The photograph shows a rotary carousel, which is a cylindrical drum rotating around a vertical axis with a frequency ν = 33 revolutions per minute. People who initially stand with their backs against the inner vertical wall of the drum move with a centripetal acceleration of 3g ( g = 10 m/s 2). As a result, they “stick” to the drum wall. For greater effect, at some point the floor automatically lowers. Assuming people are thin enough, estimate the radius of this carousel's drum, as well as the minimum coefficient of friction between people and the wall of the carousel drum that is sufficient to prevent people from sliding down.
Possible Solution
3g = ω 2 ∙R = 4∙π 2 ∙ν 2 ∙R, where ω = 2∙π∙ν.
R = 3∙g/4∙π 2 ∙ν 2 ≅ 2.5 m.
To answer the second question, let’s write down Newton’s second law for human motion in a circle in projection onto the vertical axis and in the radial direction (m is the mass of the person, N is the reaction force of the drum wall, F tr. is the modulus of the friction force): m∙g = F tr. , 3∙m∙g = N.
Let us take into account that if the friction coefficient is minimal, then F tr. = µ∙N. Then from the written equations we find: µ = 1/3.
Evaluation criteria
Problem 2
A piece of ice weighing 1 kg floats in a vertical cylindrical vessel partially filled with carbon tetrachloride, which has a density of 1600 kg/m3 and is immiscible with water. How and by how much will the carbon tetrachloride level change after all the ice has melted? The area of the bottom of the vessel is 200 cm2.
Possible Solution
Let h 1 be the initial height of the carbon tetrachloride level. Then the pressure at the bottom of the vessel is equal to
ρ T ∙g∙h 1 ,
where ρ T is the density of carbon tetrachloride.
After the ice melts, the pressure at the bottom of the vessel is equal to:
ρ T ∙g∙h 2 + ρ∙g∙H = ρ T ∙g∙h 2 + m∙g/S,
where h 2 is the final height of the carbon tetrachloride column, ρ is the density of water, H is the height of the water column. The mass of the contents of the vessel has not changed, therefore, the pressure on the bottom in the initial and final states is equal, that is:
Thus, the height of the carbon tetrachloride level will decrease by ∆h = 3.125 cm.
Evaluation criteria
Problem 3
The graphs show the dependence of pressure p and volume V of one mole of a monatomic ideal gas on time t. Determine how the heat capacity of a given amount of gas changed over time. Plot this heat capacity as a function of time.
Possible Solution
During the first 15 minutes, the dependence of gas pressure on its volume has the form
Let at some arbitrary moment of time (in the interval from 0 min. to 15 min.) the gas pressure is equal to p 1, and the volume occupied by it is equal to V 1. Let us write down the first law of thermodynamics for the process of transition from the state (p 0, V 0) to the state (p 1, V 1):
Here C is the heat capacity of one mole of gas in the process under consideration, ∆T is the change in gas temperature, ∆A is the work done by the gas. It is numerically equal to the area of the figure under the p(V) dependence graph, and this figure is a trapezoid.
Let's rewrite the last expression using the equation of state p∙V = R∙T for one mole of an ideal gas:
Let's take into account that
whence follows
that is, C = 2∙R.
Note that the pressure p 1 and volume V 1 taken at an arbitrary point in time are reduced during calculations. This is true, including for two arbitrary states of a gas separated by a very short period of time. This proves that the heat capacity in the process under consideration is a constant value, that is, it will be equal to 2∙R at any time during the first 15 minutes.
After the first fifteen minutes the process becomes isobaric.
Therefore, in this case C = 5/2∙R.
The corresponding graph of the heat capacity of one mole of a monatomic ideal gas versus time is shown in the figure.
Evaluation criteria
| The dependence of pressure on volume for the first process was obtained | 1 point |
| The first law of thermodynamics was recorded for the change in gas temperature upon transition to an arbitrary intermediate state (in the range from 0 min. to 15 min.) | 1 point |
| An expression has been written for the work of a gas during the transition to an intermediate state | 1 point |
| The heat capacity in the first process was found and it was proven that it is a constant value (if there is no justification for the constancy of the heat capacity, then 2 points are given for this point) | 3 points |
| It is indicated that the second process is isobaric | 1 point |
| The heat capacity in the second process is indicated | 1 point |
| A graph has been constructed showing characteristic values | 2 points |
Problem 4
The first point charge was placed at point A, and it created a potential of 2 V at point B. Then the first charge was removed, and a second point charge was placed at point B. He created a potential of 9 V at point A. Then the first charge was returned back to point A. With what force do these charges interact?
Possible Solution
Let the moduli of the charges that were placed at points A and B be equal to q 1 and q 2, respectively, and the distance between them equal to R. Writing the formulas for the potentials created by point charges at points B and A, we obtain:
According to Coulomb's law, the required force of interaction of charges is equal to:
Taking into account the written expressions for potentials, we obtain:
Answer: F = 2 nN
Evaluation criteria
Problem 5
Determine the reading of an ideal ammeter in the circuit, the diagram of which is shown in the figure (Fig. 5.1).
The dependence of the current I flowing through diode D on the voltage U across it is described by the expression: I = α∙U 2, where α = 0.02 A/V 2. EMF of the source E = 50 V. The internal resistance of the voltage source and resistor are equal to r = 1 Ohm and R = 19 Ohm, respectively.
Possible Solution
Let's write Ohm's law for a section of a circuit that includes a resistor, a voltage source and an ammeter:
I(R + r) = E – U,
where I is the current flowing through the diode (and through the ammeter), U is the voltage across the diode.
Using the current-voltage characteristic of the diode, we obtain:
Solving the quadratic equation, we find:
The second root of the quadratic equation, corresponding to the “+” sign in front of the square root (3.125 A), is not the root of the original equation. This can be established either by direct substitution into the given original equation, or by noting that the current flowing through the ammeter in a given circuit cannot exceed
I max = E/(R+r) = 2.5 A.
The solution to the problem looks somewhat simpler if you immediately substitute numbers into the resulting equations. For example, let's rewrite Ohm's law as:
α∙U 2 (R +r) = E – U
The root of this equation corresponds to the intersection of the parabola
y 1 (U) = α∙U 2 (R + r) = 0.4∙U 2
and graph of a linear function
y 2 (U) = E – U = 50 – U.
The intersection occurs at the point with the abscissa U 0 = 10 V (this can be established either analytically by solving the corresponding quadratic equation, or graphically). At this voltage on the diode, the current flowing through it is equal to:
Answer: I 0 = 2A
- Points for each correct action fold up.
- In case of an arithmetic error (including an error when converting units of measurement), the assessment decreases by 1 point.
- Maximum for 1 task – 10 points.
- A total of 50 points for the work.
Transcript
1 Solutions and assessment system Problem 1 The photograph shows a rotary carousel, which is a cylindrical drum rotating around a vertical axis at a frequency of 33 revolutions per minute. People who initially stand with their backs against the inner vertical wall of the drum move with a centripetal acceleration of 3 (10 m/s 2). As a result, they “stick” to the drum wall. For greater effect, at some point the floor automatically lowers. Assuming people are thin enough, estimate the radius of this carousel's drum, as well as the minimum coefficient of friction between people and the wall of the carousel drum that is sufficient to prevent people from sliding down. We will assume that people are thin enough, and in order to make the necessary estimates, we will neglect their thickness. Then from the formula for centripetal acceleration, assuming its module equal to 3g, we obtain: where 2. Hence 3 4,. Frequency is the reciprocal of the revolution period, which in this case is 60/33 s. Therefore, the frequency is 33/60 Hz. Finally 2.5 m. To answer the second question, we write down Newton’s second law for human motion in a circle in projection onto the vertical axis and in the radial direction (m is the mass of a person, N is the reaction force of the drum wall, Ftr. modulus of the friction force): mg = Ftr ., 3mg = N. Let us take into account that if the friction coefficient is minimal, then Ftr. = µn. Then from the written equations we find: µ = 1/3. 1
2 The formula for centripetal acceleration is written... 1 point The radius of the drum is expressed... 1 point The frequency of revolution is expressed in SI units... 1 point The numerical value of the radius of the drum is found... 1 point Newton's second law is written in projection onto the radial direction. .. 2 points Newton’s second law is written in projection onto the vertical axis... 2 points The friction coefficient is expressed and its numerical value is found... 2 measurement points) the score is reduced by 1 point. Maximum 10 points for the task. Problem 2 A piece of ice weighing 1 kg floats in a vertical cylindrical vessel partially filled with carbon tetrachloride, which has a density of 1600 kg/m3 and is immiscible with water. How and by how much will the carbon tetrachloride level change after all the ice has melted? The area of the bottom of the vessel is 200 cm2. Let the initial height of the level be carbon tetrachloride. Then the pressure at the bottom of the vessel is equal to m, where m is the density of carbon tetrachloride. After the ice melts, the pressure at the bottom of the vessel is equal to: t t, where is the final height of the carbon tetrachloride column, the density of water, and the height of the water column. The mass of the contents of the vessel has not changed, therefore, the pressure on the bottom in the initial and final states is equal, that is: t t 3.125 cm. t Thus, the height of the carbon tetrachloride level will decrease by 3.125 cm. The idea of equality of pressures/pressure forces at the bottom of the vessel is used. .. 2 points Formulas were written for the pressure on the bottom before and after ice melting (2 points each)... 4 points Water pressure is expressed through its mass... 1 point An expression was obtained for changing the height of the carbon tetrachloride level... 2 points 2
3 The numerical value of the change in the height of the carbon tetrachloride level was found and a conclusion was made about its decrease... 1 point of measurement) the score is reduced by 1 point. Maximum 10 points for the task. Problem 3 The graphs show the dependence of pressure p and volume V of one mole of a monatomic ideal gas on time t. Determine how the heat capacity of a given amount of gas changed over time. Plot this heat capacity as a function of time. p V 2p0 2V0 p0 V t, min t, min. During the first 15 minutes, the dependence of gas pressure on its volume looks like this: Let at some arbitrary moment of time (in the interval from 0 min. to 15 min.) the gas pressure is equal to p1, and the volume occupied by it is equal to V1. Let us write down the first law of thermodynamics for the process of transition from state (p0, V0) to state (p1, V1): Here C is the heat capacity of one mole of gas in the process under consideration, the change in gas temperature, and the work performed by the gas. It is numerically equal to the area of the figure under the p(v) dependence graph, and this figure is a trapezoid. Let's rewrite the last expression using the equation of state for one mole of an ideal gas: Δ 3 Δ Δ 2 3
4 or All-Russian Olympiad for schoolchildren in physics. g. Δ. Let's take into account that. Then it follows that 2. Note that the pressure p1 and volume V1, taken at an arbitrary moment in time, are reduced during calculations. This is true, including for two arbitrary states of a gas separated by a very short period of time. This proves that the heat capacity C 2.5R in the process under consideration is 2R a constant value, that is, it will be equal to 2R at any time during the first 15 minutes t, min. After the first fifteen minutes the process becomes isobaric. Therefore, at the same time. The corresponding graph of the heat capacity of one mole of a monatomic ideal gas versus time is shown in the figure. The dependence of pressure on volume for the first process is obtained... 1 point The first law of thermodynamics is written for the change in gas temperature during the transition to an arbitrary intermediate state (in the range from 0 min. to 15 min.)... 1 point An expression is written for the work of gas at transition to an intermediate state... 1 point The heat capacity in the first process is found and it is proven that it is a constant value (if there is no justification for the constancy of the heat capacity, then 2 points are given for this point)... 3 points It is indicated that the second process is isobaric.. 1 point The heat capacity in the second process is indicated... 1 point A graph is constructed showing the characteristic values... 2 points 4
5 dimensions) the score is reduced by 1 point. Maximum 10 points for the task. Problem 4 The first point charge was placed at point A, and it created a potential of 2 V at point B. Then the first charge was removed, and the second point charge was placed at point B. He created a potential of 9 V at point A. Then the first charge was returned back to point A. With what force do these charges interact? Let the moduli of the charges that were placed at points A and B be equal to q1 and q2, respectively, and the distance between them equal to R. Writing the formulas for the potentials created by point charges at points B and A, we obtain: q1 B k, R q2 A k. R According to Coulomb's law, the required force of interaction between charges is equal to: q1q2 F k. 2 R Taking into account the written expressions for potentials, we obtain: F A B k Н = 2 nn. Formulas for the potentials of point charges are written (2 points each)... 4 points Coulomb's law is written... 2 points An expression for the force of interaction of charges is obtained... 2 points The numerical value of the force is found... 2 measurement points) the score is reduced by 1 point. Maximum 10 points for the task. 5
6 Task 5 Determine the reading of an ideal ammeter in the circuit whose diagram is shown in the figure. The dependence of the current I flowing through diode D on the voltage U across it is described by the expression: where 0.02 A/V 2. The emf of the source is 50 V. The internal resistance of the voltage source and resistor are 1 Ohm and 19 Ohm, respectively. Let's write Ohm's law for a section of a circuit that includes a resistor, a voltage source and an ammeter: where is the current flowing through the diode (and through the ammeter), U is the voltage across the diode. Using the current-voltage characteristic of the diode, we obtain: Solving the quadratic equation, we find: 2 A. The second root of the quadratic equation, corresponding to the “+” sign in front of the square root (3.125 A), is not the root of the original equation. This can be established either by direct substitution into the specified initial equation, or by noting that the current flowing through the ammeter in a given circuit cannot exceed 2.5 A. The solution to the problem looks somewhat simpler if you immediately substitute numbers into the resulting equations. For example, let's rewrite Ohm's law in the form:. The root of this equation corresponds to the intersection of the parabola 0.4 6
7 and the graph of the linear function 50. The intersection occurs at the point with the abscissa U0 = 10 V (this can be established either analytically by solving the corresponding quadratic equation, or graphically). At this voltage on the diode, the strength of the current flowing through it is equal to: 2 A. Ohm's law is written for a section of the circuit (or for the complete circuit)... 2 points A quadratic equation regarding the current or voltage is obtained... 2 points A solution to the quadratic equation is obtained ( in any way) and, if necessary, the extra root is reasonably excluded... 4 points The numerical value of the current strength is found... 2 measurement points) the score is reduced by 1 point. Maximum 10 points for the task. A total of 50 points for the work. 7
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All-Russian Olympiad for schoolchildren in physics 2016–2017 academic year. G.
School tour. Grade 11
Solutions and grading system
The photo shows a rotary
carousel, which is
cylindrical drum rotating around a vertical axis
33 rpm.
with frequency
People who initially stand
leaning their backs against the inner vertical wall of the drum,
move with centripetal
acceleration (As a result of this they “stick”
to the drum wall. For greater effect, at some point the floor automatically lowers. Assuming people are thin enough, estimate the radius of this carousel's drum, as well as the minimum coefficient of friction between people and the wall of the carousel drum that is sufficient to prevent people from sliding down.
Then from the formula for centripetal acceleration, assuming its module equal to 3g, we obtain:
3 4 where. From here.
Frequency is the reciprocal of the revolution period, which in this case is 33/60 Hz. Final equal to 60/33 s. Therefore, the frequency is 2.5 m.
To answer the second question, let’s write down Newton’s second law for human motion in a circle in projection onto the vertical axis and in the radial direction (m is the mass of the person, N is the reaction force of the drum wall, Ftr. is the modulus of the friction force): mg = Ftr., 3mg = N.
Let us take into account that if the friction coefficient is minimal, then Ftr. = µN. Then from the written equations we find: µ = 1/3.
1 All-Russian Olympiad for schoolchildren in physics 2016–2017 academic year. G.
School tour. Grade 11 Assessment criteria The formula for centripetal acceleration is written
Drum radius expressed
Frequency of circulation is expressed in SI units
Found the numerical value of the drum radius
Newton's second law is written in projection onto the radial direction...... 2 points Newton's second law is written in projection onto the vertical axis...... 2 points The friction coefficient is expressed and its numerical value is found. .......... 2 points
– – –
Evaluation criteria The idea of equality of pressure/pressure forces at the bottom of the vessel is used...... 2 points Formulas are written for the pressure on the bottom before and after ice melting (2 points each)
Water pressure is expressed through its mass
An expression was obtained for changing the height of the carbon tetrachloride level.... 2 points
– – –
Problem 3 The graphs show the dependence of pressure p and volume V of one mole of a monatomic ideal gas on time t. Determine how the heat capacity of a given amount of gas changed over time. Plot this heat capacity as a function of time.
– – –
Possible solution During the first 15 minutes, the dependence of gas pressure on its volume looks like this: Let at some arbitrary moment of time (in the interval from 0 min. to 15 min.) the gas pressure is equal to p1, and the volume occupied by it is equal to V1.
Let us write down the first law of thermodynamics for the process of transition from state (p0, V0) to state (p1, V1):
Here C is the heat capacity of one mole of gas in the process under consideration, is the change in gas temperature, and is the work performed by the gas. It is numerically equal to the area of the figure under the p(V) dependence graph, and this figure is a trapezoid.
Let's rewrite the last expression using the equation of state for one mole of an ideal gas:
– – –
The corresponding graph of the heat capacity of one mole of a monatomic ideal gas versus time is shown in the figure.
Evaluation criteria The dependence of pressure on volume for the first process was obtained............... 1 point The first law of thermodynamics was recorded for the change in gas temperature during the transition to an arbitrary intermediate state (in the range from 0 min. to 15 min.)
An expression has been written for the work of a gas during the transition to an intermediate state
The heat capacity in the first process was found and it was proven that it is a constant value (if there is no justification for the constancy of the heat capacity, then 2 points are given for this point)
It is indicated that the second process is isobaric
The heat capacity in the second process is indicated
A graph has been constructed showing characteristic values
4 All-Russian Olympiad for schoolchildren in physics 2016–2017 academic year. G.
School tour. Grade 11 For each correctly performed action, points are added up.
– – –
Evaluation criteria Formulas for the potentials of point charges are written (2 points each)........ 4 points Coulomb's law is written
An expression for the force of interaction of charges is obtained
The numerical value of the force is found
For each correctly performed action, points are added up.
In case of an arithmetic error (including an error when converting units of measurement), the score is reduced by 1 point.
The maximum score for the task is 10 points.
– – –
Determine the reading of an ideal ammeter in the circuit whose diagram is shown in the figure. The dependence of the current I flowing through diode D on the voltage U across it is described by the expression: where 0.02 A/V2. The emf of the source is 50 V. The internal resistance of the voltage source and resistor is 1 Ohm and 19 Ohm, respectively.
are equal Possible solution Let's write Ohm's law for a section of the circuit that includes a resistor, a voltage source and an ammeter:
Where is the current flowing through the diode (and through the ammeter), U is the voltage across the diode.
Using the current-voltage characteristic of the diode, we obtain:
Solving the quadratic equation, we find:
The second root of the quadratic equation, corresponding to the “+” sign in front of the square root (3.125 A), is not the root of the original equation. This can be established either by direct substitution into the given original equation, or by noting that the current flowing is 2.5 A.
through an ammeter in a given circuit, cannot exceed
– – –
Evaluation criteria Ohm's law is written for a section of a circuit (or for a complete circuit)
A quadratic equation was obtained for current or voltage... 2 points A solution to the quadratic equation was obtained (by any method) and, if necessary, an extra root was reasonably excluded
The numerical value of the current is found
For each correctly performed action, points are added up.
In case of an arithmetic error (including an error when converting units of measurement), the score is reduced by 1 point. The maximum score for the task is 10 points.
– – –
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St. Petersburg, Russia, 1910s.
Led Zeppelin, 1969.
Khevsurs (tribe of Georgian mountaineers), Russia, 1890.
Civilians digging an anti-tank ditch near Moscow, 1941.
Consolidated PBY Catalina maritime patrol bombers at Lake Worth Air Station, 1940s.
Remains of concentration camp prisoners, Pomerania, 1945.
Charlie Chaplin, 1912.
The boy met wrestler Andre the Giant, 1970s.
We started thinking about three people at least 100 years ago. Russia, end of the 19th century.
Street sale of household appliances, Russia, 1990s.
Another sketch from the life of Russia in the 1990s. It’s hard to imagine now, but in those days they sold household appliances on the street, bringing them in on trucks. Those. straight from the wheels.
Construction of the Hindenburg airship, 1932.
Mel Gibson and Sigourney Weaver, 1983.
The first aerial explosion of a hydrogen bomb at Bikini Atoll in the Pacific Ocean, May 20/21, 1956.
Twin dancers Alice and Ellen Kessler, 1958.
Young Steven Seagal, USA, 1960s.
His paternal grandparents came to America as children from St. Petersburg.
Sometimes it is very important to rest your soul and body... Idi Amin, dictator of Uganda, Africa, 1972.
British soldiers test a special crane to extract wounded tank crews, World War II.
The device is mounted on the turret of the Mk.II Matilda II infantry tank
Rotary carousel. USA, 1950s.
It accelerated to 33 rpm, creating a centrifugal force of almost 3G. When people “stuck” to the wall of the drum from this force, the floor was automatically removed for greater effect.
Captured Soviet soldiers try to drink from a frozen river, 1941.
"Pobeda-Sport", USSR, 1950.
The famous “Sunny Clown” Oleg Popov, USSR, 1944.
Prisoner in a French prison, 1900s. Mustaches were tattooed as a sign of protest against the administration.
Cosmonauts Andriyan Nikolaev and Valentina Tereshkova, Japan, 1965.
Morning in the apartment of Vladimir Mayakovsky and the Brikovs on Gendrikov Lane, 1926. From left to right: Vladimir Mayakovsky, Varvara Stepanova, Osip Beskin, Lilya Brik.
Festive decoration on Gorky Street in Moscow on International Workers' Day, 1969.
Car traffic on Red Square in Moscow, USSR, 1960.
Until 1963, there was vehicular traffic on Red Square in Moscow. And then it was decided to make it pedestrian.
Michael Jackson in 2000 according to Ebony Magazine, 1985.
In 1985, Ebony magazine predicted what Michael Jackson would look like in 2000: "At 40, Michael will age gracefully, looking more mature and attractive. And his fan base will increase 10-fold."
Destruction of the Cathedral of Christ the Savior. Remains of a sculptural group. Moscow, USSR, 1931.
Championship for the game Space Invaders, 1980.
All ages are submissive to football, USSR.
Elizabeth Taylor in Iran, 1976.
Martin Scorsese and Robert De Niro, 1970s.
A crashed zeppelin in a field, France, 1917.
Matthias Rust (left), the 18-year-old German amateur pilot who amazed the world by landing his plane on Vasilyevsky Spusk in May 1987, has lunch in court, 1987.
Blessing of an airplane, France, 1915.
If not Taylor, then who?
In 1997, presidential elections were held in Liberia. Leading candidate Charles Taylor's campaign slogan was: "Taylor killed my father, killed my mother, but I'll still vote for him."
Civilians shot by the Nazis, 1942.
Ivan Pavlov's dogs with their "servants", Imperial Institute of Experimental Medicine, St. Petersburg, 1904.
Swimming pool Moscow on the site of the Cathedral of Christ the Savior. Moscow, USSR, 1960s.
That's who naturally rode on Bill Clinton's neck - the presidential cat Sox, USA, March 7, 1995.
Apple clothing line, 1986.
Japanese soldiers bury Chinese prisoners of war alive. Nanjing, China, Sino-Japanese War, 1937.
Children in kindergarten draw a poster for the celebration of the 12th anniversary of the October Revolution, October 1, 1929.
Assembly of the I-15 fighter designed by N. Polikarpov Design Bureau at the Spanish SAF-3 plant in Reus, Spain, 1937.
Boxing match between American boxer Gus Waldorf and a real bear, March 1949.
Ukrainian politicians Yulia Tymoshenko, Alexander Turchynov, Pavel Lazarenko, 1996.
Airplane over Manhattan, USA, 1939.
Boxers, 1890s.
Prisoners await trial in the overcrowded Butyrka prison, 1995.
Mick Jagger, 1967.
A column of Tiger I heavy tanks and a MAN ML 4500 truck of the 1st SS Panzer Division "Leibstandarte SS Adolf Hitler" in the Vinnitsa region of Ukraine, 1943.
Jean-Paul Belmondo and Alain Delon, 1997.
One of the last photographs of the icebreaker Ermak, 1960s.
New York taxi, 1905.
Hitler inspects the new Ferdinand self-propelled gun. To his left is Ferdinand Porsche.
Donald Trump and his sons Donald Jr. and Eric Trump with Hillary Clinton at the White House in 1997, Photo: Sarah Merians.
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