can assume that there is an insignificant heat transfer in that time, therefore for adiabatic conditions:

P*V^k = constant = C, k= adiabatic coefficient for air = 1.4

then the work will be

W = ∫ P dV = ∫ C*V^(-k) dV = C*[((V2^(-k+1)-V1^(-k+1)]/( -k +1) = (P2*V2 - P1*V1)/(1-k)= nR(T2-T1)/(1-k) = (P1*V1/T1)*(T2-T1)/(1-k)

W = (P1*V1/T1)*(T2-T1)/(1-k)

T2 = (1-k)W* T1/(P1*V1) +T1

replacing values (W=-450 J since it is the work done by the gas to the piston)

T2 = (1-1.4)*(-450J) *308K/(101325 Pa*650*10^-6 m³) + 308 K= 1149.76 K = 876.76°C

the final volume is

TV^(k-1)= constant

therefore

T2/T1= (V2/V1)^(1-k)

V2 = V1* (T2/T1)^(1/(1-k)) = 650 cm³ * (1149.76K/308K)^(1/(1-1.4)) = 24.14 cm³

3 0

3 years ago

ASAP correct answer plss When you are driving, if you see this traffic sign it means

vlabodo [156]

Answer:

C

Explanation:

5 0

3 years ago

How many sets of equations (V and M equations) would you need to describe shear and moment as functions of x for this beam? In o

den301095 [7]

Shear and moment as functions of x is described below .

Explanation:

1. Beam is the slender bar that carries transverse

loading; that is, the applied force are perpendicular to the bar.

2. In a beam, the internal force system consist of a shear force and

a bending moment acting on the cross section of the bar.

3. The shear force and the bending moment usually vary continuously

along the length of the beam.

4. The internal forces give rise to two kinds of stresses on a

transverse section of a beam:

(1) normal stress that is caused by

bending moment and

(2) shear stress due to the shear force.

Knowing the distribution of the shear force and the bending

moment in a beam is essential for the computation of stresses

and deformations.

Shear- Moment Equations

The determination of the internal force system acting at a given

section of a beam : draw a free-body diagram that expose these

forces and then compute the forces using equilibrium equations.

The goal of the beam analysis －determine the shear force V and the bending moment M at every cross section of the beam.

To derive the expressions for V and M in terms of the distance x

measured along the beam. By plotting these expressions to scale,

obtain the shear force and bending moment diagrams for the

beam.

The shear force and bending moment diagrams are convenient

visual references to the internal forces in a beam; in particular, they identify the maximum values of V and M

5 0

2 years ago

A steel bar is 150 mm square and has a hot-rolled finish. It will be used in a fully reversed bending application. Sut for the s

Xelga [282]

Answer:

See explanation

Explanation:

Given The bar is square and has a hot-rolled finish. The loading is fully reversed bending.

Tensile Strength

Sut: 600 MPa

Maximum temperature

Tmax: 500 °C

Bar side dimension

b: 150 mm

Alternating stress

σa: 100 MPa

Reliability

R: 0.999 Note 1.

Assumptions Infinite life is required and is obtainable since this ductile steel will have an endurance limit. A reliability factor of 99.9% will be used.

Solution See Excel file Ex06-01.xls.

1 Since no endurance-limit or fatigue strength information is given, we will estimate S'e based on the ultimate tensile strength using equation 6.5a.

S'e: 300 MPa = 0.5 * Sut

2 The loading is bending so the load factor from equation 6.7a is

Cload: 1

3 The part size is greater than the test specimen and the part is not round, so an equivalent diameter based on its 95% stressed area must be determined and used to find the size factor. For a rectangular section in nonrotating bending, the A95 area is defined in Figure 6-25c and the equivalent diameter is found from equation 6.7d

A95: 1125 mm2 = 0.05 * b * b Note 2.

dequiv: 121.2 mm = SQRT(A95val / 0.0766)

and the size factor is found for this equivalent diameter from equation 6.7b, to be

Csize: 0.747 = 1.189 * dequiv^-0.097

4 The surface factor is found from equation 6.7e and the data in Table 6-3 for the specified hot-rolled finish.

Table 6-3 constants

A: 57.7

b: -0.718 Note 3.

Csurf: 0.584 = Acoeff * Sut^bCoeff

5 The temperature factor is found from equation 6.7f :

Ctemp: 0.710 = 1 - 0.0058 * (Tmax - 450)

6 The reliability factor is taken from Table 6-4 for R = 0.999 and is

Creliab: 0.753

7 The corrected endurance limit Se can now be calculated from equation 6.6:

Se: 69.94 MPa = Cload * Csize * Csurf * Ctemp *

Creliab * Sprme

Let

Se: 70 MPa

8 To create the S-N diagram, we also need a value for the estimated strength Sm at 103 cycles based on equation 6.9 for bending loading.

Sm: 540 MPa = 0.9 * Sut

9 The estimated S-N diagram is shown in Figure 6-34 with the above values of Sm and Se. The expressions of the two lines are found from equations 6.10a through 6.10c assuming that Se begins at 106 cycles.

b: -0.2958 Note 4.

a: 4165.7

Plotting Sn as a function of N from equation 6.10a

N Sn (MPa)

1000 540 =aa*B73^bb

2000 440

4000 358

8000 292

16000 238

32000 194

64000 158

128000 129

256000 105

512000 85

1000000 70

FIGURE 6-34. S-N Diagram and Alternating Stress Line Showing Failure Point

10 The number of cycles of life for any alternating stress level can now be found from equation 6.10a by replacing σa for Sn.

At N = 103 cycles,

Sn3: 540 MPa = aa * 1000^bb

At N = 106 cycles,

Sn6: 70 MPa = aa * 1000000^bb

The figure above shows the intersection of the alternating stress line (σa = 100 MPa) with the failure line at N = 3.0 x 105 cycles.

8 0

2 years ago