Review of Physics 2 - Exam

Solution:

Let us denote the required quantities \(v\) and \(p\).

1. From Bernulli equation we get from comparison of points at the level and at the end of the tube $$\rho g h = {1 \over 2} \rho v^2 \to v = \sqrt{2 g h}; $$ $$v = \sqrt{2 \cdot 10\text{ m}\,\text{s}^{−2}\cdot 0.5\text{ m}} = \sqrt{10} \text{ m}\,\text{s}^{−1}\approx 3\text{ m}\,\text{s}^{−1}.$$
2. Let's mark the velocity in the first part of the tube as \(v_1\). From Bernulli equation we get $$\rho g h = p + {1 \over 2} \rho ^2 v_1^2;$$ relationship between velocities \(v_1\) and \(v\) in the first and second part of the horisontal tube is given by continuity equation \(v_1 S_1 = v S_2\), so we have $$v_1 = v \left({S_2 \over S_1}\right),$$ let's substitute into the first equation and express and use the equation from the point 1. and we get: $$p = \rho g h \left(1 - {S_2 \over S_1}\right)$$ and numericaly we get $$p = 1\,000\text{ kg}\,\text{m}^{-3} \cdot 10\text{ m}\,\text{s}^{−2} \cdot 0.5\text{ m} \left(1 - {0.5 \over 1}\right) = 2.5\text{ kPa}.$$

Solution:

1. $$\omega^2={\left({2\pi \over T}\right)^2}={g \over l} \to l={g T^2 \over 4\pi^2} \approx {10\cdot2^2 \over 4\cdot10}=1\text{ m}.$$
2. $$E_\text{tot}={1 \over 2} m \omega^2 x_0^2 = {1 \over 2} m {g \over l} \left( l \alpha_0 \right)^2 = {1 \over 2} m g l \alpha_0^2 = {{m g^2 T^2 \alpha_0^2} \over {8\pi^2}}$$ numerical value can be more effectively calculateble from $$E_\text{tot}={1 \over 2} m g l \alpha_0^2 = {2\text{ kg}\cdot 10\text{ m}\,\text{s}^{-2}\cdot 1\text{ m} \over 2} \left(2{2 \pi \over 360}\right)^2;$$ and the final numerical result is $$E_\text{tot}=2\left({3.14 \over 90}\right)^2 \text{ J} \approx 0.0122\text{ J} \approx 10\text{ mJ}.$$
3. $${\require{cancel} l_\text{E} \over l_\text{M}} = {g_\text{E} \cancel{T^2} \cancel{4\pi^2} \over g_\text{M} \cancel{T^2} \cancel{4\pi^2} } = 6;$$ $${E_\text{E} \over E_\text{M}} = {\cancel{2} \cancel{m} g_\text{E} l_ \text{E} \cancel{\alpha_0^2} \over \cancel{2} \cancel{m} g_\text{M} l_\text{M} \cancel{\alpha_0^2}}=36.$$

Solution:

1. $$C=C_1 + {1 \over {1 \over C_2} + {1 \over C_3} } = C_1 + {{C_2 C_3} \over {C_1 + C_3}};$$ $$C=\left(5 + {{3 \cdot 2} \over {3 + 2}} \right)\text{ }\mu\text{F} = {31 \over 5}\text{ }\mu\text{F} \approx 6\text{ }\mu\text{F}.$$
2. Capacities and charges satisfy the following set of equations (from definition of the capacity) $$Q_1=C_1 U_1; Q_2=C_2 U_2; Q_3=C_3 U_3$$ and because the capacities \(C_2\) and \(C_3\) are connected serial and the charge that flowed through both is the same, $$Q_2 = Q_3 \to C_2 U_2 = C_3 U_3 ; $$ $$U = U_1 = U_2 + U_3;$$ and from last two equations we get $$U_3 = U {C_2 \over {C_2 + C_3}} = 10\text{ V} \cdot {3 \over {3 + 2}} = 6\text{ V}.$$
3. $$Q=Q_1 + Q_2 + Q_3 = Q_1 + 2 Q_3 = U \left(C_1 + 2 C_3{{C_2 } \over {C_2 + C_3}} \right);$$ $$Q=10\text{ V}\left(5 + 2\cdot 3{2 \over {3 + 2}} \right)\text{ }\mu\text{F} = 74\text{ }\mu\text{C}.$$

Solution:

1. $$F=mg \to m = {F \over g} = {pS \over g} = {p \cdot 4\pi R^2 \over g} $$ $$m = {10^5\text{ Pa} \over {10\text{ m}\,\text{s}^{−2}}}\cdot 4 \cdot 3.14 \cdot 64^2 \cdot 10^{10}\text{ m}^2 \approx 5\cdot10^{18}\text{ kg}.$$
2. Molar mass of the ear is the weighted mean value of molar masses of individual components, $$M={{4 M_\text{N_2} + 1 \cdot M_\text{O_2}} \over {4+1} } = {{4\cdot28 + 32} \over 5 }=28.8\text{ g}\,\text{mol}^{-1}$$ $$M={m\over s} \to s ={m \over M} \approx {{5\cdot10^{18}\text{ kg}} \over 28.8\text{ g}\,\text{mol}^{-1}} \approx 179\cdot 10^{20}\text{ mol} \; \dot{=} \; 2\cdot 10^{22}\text{ mol}.$$
3. $$N=s M_A = 179\cdot 10^{20}\text{mol} \cdot 6.6\cdot10^{23}\text{ mol}^{-1} \approx 1.18\cdot10^{46} \approx 1\cdot10^{46}\text{ molecules.}$$
4. $$V = S h \to h = {V \over S} \underset{p V = s R_\text{m} T}{\underset{\uparrow}=} {s R_\text{m} T \over p S } = {\cancel{p} \cdot \cancel{4\pi R^2} R_\text{m} T \over \cancel{p} g \cdot \cancel{4\pi R^2} M} = {R_\text{m} T \over g M};$$ $$h = { 8.3\text{ J}\,\text{K}^{-1}\,\text{mol}^{-1} \cdot 293\text{ K} \over 10\text{ m}\,\text{s}^{−2} \cdot 28.8\text{ g}\,\text{mol}^{-1} } \approx 8\text{ km}.$$