1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
elena-s [515]
3 years ago
14

A plane, opaque, surface M has the following properties: gray, diffuse, absorptivity = 0.7, surface area = 0.5 m2 , temperature

maintained at 500°C. Surface M experiences incoming radiant energy at 10,000 W/m2 . Find the energy absorbed per unit time. Parts d – f will also refer to surface M.
Engineering
1 answer:
BaLLatris [955]3 years ago
7 0

Answer:

The rate of energy absorbed per unit time is 3500W.

Explanation:

From the question, we were given the following parameters;

Plane, opaque, gray, diffuse surface

â = 0.7

Surface area, A = 0.5m²

Incoming radiant energy, G = 10000w/m²

T = 500°C

Rate of energy absorbed is âAG;

âAG = 0.7 × 0.5 × 10000

âAG = 3500W.

The energy absorbed is measured in watts and denoted by the symbol W.

You might be interested in
A rotor in a compressor stage has a mean blade radius of 0.285 m and an angular rotor velocity of 8500 RPMs. The static temperat
pantera1 [17]

Answer:

0,285 is the answer

Explanation:

6 0
3 years ago
Pls help me answer my module
Otrada [13]

Answer:

Hand tools based on job requirement and its importance and the classification of hand tools according to its function and its importance are discussed below in details.

Explanation:

Hand tools based on work requirement is essential because Every tool is specifically invented for a particular purpose, so picking the accurate tool will also reduce the amount of energy needed to get work done right without causing injury or harm to either the tools or the exterior being worked on.

classifying of hand tools: wrenches, screwdrivers, cutters, striking tools, hammer tool or struck, pliers, vise, clamps, snips, saws, drills, and knives.

4 0
3 years ago
A cylindrical specimen of some metal alloy having an elastic modulus of 124 GPa and an original cross-sectional diameter of 4.2
IrinaVladis [17]

Answer:

the maximum length of the specimen before deformation is 0.4366 m

Explanation:

Given the data in the question;

Elastic modulus E = 124 GPa = 124 × 10⁹ Nm⁻²

cross-sectional diameter D = 4.2 mm = 4.2 × 10⁻³ m

tensile load F = 1810 N

maximum allowable elongation Δl = 0.46 mm = 0.46 × 10⁻³ m

Now to calculate the maximum length l for the deformation, we use the following relation;

l = [ Δl × E × π × D² ] / 4F

so we substitute our values into the formula

l = [ (0.46 × 10⁻³) × (124 × 10⁹) × π × (4.2 × 10⁻³)² ] / ( 4 × 1810 )

l = 3161.025289 / 7240

l = 0.4366 m

Therefore, the maximum length of the specimen before deformation is 0.4366 m

5 0
3 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
3 years ago
PLS HELP ME
Oksana_A [137]

Answer:

The Euler buckling load of a 160-cm-long column will be 1.33 times the Euler buckling load of an equivalent 120-cm-long column.

Explanation:

160 - 120 = 40

120 = 100

40 = X

40 x 100 / 120 = X

4000 / 120 = X

33.333 = X

120 = 100

160 = X

160 x 100 /120 = X

16000 / 120 = X

133.333 = X

4 0
3 years ago
Other questions:
  • Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm
    5·1 answer
  • In which of the following states are you most likely to find a home with a basement and why?
    13·1 answer
  • Write a static method named fixSpacing that accepts a Scanner representing a file as a parameter and writes that file's text to
    5·1 answer
  • 1.The moist unit weights and degrees of saturation of a soil are given: moist unit weight (1) = 16.62 kN/m^3, degree of saturati
    11·1 answer
  • The shaft is hollow from A to B and solid from B to C. The shaft has an outer diameter of 79 mm, and the thickness of the wall o
    6·1 answer
  • Hot carbon dioxide exhaust gas at 1 atm is being cooled by flat plates. The gas at 220 °C flows in parallel over the upper and l
    15·1 answer
  • A 250 kilo ohms and a 750 kilo ohms resistor are connected in series across a 75-volt source. Determine the error in measuring t
    10·1 answer
  • Problem 2
    11·1 answer
  • The art of manipulating, influencing, or deceiving you into taking some action that isn’t in your best interest or in the best i
    5·1 answer
  • A company intends to market a new product and it estimates that there is a 20% chance that it will be first in the market
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!