Review of Physics 2 - Exam, muster
Semestr: sommer, 20XX/XX, Tutor: Martin Žáček, Date: 20XX-XX-XX
For every task is for correct general result 1 point, for correct numerical result 1 point and correct way of solution for 3 points, i.e. maximum of possible points is 5 per task and maximum 20 points for the test. Numerical results estimate with the 1-digit of precision.
\(\)Calculate the difference in the river levels before and after the hydroelectric power station, where the current flow rate is \(I = 400\text{ m}^3\text{s}^{−1},\) the output power is \(P = 2\text{ MW,}\) and efficiency of the power plant is \(\eta=75\text{ }\%.\) Assume gravitational acceleration as \(g = 10\text{ m}\,\text{s}^{−2}.\)
Solution:
- $$\Delta W=mg\Delta h;$$ $$P_\text{water}={{\Delta W}\over t}={mg{\Delta h}\over t}=\rho Ig\Delta h;$$ $$P_\text{out}=\eta P_\text{water};$$ $$\Delta h={{P_\text{out}}\over{I\rho g\eta}}={{2\cdot 10^6}\over{400\cdot10\cdot1\,000\cdot0.75}}={{2}\over{3}}\text{ m}.$$
The linear harmonic oscillator with the amplitude of displacement \(y_0 = 12\text{ cm}\) has a period of movement \(T = 40\text{ ms}\). The rigidity of the oscillating system is \(k = 6\text{ N}\text{m}^{−1}.\) Calculate the total mass, maximal velocity, maximal acceleration, and total energy of the oscillator.
Solution:
- $$m={k \over {\omega^2}}={{T^2k}\over{4\pi^2}}={{0.04\cdot0.04\cdot6}\over 40}=24\cdot10^{-5}\text{ kg}=210\text{ mg};$$ $$v_\text{max}=\omega y_0={{2\pi}\over T}y_0\approx{{2\cdot3.14}\over{0.04}}\cdot 0.12=2\cdot 3.14\cdot3\approx19\text{ m}\,\text{s}^{-1};$$ $$a_\text{max}=\omega v_\text{max}=\omega^2 y_0={{4\pi^2}\over T^2}y_0\approx{{2\cdot3.14}\over{0.04}}\cdot 19\approx3\,000\text{ m}\,\text{s}^{-2};$$ $$E_\text{tot}=E_\text{k max}={1\over2}{k\over{\omega^2}}\left(\omega y_0\right)^2={1\over2}k y_0^2={1\over2}\cdot 6\cdot0.12^2\approx 3\cdot0.014\,4\approx0.04\text{ J}.$$
Three capacitors with capacities \(C_1 = 1\text{ }\mu\text{F},\) \(C_2 = 3\text{ }\mu\text{F}\) and \(C_3 = 20\text{ }\mu\text{F}\) are connected serial. What is the total capacity? What is the total bound electric charge, if the capacities are charged to voltage \(U = 200\text{ V}\)? how is the voltage divided into individual capacities?
Solution:
- $$C={1 \over {{1\over{C_1}}+{1\over{C_2}}+{1\over{C_3}}}}={1 \over {{1\over{\mu F}}\left({1\over1}+{1\over3}+{1\over{20}}\right)}}={{1 \mu F}\over{{{60+20+3}\over{60}}}}={{60}\over{83}}\text{ }\mu{F};$$ $$Q=CU=\left({{60}\over{83}}\text{ }\mu{F}\right)\cdot200\text{ V}\approx145\text{ }\mu{C};$$ $$U_1={Q\over{C_1}}={{UC}\over{C_1}}=U{{1\over {{1\over{C_1}}+{1\over{C_2}}+{1\over{C_3}}}}\over {C_1}}=U{1\over {{{C_1}\over{C_1}}+{{C_1}\over{C_2}}+{{C_1}\over{C_3}}}}=200{1\over{1+{1\over3}+{1\over{20}}}}=200{{60}\over{83}}\approx145\text{ V};$$ $$U_2=U{1\over {{{C_2}\over{C_1}}+{{C_2}\over{C_2}}+{{C_2}\over{C_3}}}}=200{1\over{3+1+{3\over{20}}}}=200{{20}\over{83}}\approx48\text{ V};$$ $$U_3=U{1\over {{{C_3}\over{C_1}}+{{C_3}\over{C_2}}+{{C_3}\over{C_3}}}}=200{1\over{20+{{20}\over3}+1}}=200{3\over{81}}\approx7\text{ V};$$
Inside the closed glass tube with a volume of \(V = 3\text{ cm}^3\) is located pure water vapor with the pressure \(p = 2\,000\text{ Pa}\) and the temperature \(ϑ = 20\text{ °C}.\) Calculate their total mass, molar mass, density and number of molecules of the vapor. Relative atomic mass let's assume as 1 for hydrogen and 16 for oxygen; the molar gas constant is \(R = 8.3\text{ J}\,\text{K}^{-1}\,\text{mol}^{-1},\) Avogadro constant is \(N_\text{A} = 6.6\cdot10^{23}\text{ mol}^{-1},\) let's assume the behavior of the vapor as an ideal gas.
- $$M={m\over n}={{Mm_\text{molecule}}\over{N\over{N_\text{A}}}}=N_\text{A}u\left(2A_\text{H}+A_\text{O}\right)={{1\text{ g}}\over{1\text{ mol}}}\left(2\cdot 1+16\right)=18\text{ g}\,\text{mol}^{-1};$$ $$pV=nRT={m\over M}RT; T=\left(ϑ+273.15\right)\text{ K}\approx293\text{ K};$$ $$m={{pVM}\over{RT}}\approx{{2\,000\cdot3\cdot10^{-6}\cdot18\text{ g}\,\text{mol}^{-1}}\over{8.3\cdot293}}\approx{9\over2}\cdot10^{-5}\text{ g}\approx45\text{ ng};$$ $$\rho={m\over V}={{pM}\over{RT}}\approx{{2\,000\cdot18\text{ g}\,\text{mol}^{-1}}\over{8.3\cdot293}}\approx{30\over2}\text{ g}\,\text{m}^{-3}=15\text{ g}\,\text{m}^{-3};$$ $$N=nN_\text{A}={{pVM}\over{RT}}N_\text{A}\approx{{{2\,000\cdot3\cdot10^{-6}}}\over{8.3\cdot293}}\cdot6\cdot10^{23}\approx 1.5\cdot10^{18}.$$