Russia 2020

Problem 1. Three ice cubes are floating in each glass full of water. The ice cube in glass 1 has an air bubble in it. The ice cube in glass 2 has a cavity inside and the third ice cube floats in glass 3 with an aluminum core. What can be said about the water level in the glasses directly?

A. The water in glass 1 has risen, the other glasses are unchanged.
B. The water level in glass 3 has got down, the other glasses are unchanged.
C. The water level in glass 1 and 3 has risen, that in glass 2 is unchanged.
D. The water level in all glass is unchanged.

Topic: Fluid Mechanics Metodi: Hydrostatic Equilibrium, Physical Modeling Competenze: Physical Reasoning Objects: Container, Bubble Fonte: Testo (PDF) — p.1

Il problema 1. Tre cubetti di ghiaccio galleggiano in ogni bicchiere pieno di acqua. Il cubo di ghiaccio in vetro 1 ha un’aria bollata in it. Il cubo di ghiaccio in vetro 2 ha una cavità all’interno e il terzo cubo di ghiaccio galleggia in vetro 3 con un nucleo di alluminio. Cosa si può dire direttamente del livello dell’acqua nei bicchieri?

A. L’acqua in vetro 1 ha risciuto, gli altri bicchieri sono invariati.
B. Il livello di acqua in vetro 3 ha diminuito, gli altri bicchieri sono invariati.
C. Il livello di acqua in vetro 1 e 3 è aumentato, che in vetro 2 è invariato.
D. Il livello di acqua in tutto il vetro è invariato.

Topic: Fluid Mechanics Metodi: Hydrostatic Equilibrium, Physical Modeling Competenze: Physical Reasoning Objects: Container, Bubble Fonte: Testo (PDF) — p.1

Problem number one. Three ice cubes are floating in each glass full of water. The ice cube in glass 1 has an air bubble in it. The ice cube in glass 2 has a cavity inside and the third ice cube floats in glass 3 with an aluminum core. What can be said about the water level in the glasses directly?

A. The water in glass 1 has risen, the other glasses are unchanged. The water level in glass 3 has got down, the other glasses are unchanged.
C. The water level in glass 1 and 3 has risen, that in glass 2 is unchanged.
D. The water level in all glass is unchanged.

Topic: Fluid Mechanics Metodi: Hydrostatic Equilibrium, Physical Modeling Competenze: Physical Reasoning Objects: Container, Bubble Fonte: Testo (PDF) — p.1

Problem 2. Two glass slabs are kept in a position as shown in the figure. A beam of light hits the left slab at an angle of $4 5^{\circ}$ . The path the light beam would have taken if there was no glass is shown by the dotted line. The refractive index of the glass is 1.5 and the surrounding is air. Which of the following sections of the option shows the course of the refracted light beam after the exit from the second cuboid?

German Physics Olympiad 2021 11th Final Round 2

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing Competenze: Diagrammatic Reasoning, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.1

Il problema 2. Due lastre di vetro sono tenute in posizione come mostrato in Il Figuro. Un fascio di luce colpisce la lastra sinistra ad un angolo di $4 5^{\circ}$ . Il percorso che il raggio di luce avrebbe intrapreso se non ci fosse alcun vetro mostrato dalla linea puntata. L’indice di refraczione del vetro è di 1,5 e l’aria circostante è aria. Qual è la sezione di opzione che mostra il Il corso del fascio di luce refratto dopo l’uscita dal secondo cuboide?

L’Olimpiade di Fisica tedesca 2021 11° Final Round 2

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing Competenze: Diagrammatic Reasoning, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.1

Problem two. Two glass slabs are kept in a position as shown in The figure. A beam of light hits the left slab at an angle of $4 5^{\circ}$ . The path the light beam would have taken if there was no glass is shown by the dotted line. The refractive index of the glass is 1.5 and the surrounding is air. Which of the following sections of the option shows the The course of the refracted light beam after the exit from the Second cuboid?

German Physics Olympiad 2021 11th Final Round 2

Topic: Geometric Optics Metodi: Snell’s Law, Ray Tracing Competenze: Diagrammatic Reasoning, Physical Reasoning Objects: — Fonte: Testo (PDF) — p.1

Problem 3. A small metal ball hangs from a pivot, with the string length $L$ . The period of oscillation about this pivot is $T = 1.0$ Now, a nail at a distance $\frac{3}{4} L$ from the pivot is hammered down. When swinging to the right the pendulum hits the nail, it’s path of motion is hindered by it. The ball will now take a motion like the 2nd figure. Which of the following figures shows the position of the ball after $t = 1.5$ seconds of letting go?

German Physics Olympiad 2021 11th Final Round 2

Topic: Oscillations & Waves Metodi: Simple Harmonic Motion Analysis, Physical Modeling Competenze: Physical Reasoning, Mathematical Modeling Objects: Pendulum Fonte: Testo (PDF) — p.2

Il problema 3. A small metal ball hangs from a pivot, with the string length $L$ . Il periodo di oscillazione circa This pivot is $T = 1.0$ Now, a nail at a distance $\frac{3}{4} L$ from the pivot is hammered down. Quando svinghi a destra il pendolo colpisce Il naso, il suo percorso di movimento è ostacolato da esso. La palla prenderà una mossa come la seconda figura. Which of the following figures shows the position of the ball after $t = 1.5$ seconds of letting go?

L’Olimpiade di Fisica tedesca 2021 11° Final Round 2

Problem three. A small metal ball hangs from a pivot, with the string length $L$ . The period of oscillation about This pivot is $T = 1.0$ Now, a nail at a distance $\frac{3}{4} L$ from the pivot is hammered down. When swinging to the right the pendulum hits The nail, its path of motion is hindered by it. The ball will now take a motion like the second figure. Which of the following figures shows the position of the ball after $t = 1.5$ seconds of letting go?

German Physics Olympiad 2021 11th Final Round 2

Problem 4. In one experiment, monochromatic laser light falls perpendicularly onto an optical grating with 300 lines each mm. The interference pattern is observed on a screen behind the grating. The distance of the screen to the grating is very large compared to the extent of the interference pattern. The adjacent Grayscale images demonstrate when using two lasers with different wavelengths but otherwise the same Experimental set-up, arising from the screen Interference pattern. The wavelength of from the first laser emitted Light is 650 nm. How big is the wavelength of the laser light emitted by the second laser?

Topic: Wave Optics, Oscillations & Waves Metodi: Interference & Diffraction Analysis, Superposition Principle Competenze: Diagrammatic Reasoning, Mathematical Modeling Objects: Diffraction Grating, Screen Fonte: Testo (PDF) — p.3

Problema 4. In un esperimento, la luce laser monocromatica cade perpendicolare su una griglia ottica con 300 linee di ogni mm. Il pattern di interferenza è osservato su uno schermo dietro la griglia. La distanza dello schermo la rete è molto grande rispetto all’entità del modello di interferenza. Le immagini adiacenti a scala di grigio dimostrano quando si utilizzano due laser con lunghezze d’onda diverse ma altrimenti il la stessa configurazione sperimentale, derivante dal modello di interferenza dello schermo. La lunghezza d’onda della prima luce emessa dal laser è di 650 nm. Quanto è grande la lunghezza d’onda della luce laser emessa dal secondo laser?

Problem four. In one experiment, monochromatic laser light falls perpendicularly onto an optical grating with 300 lines each mm. The interference pattern is observed on a screen behind the grating. The distance of the screen to the grating is very large compared to the extent of the interference pattern. The adjacent grayscale images demonstrate when using two lasers with different wavelengths but otherwise the same experimental setup, arising from the screen interference pattern. The wavelength of the light emitted by the first laser is 650 nm. How big is the wavelength of the laser light emitted by the second laser?

Problem 5. Three radioactive preparations are considered below. Initially, at time $t = 0$ , they consist of 100% of a single radioactive isotope, the respective parent nuclide. The initial activity of the preparations is designated $A_{0}$ in each case. The direct products of decay, the daughter nuclides, are also radioactive again and decay. Further subsequent decays are no longer considered. The mother and daughter nuclides of the three preparations are: Preparation 1: $^{226}$ Ra ( $T_{mother} = 1600$ a) $\to^{222}$ Rn ( $T_{daughter} = 3.8$ d) Preparation 2: $^{211}$ Pb ( $T_{mother} = 36.1$ min) $\to^{211}$ Bi ( $T_{daughter} = 2.14$ min) Preparation 3: $^{214}$ Pb ( $T_{mother} = 26.8$ min) $\to^{214}$ Bi ( $T_{daughter} = 19.9$ min)

German Physics Olympiad 2021 11th Final Round 2 With $T_{mother}$ or $T_{daughter}$ the half-lives of the respective nuclides are indicated. The following graphs show the time courses of the activities A of both the mother nuclide and of the daughter nuclide and the total activity for the three preparations. Which of the three nuclide pairs belongs to which diagram?

Topic: Nuclear & Particle Physics Metodi: Radioactive Decay Law, Experimental Data Analysis Competenze: Diagrammatic Reasoning, Physical Reasoning Objects: Nucleus Fonte: Testo (PDF) — p.3

Problema 5. Tre preparati radioattivi sono considerati di seguito. Inizialmente, a tempo $t = 0$ , consistono di 100% di un singolo isotopo radioattivo, il rispettivo nucleide genitore. L’attività iniziale dei preparati è designata $A_{0}$ in ogni caso. I prodotti diretti del decadimento, i daughter nuclides, sono radioattivi di nuovo e decadono. Ulteriori Le decise successive non sono più considerate. I nuclidi madre e figlia dei tre preparati sono: Preparazione 1: $^{226}$ Ra ( $T_{mother} = 1600$ a) $\to^{222}$ Rn ( $T_{daughter} = 3.8$ d) Preparazione 2: $^{211}$ Pb ( $T_{mother} = 36.1$ min) $\to^{211}$ Bi ( $T_{daughter} = 2.14$ min) Preparazione 3: $^{214}$ Pb ( $T_{mother} = 26.8$ min) $\to^{214}$ Bi ( $T_{daughter} = 19.9$ min)

L’Olimpiade di Fisica tedesca 2021 11° Final Round 2 Con $T_{mother}$ o $T_{daughter}$ sono indicate le half-lives dei rispettivi nuclidi. Le seguenti grafiche mostrano il A di entrambi i nuclidi madre e della figlia e l’attività totale per Le tre preparazioni. Quale delle tre coppie di nuclidi appartiene a quale diagramma?

Problem five. Three radioactive preparations are considered below. Initially, at time $t = 0$ , they consist of 100% of a single radioactive isotope, the respective parent nucleid. The initial activity of the preparations is designated $A_{0}$ in each case. The direct products of decay, the daughter nuclides, are also radioactive again and decay. Further subsequent decays are no longer considered. The mother and daughter nuclides of the three preparations are: Preparation 1: $^{226}$ Ra ( $T_{mother} = 1600$ a) $\to^{222}$ Rn ( $T_{daughter} = 3.8$ d) Preparation 2: $^{211}$ Pb ( $T_{mother} = 36.1$ min) $\to^{211}$ Bi ( $T_{daughter} = 2.14$ min) Preparation 3: $^{214}$ Pb ( $T_{mother} = 26.8$ min) $\to^{214}$ Bi ( $T_{daughter} = 19.9$ min)

German Physics Olympiad 2021 11th Final Round 2 With $T_{mother}$ or $T_{daughter}$ the half-lives of the respective nuclides are indicated. The following graphs show the time courses of the activities A of both the mother nucleid and of the daughter nucleid and the total activity for The three preparations. Which of the three nuclide pairs belongs to which diagram?

Problem 6. A diode is an electronic component which, in simplified form, has a completely isolating effect in one direction, the reverse direction. In the opposite direction, the diode hardly allows any current to pass up to a certain voltage. But after this voltage limit is crossed, however, it behaves approximately like an ideal conductor. In the circuit shown below are a diode and two resistors installed with resistance values $R_{1}$ and $R_{2}$ . In the adjacent graph are measured values of the current I in the circuit as a function of the applied voltage U shown. Which resistance values best match the displayed measured values?

A. $R_{1} = 220 Ω$ and $R_{2} = 670 Ω$
B. $R_{1} = 470 Ω$ and $R_{2} = 220 Ω$
C. $R_{1} = 470 Ω$ and $R_{2} = 150 Ω$

Topic: Circuits Metodi: Kirchhoff’s Laws, Equivalent Circuit Reduction Competenze: Diagrammatic Reasoning, Mathematical Modeling Objects: Resistor Fonte: Testo (PDF) — p.4

Problema 6. Un diodo è un componente elettronico che, in forma semplificata, ha un effetto completamente isolante in una direzione, il direzione inversa. In direzione opposta, il diodo difficilmente permette a qualsiasi corrente di passare fino a una certa tensione. Ma dopo questo voltage limit is crossed, however, it maintains approximately like an

Il conducente ideale. In questo circuito sono installati un diodo e due resistori con valori di resistenza $R_{1}$ e $R_{2}$ . In the adjacent graph are measured values of the current I in the circuit as a function of the applied voltage U shown. Quali valori di resistenza corrispondono meglio ai valori misurati mostrati?
A. $R_{1} = 220 Ω$ and $R_{2} = 670 Ω$
B. $R_{1} = 470 Ω$ and $R_{2} = 220 Ω$
C. $R_{1} = 470 Ω$ and $R_{2} = 150 Ω$

Topic: Circuits Metodi: Kirchhoff’s Laws, Equivalent Circuit Reduction Competenze: Diagrammatic Reasoning, Mathematical Modeling Objects: Resistor Fonte: Testo (PDF) — p.4

Problem number six. A diode is an electronic component which, in simplified form, has a completely isolating effect in one direction, the Reverse direction. In the opposite direction, the diode hardly allows any current to pass up to a certain voltage. But after this The voltage limit is crossed, however, it maintains approximately like an It’s an ideal conductor. In the circuit shown below are a diode and two resistors installed with resistance values $R_{1}$ and $R_{2}$ . In the adjacent graph are measured values of the current I in the circuit as a function of the applied voltage U shown. Which resistance values best match the measured values displayed?

A. $R_{1} = 220 Ω$ and $R_{2} = 670 Ω$
**B ** $R_{1} = 470 Ω$ and $R_{2} = 220 Ω$
**C ** $R_{1} = 470 Ω$ and $R_{2} = 150 Ω$

Topic: Circuits Metodi: Kirchhoff’s Laws, Equivalent Circuit Reduction Competenze: Diagrammatic Reasoning, Mathematical Modeling Objects: Resistor Fonte: Testo (PDF) — p.4

Problem 7. The ends of three round metal rods made of identical material are each held on constant temperature. The following data are known for the members:

A. Rod I - diameter: 2.0 cm, length: 20 cm, temperatures of the rod ends: $50^{\circ} C$ and $20^{\circ} C$
B. Rod II - diameter: 3.0 cm, length: 50 cm, temperatures of the rod ends: $60^{\circ} C$ and $30^{\circ} C$
C. Rod III - diameter: 4.0 cm, length: 80 cm, temperatures of the rod ends: $70^{\circ} C$ and $40^{\circ} C$ How do the thermal powers transferred through the rods due to thermal conduction behave $P_{I}$ , $P_{II}$ and $P_{III}$ to each other (the services can all be accepted as positive)?

German Physics Olympiad 2021 11th Final Round 2

Topic: Thermodynamics Metodi: Physical Modeling, Dimensional Analysis Competenze: Mathematical Modeling, Physical Reasoning Objects: Rod Fonte: Testo (PDF) — p.4

Problema 7. Le estremità di tre barre di metallo rotondo fatte di materiale identico sono ciascuna tenuta a temperatura costante. I seguenti dati sono noti per i membri:

A. Rod I - diametro: 2,0 cm, lunghezza: 20 cm, temperature delle estremità della canna: $50^{\circ} C$ e $20^{\circ} C$
B. Rod II - diametro: 3,0 cm, lunghezza: 50 cm, temperature delle estremità della canna: $60^{\circ} C$ e $30^{\circ} C$
C. Rod III - diametro: 4.0 cm, lunghezza: 80 cm, temperature delle estremità della canna: $70^{\circ} C$ e $40^{\circ} C$ How do the thermal powers transferred through the rods due to thermal conduction behave $P_{I}$ , $P_{II}$ e $P_{III}$ a vicenda (i servizi possono essere tutti accettati come positivi)?

L’Olimpiade di Fisica tedesca 2021 11° Final Round 2

Topic: Thermodynamics Metodi: Physical Modeling, Dimensional Analysis Competenze: Mathematical Modeling, Physical Reasoning Objects: Rod Fonte: Testo (PDF) — p.4

Problem seven. The ends of three round metal rods made of identical material are each held at constant temperature. The following data are known for the members:

A. Rod I - diameter: 2.0 cm, length: 20 cm, temperatures of the rod ends: $50^{\circ} C$ and $20^{\circ} C$
B. Rod II - diameter: 3.0 cm, length: 50 cm, temperatures of the rod ends: $60^{\circ} C$ and $30^{\circ} C$
C. Rod III - diameter: 4.0 cm, length: 80 cm, temperatures of the rod ends: $70^{\circ} C$ and $40^{\circ} C$ How do the thermal powers transferred through the rods due to thermal conduction behave $P_{I}$ , $P_{II}$ and $P_{III}$ to each other (the services can all be accepted as positive)?

German Physics Olympiad 2021 11th Final Round 2

Topic: Thermodynamics Metodi: Physical Modeling, Dimensional Analysis Competenze: Mathematical Modeling, Physical Reasoning Objects: Rod Fonte: Testo (PDF) — p.4

Problem 8. In the manufacturing of optical lenses, they are often provided with a very thin layer of transparent material in order of reduce the reflections made at certain wavelength ranges. Consider a lens made from a material of refractive index 1.40. This should be given a thinnest possible layer of transparent material with index 1.24, to minimize reflections at normal incidence of light with a wavelength of 500 nm. (A) Determine how thick this layer should be in order to determine the intensity of the reflected light minimize. When light is perpendicular to a transition from a medium with a refractive index $n_{A}$ to one with a refractive index $n_{B}$ , a portion of the incident light intensity is reflected. Since the reflected components are very small under the present conditions, it is sufficient to only consider simple reflections. The portion of light reflected, $R_{A \to B} = (\frac{n _{B} - n _{A}}{n _{B} + n _{A}})^{2}$ (B) Compare the intensity of the on the coated lens at the specified wavelength total reflected light with the intensity of the light that passes through an uncoated Lens is reflected. To do this, calculate the ratio of these intensities. Despite the coating described, the intensity of the reflected light is low, however not zero. (C) State and explain how the coating would need to be changed to make it clear to achieve a better anti-reflection effect at the wavelength in question.

Topic: Wave Optics, Geometric Optics Metodi: Interference & Diffraction Analysis, Superposition Principle, Snell’s Law Competenze: Mathematical Modeling, Physical Reasoning Objects: Lens Fonte: Testo (PDF) — p.5

Problema 8. Nella produzione di lenti ottiche, sono spesso fornite con un strato molto sottile di trasparenza. Il materiale in ordine di ridurre le riflessioni fatte a determinate lunghezze d’onda. Considerate un obiettivo fatto con un materiale con un indice di refraczione 1.40. Questo dovrebbe essere dato un livello più sottile possibile di materiale trasparente con indice 1.24, per minimizzare le riflessioni a normale incidenza di luce con una lunghezza d’onda di 500 nm. (A) Determina come deve essere spessa questa strata per determinare l’intensità del

Minimizzare la luce. Quando la luce è perpendicolare a una transizione da un mezzo con un indice di refraczione $n_{A}$ a uno con un refractivo Indice $n_{B}$ , una porzione dell’intensità luminosa incidente è riflessa. Poiché i componenti riflessi sono molto piccoli In queste condizioni, è sufficiente considerare solo le semplici riflessioni. La porzione di luce riflessa, $R_{A \to B} = (\frac{n _{B} - n _{A}}{n _{B} + n _{A}})^{2}$ (B) Compare l’intensità dell’on the coated lens at the specified wavelength total reflected la luce con l’intensità della luce che passa attraverso un obiettivo non rivestito è riflessa. To do Questo, calcola il rapporto di queste intensità. Nonostante il rivestimento descritto, l’intensità della luce riflessa è bassa, tuttavia non zero. (c) State and explain how the coating would need to be changed to make it clear to achieve a migliore effetto anti-riflessione alla lunghezza d’onda in questione.

Problem eight. In the manufacture of optical lenses, they are often provided with a very thin layer of transparent The material in order to reduce the reflections made at certain wavelength ranges. Consider a lens made from a material of refractive index 1.40. This should be given a thinnest possible layer of transparent material with index 1.24, to minimize reflections at normal incidence of light with a wavelength of 500 nm. (A) Determine how thick this layer should be in order to determine the intensity of the reflected Light minimize. When light is perpendicular to a transition from a medium with a refractive index $n_{A}$ to one with a refractive The index $n_{B}$ , a portion of the incident light intensity is reflected. Since the reflected components are very small In the present conditions, it is sufficient to consider only simple reflections. The portion of light reflected, $R_{A \to B} = (\frac{n _{B} - n _{A}}{n _{B} + n _{A}})^{2}$ (B) Compare the intensity of the on the coated lens at the specified wavelength total reflected light with the intensity of the light that passes through an uncoated lens is reflected. To do This, calculate the ratio of these intensities. Despite the coating described, the intensity of the reflected light is low, however not zero. (c) State and explain how the coating would need to be changed to make it clear to achieve a The effect of the test is to improve the anti-reflection effect at the wavelength in question.

Problem 9. The photo shows the test vehicle called Blackbird. The vehicle does not have its own Energy storage devices such as batteries or fuels, it is driven solely by the wind. In the vehicle there is only a transmission for the drive, which can transfer energy between the wheels and the propeller. With the Blackbird, test drives were done on level ground with constant wind direction and constant wind speed

both along and against the wind. The speed of the vehicle is say $v$ . The vehicle was in parallel the whole time or anti-parallel to the wind velocity $v_{w}$ . You can assume that during the test drives a constant speed was reached. The Blackbird’s designers claimed that when traveling in the direction of the wind, the vehicle speed was faster than the wind, i.e. with a constant speed $v$ , for which $∥ v ∥>∥ v_{w} ∥$ . Some people criticized this as being unphysical and therefore impossible. But is that it? (A) Explain why it is possible with constant wind speed $v_{w}$ a constant speed $v > v_{w}$ in the direction of the wind. Indicate if energy is transferred from the propeller to the wheels or vice versa. To estimate the achievable speed, assume that when transmitting Energy between the surrounding air and the ground and vice versa, a proportion $α$ of the available performance for further use is lost. So if, for example, energy from the surrounding air is transferred to the ground via the propeller, the gearbox and the wheels of the energy transmitted to the vehicle by the wind, only a portion $1 - α$ for the drive be used. (B) Determine the speed $v$ that the vehicle can reach when driving in the direction of the wind can. Express your result in terms of $v_{w}$ and $α$ .

German Physics Olympiad 2021 11th Final Round 2 (C) Determine the achievable speed for driving directly against the wind. Show this in terms of this by $v_{w}$ and $α$ and justify whether it is also possible in this case, faster to be than the wind.

Topic: Conservation of Energy, Newtonian Mechanics Metodi: Energy Conservation Method, Physical Modeling Competenze: Physical Reasoning, Mathematical Modeling Objects: Cart, Wheel Fonte: Testo (PDF) — p.5

Problema 9. La foto mostra il veicolo di prova chiamato Blackbird. Il veicolo non ha il suo spazio di stoccaggio dell’energia dispositivi come le batterie o i combustibili, è guidato solo dal vento. In the vehicle there is only a transmission for il propulsore, che può trasferire energia tra le ruote e la propeller. Con il Blackbird, i test drive sono stati effettuati su terreno di livello con costante direzione del vento e velocità del vento costante

sia lungo che contro il vento. La velocità del veicolo è dire $v$ . The vehicle was in parallel the whole time or anti-parallel to the wind velocity $v_{w}$ . Si può supporre che durante il test guida una velocità costante di ciò che è stato raggiunto. I progettisti di Blackbird hanno affermato che quando si viaggia in direzione del vento, la velocità del veicolo era più veloce che il vento, cioè con una velocità costante $v$ , per la quale $∥ v ∥>∥ v_{w} ∥$ . Alcuni hanno criticato questo come non fisico. E quindi impossibile. Ma è così? (A) Spiegare perché è possibile con velocità di vento costante $v_{w}$ a velocità costante $v > v_{w}$ in direzione del vento. Indicare se l’energia è trasferita dal propellore alle ruote o viceversa. Per stimare la velocità raggiungibile, supponiamo che quando trasmettiamo energia tra l’aria circostante e Il terreno e viceversa, una percentuale $α$ del rendimento disponibile per ulteriore utilizzo è persa. Quindi se, per Per esempio, l’energia proveniente dall’aria circostante viene trasferita al suolo tramite il propeller, la gearbox e il le ruote dell’energia trasmessa al veicolo dal vento, solo una porzione $1 - α$ per la guida deve essere utilizzata. (B) Determina la velocità $v$ che il veicolo può raggiungere quando guida nella direzione del vento
Sì, si può. Esprimere il tuo risultato in termini di $v_{w}$ e $α$ .

L’Olimpiade di Fisica tedesca 2021 11° Final Round 2 (C) Determine la velocità raggiungibile per la guida diretta contro il vento. Show this in terms of Questo è stato fatto da $v_{w}$ e $α$ e giustificare se è anche possibile in questo caso, più veloce che il vento.

Problem nine. The photo shows the test vehicle called Blackbird. The vehicle does not have its own energy storage devices like batteries or fuels, it’s driven solely by the wind. In the vehicle there is only a transmission for The drive, which can transfer energy between the wheels and the propeller. With the Blackbird, test drives were done on level ground with constant wind direction and constant wind speed

both along and against the wind. The speed of the vehicle is say $v$ . The vehicle was in parallel the whole time or anti-parallel to the wind speed $v_{w}$ . You can assume that during The test drives a constant speed of what is reached. The Blackbird’s designers claimed that when traveling in the direction of the wind, the vehicle speed was faster than the wind, i.e. with a constant speed $v$ , for which $∥ v ∥>∥ v_{w} ∥$ . Some people criticized this as being unphysical. and therefore impossible. But is that it? (A) Explain why it is possible with constant wind speed $v_{w}$ a constant speed $v > v_{w}$ in the direction of the wind. Indicate whether energy is transferred from the propeller to the wheels or vice versa. To estimate the achievable speed, assume that when transmitting energy between the surrounding air and the the ground and vice versa, a proportion $α$ of the available performance for further use is lost. So if, for For example, energy from the surrounding air is transferred to the ground via the propeller, the gearbox and the wheels of the energy transmitted to the vehicle by the wind, only a portion $1 - α$ for the drive be used. (B) Determine the speed $v$ that the vehicle can reach when driving in the direction of the wind Can. Express your result in terms of $v_{w}$ and $α$ .

German Physics Olympiad 2021 11th Final Round 2 (C) Determine the achievable speed for driving directly against the wind. Show this in terms of This by $v_{w}$ and $α$ and justify whether it is also possible in this case, faster to be than the wind.

Problem 10. Controlled nuclear fusion could make an important contribution to energy supply in the future and is therefore intensively researched in plasma physics. In the following, you should see the investigation of the fusion of the two hydrogen isotopes, Deuterium ( $^{2}$ H ) and Tritium, ( $^{3}$ H), because of the fusion, a Helium Core is created. $^{4}$ He and a neutron $^{1}$ n. The response can be represented as, $^{2} H +^{3} H \to^{4} He +^{1} n + E$ E denotes the energy released during the fusion in the form of kinetic energy. the Rest masses of the particles and nuclei are: Deuterium nucleus: $m_{0} = 3, 344494 \cdot 1 0^{- 27}$ kg Tritium core: $m_{T} = 5.008 268 \cdot 1 0^{- 27}$ kg Helium core: $m_{H e} = 6.646 477 \cdot 1 0^{- 27}$ kg Neutron: $m_{n} = 1, 674927 \cdot 1 0^{- 27}$ kg (A) Calculate the energy E released during a single nuclear fusion by the above reaction. Determine the proportions of energy attributable to the helium nucleus and the neutron $E_{H e}$ and $E_{n}$ for the case that the initial kinetic energy of the hydrogen isotopes is negligible. Give your results in the unit MeV with $1 MeV \approx 1.602 177 \cdot 1 0^{- 13}$ J at. For the fusion reaction ot take place, the hydrogen nuclei have to come close enough together. This can be achieved by putting an electrically neutral gas of isotopes to a very high temperature. The gas then becomes completely ionized and becomes plasma. (B) Estimate how high the temperature of the plasma must be at least so that the hydrogen nuclei can come together at a distance of less than $1 0^{- 14}$ m and thus one Fusion reaction becomes possible. Assume that all the nuclei of an isotope correspond to move with the same amount of speed. In fact, the amount of speed of the nucleus is not the same for all. As a result of that, some nuclei have more kinetic energy than others, the nuclear fusion can also initiate with lower temperatures. In the following, consider a plasma that is divided equally into deuterium and tritium nuclei with a particle density of $n = 1.0 \cdot 1 0^{20} m^{- 3}$ . The temperature of the plasma is $T = 1.0 \cdot 1 0^{8}$ K. In order to keep the plasma at these high temperatures and thus to maintain the nuclear fusion over a longer period of time, the energy losses of the plasma must be compensated. The electrically neutral neutrons released during the fusion leave the plasma very quickly and its kinetic energy is no longer available to the plasma. In addition, there is a loss of energy through radiation and transport. The total resulting power loss $P_{V}$ can be expressed with the help of the internal thermal energy of the plasma U and the so-called Express energy containment time $τ$ by $P_{V} = \frac{U}{τ}$ You can assume that the plasma behaves in a good approximation like an ideal gas and for use the value $τ = 1.0$ s for the energy containment time. The helium nuclei generated during the fusion, on the other

German Physics Olympiad 2021 11th Final Round 2 hand, remain in the plasma and their kinetic energy heats the plasma. The heating power $P_{H}$ is a function of the nuclear reaction rate density r, i.e. the mean number of fusion reactions per unit of time and volume, the plasma volume V and the kinetic energy $E_{H e}$ of the helium nuclei and amounts to $P_{H} = r E_{H e} V$ (C) Determine the mean nuclear reaction rate density r for the plasma at constant temperature. Note that the plasma, which is electrically neutral as a whole, is not only ions but also contains electrons. For the technical use of nuclear fusion, it is important to maintain the temperature. It is also necessary to spatially enclose the plasma for as long as possible. To compensate for the Plasma pressure and thus to contain the plasma, a magnetic field is applied. Of the The pressure generated by this field can be estimated in a simplified manner using the magnetic express the table flux density B and $μ_{0}$ (occurring numerical factors can be set to 1 will). (D) Estimate how large the magnetic flux density B has to be in order to produce the described plasma.

Topic: Nuclear & Particle Physics, Thermodynamics, Magnetism Metodi: Mass-Energy Equivalence, Kinetic Theory of Gases, Conservation of Momentum, Order-of-Magnitude Estimation Competenze: Mathematical Modeling, Estimation & Approximation, Physical Reasoning Objects: Nucleus Fonte: Testo (PDF) — p.6

Problema 10. La fusione nucleare controllata potrebbe contribuire in modo significativo all’approvvigionamento energetico nel futuro e è quindi molto studiato in fisica del plasma. In the following, you should see the investigation of the fusion of the two hydrogen isotopes, deuterium ( $^{2}$ H) e tritio, ( $^{3}$ H), a causa della fusione, un nucleo di elio è creato. $^{4}$ He and a neutron $^{1}$ n. La risposta può essere rappresentata come, $^{2} H +^{3} H \to^{4} He +^{1} n + E$ E indica l’energia rilasciata durante la fusione sotto forma di energia cinetica. il resto delle masse delle particelle e nuclei sono: Deuterium nucleus: $m_{0} = 3, 344494 \cdot 1 0^{- 27}$ kg Tritico core: $m_{T} = 5.008 268 \cdot 1 0^{- 27}$ kg Cori di elio: $m_{H e} = 6.646 477 \cdot 1 0^{- 27}$ kg Neutron: $m_{n} = 1, 674927 \cdot 1 0^{- 27}$ kg (A) Calcolare l’energia E rilasciata durante una singola fusione nucleare dalla reazione sopra. Determinazione le proporzioni di energia attribuibili al nucleo di elio e al neutrone $E_{H e}$ e $E_{n}$ per il caso che L’energia cinetica iniziale degli isotopi di idrogeno è trascurabile. Give your results in the unit MeV with $1 MeV \approx 1.602 177 \cdot 1 0^{- 13}$ J at. Per la reazione di fusione che ha luogo, i nuclei di idrogeno devono venire abbastanza vicino insieme. Questo può essere raggiunto mettendo un gas elettricamente neutro di isotopi a una temperatura molto alta. Il gas diventa completamente ionizzato e diventa plasma. (B) Estimare quanto alta la temperatura del plasma deve essere almeno in modo che i nuclei di idrogeno può venire insieme a una distanza di meno di $1 0^{- 14}$ m e quindi una reazione di fusione diventa

Potrebbe essere. Supponiamo che tutti i nuclei di un isotopo corrispondano a muoversi con la stessa quantità di velocità. Infatti, la velocità del nucleo non è la stessa per tutti. Come risultato di ciò, alcuni nuclei Se i nuclei hanno più energia cinetica di altri, la fusione nucleare può anche iniziare con temperature più basse. In following, considerate un plasma che è diviso equamente in deuterium e tritium nuclei with a particle density of $n = 1.0 \cdot 1 0^{20} m^{- 3}$ . La temperatura del plasma è $T = 1.0 \cdot 1 0^{8}$ K. Per mantenere il plasma a queste temperature elevate E quindi per mantenere la fusione nucleare per un periodo di tempo più lungo, le perdite energetiche del plasma devono essere
il pagamento di un’importo di circa il 50%. I neutroni elettricamente neutri rilasciati durante la fusione lasciano il plasma molto rapidamente e il suo kinetico energia non è più disponibile per il plasma. Inoltre, c’è una perdita di energia attraverso la radiazione e Trasporti. Il totale di perdita di potenza risultante $P_{V}$ può essere espresso con l’aiuto dell’energia termica interna di plasma U e il cosiddetto tempo di contenimento di energia espresso $τ$ $P_{V} = \frac{U}{τ}$ Si può supporre che il plasma si comporta in una buona approssimazione come un gas ideale e per utilizzare il valore $τ = 1.0$ s per il tempo di contenimento dell’energia. Gli elio nuclei generati durante la fusione, dall’altro

L’Olimpiade di Fisica tedesca 2021 11° Final Round 2 La loro energia cinetica riscalda il plasma. Il calore $P_{H}$ è una funzione di densità r del tasso di reazione nucleare, cioè il numero medio di reazioni di fusione per unità di tempo e volume, il volume di plasma V e l’energia cinetica $E_{H e}$ dei nuclei di elio e quantità di $P_{H} = r E_{H e} V$ (C) Determina la densità media di reazione nucleare r per il plasma a temperatura costante. Si noti che il plasma, che è elettricamente neutro nel suo complesso, non è solo ionico ma contiene anche elettroni. Per l’uso tecnico della fusione nucleare, è importante mantenere la temperatura. È necessario In questo modo, si può isolare il plasma per il più lungo possibile. Per compensare la pressione plasmatica e quindi per contenere il plasma, un campo magnetico è applicato. Of the The pressure generated by this field can be estimated in a simplified manner using the magnetic express the table flux density B and $μ_{0}$ (occurring numerical (Factors can be set to 1 will) (D) Estimare quanto grande la densità del flusso magnetico B deve essere in modo da produrre il descritto plasma.

Problem number ten. Controlled nuclear fusion could make an important contribution to energy supply in the future and is therefore intensively researched in plasma physics. In the following, you should see the investigation of the fusion of the two hydrogen isotopes, deuterium ( $^{2}$ H) and tritium, ( $^{3}$ H), because of the fusion, a helium core is created. $^{4}$ He and a neutron $^{1}$ n. The response can be represented as, $^{2} H +^{3} H \to^{4} He +^{1} n + E$ E denotes the energy released during the fusion in the form of kinetic energy. The rest of the particles and nuclei are: Deuterium nucleus: $m_{0} = 3, 344494 \cdot 1 0^{- 27}$ kg Tritium core: $m_{T} = 5.008 268 \cdot 1 0^{- 27}$ kg Helium core: $m_{H e} = 6.646 477 \cdot 1 0^{- 27}$ kg Neutron: $m_{n} = 1, 674927 \cdot 1 0^{- 27}$ kg (A) Calculate the energy E released during a single nuclear fusion by the above reaction. Determine the proportions of energy attributable to the helium nucleus and the neutron $E_{H e}$ and $E_{n}$ for the case that The initial kinetic energy of the hydrogen isotopes is negligible. Give your results in the unit MeV with $1 MeV \approx 1.602 177 \cdot 1 0^{- 13}$ J at. For the fusion reaction to take place, the hydrogen nuclei have to come close enough together. This can be achieved by putting an electrically neutral gas of isotopes to a very high temperature. The gas then becomes completely ionized and becomes plasma. (B) Estimate how high the temperature of the plasma must be at least so that the hydrogen nuclei can come together at a distance of less than $1 0^{- 14}$ m and thus one fusion reaction becomes It is possible. Assume that all the nuclei of an isotope correspond to move with the same amount of speed. In fact, the amount of speed of the nucleus is not the same for all. As a result of that, some nuclei If nuclear fusion can also start at lower temperatures, it can also start at lower temperatures. In the following, consider a plasma that is divided equally into deuterium and tritium nuclei with a particle density of $n = 1.0 \cdot 1 0^{20} m^{- 3}$ . The temperature of the plasma is $T = 1.0 \cdot 1 0^{8}$ K. In order to keep the plasma at these high temperatures The energy losses of the plasma must be maintained over a longer period of time. The Commission will be compensated. The electrically neutral neutrons released during the fusion leave the plasma very quickly and its kinetic The plasma is no longer available. In addition, there is a loss of energy through radiation and The transportation system. The total resulting power loss $P_{V}$ can be expressed with the help of the internal thermal energy of the plasma U and the so-called Express energy containment time $τ$ by $P_{V} = \frac{U}{τ}$ You can assume that the plasma behaves in a good approximation like an ideal gas and for use the value $τ = 1.0$ s for the energy containment time. The helium nuclei generated during the fusion, on the other hand,

German Physics Olympiad 2021 11th Final Round 2 Hand, remain in the plasma and their kinetic energy heats the plasma. The heating power $P_{H}$ is a function of the nuclear reaction rate density r, i.e. the mean number of fusion reactions per unit of time and volume, the plasma volume V and the kinetic energy $E_{H e}$ of the helium nuclei and amounts to $P_{H} = r E_{H e} V$ (C) Determine the mean nuclear reaction rate density r for the plasma at constant temperature. Note that the plasma, which is electrically neutral as a whole, is not only ionic but also contains electrons. For the technical use of nuclear fusion, it is important to maintain the temperature. It is also necessary to Spatially enclose the plasma for as long as possible. To compensate for the plasma pressure and thus to contain the plasma, a magnetic field is applied. Of the pressure generated by this field can be estimated in a simplified manner using the magnetic express the table flux density B and $μ_{0}$ (occurring numerical factors can be set to 1 will). (D) Estimate how large the magnetic flux density B has to be in order to produce the described The plasma.

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