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1
Correct effective cross section area $$A = \pi R_E^2$$ |
0.10 |
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2
Correct final answer $$P_0 = (1-a) \pi R_E^2 F_s$$
If a different cross sectional area is used, 0.1 points can be given here, provided it is the only mistake. |
0.10 |
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| 1 Energy balance | 0.10 |
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| 2 Correct explicit blackbody radiation formula, using the surface area of a sphere | 0.10 |
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3
Correct numerical value $$T_{g0} \approx 255~К \approx -18^\circ \mbox{C}$$ |
0.10 |
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| 1 Statement on radiation balance in the region outside the atmosphere | 0.10 |
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| 2 Statement on radiation balance in the region between the atmosphere and Earth | 0.10 |
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| 3 Using $t_{sw}$ correctly | 0.10 |
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| 4 Using $t_{lw}$ correctly | 0.10 |
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5
Correct numerical result $$T_g = \left(\frac{1+t_{sw}}{1+t_{lw}}\right)^{1/4} T_{g0} \approx 286~К \approx 13^\circ \mbox{C}$$ |
0.30 |
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| 6 Only analytical result | 0.20 |
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1
Writing down correct equations of motion for $A$ and $B$
Points are given automatically in case the answer is correct |
2 × 0.10 |
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2
Studying the equation of motion for $x_A - x_B$
Points are given automatically in case the answer is correct |
0.10 |
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3
Correct answer $$\omega_d = \sqrt{k \frac{m_A + m_B}{m_A m_B}}$$
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0.20 |
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1
Correct answer $$E = \hbar \omega_d$$ |
0.20 |
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| 2 $h$ is used instead of $\hbar$ | 0.10 |
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| 1 Writing down an expression for Doppler effect (even if incorrect) | 0.10 |
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2
Correct answer $$f - f_0 = \frac{v}{c} f_0 \\ \text{or} \\ f - f_0 = -\frac{v}{c} f_0 \\ \text{or} \\ f - f_0 = f_0 \left(\sqrt{\frac{1+v/c}{1-v/c}}-1\right)$$ |
0.10 |
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| 1 Normalization condition (even if done incorrectly from 0 to $\infty$) | 0.10 |
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2
Correct answer $$C = \sqrt{\frac{m}{2 \pi k_B T}}$$ |
0.10 |
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1
$v$ is expressed in terms of $f$
Points are given even if the incorrect formula for Doppler effect from B3 is used. |
0.10 |
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2
Correct answer $$p(f) \propto \exp \left[-\frac{mc^2}{2k_B T} \left(\frac{f-f_0}{f_0}\right)^2\right]$$
Points are not given if the incorrect formula for Doppler effect from B3 is used. |
0.20 |
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| 1 The distribution is has a single peak at zero | 0.10 |
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| 2 The distribution is symmetric | 0.10 |
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| 3 The distribution decays to zero for $f-f_0 \rightarrow \pm \infty$ | 0.10 |
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4
Correct answer $$f^* - f_0 = f_0 \sqrt{\frac{2 k_B T}{mc^2}}$$ |
0.10 |
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| 1 Sum of forces equals zero | 0.10 |
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| 2 Correct pressure force above and below | 0.10 |
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3
Correct answer $$\frac{dp}{dz} = -\rho(z) g$$ |
0.10 |
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| 1 Ideal gas law | 0.10 |
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2
Correct answer $$\frac{dp}{dz} = -\frac{\mu_{air} p(z)}{RT(z)}g$$ |
0.10 |
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| 1 Recognizing a separable differential equation | 0.10 |
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2
Correct answer $$p(z) = p_0 \exp \left(-\frac{\mu_{air} g}{RT} z\right)$$ |
0.10 |
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| 1 Writing the adiabatic relation in any form | 0.10 |
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| 2 Relating $dT/dz$ to $dP/dz$ | 0.30 |
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3
Correct answer $$\Gamma_a = \frac{\mu_{air}}{c_p} g$$ |
0.20 |
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| 1 Inclusion of gravitational force with parcel density | 0.20 |
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| 2 Inclusion of buoyancy force with air density | 0.20 |
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| 3 Correct equation of motion | 0.20 |
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4
Relating density to temperature $$\rho \propto \frac{1}{T}$$ |
0.20 |
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| 5 Smallness of $\delta z$ is used | 0.20 |
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6
Correct stability requirements $$\mu_{air} g / c_p > \Gamma$$ |
0.20 |
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7
Correct answer $$\omega = \sqrt{\frac{\mu_{air} g / c_p - \Gamma}{T}g}$$ |
0.20 |
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1
Correct entropy change $$\Delta S = \frac{Lm}{T}$$ |
0.20 |
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| 2 Volume of liquid is much less than volume of vapor | 0.10 |
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3
Correct answer $$\frac{dp_s}{dT} = \frac{\mu_{\mbox{H}_2\mbox{O}} L p_s}{RT^2}$$ |
0.20 |
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| 1 Recognizing a separable differential equation | 0.10 |
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2
Correct answer $$p_s(T) = p_{so} \exp \left[ -\frac{\mu_{\mathrm{H_2O}} L}{R} \left( \frac{1}{T} - \frac{1}{T_o} \right) \right]$$ |
0.10 |
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1
According to Dalton's law obtained $$p_w = \phi \frac{\mu_{air}}{\mu_{\mbox{H}_2\mbox{O}}} p$$
If the formula $p_w = \phi p$ is used further in this task, the maximum score for D3 is 1.2 |
0.40 |
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| 2 The dependence $p(T)$ for an adiabatic process | 0.20 |
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| 3 Understanding that partial pressure of water needs to reach saturation for condensation to start | 0.40 |
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4
An equation in a convenient form for numerical solving is obtained |
0.40 |
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| 5 A method for numerical solving is suggested | 0.20 |
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6
Correct numerical answer $$T \approx 286.8~К \approx 13.7^\circ \mbox{С}$$ |
0.40 |
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| 1 Writing Snell’s law for the two refractions | 2 × 0.20 |
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| 2 A dependence between angles is used | 0.10 |
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| 3 | 0.00 |
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| 4 Correct calculations for $\delta$ | 0.00 |
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5
Correct answer $$\delta = \alpha + \arcsin \left\{ n \sin \left[ \varphi - \arcsin \left( \frac{\sin \alpha}{n} \right) \right] \right\} - \varphi$$ |
0.30 |
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| 1 Calculation of $\delta$ for all values of $\alpha \in [20^\circ,70^\circ]$ with a step of $5^\circ$ | 0.20 |
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| 2 Calculation of $\delta$ for 6 values of $\alpha$ | 0.10 |
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| 3 A graph is plotted | 0.20 |
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| 4 The graph has a local minimum | 0.20 |
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1
The minimal value of $\delta$ is obtained $$\delta_{min} \approx 21.8^\circ$$ |
0.10 |
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| 2 Concluding that the angular size of halo corresponds to the minimal value of $\delta$ | 0.10 |
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