Practice Questions

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.

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}

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.

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}

\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}

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}

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}

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}