Practice Questions

The First step is to divide the object in to three parts: $$\\$$ $$\\$$ Determine A and \(\bar{y}\) for each shape: \begin{matrix} A_1=1(4.5) & A_2=8(2) & A_3=.5(5.5) \\ \bar{y_1}=.5+8+.5=9 & \bar{y_2}=.5+4=4.5 & \bar{y_2}=.25 \\ A_1\bar{y_1}=40.5 & A_2\bar{y_2}=72 & A_3\bar{y_3}=.69 \\ \end{matrix}\\ Plug in the values to determine} \(\bar{y}_{composite}\): \begin{aligned} \bar{y}_{composite}=\frac{\sum{A\bar{y}}}{\sum{A}}=\frac{40.5+72+.69}{4.5+16+2.75}=4.87 \end{aligned} Determine the Centroidal Moment of Inertia for each shape and the distance from \(\bar{y}_{composite}\) for each shape noting that \( I_{rec}=\frac{bh^3}{12}\) : \begin{matrix} I_{c1}=\frac{4.5(1)^3}{12}=.375 & d_1=4.87-9=-4.13 \\ I_{c2}=\frac{2(8)^3}{12}=85.33 & d_2=.37 \\ I_{c3}=\frac{5.5(.5)^3}{12}=.057 & d_3=4.18 \\ \end{matrix} Utilizing the Parallel Axis Theorem to get the Centroidal Moment of Inertia of the complete object we get: \begin{aligned} I=.375+4.5(-4.13)^2+85.33+16(.37)^2+.57+2.75(4.18)^2=213.3 \end{aligned}

Given: \begin{aligned} &W=200kg(9.81 \frac{m}{s^2})=1,962N \\ &a_x=3 \frac{m}{s^2} \\ &a_y=0 \\ &m=200kg \\ \end{aligned} First compose the Free Body Diagram of the Container: $$\\$$ $$\\$$ Using Newtons's Law, we know: \begin{aligned} &\sum F_x=ma_x \\ &\sum F_y=ma_y \\ &\sum F_y=ma_y=0 \;\;\;\;\text{where } a_y=0 \text{ so } \sum F_y=0 \\ &\sum F_y=0=N+Tsin(12)-Wcos(20) \\ \end{aligned} \begin{equation} \label{eq:erl} N=Wcos(20)-Tsin(12) \end{equation} \begin{aligned} \sum F_x=ma_x \Rightarrow Tcos(12)+T-Wsin(20)-F_{\mu}=ma_x \end{aligned} Use Equation 1 to determine \(F_{\mu}\): \begin{aligned} F_{\mu}=\mu_{k}N=.25(Wcos(20)-Tsin(12)) \end{aligned} Plugging this result in to \(\sum F_x\): \begin{equation} \label{eq:er2} Tcos(12)+T-Wsin(20)-.25(Wcos(20)-Tsin(12))=200kg(\frac{m}{s^2}) \end{equation} We can now rearrange and solve for \(T\): \begin{aligned} T(1.98)-671N-461N+.052T=600N \end{aligned} ANSWER: \begin{aligned} T=853.2N \end{aligned}

1. Rearrange the equation so that it is in the form: \begin{aligned} &\frac{dy}{dt}+p(t)y=g(t) \\ &\frac{dy}{dt}+5y=10 \end{aligned} 2. Determine the Integrating Factor, such that: \begin{aligned} &\mu(t)=e^{\int p(t)dt} \\ &p(t)=5 \text{ so } \mu(t)=e^{\int5dt} \\ &\mu(t)=e^{5t} \\ \end{aligned} 3. Multiply all terms of the Differential Equation by the Integrating Factor: \begin{aligned} &\mu(t) \frac{dy}{dt}+\mu(t)p(t)y=\mu(t)g(t) \\ &e^{5t} \frac{dy}{dt}+e^{5t}5y=e^{5t}(10) \\ \end{aligned} 4. Notice that the left side simply becomes the product rule and rewrite such that: \begin{aligned} &(\mu(t)y(t))'=\mu(t)g(t) \\ &(e^{5t}y(t))'=e^{5t}(10) \\ \end{aligned} 5. Integrate both sides and combine the constant of integration on the right side of the equation: \begin{aligned} &\int(e^{5t}y(t))'dt=\int e^{5t}(10)dt \\ &e^{5t}y(t)=2e^{5t}+C \\ \end{aligned} 6. Rearrange to form the equation for y(t): \[ y(t)=2+\frac{C}{e^{5t}} \] 7. Plug in the initial value (if given) to determine the Constant C: \begin{aligned} y(0)&=1=2+Ce^{-5(0)} \\ C&=-3 \end{aligned} ANSWER: \[ y(t)=2-3e^{-5t} \]

Kircoff's Second Law states that for any loop in a circuit, the voltage rises must equal the voltage drops, or in other words, whatever energy a charge starts with in a circuit loop, it must end up losing all that energy by the time it gets to the end so that: \begin{aligned} \sum V=0 \end{aligned} Knowing \(V=IR\), we can annotate the given circuit as such: \begin{aligned} \sum {V}=V_1+I(R_{eq1})+I(R_{2})+I(R_{eq3})-V_2=0 \end{aligned} The equivalent of the resistors in parallel is: \begin{aligned} \frac{1}{R_{eq1}}=\frac{1}{10}+\frac{1}{10}=\frac{2}{10}\Rightarrow R_{eq1}=5 \\ \end{aligned} And \begin{aligned} \frac{1}{R_{eq3}}=\frac{1}{10}+\frac{1}{5}=\frac{3}{10}\Rightarrow R_{eq3}=3.33 \\ \end{aligned} So \begin{aligned} \sum V=48+I(5)+I(20)+I(3.33)-24=0 \end{aligned} Solving for I: \begin{aligned} I=\frac{-24}{28.33} \\ \end{aligned} ANSWER: \begin{aligned} I=.85\;amps \end{aligned}

Using the distance formula, we can formulate an equation that represents the distance, R, from each point to the origin, O: \begin{aligned} &| AO |=\sqrt{(h+5)^2+(k)^2} \\ &| BO |=\sqrt{(h-1)^2+(k)^2} \\ &| CO |=\sqrt{(h+2)^2+(k+3)^2} \\ \end{aligned} At this point we have 3 equations and 3 unknowns (h,k,and R) Knowing that the distance from each point is equal to R, we can set the equations equal to one another to eliminate the variable R for now: \begin{aligned} &\sqrt{(h+5)^2+(k)^2}=\sqrt{(h-1)^2+(k)^2} \\ &\sqrt{(h-1)^2+(k)^2}=\sqrt{(h+2)^2+(k+3)^2} \\ \end{aligned} Squaring both sides and expanding we get: \begin{aligned} &(h+5)^2+(k)^2=(h-1)^2+(k)^2\Rightarrow h^2+10h+25+k^2=h^2-2h+1+k^2 \\ &(h-1)^2+(k)^2=(h+2)^2+(k+3)^2\Rightarrow h^2-2h+1+k^2=h^2+4h+4+k^2+6k+9 \\ \end{aligned} Combining the terms in the first series: \begin{aligned} 12h=-24 \text{ which gives} h=-2 \end{aligned} Combining the terms in the second series: \begin{aligned} -6h-6k=12 \end{aligned} Plugging in the value for h: \begin{aligned} -6(-2)-6k=12 \text{ which gives} k=0 \end{aligned} Find R using the values of h and k and any of the original Distance equations you formulated in the first step: \begin{aligned} R^2=(-2+5)^2+(0)^2=9 \end{aligned} Finally, insert all the derived values in to the general equation of a circle: \begin{aligned} (x+2)^2+(y-0)^2=9 \end{aligned} ANSWER: \begin{aligned} (x+2)^2+y^2=9 \end{aligned}

The Voltage across the Capacitor is \(12V\). After a source is connected to a Capacitor for "a long time", the Capacitor becomes fully charged and no more current flows through the terminals at that point, it essentially becomes an open circuit.

\begin{aligned} P=-12,000+3,500(\frac{P}{A},9\%,8)(1-.35)-200(\frac{P}{A},9\%,8)+500(\frac{P}{F},9\%,8) \end{aligned} Using the Present Value Tables, determine the values of the unknowns and plug them in: \begin{aligned} P=-12,000+3,500(5.5348)(.65)-200(5.5348)+500(.5019) \end{aligned} ANSWER: \begin{aligned} P=($63.58) \end{aligned}

The equation for the stress intensity factor is: \begin{aligned} K_1=Y\sigma\sqrt{\pi a} \end{aligned} Given: \begin{aligned} &\sigma = 145 MPa \\ &a=2.5\times10^{-3} m \\ &Y=1.25 \\ \end{aligned} Plugin the information that is given to determine the stress intensity factor: \begin{aligned} K_1=(1.25)(145 MPa)\sqrt{\pi (2.5\times10^{-3}m)}=16.06MPa\sqrt{m} \end{aligned} ANSWER: \begin{aligned} K_1=16.06\;MPa\sqrt{m} \end{aligned}

The efficiency of an Ideal Compression Cycle is \( \eta_t= \frac{W_{cyc}}{Q_{in}} \). If it was the way it is relayed in the option above, the efficiency would be \( \eta_t=\frac{Q_{23}-Q_{41}}{Q_{23}-Q_{41}} \) which would make for a \(100\% \) efficient cycle, which is not possible.

Gibbs Phase rule states that: \begin{aligned} P+F=C+2 \end{aligned} Where: \begin{aligned} &\text{P: Number of phases (solid, liquid, or gas)}\\ &\text{C: Number of components}\\ &\text{F: Degrees of Freedom}\\ \end{aligned} At the triple point of water: \begin{aligned} &\text{P = 3 (solid, liquid, and gas)} \\ &\text{C = 1 (water)} \end{aligned} Plugging the values in to the equation and rearranging: \begin{aligned} 3+F=1+2 \end{aligned} ANSWER: \begin{aligned} F=0 \text{ there are zero degrees of freedom at the triple point of water} \end{aligned}