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dexar [7]
1 year ago
10

Question 2

Engineering
1 answer:
Neko [114]1 year ago
6 0

A differential equation that represents the system is

using Newtons law of motion

Mx: - (-cx· - kx + F)

Mx: + cx· + kx - F

250x: + 100x· + 640x = 10 sin (2t)

<h3>What is a differential equation?</h3>

An equation that connects one or more unknown functions and their derivatives is known as a differential equation in mathematics. In practical settings, functions typically represent physical quantities, derivatives their rates of change, and the differential equation establishes a connection between the two.

Differential equations are widely used in many fields, including engineering, physics, economics, and biology, as a result of the prevalence of these relationships.

Studying differential equations primarily entails learning about the solutions—the collection of functions that satisfy each equation—and the solutions' characteristics. The only differential equations that can be solved with explicit formulas are the simplest; however, many properties of a given differential equation's solutions can be determined without performing an exact calculation.

Learn more about differential equations

brainly.com/question/1164377

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Answer:

a). 139498.24 kg

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Explanation:

Given :

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Young's modulus, E = 30 GPa

Gauge factor, k = 6.9

Gauge resistance, R = 340 Ω

a). Maximum truck weight

σ = Eε

σ = $0.03 \times 30 \times 10^9$

$\frac{P}{A} =0.03 \times 30 \times 10^9$

$P = 0.03 \times 30 \times 10^9\times \frac{\pi}{4}\times (0.022)^2$

 = 342119.44 N

For the four sensors,

Maximum weight = 4 x P

                            =  4 x 342119.44

                            = 1368477.76 N

Therefore, weight in kg is $m=\frac{W}{g}=\frac{1368477.76}{9.81}$

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b). Change in resistance

k=\frac{\Delta R/R}{\Delta L/L}

$\Delta R = k. \epsilon R$    , since $\epsilon= \Delta L/ L$

$\Delta R = 6.9 \times 0.03 \times 340$

$\Delta R = 70.38 $ Ω

For 4 resistance of the sensors,

$\Delta R = 70.38 \times 4 = 281.52$ Ω

c). $k=\frac{\Delta R/R}{\epsilon}$

If linear strain,

$\frac{\Delta R}{R} \approx \frac{\Delta L}{L}$  , where k = 1

$\Delta R = \frac{\Delta L}{L} \times R$

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The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate t
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Q)The shear flow at a point is given by q = 1 of 2 An I-beam has a flange width b = 200 mm , height h = 200 mm , web thickness tw = 8 mm, and flange thickness t = 12 mm, For the given point The shear flow at the point depends on the value of Q for the portion of the upper flange to the right of the point. Calculate the value of Q. Express your answer with appropriate units to three significant figures.( <u>I have also attached the diagram for better understanding)</u>

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