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Aneli [31]
3 years ago
11

What are the well-known effects of electricity​

Engineering
2 answers:
zavuch27 [327]3 years ago
4 0
Answer:the facts are electric current, circuits
Explanation:
Sever21 [200]3 years ago
3 0

Answer:

Hence, the three effects of electric current are heating effect, magnetic effect and chemical effect.

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Does anybody want 20 points? They're free get 'em while ya can...
DedPeter [7]

Answer:

hi

Explanation:

4 0
3 years ago
Read 2 more answers
A 0.66 ft steel bar undergoes a stretch of 0.75 in in an axial direction. What is the strain? Use three significant figures (fou
Lapatulllka [165]

Answer:

The strain is 0.0947.

Explanation:

Strain is the ration of change in dimension to the original dimension under application of load. Depending upon load, the bar may stretch or compress. So, the strain can be negative in case of compression or positive in case of tension.

Step1

Given:

Original length of the steel bar is 0.66 ft or 7.92 in.

Stretch in the bar is of 0.75 in.

Calculation:

Step2

Strain is calculated as follows:

e=\frac{\bigtriangleup l}{l_{o}}

e=\frac{0.75}{7.92}

e = 0.0947

Positive sign is for stretching condition of the bar.

Thus, the strain is 0.0947.

8 0
3 years ago
Convert the following indoor air quality standards, established by the U.S. Occupational Safety and Health Administration (OSHA)
sp2606 [1]
Wouldn’t this be science not engineering?
3 0
3 years ago
Consider a voltage v = Vdc + vac where Vdc = a constant and the average value of vac = 0. Apply the integral definition of RMS t
Anna11 [10]

Answer:

Proof is as follows

Proof:

Given that , V = V_{ac} + V_{dc}

<u>for any function f with period T, RMS is given by</u>

<u />RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[f(t)]^{2} } \, dt  }<u />

In our case, function is V = V_{ac} + V_{dc}

RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac} + V_{dc}]^{2} } \, dt  }

Now open the square term as follows

RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {[V_{ac}^{2} + V_{dc}^{2} + 2V_{dc}V_{ac}] } \, dt  }

Rearranging  terms

RMS = \sqrt{\frac{1}{T}\int\limits^T_0 {V_{dc}^{2}  } \, dt  + \frac{1}{T}\int\limits^T_0 {V_{ac}^{2}  } \, dt  + \frac{1}{T}\int\limits^T_0 {2V_{dc}V_{ac}  } \, dt  }

You can see that

  • second term is square of RMS value of Vac
  • Third terms is average of VdcVac and given is that                      average of  V_{ac}V_{dc} = 0

so

RMS = \sqrt{\frac{1}{T}TV_{dc}^{2}   + [RMS~~ of~~ V_{ac}]^2 }

RMS = \sqrt{V_{dc}^{2}   + [RMS~~ of~~ V_{ac}]^2 }

So it has been proved that given expression for root mean square (RMS) is valid

7 0
3 years ago
A basketball has a 300-mm outer diameter and a 3-mm wall thickness. Determine the normal stress in the wall when the basketball
faltersainse [42]

Answer:

2.65 MPa

Explanation:

To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

\sigma = \frac{p*r}{2t}

<u>Where:</u>

p: is the gage pressure = 108 kPa

r: is the inner radius of the ball

t: is the thickness = 3 mm  

Hence, we need to find r, as follows:

r_{inner} = r_{outer} - t    

r_{inner} = \frac{d}{2} - t

<u>Where:</u>

d: is the outer diameter = 300 mm

r_{inner} = \frac{300 mm}{2} - 3 mm = 147 mm

Now, we can find the normal stress (σ) in the wall of the basketball:

\sigma = \frac{p*r}{2t} = \frac{108 kPa*147 mm}{2*3 mm} = 2646 kPa = 2.65 MPa

Therefore, the normal stress is 2.65 MPa.

I hope it helps you!

3 0
3 years ago
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