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Two concepts are in play here: Calculation of moment of inertia for a composite section. Calculation of elastic beam stresses. First, the calculation of $I_{total}$ for the wide flange section. Your equation for $I_{total}$ via the parallel axis theorem is correct, but the execution went awry. $$I_{total} = \sum (I + A \cdot d^2)$$ Since the section is ...


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As TimWescott indicated, this problem is more suitable for control theory for a fuller understanding than for digital signal processing, yet there's no control.se as far as I know. Hence you can give it a try here too since such damped-spring systems is not uncommon subject for audio / acoustic engineers etc. First of all you have a programming error: Your ...


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Well, material limits tend to restrict edge-of-disk speeds to less than 1000 m/s. So what happens to a tennis ball if it is accelerated to Mach 2.9? The stagnation temperature is about 540 F, so it will begin to melt, then explode. Plus the shock pressures would probably cause, um, spontaneous self disassembly. So maybe take Mach 0.95 as a practical upper ...


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The trajectory of a tennis ball ejected from a machine gets a bit complicated due to the air resistance drag which is of second order. The typical speeds are in the range of 30-50m/s. Also, some machines' wheels rotate at different speeds spinning the ball that makes a difference. Let's say your turning wheels are shooting the balls out at a speed of 30m/s. ...


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Assuming equilibrium, $\vec{F_1} + \vec{F_2} + \vec{F_3} = 0$ . We have: $$\vec{F_1} = \vec{{F_1}_u} + \vec{{F_1}_v}$$ $$\vec{F_2} = \vec{{F_2}_v} + \vec{{F_2}_w}$$ $$\vec{F_3} = \vec{{F_3}_w} + \vec{{F_3}_u}$$ And the problem states that: $$\color{red}{\vec{{F_1}_v}} = \color{red}{\vec{{F_2}_v}}$$ $$\color{blue}{\vec{{F_2}_w}} = \color{blue}{\vec{{F_3}_w}}$$...


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You are on the right track but your equation is wrong on some places and there's some misinterpretation. First error your drawing doesn't show that the midle piece thickness is 5mm (for this type of beam usualy the web and flange thickness is different). Just note some numbers aren't fully in cientific notation. I've done this to stick with your precision ...


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Entropy change is $$ \Delta S = \int \frac{\delta q}{T} $$ For a phase transformation this becomes $$ \Delta_{pt} S = \frac{\Delta_{pt} H}{T_{pt}}$$ Enthalpy and entropy depend on temperature as $$ \Delta H(T_2) - \Delta H(T_1) = \int C_p dT $$ $$ \Delta S(T_2) - \Delta S(T_1) = \int \frac{C_p}{T} dT $$ The Gibbs energy change is $$ \Delta G = \...


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Because the specific value are not given, I will give a general overview of the method. When applying the direct stiffness method for trusses, the following steps must be used: Find the local stiffness matrix for each individual element as following: $\begin{bmatrix}AE/L & 0 & -AE/L & 0 \\0 & 0 & 0 & 0\\-AE/L & 0 & AE/L &...


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