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WARRIOR [948]
3 years ago
7

A non-linear analog force sensor outputs the following voltages for different forces.

Engineering
1 answer:
Serjik [45]3 years ago
4 0

Answer:

see explaination

Explanation:

Matlab code

Clc

F=[10, 15, 20, 25, 30];

V=[10, 20, 35, 55, 80];

figure(1) see attachment

subplot (2,1,1)

plot(F,V, 'o')

figure(1) see attachment

subplot (2,1,2)

plot(F,V, 'o')

Equation relating voltage to force is shown in a graph

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True or false? if i were to hook up an ac voltage source to a resistor, the voltage drop across the resistor would be in phase w
hodyreva [135]

Answer: True

Explanation:

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2 years ago
Draw the ipo chart for a program that reads a number from the user and display the square of that number ???Anyone please
kompoz [17]

Answer:

See attachment for chart

Explanation:

The IPO chart implements he following algorithm

The expressions in bracket are typical examples

<u>Input</u>

Input Number (5, 4.2 or -1.2) --- This will be passed to the Processing module

<u>Processing</u>

Assign variable to the input number (x)

Calculate the square (x = 5 * 5)

Display the result (25) ----> This will be passed to the output module

<u>Output</u>

Display 25

5 0
2 years ago
Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?
Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

         x_average = ∑ x_{i} / n

The tolerance or error is the current value over the mean value per 100

         Δx₁ = x₁ / x_average

         tolerance = | 100 -Δx₁  100 |

bars indicate absolute value

let's look for these values ​​for each case

a)

    x_average = (2.1700000+ 2.258571429) / 2

    x_average = 2.2142857145

fluctuation for x₁

        Δx₁ = 2.17000 / 2.2142857145

        Tolerance = 100 - 97.999999991

        Tolerance = 2.000000001%

fluctuation x₂

        Δx₂ = 2.258571429 / 2.2142857145

        Δx2 = 1.02

        tolerance = 100 - 102.000000009

        tolerance 2.000000001%

b)

    x_average = (2.2 + 2.29) / 2

    x_average = 2,245

fluctuation x₁

         Δx₁ = 2.2 / 2.245

         Δx₁ = 0.9799554

         tolerance = 100 - 97,999

         Tolerance = 2.00446%

fluctuation x₂

          Δx₂ = 2.29 / 2.245

          Δx₂ = 1.0200445

          Tolerance = 2.00445%

c)

   x_average = (2.211445 +2.3) / 2

   x_average = 2.2557225

       Δx₁ = 2.211445 / 2.2557225 = 0.9803710

       tolerance = 100 - 98.0371

       tolerance = 1.96%

       Δx₂ = 2.3 / 2.2557225 = 1.024624

       tolerance = 100 -101.962896

       tolerance = 1.96%

d)

   x_average = (2.20144927 + 2.29130435) / 2

   x_average = 2.24637681

       Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

       tolerance = 100 - 98.000043

       tolerance = 2.000002%

       Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

       tolerance = 2.0000002%

e)

   x_average = (2 +2,3) / 2

   x_average = 2.15

   Δx₁ = 2 / 2.15 = 0.93023

   tolerance = 100 -93.023

   tolerance = 6.98%

   Δx₂ = 2.3 / 2.15 = 1.0698

   tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values ​​may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0
3 years ago
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Consider a cubic workpiece of rigid perfect plastic material with side length lo. The cube is deformed plastically to the shape
Taya2010 [7]

Answer:  ε₁+ε₂+ε₃ = 0

Explanation: Considering the initial and final volume to be constant which gives rise to the relation:-

                         l₀l₀l₀=l₁l₂l₃

                        \frac{lo*lo*lo}{l1*l2*l3}=1.0

                      taking natural log on both sides

                              ln(\frac{(lo*lo*lo)}{l1*l2*l3})=ln(1)

Considering the logarithmic Laws of division and multiplication :

                                ln(AB) = ln(A)+ln(B)

                                ln(A/B) = ln(A)-ln(B)

                           ln(\frac{(l1)}{lo})*ln(\frac{(l2)}{lo})*ln(\frac{(l3)}{lo}) = 0

Use the image attached to see the definition of true strain defined as

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which then proves that ε₁+ε₂+ε₃ = 0

8 0
3 years ago
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