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lorasvet [3.4K]
2 years ago
15

Disconnecting a circuit while in operation can create a voltage blank

Engineering
1 answer:
zlopas [31]2 years ago
5 0

Answer:

what is the question

Explanation:

confused

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For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
saveliy_v [14]

Complete Question

For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.

Answer:

The elongation is =21.29mm

Explanation:

In order to gain a good understanding of this solution let define some terms

True Stress

       A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as \sigma_T.

True Strain

     A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as \epsilon_T.

The mathematical relation between stress to strain on the plastic region of deformation is

              \sigma _T =K\epsilon^n_T

Where K is a constant

          n is known as the strain hardening exponent

           This constant K can be obtained as follows

                        K = \frac{\sigma_T}{(\epsilon_T)^n}

No substituting  345MPa \ for  \ \sigma_T, \ 0.02 \ for \ \epsilon_T , \ and  \ 0.22 \ for  \ n from the question we have

                     K = \frac{345}{(0.02)^{0.22}}

                          = 815.82MPa

Making \epsilon_T the subject from the equation above

              \epsilon_T = (\frac{\sigma_T}{K} )^{\frac{1}{n} }

Substituting \ 411MPa \ for \ \sigma_T \ 815.82MPa \ for \ K  \ and  \  0.22 \ for \ n

       \epsilon_T = (\frac{411MPa}{815.82MPa} )^{\frac{1}{0.22} }

            =0.0443

       

From the definition we mentioned instantaneous length and this can be  obtained mathematically as follows

           l_i = l_o e^{\epsilon_T}

Where

       l_i is the instantaneous length

      l_o is the original length

Substituting  \ 470mm \ for \ l_o \ and \ 0.0443 \ for  \ \epsilon_T

             l_i = 470 * e^{0.0443}

                =491.28mm

We can also obtain the elongated length mathematically as follows

            Elongated \ Length =l_i - l_o

Substituting \ 470mm \ for l_o and \ 491.28 \ for \ l_i

          Elongated \ Length = 491.28 - 470

                                       =21.29mm

4 0
3 years ago
Technician A says high resistance causes an increase in current flow. Technician B says a higher than
Kryger [21]

The correct statement is: a higher than a normal voltage drop could indicate high resistance. Technician B is correct.

<h3>Ohm's law</h3>

Ohm's law states that the current flowing through a metallic conductor is directly proportional to the voltage provided all physical conditions are constant. Mathematically, it is expressed as

V = IR

Where

V is the potential difference

I is the current

R is the resistance

<h3>Technician A</h3>

High resistance causes an increase in current flow

V = IR

Divide both side by I

R = V / I

Thus, technician A is wrong as high resistance suggest low current flow

<h3>Technician B</h3>

Higher than normal voltage drop could indicate high resistance

V = IR

Thus, technician B is correct as high voltage indicates high resistance

<h3>Conclusion </h3>

From the above illustration, we can see that technician B is correct

Learn more about Ohm's law:

brainly.com/question/796939

8 0
2 years ago
Depending on the environmental demands, there are different types of organizational structures, including __________.
Andrews [41]

Answer:

  • Functional (Centralized) Organization
  • Divisional Organization
  • Team-Based Organization
  • Product-Based Organization
  • Modular Organization
  • Matrix Organization

Explanation:

Organization structure:

refers to the idea of how people are supposed to work and coordinate in an organization to maintain a healthy and effective work environment.

There are various types of organizational structures which depends on several factors. There is no single best organization structure. Each structure has its own advantages and disadvantages. In order to select a structure the organization's vision, mission, culture values, goals are to be identified first.

6 0
3 years ago
water at 20c discharges from a stroke tank through a 150m length of horizontal pipe. The pipe is smooth and has a diameter of 75
inessss [21]

Answer:

The dept of the water needed is 26.11m

Explanation:

Given the following parameters:

Length of the pipe = 150m

Diameter of the pipe = 75mm = 0.75m

Volumetric flow rate 1m^{3}/s

Knowing that:

Volumetric flowrate = area (A) * Velocity(V)

==> 0.1 = A*V                              *Knowing that V = \sqrt{2gh}

==> 0.1  = \frac{\pi }{4}* (0.075)^{2} * \sqrt{2gh}

==> h = 26.11m

Hence, the dept of water needed to produce a volumetric flow rate of 1m^{3}/s is 26.11m.

7 0
3 years ago
Determine the flow velocities at the inlet and exit sections of an
VikaD [51]

Answer:

The inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

Explanation:

Using Bernoulli's equation

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂ + 1/2ρv₂²

P₁ - P₂ + ρgh₁ -  ρgh₂ = 1/2ρv₂² - 1/2ρv₁²

ΔP + ρgΔh = 1/2ρ(v₂² - v₁²)  (1)

where ΔP = pressure difference = 12.35 kPa = 12350 Pa

Δh = height difference = 1.35 m

From the flow rate equation Q = A₁v₁ = A₂v₂ and v₁ = A₂v₂/A₁ = d₂²v₂/d₁² where v₁ and v₂ represent the inlet and exit velocities from the pipe and d₁ = 0.95 m and d₂ = 0.44 m represent the diameters at the top end and lower end of the pipe respectively.

Substituting v₁ into (1), we have

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂²v₂/d₁² )²)

ΔP + ρgΔh = 1/2ρ(v₂² - (d₂/d₁)⁴v₂²)

v₂ = √[2(ΔP + ρgΔh)/ρ(1 - (d₂/d₁)⁴)}

substituting the values of the variables, we have

v₂ = √[2(12350 Pa + 1000 kg/m³ × 9.8 m/s² × 1.35 m)/(1000 kg/m³ (1 - (0.44 m/0.95 m)⁴))}

= √[2(12350 Pa + 13230 Pa)/(1000 kg/m³ × 0.954)]

= √[2(25580 Pa)/954 kg/m³]

= √[51160 Pa/954 kg/m³]

= √53.627

= 7.32 m/s

v₁ = d₂²v₂/d₁²

  = (0.44 m/0.95 m)² × 7.32 m/s

  = (0.954)² × 7.32 m/s

  = 6.66 m/s

The volume flow rate Q = A₁v₁

= πd₁²v₁/4

= π(0.95 m)² × 6.66 m/s ÷ 4

= 18.883 m³/s ÷ 4

= 4.72 m³/s

So, the inlet velocity v₁ = 6.66 m/s, the exit velocity v₂ = 7.32 m/s and the volume flow rate Q = 4.72 m³/s

7 0
3 years ago
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