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marta [7]
3 years ago
8

Fire andwater ______(do/does) not agree​

Engineering
1 answer:
Elina [12.6K]3 years ago
3 0

Answer:

<em><u>Hii</u></em><em><u> </u></em><em><u>Dear</u></em><em><u> </u></em><em><u>!</u></em><em><u>!</u></em><em><u>!</u></em><em><u> </u></em>

<em><u>How</u></em><em><u> are</u></em><em><u> you</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

<em><u>I</u></em><em><u> </u></em><em><u>hop</u></em><em><u>e</u></em><em><u> </u></em><em><u>That</u></em><em><u> </u></em><em><u>you</u></em><em><u> are</u></em><em><u> </u></em><em><u>Fine</u></em><em><u> </u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

Explanation:

<em><u>So</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>answer</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>_</u></em><em><u>_</u></em><em><u>d</u></em><em><u>o</u></em><em><u>e</u></em><em><u>s</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em><em><u>_</u></em>

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A three-point bending test was performed on an aluminum oxide specimen having a circular cross section of radius 5.6 mm; the spe
ankoles [38]

Answer:

F =  8849 N

Explanation:

Given:

Load at a given point = F =  4250 N

Support span = L = 44 mm

Radius = R = 5.6 mm

length thickness of tested material = 12 mm

First compute the flexural strength for circular cross section using the formula below:

σ_{fs} = F_{f} L / \pi  R^{3}

σ = FL / π R³

Putting the given values in the above formula:

σ = 4250 ( 44 x 10⁻³ ) / π  ( 5.6 x 10⁻³ ) ³

  = 4250 ( 44 x 10⁻³ )  / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 (44 x 1 /1000 )) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 4250 ( 11 / 250  ) / 3.141593 ( 5.6 x 10⁻³ ) ³

  = 187 / 3.141593 ( 5.6 x 1 / 1000 ) ³

  = 187 / 3.141593 (0.0056)³

  = 338943767.745358

  = 338.943768 x 10⁶

σ = 338 x 10⁶ N/m²

Now we compute the load i.e. F from the following formula:

F_{f} = 2 σ_{fs} d³/3 L

F = 2σd³/3L

  = 2(338 x 10⁶)(12 x 10⁻³)³ / 3(44 x 10⁻³)

  = 2 ( 338 x 1000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 2 ( 338000000 ) ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12 x 10⁻³)³ / 3 ( 44 x 10⁻³)

  = 676000000 ( 12  x  1/1000  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  3  / 250  )³ / 3 ( 44 x 10⁻³)

  = 676000000 (  27  / 15625000 )  / 3 ( 44 x 10⁻³)

  = 146016  / 125 / 3 ( 44 x 1 / 1000  )

  = ( 146016  / 125 ) /  (3 ( 11 /  250 ))

  =  97344  / 11

F =  8849 N

4 0
3 years ago
_____ are used to control the flow of electricity in a circuit.
Travka [436]

Answer:

Switches control the flow of electricity in a circuit.

8 0
3 years ago
Read 2 more answers
A robot was able to detect a burning smell at a shopping mall and prevent a major disaster. Which function enabled the robot to
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Answer:

Smoke detectors on the robot.

Plz rate as Brainliest. I need it to get to the next rank.

3 0
3 years ago
Technician A says that a radio may be able to receive AM signals, but not FM signals if the antenna is defective. Technician B s
DIA [1.3K]

The response to whether the statements made by both technicians are correct is that;

D: Neither Technician A nor Technician B are correct.

<h3>Radio Antennas</h3>

In radios, antennas are the means by which signals to the sought frequency be it AM or FM are received.

Now, if the antenna is bad, it means it cannot pick any radio frequency at all and so Technician A is wrong.

Now, most commercial antennas usually come around a resistance of 60 ohms and so it is not required for a good antenna to have as much as 500 ohms resistance and so Technician B is wrong.

Read more about Antennas at; brainly.com/question/25789224

3 0
2 years ago
BTSAUDY5 NAME STARTS FROM THIS
otez555 [7]

Answer:

What's your question?

Explanation:

I'm not able to understand it....

Sorry.

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3 years ago
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