Select four types of engineers who might be involved in the development of a product such as an iPhone.

As the temperature of a thermal radiator is increased Group of answer choices the object appears redder. the object appears blue

Elanso [62]

Answer:

As the temperature of a thermal radiator is increased

Group of answer choices

the object appears redder.
the object appears bluer.
the object emits more power for the same area.
the object emits less power for the same area.
the object expands to keep the same power per units area.

<em>When the temperature of a thermal radiator increases ;</em>

<em>the object emits more power for the same are</em>
<em>the object becomes bluer</em>

Explanation:

Thermal radiation involves the transfer of heat between molecules of two substances without direct contact with each other. When a body is heated to a given temperature it begins to emit light which is transferred to nearby objects as thermal radiation. The medium through which the heat is transferred could be liquid, solid, or in a vacuum.

<h3>How temperature affects thermal radiation.</h3>

Temperature determines the amount of heat that is been radiated from a body. An increase in temperature would increase the thermal radiation of the body. The increase in the heat radiation results to increase in the thermal energy of the body. Also when a body is heated it tends to be bluer than a cool object, this is caused by the rapid movement of the molecules.

Therefore When the temperature of a thermal radiator increases ;

the object emits more power for the same are
the object becomes bluer

4 0

4 years ago

Illinois furniture , Inc produces all types of coffee furniture the executive secretary is a chair that has been designed using

o-na [289]

This question is incomplete, the complete question is;

Illinois furniture , Inc produces all types of office furniture. The "Executive Secretary" is a chair that has been designed using ergonomics to provide comfort during long work hours the chair sells for $130. There are 480 minutes available during the day and the average daily demand has been 50 chairs. There are eight tasks.

TASK PERFORMANCE TIME( MIN ) TASK MUST FOLLOW TASK LISTED

A 4 ---------

B 7 ----------

C 6 A,B

D 5 C

E 6 D

F 7 E

G 8 E

H 8 F,G

1) What is the cycle time for operation?

2) What is the theoretical minimum number of workstation?

Answer:

1) the cycle time for operation is 9.6 min

2) the theoretical minimum number of workstation is 5

Explanation:

Given the data in the question;

production time per day = 480 minutes

average daily demand = 50

Given the data in the question;

1) cycle time for operation

this is simply referred to as the total time for the process from start to finish.

cycle time = production time per day / units demand per day

we substitute

cycle time = 480 min / 50

cycle time = 9.6 min

Therefore, the cycle time for operation is 9.6 min

2) theoretical minimum number of workstation.

theoretical minimum number of workstation = total task time / cycle time

Total task time = ( 4 + 7 + 6 + 5 + 6 + 7 + 8 + 6 ) = 49 min

∴ theoretical minimum number of workstation = 49 min / 9.6 min

theoretical minimum number of workstation = 5.104 ≈ 5

Therefore, the theoretical minimum number of workstation is 5

7 0

3 years ago

What is the difference between a refrigeration cycle and a heat pump cycle?

sukhopar [10]

Answer:

In refrigeration cycle heat transfer from inside refrigeration

In heat pump cycle heat transfer from environment

Explanation:

heat cycle is mechanical process use for cool the temperature but

In refrigeration heat transfer from inside of refrigeration that decrease temperature of refrigerator and in heat pump it decrease temperature negligible as compare to refrigerator

5 0

3 years ago

Calculate KI for a rectangular bar containing an edge crack loaded in three-point bending where P=35.0 kN, W=50.8 mm, B=25 mm, a

Katyanochek1 [597]

Answer:

$K_{I}=5.21 MPa\sqrt{m}$

Explanation:

given data

Load P = 35 kN

Width of bar W = 50.8 mm

Breadth of bar B = 25 mm

Ratio of crack length to width α = a/W = 0.2

solution

we get here KI for a rectangular bar that is express as

$K_{I} = \frac{6P}{BW}Y\sqrt{\pi a}$ ................................1

here Y is the geometrical function

so

Y = $\frac{1.12+\alpha (3.43\alpha -1.89)}{1-0.55\alpha}$

Y = $\frac{1.12+0.2(3.43\times 0.2-1.89)}{1-0.55\times 0.2}$

Y = $\frac{0.8792}{0.89}$

Y = 0.9878

so put here value in equation 1

$K_{I} = \frac{6\times 35\times 10^{3}}{0.025\times 0.0508}\times 0.9878\times \sqrt{3.1415\times (0.2\times 0.0508)}$

$K_{I} = 165354.33\times 10^{3}\times 0.9878\times 0.0319$

$K_{I}$ = 5210.45 × 10³ $Pa\sqrt{m}$

$K_{I}$ = 5.21 MPa $\sqrt{m}$

5 0

3 years ago

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$