5. Find the inductance of the circuit shown in the figure. The inductances of all coils in the circuit are the same and equal to L, neglect the inductances of the connecting wires and the influence of the coils on each other. Solution. If an alternating sinusoidal voltage with a circular frequency ω is applied to the input terminals of this circuit, then the inductive reactances of all coils will be the same and equal to ωL. In this case, the rules for adding these resistances will be the same as for a circuit consisting of resistors, that is, with a series connection, the inductances are added, and with a parallel connection, their inverse values are added. To calculate the inductance, we first redraw the circuit, for example, as shown in the figure on the left. Since all inductances are the same, it follows from symmetry considerations that the potentials of points A and B at any time will coincide. Therefore, you can short them with a conductor. We find that inductors L1 and L4, L2 and L3, L5 and L7 are connected in parallel. This means that the circuit can now be redrawn as shown in the figure on the right, and it will consist of coils with inductances L/2 and L connected in series and parallel. The total inductance of such a circuit will be equal to 1 1 7 Ltot = = = L. 1 1 1 1 15 + + L L 1 L L 3L + + 2 1 1 2 8 + L/2 1 L+ L/2 11 Urban stage. The first theoretical round took place on February 19, 2006. 7th grade 3 astronomical hours were allotted to complete the task. 1. Find the approximate pressure value at the center of the Earth, assuming that average density substances globe equal to ρ = 5000 kg/m3. The radius of the Earth RЗ = 6400 km. Acceleration free fall on the Earth's surface g = 10 m/s2. Solution. At a depth h below the surface of the liquid, the pressure is p = ρgh, where ρ is its density, and g is the acceleration of gravity. But we cannot use this formula to find the pressure at the center of the Earth, since g does not remain constant as we move deeper into the Earth. Indeed, let's imagine that we managed to drill a well to the center of the Earth. It is clear that a body lowered into it to this center will be equally attracted from all sides by the substance of the Earth and will be in a state of weightlessness, that is, the acceleration of free fall gradually decreases from a value of 10 m/s2 on the surface of the Earth to zero at its center. Therefore, the average value of the acceleration of free fall, equal to g/2, must be substituted into the formula for pressure. This means that the pressure in the center of the Earth is approximately equal to p = ρgRЗ /2 ≈ 1.6 1011 Pa = 1.6 million atmospheres! Comment. By modern ideas, The Earth consists of three main layers of a thin crust, a rather thick mantle (about 3000 km), composed of rocks of relatively low density, and a heavy (iron) core. The acceleration of free fall is also quite in a complex way depends on the depth (see task No. 3 of the district stage, page 7). Taking this into account, the calculation gives an even larger value for the pressure in the center of the Earth: pc ≈ 3.6 million atmospheres! o 2. The schoolchildren visited the museum-estate of L. N. Tolstoy Yasnaya Polyana and returned to Ryazan on buses that traveled at a speed of v1 = 70 km/h. It began to rain, and the drivers reduced their speed to v2 = 60 km/h. When the rain stopped, S = 40 km remained to travel to Ryazan. The buses drove at a speed v3 = 75 km/h and entered Ryazan at exactly the planned time. How long did it rain? What is it equal to average speed bus? To simplify, assume that the buses did not stop along the way. 12 Solution. The average speed of a bus is the ratio of the distance traveled to the time taken. Since the distance from Yasnaya Polyana to Ryazan did not change due to the rain, and the time spent by schoolchildren on the bus also did not change (because the buses entered Ryazan at exactly the planned time), then the average speed coincides with initial speed vav = 70 km/h. Let it rain for time t. Then the path traveled during this time was v2 t. The time during which the buses covered the remaining distance after the rain is equal to S/v3. It is clear that the time spent by buses from the moment the rain begins until their arrival in Ryazan must be equal to the time that would be required to cover the same distance with an initial speed v1: S v2 t + S t+ = . v3 v1 From here we find the time during which it rained: v1 S S S(v3 − v1) t= − = = 16 minutes. v1 − v2 v1 v3 v3 (v1 − v2) 3. In the ice of the Arctic, in the center of a small flat ice floe with an area of S = 70 m2, there is a polar bear with a mass of m = 700 kg. In this case, the surface part of the ice floe protrudes above the water surface to a height of h = 10 cm. At what depth under water is it located? bottom surface ice floes? Water density ρw = 1000 kg/m3, ice density ρl = 900 kg/m3. Solution. Let us denote the desired depth by x. The force of gravity acting on the ice floe with the bear is obviously equal to g(m + ρл S(h + x)). It should be equal to the force of water pressure on the lower surface of the ice floe, located at depth x, that is, ρв gxS, since the ice floe is in a state of equilibrium. From here we get: x = (m + ρл hS)/((ρв − ρл)S) = 1 m. 4. The wires above the railway supplying electric trains with current are tensioned using the system shown in the figure. It is attached to a pole and consists of cables, blocks with insulators and a square steel weight with side a = 20 cm. The tension force of the thick cable that goes from the outer block to the wire holder is T = 8 kN. What is the height h of the steel weight? The density of steel is ρс = 7800 kg/m3. Gravity acceleration g = 10 m/s2. 13 Solution. It is easy to see that each block covered by two horizontal sections of cables gives a 2-fold gain in strength. This means that three such blocks shown in the figure will give a winning of 23 = 8 times. The force of gravity acting on the load is equal to ρс gV, where V = a2 h volume of the load. This means that the tension force of a thick cable will be 8 times greater: T = 8ρс gV. From here we find that the volume of the steel load is V = T /(8ρс g), and its length is equal to h = T /(8ρс ga2) ≈ 0.32 m = 32 cm. 8th grade 3 astronomical hours were allotted to complete the task. 1. The schoolchildren visited the village of Konstantinovo, the homeland of Sergei Yesenin, and returned to their home in Ryazan by bus. The buses were traveling at a speed v1 = 70 km/h. It started to rain, and the drivers reduced their speed to v2 = 50 km/h. When the rain stopped, the buses moved again at the same speed and entered Ryazan 10 minutes later than planned. How long did it rain? Solution. Let's make a drawing and introduce the following notation on it: To Konstantinovo; R Ryazan; AB is the section that the bus covered in the rain during the required time t; AC is the section that the bus would cover in the same time t if it were not raining. It is clear that BC = AC − AB = (v1 − v2)t. On the other hand, the bus covered the path KA + AB + CR in the same time as it was planned to cover the entire path KR. This means BC = v1 ∆t, where ∆t = 10 minutes is the time by which the buses were late. Equating the resulting expressions, we have: (v1 − v2)t = v1 ∆t, whence t = v1 ∆t/(v1 − v2). 2. Air is pumped into a two-liter plastic bottle through a short hose to a pressure of 2 atm. The hose is pinched, and a sealed thin-walled plastic bag of large capacity (more than 10 liters) without air inside is attached to it. The bottle and the package are placed on one pan of the scale and balanced with weights that are placed on the other pan, and then the clamp is loosened. Air from the bottle flows into the bag, and the balance of the scales is disrupted. What mass must be placed on what scale to restore balance? The air density is 1.3 kg/m3, the acceleration of free fall is assumed to be 10 m/s2. Solution. The total mass of air inside the bottle and bag did not change after the air flowed from the bottle into the bag. Consequently, the total force of gravity acting on both shells and the air inside them remains the same. However, the total volume occupied by the bottle and the bag together has changed, since after the clamp was loosened, some of the air from the bottle passed into the bag. The pressure in the bag became equal to 1 atm, which means that the same pressure was established in the bottle. Air, which in the bottle occupied a volume of 2 liters at a pressure of 2 atm, now occupies a volume of 4 liters at a pressure of 1 atm. Thus, there were 2 liters of air in the bag, and the total volume increased by 2 liters. A buoyant (Archimedean) force acts on the bottle and the bag from the air. The increment of this force is equal to: ∆FA = 0.002 m3 · (1.3 kg/m3) · (10 m/s2) = 0.026 N. Thus, in order for the balance of the scales to be restored, it is necessary to move to the same cup where the bottle is located and a bag, add weights with a total mass M = ∆FA /g = 2.6 g. 3. The calorimeter contains m = 100 g of molten gallium metal at its melting temperature tmelt = 29.8 ◦ C. They began to slowly cool it, protecting from external influences, and as a result, the temperature dropped to t = 19.8 ◦ C, and gallium remained liquid. When liquid gallium supercooled in this way was stirred with a stick, it partially turned into solid state. Find the mass of solidified gallium and the temperature established in the calorimeter. Specific heat of fusion of gallium λ = 80 kJ/kg, specific heat liquid gallium c = 410 J/(kg ◦ C). Neglect the heat capacity of the calorimeter and the stick. 15 Solution. When gallium solidifies, the heat of crystallization is released, which leads to heating of the system to the melting temperature of gallium tmelt = 29.8 ◦ C, since only at this temperature will liquid and solid gallium be in equilibrium. The amount of heat released during the solidification of the mass m1 of gallium is equal to λm1. It is used to heat all the gallium to the melting point; this requires an amount of heat cm(tmel − t). Consequently, m1 = cm(tmelt − t)/λ ≈ 5.1 g. Note that if the supercooling were very strong, then the heat of crystallization might not be enough to heat the entire mass of gallium to the melting temperature. However, since m1< m, то в
нашем случае галлий действительно нагреется до этой температуры.
9 класс
На выполнение задания отводилось 4 астрономических часа.
1. Цилиндр массой M поместили на рельсы,
наклоненные под углом α к горизонту (вид
сбоку показан на рисунке). Груз какой мини-
мальной массы m нужно прикрепить к намо-
танной на цилиндр нити, чтобы он покатился
вверх? Проскальзывание отсутствует.
Решение. На цилиндр действуют при-
ложенная к его центру сила тяжести M g
и приложенная к его краю сила натяже-
ния нити, равная mg. Цилиндр покатится
вверх, если момент силы тяжести относи-
тельно оси, проходящей через точку А пер-
пендикулярно плоскости рисунка, будет
меньше момента силы натяжения нити.
Поскольку плечи сил тяжести и натяже-
ния нити равны R sin α и R(1 − sin α), то искомое условие имеет вид:
M gR sin α < mgR(1 − sin α), или m >(M sin α)/(1 − sin α). 2. Aluminum wire diameter d = 2.5 mm, not too bent, covered with ice. The total diameter of the wire with ice is D = 3.5 mm. The temperature of the ice and the wire is t = 0 ◦ C. A current of force I = 15 A is passed through the wire. How long will it take for the ice to melt? Ice density ρl = 0.9 g/cm3, and its specific heat melting λ = 340 kJ/kg. Resistivity of aluminum ρ = 2.8 · 10−8 Ohm · m. 16 Solution. When a current passes through a wire, heat is released in it, equal to the Joule-Lenz law Q = I 2 Rτ, where τ is the desired time for ice melting, and R is the resistance of the wire. This resistance, according to known formula, is equal to R = ρl/S = 4ρl/πd2 (here l is the length of the wire, S is its area cross section). This amount of heat is spent on melting ice: Q = λm. The mass of ice m is equal to the product of its density and volume: m = ρl V = ρl (1/4)π(D2 − d2)l. Equating the obtained expressions for the amounts of heat, we finally obtain: τ = λρл π 2 d2 (D2 − d2)/(16I 2 ρ) ≈ 19 min. 3. The electrical circuit consists of three resistors with known resistances R1 = 20 Ohm, R2 = 30 Ohm, R4 = 60 Ohm, one resistor with unknown resistance R3 and one variable resistor (see figure) When measuring resistance RAB between points A and B of this electrical circuit It turned out that it does not depend on the resistance of the variable resistor. Find the resistance values of the unknown resistor R3 and the entire circuit RAB. Solution. The idea of the solution is that, under the conditions of the problem, no current flows through the variable resistor, and the voltage across it is zero (otherwise, a change in the resistance of this resistor would inevitably lead to a change in the value of RAB). It follows that the voltages U1 and U3 on resistors R1 and R3 are the same. Since R1 R3 U1 = UAB · , U3 = UAB · , R1 + R2 R3 + R4 then hence R1 R4 = R2 R3 , and the resistance of the unknown resistor R3 = R1 R4 /R2 = 40 Ohms. The resistance of the entire circuit can be found using the formula for parallel connection resistors: 1 1 1 (R1 + R2)R4 = + , whence RAB = ≈ 33 Ohm. RAB R1 + R2 R3 + R4 R2 + R4 4. In a sextant, which allows you to determine the angle ϕ of the elevation of the Sun above the horizon at noon and, thus, the latitude of the area, two flat mirrors a, from which light is alternately reflected and the angle α between which is adjusted. When measuring with a sextant, the image of the Sun in these mirrors must be aligned with the horizon line, selecting the angle α. Find the connection between angle α and 17 angle ϕ and explain why using a sextant greatly simplifies the task of finding angle ϕ, especially when the ship is rocking. Solution. Let us construct the path of a ray of light from the Sun in a sextant with two reflections of light from flat mirrors, the angle between which is equal to α (see figure). Let us denote the vertex of the angle α by the point O, the points of incidence of the ray on the first and second mirrors A and B, the point of intersection of the perpendiculars raised to the mirrors at points A and B by C, the point of intersection of the rays entering and exiting the device by D. At the moment of removal indications for correct position mirrors, the straight line BD is horizontal, and the angles of incidence of light on the mirrors are equal to i1 and i2, respectively. In quadrilateral AOBC, two angles OBC and OAC are right angles, so angle BCA is equal to (π − α), and its adjacent angle is equal to α. But this same angle is outer corner triangle ABC, so α = i1 + i2. In turn, the angle ϕ of the Sun’s elevation above the horizon equal to angle BDA in triangle ABD, the remaining angles of which are equal to 2i1 and 2i2, respectively. Therefore ϕ = π − 2(i1 + i2) = π − 2α. Thus, α = (π − ϕ)/2 and does not depend on the angle of incidence of light on the mirrors. Therefore, even when the ship rocks and the angle i1 changes, the ray of light from the Sun emerging from the sextant retains its direction (horizontal when correct selection angle α). At the same time, it is much easier to combine the image of the Sun with the horizon than to sight two directions to the Sun and the horizon at once on a goniometer instrument, and even if everything is swaying! 18 10th grade 5 astronomical hours were allotted to complete the task. 1. One summer morning, a grasshopper was sitting on the asphalt. When the Sun rose at an angle ϕ above the horizon, he jumped towards the Sun with an initial speed v0 at an angle α to the horizon. At what speed does the grasshopper’s shadow move on the asphalt after time t after the jump? Solution. Let's direct the X axis horizontally towards the Sun, the Y axis vertically upward, and place the origin of coordinates at the point where the grasshopper was sitting. The law of motion of the grasshopper has the form: gt2 2v0 sin α x(t) = v0 cos α · t, y(t) = v0 sin α · t − , with 0
Huge frozen hat sea water on the surface of the North Arctic Ocean and neighboring seas last decades suffered a double blow: its area was reduced, the oldest and thickest ice either thinned or melted completely, leaving the ice sheet more vulnerable to warming oceans and the atmosphere.
Old ice is shown in white, young ice is in gray.
"We've been seeing the old ice disappear for many years," says Walt Meyer of the Center. space flights NASA Goddard in Greenbelt. “This thick cover serves as a protective support for the entire cap: summer, with its warmth, can melt thin ice, but it cannot completely get rid of old ice. However, it is becoming less and less, and the remaining one is thinning, and the resistance to above-zero temperatures is no longer as stable as before.”
Direct thickness measurements sea ice in the Arctic are sporadic and incomplete, so scientists have developed a model of the age evolution of sea ice from 1984 to the present. A new NASA visualization shows how sea ice has grown and shrunk, melted and drifted over the past three decades.
Scientists say the age of ice is a great indicator of its thickness because as ice gets older, it gets thicker. This is due to the fact that, as a rule, in winter it increases more ice than it manages to melt over the summer.
In the early 2000s, scientists at the University of Colorado developed a way to monitor Arctic sea ice movement and age changes using data from various sources, mainly based on passive satellite microwave instruments. These instruments estimate brightness temperature, a measure of the microwave energy emitted by sea ice that depends on temperature, salinity, ice surface texture and the layer of snow on the sea ice. Each ice floe has a characteristic brightness temperature, and scientists identified and tracked them using serial passive microwave images. The system also uses information from drifting buoys and meteorological data.
“It’s like an accounting book, we track what happens to sea ice, how it moves, grows and retreats until it either melts in place or goes beyond the Arctic,” Meyer explains.
Every year, sea ice forms in winter and melts in summer. The one that survives the melting season thickens every year. Newly formed ice grows to 1-2 meters thick within the first year, while the thickness multi-year ice, which has survived several warm seasons, is approximately 3-4 meters. The older and thicker the ice, the more resistant it is to melting and the less susceptible to the influence of wind and storm waves.
The movement of sea ice is not limited to its seasonal expansion and retreat: except for coastal areas, the sea ice cover is almost constant movement. Home driving force The wind acts in this process. There are two main circulations in the Arctic air masses: the Teaufort Gyre, where ice rotates like a wheel clockwise in the Beaufort Sea north of Alaska, and the Transpolar Drift Current, which moves ice from the coast of Siberia towards the Fram Strait east of Greenland, where it emerges from the Arctic basin and melts into warm waters Atlantic Ocean.
However, approximately every week, weather systems follow this same trajectory and affect these flows. So the speed of ice movement is not constant. When spring comes and the ice begins to melt, it disappears from the peripheral seas.
New video shows two major losses of thick ice. The first, which began in 1989 and lasted for several years, was associated with a change in the Arctic Oscillation, which weakened the Trough Gyre and strengthened the Transpolar Drift Current, which washed more sea ice out of the Arctic than usual. The second melting peak began in the mid-2000s.
“Unlike the 1980s, old ice is now melting within the Arctic Ocean in summer time. One reason is that multi-year ice tends to clump together, and we now see relatively smaller pieces of it mixed in with younger ice. These isolated floes of thick ice are much easier to melt,” says Meyer.
The loss of thick ice, according to scientists, is colossal. In the 1980s it made up 20% of sea ice cover. Now it’s only about 3%. It is possible that very soon the summer Arctic will be completely ice-free. GisMeteo
Physics Olympiad for schoolchildren, 8th grade, 2010-2011 school year. year
Schoolchildren's Olympics
in physics
2010–2011 academic year
8th grade
Dear friend! We wish you success!
Quests ( maximum score for all work - 40)
1 The calorimeter contains m = 100 g of molten gallium metal at its melting temperature t pl = 29.8 °C. They began to slowly cool it, protecting it from external influences, and as a result the temperature dropped to t = 19.8 C, and gallium remained liquid. When liquid gallium supercooled in this way was stirred with a stick, it partially turned into a solid state. Find the mass of solidified gallium and the temperature established in the calorimeter. Specific heat of fusion of gallium λ = 80 kJ/kg, specific heat capacity of liquid gallium c = 410 J / (kg∙°C). Neglect the heat capacity of the calorimeter and the stick.
10 points
2 The schoolchildren visited the museum-estate of L. N. Tolstoy " Yasnaya Polyana"and returned to Ryazan on buses that traveled at a speed v 1 = 70 km/h. It began to rain, and the drivers reduced their speed to v 2 = 60 km/h. When the rain stopped, all that was left was to drive to Ryazan S= 40 km. The buses drove at a speed v 3 = 75 km/h and entered Ryazan exactly at the planned time. How long did it rain? What is the average speed of the bus? To simplify, assume that the buses did not stop along the way.
10 points
3. In the Arctic ice in the center of a small flat ice floe with an area of S= 70 m 2 sits a polar bear with mass m= 700 kg. In this case, the surface part of the ice floe protrudes above the water surface to a height h= 10 cm. At what depth under water is the lower surface of the ice floe? Density of water in = 1000 kg/m 3, density of ice l = 900 kg/m 3.
10 points
4. The wires above the railway supplying electric trains with current are tensioned using the system shown in the figure. It is attached to a pole and consists of cables, blocks with insulators and a square steel weight with a side a= 20 cm. The tension force of the thick cable that goes from the outermost block to the wire holder is equal to T = 8 kN. What is the height h steel load? The density of steel is c = 7800 kg/m 3. Acceleration of gravity g= 10 m/s 2 .
10 points
8th grade
Possible solutions to problems
1. Solution. When gallium solidifies, the heat of crystallization is released, which leads to heating of the system to the melting temperature of gallium t pl = 29.8 ◦ C, since only at this temperature will liquid and solid gallium be in equilibrium.
The amount of heat released during solidification of the mass m1 of gallium is equal to λm 1.
It is used to heat all the gallium to its melting point; this requires
amount of heat cm(t pl − t). Consequently, m1 = cm(t pl − t)/λ ≈ 5.1 g.
Note that if the supercooling were very strong, then the heat of crystallization might not be enough to heat the entire mass of gallium to the melting temperature.
However, since m 1
2. Solution. The average speed of a bus is the ratio of the distance traveled to the time taken. Since the distance from Yasnaya Polyana to Ryazan did not change due to the rain, and the time spent by schoolchildren on the bus also did not change (because the buses entered Ryazan at exactly the planned time), then the average speed coincides with the initial speed v av = 70 km/h Let it rain for time t. Then the path traveled during this time was v 2 t. The time it takes for the buses to cover the remaining distance after the rain is S/v3. It is clear that the time taken by buses from the moment the rain begins to arrival in Ryazan must be equal to the time that would be required to cover the same distance with an initial speed v 1:
From here we find the time during which it rained:
3. Solution. Let us denote by x the desired depth. The force of gravity acting on the ice floe with the bear is obviously equal to g(m +ρлS(h + x)). It must be equal to the force of water pressure on the lower surface of the ice floe, located at depth x, that is, ρ in gxS, since the ice floe is in a state of equilibrium. From here we get:
x = (m+ ρ l hS)/((ρ in − ρ l)S) = 1 m.
4. Solution. It is easy to see that each block covered by two horizontal sections of cables gives a 2-fold gain in strength. This means that three such blocks shown in the figure will give a winning of 2 3 = 8 times. The force of gravity acting on the load is equal to ρ with gV, where V = a 2 h is the volume of the load.
This means that the tension force of a thick cable will be 8 times greater: T = 8ρ with gV.
From this we obtain that the volume of the steel load is V = T/(8ρ with g),
and its length is equal to h = T/(8ρ with g 2) ≈ 0.32 m = 32 cm.
/ Myths about the Arctic /
Myths about the Arctic
It seems that the 21st century has left no chance for unenlightened people, but surprisingly, even those who have higher education know very little about North Country Russia either has little idea of what the Arctic is.
The Arctic is a land of poets, historical New Earth, sacred Vaigache, the unattainable North Pole, Franz Josef Land, glorified by Vizbor. We seem to be listening with half an ear to news from distant regions - the Arctic scientific stations, melting ice, global warming, polar year, Chilingarov, - but for some reason we cannot answer with complete confidence why bears do not eat penguins, why an iceberg does not sink, or what the eighty-fourth parallel is. So, let's try to dispel some myths about the Arctic.
Myth No. 1
Bears eat penguins.
This is the funniest and most absurd misconception that can often be heard from children's lips. What can I say, some adults are convinced that bears, no, no, will even snack on a couple of flightless birds.
This myth can only be debunked with one statement: penguins are at the bottom, and bears are at the top. And we are not talking about sea-land, but about the fact that penguins live only in Antarctica, that is, in the south, and polar polar bears live only in the Arctic, that is, in the north. That is why they are not destined to meet and fight for life.
The most popular answer to the riddle of why a bear won't eat a penguin is that penguins smell bad. Members of the Russian Polar Explorers Association decided to dispel people’s doubts once and for all and demonstrated the principle of non-coexistence of these animals on their symbol.
Myth No. 2
In the Arctic there are fifteen-meter thicknesses of ice and terrible frosts from which a person can die.
Sometimes people, in their skepticism and ignorance, cannot change their minds about distant regions and believe someone who knows. Some even think that expedition cruises to the North Pole and trips there on icebreakers are something beyond fantasy.
Actually in summer months(July, August), when cruise flights to the North Pole and the Arctic operate, the ice thickness is only 2–3 meters, and the air temperature fluctuates around 0 °, sometimes reaching +1–3 °C. The average winter temperature is –40 °C.
Myth No. 3
The Arctic is a kingdom of ice and desert plains of snow.
For the first time entering this land, it may seem that there is no life here at all, only snow. But the Arctic appears extinct only to the uninitiated. In fact, it is diverse, rich in animals and plants, flowers and colors.
What is Wrangel Island worth: thanks to the diversity of flora and fauna, it was included in the UNESCO World Heritage List. The reserve organized on the island protects walruses, musk oxen, polar bears and many Arctic bird species from human encroachment.
The bears fell in love with this undoubted symbol of the Arctic so much that they made it a “maternity hospital.” Bears from all over the eastern Arctic regularly come to Wrangel Island and the nearby Herald Island. In the snow they make dens for themselves and lie down in them.
The Franz Josef Land archipelago can also be considered a country of polar bears, in the area of which passengers on expedition cruises to the North Pole and the Arctic most often get to see the bears in their natural habitat. And don’t think that they will run away in all directions as soon as they see the bulk of the icebreaker. They behave befitting the owners of the Arctic: they come up to the board of the nuclear-powered ship and look at their dear guests with interest. There are both lonely bears and entire families: for example, a mother with children.
Birds, of which there are a great many here, are not so sociable. If a visit to the island where they nest falls during the period of hatching chicks and caring for the newly born, maximum caution will have to be taken. With any rash movement on the part of a person, restless parents can attack uninvited visitors.
The majestic walruses are very impressive and unusual. Often, during a trip on special inflatable boats, they can float nearby to the delight of photographers and even accompany “Arctic” tourists until the end of the trip.
Some travelers are delighted with the local flora: it is amazing here. Nature can charm even the most urban people. Bright, dense red, green, yellow carpets of lichens and mosses, polar poppies blooming on the border of thawed soil and ice - what kind of desert is this?..
Myth No. 4
Get to North Pole unrealistic, it is available only to a select few.
Many people, having read heroic stories and watched tragic dramas about the conquest of the Arctic, still remember doubts and disputes about the fate of the pioneers of the North Pole and refuse to believe that they will conquer ice world everyone can.
Today, anyone can become a passenger on an expedition cruise to the Arctic and make an amazing two-week journey on board a powerful nuclear icebreaker.
It is impossible to say exactly what exactly leaves greatest impression: conquering the “edge of the world” or enjoying the process of persistently achieving a goal. The 75 thousand horsepower of the icebreaker confidently makes its way through the thickness of the pack ice - this is perhaps the most memorable sight.
The most desperate travelers, who do not care about comfort and do not care about peace, can go off to conquer the snow on skis. And during this trip there will be everything: high (5–10 meters) ridges of hummocks, low (30°C) temperatures, crossings with ice water, amazing arctic landscapes, bright polar sun, overcoming obstacles and learning your own limits.
All this makes the journey atypical, non-trivial, because in the ocean everything moves, nothing is constant, not even the pole. And most importantly, it is only yours. You will be the first of all people on the planet to set foot on ice, because it constantly drifts and lingers in one place for several minutes. In a word, for everyone the pole is unique, unique, its own.
Myth #5
The point of the North Pole is indicated by a monument such as the “Top of the World” sign, a boundary pillar, or a stele with the inscription “90°N”.
For many, the first achievement of the world peak seems like the conquest of Everest: heroic pioneers, using their last strength, literally crawl to 90° northern latitude- your cherished goal.
And the conquest must invariably end with the installation of something material or at least a commemoration of the fact that a person was here. That is, something important and eternal. This is absolutely not true.
In fact, the North Pole is the most elusive point on the planet. At 90° north latitude there is nothing: no column with an inscription, no memorial flag, no mountain, nothing, not even land. Just a frozen ocean.
But don't be upset. Having reached geographical point poles and having caught the long-awaited 90 ° N on the GPS navigator screen, you will be able to go down onto the ice floe, and it will become the symbol of your North Pole. The ice is in constant motion, and in a few minutes your symbol will leave the “top of the planet.”
And all you have to do is break off a piece of the North Pole, carve a memorable inscription on it and store it in the freezer so that this ice souvenir warms your heart, reminding you that you too have become the chosen one - one of those who conquered the North Pole.
I think we have a lot of them...
1.. The toy bucket has the same shape as a real bucket, but the height of the toy bucket is 4 times less. How many toy buckets of water will need to be poured into a real bucket to fill it?
2. From the window you observe a crowd of people in the square who have come to the festive festivities. The square is crowded. If you mentally replace each person with a molecule, what state of matter does this resemble?
3. A fisherman was sailing along the river in a boat, caught his hat on the bridge, and it fell into the water. An hour later, the fisherman came to his senses, turned back and picked up the hat 4 km below the bridge. What is the speed of the current? The speed of the boat relative to the water remained unchanged in absolute value.
4. A cyclist left village A along a straight road. When he had traveled 16 km, a motorcyclist came after him at a speed 9 times greater than the speed of the cyclist, and caught up with him in village B. What is the distance between the villages?
5. Distance from the Earth to the nearby star(not counting the Sun) is equal to approximately 38 trillion kilometers (this number includes 14 digits). How long does it take for light from this star to reach us?
6. The most difficult thing is to get to know a person. They say that to do this you need to eat a pood (16 kg) of salt with it. How long will it take for this if the medical norm for salt consumption is 5 g per day? Eleven school years enough?
7. Masha and Pavlik each bought a portion of ice cream and brought it home. Pavlik put his ice cream in a saucer on the table, and Masha put her ice cream under the fan. Whose ice cream won't melt longer?
8. The schoolchildren visited the museum of L.N. Tolstoy’s estate “Yasnaya Polyana” and returned to Ryazan on buses that traveled at a speed of υ1 = 70 km/h. It began to rain, and the drivers reduced their speed to υ2 = 60 km/h. When the rain stopped, s = 40 km remained to travel to Ryazan. The buses drove at a speed of υ3 = 75 km/h and entered Ryazan at exactly the planned time. How long did it rain? What is the average speed of the bus? To simplify, assume that the buses did not stop along the way.
9. In the Arctic ice, in the center of a small flat ice floe with an area of S = 70 m2, there is a polar bear weighing m = 700 kg. In this case, the surface part of the ice floe protrudes above the water surface to a height of h = 10 cm. At what depth under the water was the lower surface of the ice floe? Water density rв = 1000 kg/m3, ice density rл = 900 kg/m3.
10. The wires above the railway supplying electric trains with current are tensioned using the system shown in the figure. It is attached to a pole and consists of cables, blocks with insulators and a square steel weight with side a = 20 cm. The tension force of the thick cable that goes from the outer block to the wire holder is T = 8 kN. What is the height h of the steel weight? The density of steel is rc = 7800 kg/m3. Gravity acceleration g = 10 m/s2.
11. There is a layer of snow with a thickness of h = 70 cm on the ground. The pressure of the snow on the ground (without taking into account atmospheric pressure) is equal to p = 630 Pa. The weather is frosty and the snow is made of air and ice. Determine what percentage of the snow volume is occupied by ice, and what percentage is air. The density of ice is r = 0.9 g/cm3. The free fall acceleration is assumed to be g = 10 m/s2.
