Επαναληπτικές 2019

Ενδεικτικές απαντήσεις

Θέμα Α

Α1 - $\beta$

Α2 - $\alpha$

Α3 - $\delta$

Α4 - $\beta$

Α5: $\Sigma -\Lambda -\Lambda -\Sigma -\Lambda$

Θέμα Β

B1-$(i)$

$$x_1=A\cdot \eta \mu (399\pi t)(S.I.),x_2=A\cdot \eta \mu (401\pi t)(S.I.)$$ $${\omega }_1=399\pi \frac{rad}{s}$$ $${\omega }_2=401\pi \frac{rad}{s}$$ $${\omega }_{\tau \epsilon \lambda }=\frac{{\omega }_1+{\omega }_2}{2}=400\pi \frac{rad}{s},T_{\tau \epsilon \lambda }=\frac{2\pi }{{\omega }_{\tau \epsilon \lambda }}=\frac{1}{200}sec$$ $$T_{\delta }=\frac{2\pi }{|{\omega }_1-{\omega }_2|}=\frac{2\pi }{2\pi }=1sec$$ $$\sigma \epsilon t=2\cdot T_{\delta }:N_{\tau \alpha \lambda }=\frac{t}{T_{\tau \alpha \lambda }}=\frac{2\cdot 1}{\frac{1}{200}}=400\tau \alpha \lambda \alpha \nu \tau ώ\sigma \epsilon \iota \varsigma$$

άρα σωστό το $i$

Β2 - $(iii)$

πριν

$${\Pi }_1={\Pi }_2\Rightarrow A_1\cdot {\upsilon }_1={{\rm A}}_2\cdot {\upsilon }_2\stackrel{\frac{A_1}{{{\rm A}}_2}=2}{⟹}2\cdot {\upsilon }_1={\upsilon }_2$$

Εξίσωση Bernoulli για μια ρευματική γραμμή $(1\to 2)$

$$P_1+\frac{1}{2}\rho \cdot {\upsilon }_1^2=P_2+\frac{1}{2}\rho \cdot {\upsilon }_2^2\Rightarrow P_{\alpha \tau \mu }+\rho \cdot g\cdot h_1+\frac{1}{2}\rho \cdot {\upsilon }_1^2=P_{\alpha \tau \mu }+\rho \cdot g\cdot h_2+\frac{1}{2}\rho \cdot {\upsilon }_2^2$$ $$\rho \cdot g\cdot (h_1-h_2)=\frac{3}{2}\cdot \rho \cdot {\upsilon }_1^2\Rightarrow g\cdot h=\frac{3}{2}\cdot {\upsilon }_1^2$$

μετά

με ταχύτητα ροής ${\upsilon }_1^{\prime }=2{\upsilon }_1$ στην περιοχή (1) θα έχουμε

$${\Pi }_1^{\prime }={\Pi }_2^{\prime }\Rightarrow A_1\cdot {\upsilon }_1^{\prime }={{\rm A}}_2\cdot {\upsilon }_2^{\prime }\stackrel{\frac{A_1}{{{\rm A}}_2}=2}{⟹}2\cdot {\upsilon }_1^{\prime }={\upsilon }_2^{\prime }$$

Εξίσωση Bernoulli για μια ρευματική γραμμή $(1^{\prime }\to 2)$

$$P_1+\frac{1}{2}\rho \cdot {\upsilon }_1^{\prime 2}=P_2+\frac{1}{2}\rho \cdot {\upsilon }_2^{\prime 2}\Rightarrow P_{\alpha \tau \mu }+\rho \cdot g\cdot h_1^{\prime }+\frac{1}{2}\rho \cdot {\upsilon }_1^{\prime 2}=P_{\alpha \tau \mu }+\rho \cdot g\cdot h_2^{\prime }+\frac{1}{2}\rho \cdot {\upsilon }_2^{\prime 2}$$ $$\rho \cdot g\cdot (h_1^{\prime }-h_2^{\prime })=\frac{3}{2}\cdot \rho \cdot {\upsilon }_1^{\prime 2}\Rightarrow g\cdot h^{\prime }=\frac{3}{2}\cdot {\upsilon }_1^{\prime 2}$$ $$\frac{gh}{g{h}^{\prime }}=\frac{\frac{3}{2}\cdot {\upsilon }_1^2}{\frac{3}{2}\cdot {\upsilon }_1^{\prime 2}}\Rightarrow \frac{h}{h^{\prime }}=\frac{{\upsilon }_1^2}{{\upsilon }_1^{\prime 2}}\stackrel{{\upsilon }_1^{\prime }=2{\upsilon }_1}{⟹}{h}^{\prime }=4h$$

άρα σωστό το $iii$

Β3 - $(ii)$

πρίν την κρούση

$$p_1=m\cdot \upsilon ,p_2=0$$

μετά την κρούση

$$p_1^{\prime }=m\cdot {\upsilon }_1{p}_2^{\prime }=m\cdot {\upsilon }_2$$ $$\Sigma \overrightarrow{F_{\epsilon \xi }}=0⟺{\rm A}.\Delta .{\rm O}.{\overrightarrow{p}}_{\pi \rho \iota \nu }={\overrightarrow{p}}_{\mu \epsilon \tau \alpha }$$ $$\overrightarrow{p_1}={\overrightarrow{p_1}}^{\prime }+{\overrightarrow{p_2}}^{\prime }$$ $$p_1^2=p_1^{\prime 2}+p_2^{\prime 2}+2\cdot p_1^{\prime }\cdot p_2^{\prime }\cdot \sigma \upsilon \nu \phi \Rightarrow {\upsilon }^2={\upsilon }_1^2+{\upsilon }_2^2+2\cdot {\upsilon }_1\cdot {\upsilon }_2\cdot \sigma \upsilon \nu \phi$$

Διατήρηση Κινητικής Ενέργειας

$$K_{\alpha \rho \chi }=K_{\tau \epsilon \lambda }\Rightarrow \frac{1}{2}\cdot m\cdot {\upsilon }^2+0=\frac{1}{2}\cdot m\cdot {\upsilon }_1^2+\frac{1}{2}\cdot m\cdot {\upsilon }_2^2\Rightarrow {\upsilon }^2={\upsilon }_1^2+{\upsilon }_2^2$$

οπότε από τις τελευταίες σχέσεις

$$2\cdot {\upsilon }_1\cdot {\upsilon }_2\cdot \sigma \upsilon \nu \phi =0\Rightarrow \sigma \upsilon \nu \phi =0\Rightarrow \phi ={90}^o$$

άρα σωστό το $ii$

Θέμα Γ

$$y_0=A\cdot \eta \mu \omega t$$ $$y=A\cdot \eta \mu (\frac{2\pi t}{T}-\frac{2\pi x}{\lambda })$$

t=2sec

$${\phi }_{\rho }=0\Rightarrow \frac{2\pi \cdot 2}{T}-\frac{2\pi \cdot 1}{\lambda }=0\Rightarrow T=2\lambda$$

t=4sec

$${\phi }_{\rho }=20\pi \Rightarrow \frac{2\pi \cdot 4}{T}-\frac{2\pi \cdot 1}{\lambda }=20\pi$$ $$\frac{4}{T}-\frac{1}{\lambda }=10\pi \Rightarrow \frac{4}{2\lambda }-\frac{1}{\lambda }=10\pi \Rightarrow \lambda =0,1m$$ $$T=0.2sec\omega =\frac{2\pi }{0,2}=10\pi \frac{rad}{s}$$

Γ1

$$D=\Delta m\cdot {\omega }^2\Rightarrow D=2{\pi }^2\cdot {10}^{-4}\frac{N}{m}$$ $$E_T=\frac{1}{2}\cdot D\cdot A^2\Rightarrow 16{\pi }^2\cdot {10}^{-8}=\frac{1}{2}\cdot 2{\pi }^2\cdot {10}^{-4}\cdot A^2\Rightarrow A=0.04m$$

Γ2

$$y=0.04\cdot \eta \mu (\frac{2\pi t}{0.2}-\frac{2\pi x}{0.1})\Rightarrow y=0.04\cdot \eta \mu (10\pi t-20\pi x)(S.I.)$$

Γ3

$${\upsilon }_{\delta }=\frac{\lambda }{{\rm T}}=\frac{0.1}{0.2}=0.5\frac{m}{s}$$ $$t_{\Sigma }=\frac{x_{\Sigma }}{{\upsilon }_{\delta }}=\frac{1.15}{0.5}=2.3sec$$ $$y_P=0\Rightarrow A\cdot \eta \mu {\phi }_{{\rm P}}\Rightarrow {\phi }_{{\rm P}}=2k\pi ,{\upsilon }_{{\rm P}}>0$$ $${\phi }_{{\rm P}}=10\pi t-20\pi \cdot 1.15=10\pi t-23\pi (S.I.)$$ $${\phi }_{{\rm P}}-{\phi }_{\Sigma }=(10\pi t-20\pi )-(10\pi t-23\pi )=3\pi rad\stackrel{{\phi }_{{\rm P}}=2k\pi }{⟹}{\phi }_{\Sigma }=2k\pi -3\pi$$ $${\upsilon }_{\Sigma }=\omega {\rm A}\cdot \sigma \upsilon \nu {\phi }_{\Sigma }=10\pi \cdot 0.04\cdot \sigma \upsilon \nu (2k\pi -3\pi )=0,4\pi \cdot (-1)-0.4\pi \frac{m}{s}$$

Γ4

$$y_{\Sigma }\{\begin{array}{ll}0& 0\le t<2,3s\\ 0,04\cdot \eta \mu (10\pi t-23\pi ),& t\ge 2.3sec\end{array}$$

Θέμα Δ

Δ1

$$M_1,\iota \sigma o\rho \rho o\pi ί\alpha ,\Rightarrow \Sigma \overrightarrow{F}=0$$ $$T_1={{\rm M}}_1\cdot g\Rightarrow T_1=20N$$ $$M_2,\iota \sigma o\rho \rho o\pi ί\alpha ,\Rightarrow \Sigma \overrightarrow{F}=0$$ $$T_2={{\rm M}}_2\cdot g\Rightarrow T_2=15N$$ $$M_{\Sigma },\iota \sigma o\rho \rho o\pi ί\alpha ,\Rightarrow \Sigma {\overrightarrow{\tau }}_{(K)}=0$$ $$T_2\cdot 2R=T_1\cdot R\cdot T_3\cdot R\Rightarrow T_3=20N$$ $$M_3,\iota \sigma o\rho \rho o\pi ί\alpha ,\Rightarrow \Sigma \overrightarrow{F}=0$$ $$\Sigma F_x=0\Rightarrow F_{\epsilon \lambda }-W_{3x}-T_{3x}=0\Rightarrow k\cdot \Delta l={{\rm M}}_3\cdot g\cdot \eta \mu \phi +{{\rm T}}_3\cdot \sigma \upsilon \nu \phi \Rightarrow \Delta l=0.12m$$

Δ2

$$D=k=M_3\cdot {\omega }^2\Rightarrow \omega =10\frac{rad}{s},T=\frac{\pi }{5}sec$$ $$x=A\cdot \eta \mu (\omega t+{\phi }_o)$$ $$t=0,\upsilon =0x_{\Gamma }=-{\rm A}$$ $$\Theta .{\rm I}.\Sigma F_x=0\Rightarrow F_{\epsilon \lambda }-W_{3x}=0\Rightarrow k\cdot \Delta l_3=M_3\cdot g\cdot \eta \mu \phi \Rightarrow \Delta l_3=0.08m$$ $${\rm A}=\Delta l_1-\Delta l_3\Rightarrow A=0.04m$$ $$t=0\Rightarrow -A=A\cdot \eta \mu {\phi }_o\Rightarrow \eta \mu {\phi }_o=\eta \mu \frac{3\pi }{2}\Rightarrow {\phi }_o=2k\pi +\frac{3\pi }{2}\Rightarrow {\phi }_o=\frac{3\pi }{2}\frac{rad}{s}$$ $$t_1=\frac{\pi }{15}sec{x}_1=0.04\cdot \eta \mu (10\cdot \frac{\pi }{15}+\frac{3\pi }{2})\Rightarrow x_1=0.02m$$ $$\frac{dP}{dt}=\Sigma F_x=-D\cdot x\stackrel{t=t_1}{⟹}\frac{dP}{dt}=-D\cdot x_1\Rightarrow \frac{dP}{dt}=-6N$$

Δ3

$${\upsilon }_1=\omega \cdot R\Rightarrow a_1={\alpha }_{\gamma \omega \nu }\cdot R$$ $${\upsilon }_2=\omega \cdot 2R\Rightarrow a_2={\alpha }_{\gamma \omega \nu }\cdot 2R$$ $${{\rm M}}_2\Sigma F=m\cdot a\Rightarrow W_2-T_2^{\prime }=M_2\cdot a_2\Rightarrow 15-T_2^{\prime }=1.5\cdot {\alpha }_{\gamma \omega \nu }\cdot 0.2$$ $${\rm M}\Sigma \tau ={\rm I}\cdot {\alpha }_{\gamma \omega \nu }\Rightarrow T_2^{\prime }\cdot 2R-T_1^{\prime }\cdot R=2M\cdot R^2\cdot {\alpha }_{\gamma \omega \nu }\Rightarrow T_2^{\prime }\cdot 2-T_1^{\prime }=2\cdot 1.5\cdot 0.1\cdot {\alpha }_{\gamma \omega \nu }$$ $${{\rm M}}_1\Sigma F=m\cdot a\Rightarrow {{\rm T}}_1^{\prime }-W_1=M_1\cdot a_1\Rightarrow T_1^{\prime }-10=1\cdot {\alpha }_{\gamma \omega \nu }\cdot 0.1$$

Λύση του συστήματος

$${\alpha }_{\gamma \omega \nu }=20\frac{rad}{s^2}$$

Δ4

$$t_2:h_1=\theta \cdot R{h}_2=\theta \cdot 2R\stackrel{h_1+h_2=h}{⟹}3\theta \cdot R=h\Rightarrow \theta =\frac{h}{3R}=\frac{0.48}{0.3}\Rightarrow \theta =1,6rad$$ $$\theta =\frac{1}{2}\cdot {\alpha }_{\gamma \omega \nu }{t}^2\Rightarrow 1.6=\frac{1}{2}\cdot 20\cdot t_2^2\Rightarrow t_2=0.4sec$$ $$\omega ={\alpha }_{\gamma \omega \nu }\cdot t_2\Rightarrow \omega =8\frac{rad}{s}$$ $$L=I\cdot {\alpha }_{\gamma \omega \nu }=2{\rm M}\cdot R^2\cdot {\alpha }_{\gamma \omega \nu }\Rightarrow L=0.24kg\cdot \frac{m}{s^2}$$

Δ5

$${\rm N}=\frac{{\theta }_3}{2\pi }\Rightarrow {\theta }_3=\frac{20}{\pi }\cdot 2\pi \Rightarrow {\theta }_3=40rad$$ $${\theta }_3=\frac{1}{2}{\alpha }_{\gamma \omega \nu }{t}_3^2\Rightarrow t_3=2sec$$ $${\omega }_3={\alpha }_{\gamma \omega \nu }{t}_3\Rightarrow {\omega }_3=40\frac{rad}{s}$$ $$\Sigma \tau ={\rm I}\cdot {\alpha }_{\gamma \omega \nu }=2M\cdot R^2\cdot {\alpha }_{\gamma \omega \nu }\Rightarrow \Sigma \tau =0,6N\cdot m$$ $$\Theta .{\rm M}.{\rm K}.{\rm E}.dK=dW\Rightarrow dK=\Sigma \tau \cdot d\theta \Rightarrow \frac{dK}{dt}=\Sigma \tau \cdot \frac{d\theta }{dt}\stackrel{t=t_3}{⟹}\frac{dK}{dt}=24\frac{J}{s}$$

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Published

10 September 2019

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Άσκηση

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• Βαθμολογικό 21