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tresset_1 [31]
2 years ago
7

Two units for measuring magnetic flux density are:

Engineering
1 answer:
romanna [79]2 years ago
5 0
The correct answer is A
Faraday and Weber
:)
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zubka84 [21]

Answer:

Routine

Explanation:

Loop Structures — The Method Of Repeating Routines In Statements. Repetition of code are called loops, and they are defined as statements that execute lines of code (or routines) repeatedly according to conditions or iterations. ... Take for example a routine that must write as output the string “Hello” 40 times

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3 years ago
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Solve the inequality. Then graph your solution.<br> -9v – 10 &lt; 7y +6
Cerrena [4.2K]
16

if you add 9+10 you get 18 - 7+6
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2 years ago
Dalton needs to prepare a close-out report for his project. Which part of the close-out report would describe
lys-0071 [83]

Answer:

Dalton

The part of the close-out report that would describe how he would plan and manage projects in the future is:

summary of project management effectiveness

Explanation:

The Project Close-out Report is a project management document, which identifies the variances from the baseline plans.  These variances are specified in terms of project performance, project cost, and schedule.  The project close-out report records the completion of the project and the subsequent handover of project deliverables to others.  The project management effectiveness summary details the project's objectives and the achievements recorded, including the lessons learned.

4 0
2 years ago
How can the use of local materials improve the standard of living of Filipinos?
Rudiy27
Choose a quality one, and don't use it as necessary
7 0
2 years ago
The wall of drying oven is constructed by sandwiching insulation material of thermal conductivity k = 0.05 W/m°K between thin me
masha68 [24]

Answer:

86 mm

Explanation:

From the attached thermal circuit diagram, equation for i-nodes will be

\frac {T_ \infty, i-T_{i}}{ R^{"}_{cv, i}} + \frac {T_{o}-T_{i}}{ R^{"}_{cd}} + q_{rad} = 0 Equation 1

Similarly, the equation for outer node “o” will be

\frac {T_{ i}-T_{o}}{ R^{"}_{cd}} + \frac {T_{\infty, o} -T_{o}}{ R^{"}_{cv, o}} = 0 Equation 2

The conventive thermal resistance in i-node will be

R^{"}_{cv, i}= \frac {1}{h_{i}}= \frac {1}{30}= 0.033 m^{2}K/w Equation 3

The conventive hermal resistance per unit area is

R^{"}_{cv, o}= \frac {1}{h_{o}}= \frac {1}{10}= 0.100 m^{2}K/w Equation 4

The conductive thermal resistance per unit area is

R^{"}_{cd}= \frac {L}{K}= \frac {L}{0.05} m^{2}K/w Equation 5

Since q_{rad}  is given as 100, T_{o}  is 40 T_ \infty  is 300 T_{\infty, o}  is 25  

Substituting the values in equations 3,4 and 5 into equations 1 and 2 we obtain

\frac {300-T_{i}}{0.033} +\frac {40-T_{i}}{L/0.05} +100=0  Equation 6

\frac {T_{ i}-40}{L/0.05}+ \frac {25-40}{0.100}=0

T_{i}-40= \frac {L}{0.05}*150

T_{i}-40=3000L

T_{i}=3000L+40 Equation 7

From equation 6 we can substitute wherever there’s T_{i} with 3000L+40 as seen in equation 7 hence we obtain

\frac {300- (3000L+40)}{0.033} + \frac {40- (3000L+40)}{L/0.05}+100=0

The above can be simplified to be

\frac {260-3000L}{0.033}+ \frac {(-3000L)}{L/0.05}+100=0

\frac {260-3000L}{0.033}=50

-3000L=1.665-260

L= \frac {-258.33}{-3000}=0.086*10^{-3}m= 86mm

Therefore, insulation thickness is 86mm

8 0
3 years ago
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