Physics, Chemistry, & Maths Lomonosov RussiaPhysics.Word
problems.8 cells.(Option for writing in MIEM).In all the proposed
objectives, it is desirable to bring the answer except some,
possibly convincing arguments in favor of this answer.1. a).Subway
train arrives at the station "Kirov".What is the half open door
before?b).At the moment there is a train stop push.In what
direction?2. Evaluate the tire pressure of a car and a
bicycle.Allowed to make a mistake 10 times.3. Small Chriqui thrown
from the plane.He falls to the horizontal plate and an elastic
rebound.What is the acceleration of the ball for the first time
after the bounce (for example, when he rose from the plate by 1
cm)?4. Which of the ball more thermal capacity: that is on the
table or that hangs by a thread?(Balls exactly the
same).1979.Mathematics1. How many ways are the number 1979 can be
written as the difference of two squares of natural numbers?2. The
competition involved 10 skaters.The competition is judged three
judges in the following way: each judge in his own distributes
between skaters places (from first to tenth), after which the
winner is the skater with the least amount of seats.What is the
greatest value may take this amount from the winner (winner
only)?3. At the deck of 16 cards, numbered from top to
bottom.Allowed to take some of the top deck, and then removed and
the remaining portion of the deck, without turning, "punch" into
each other.Could it be that after several such operations the cards
are numbered from the bottom up?If yes, for what is the least
number of operations it can happen?Theoretical physics1. Why is
extinguished match if it blow?2. Is it possible to cram 40 people
into a phone booth?(Weight pers. 70 lbs.).3. In the two goals
srelyayut same bullets in different places (see. Fig.).Bullets
merges into the washer.Which of them will spread to the wall
(without friction)?m rectangle?Mathematics oral 7-8 cells.(Game)1.
By placing a circle 15 minuses.Each of the two playing with his
progress can cross out one or two neighboring minus.The player who
last zacherknt minus.2. 15 stones.One divides into two piles, one
(and then alternate with passages) one pile takes himself another
divides again into two .. The player who can not make progress.3.
15 nuts.Each of the two playing with his progress can not take
yourself more than half of all nuts.The player who can not make a
move.Theoretical Physics1. To determine the density of Hg (mercury)
with an accuracy of 1%.For a while all the necessary data many
times a day were broadcast on radio and TV.2. Will there be enough
energy produced by the power plant to evaporate the water passing
through the turbines it?3. Peter 2 times wider Wani, but a 2-fold
lower.One of them weighs more?4. When testing a missile mounted in
the tail of the aircraft to protect against attacks from behind,
was discovered a surprising fact: when starting a shell he turns
around and catches up with the plane.How to explain it.Experimental
Physics1. Why does not take off?
2. The tape from the reel.ThanHis, thevwinderanymore.
3. Why is rotated in the fall and how to make to rotate in the
opposite direction?
Chemistry1. To determine the surface area of the activated
carbon.C= 4 * 103J / (kg *OC) L= 2.3 * 106J / kgC= 1 kcal / (kg
*OC) L= 500 kcal / kg2. What fundamental differences should be
expected in the composition of lunar minerals compared to the
earth.3. How to light the burner without matches at the hardware
store?4th Tournament them.Lomonosov1981.MathematicsGrade 61. From
the results ofAdivided by 2;Ais divided by 2;Ais divided by 12;Ais
divided into 24three true and one false.What is it?
2. To write the numbers 0;1;0;0. In one step is allowed to add
one to any two of them.Can I get that all the numbers become
equal?3. Several boxes together weigh 10 tons.Each of not more than
one ton.How trhtonok certainly enough to take away this
burden?Grade 71. Motorist rode the 4-wheel drive with a spare tire
5000 miles, changing the tires so that they wore out the same
way.How many kilometers passed each tire?2. Uvadratnaya area 100 m
* 100 m square is lined with slabs of 1 m * 1 m four colors.No two
plates of one color are not common side or top.How many plates can
be one of the colors?3. Several boxes together weigh 10 tons.Each
of not more than one ton.How trhtonok certainly enough to take away
this burden?8-9 classes1. Can there be a trapezoid 4-gon with
sidesL= 10, 11, 12, 13,2. Dana repeating decimals with periodN.Can
this period when multiplying the fraction by 2 or its division by
2a) increase;b) decrease?3. 50 at the round table.They give 101
ball, among them 2k+1 black.First 3 gets the ball, the other by 2.
Then the motion of the balls according to the rules: if a person
has a stroke after 3 balls of the same color - the game is over, if
different - reserve the 2 identical, and gives one a different
color neighbor on the right.What is the maximum number of steps can
be in the game ifa)k= 1;b) 1 0), but less than half.The player who
can not make the next move (this happens when the remaining two
nuts).Who wins the game with the right - beginner or the second?5.
The hour hand stands at 12. Two igrayuschihpo queue move her.You
can move the arrow to 2 or 3 hours in advance.The winner is the one
who put it on 11. Who wins the game with the right - the first or
the second?6. Written number 1000000. The first player writes
natralnyh products of two numbers greater than 1, equal to
1,000,000 Second selects any factor, and underneath it says equal
to the product of two factors.And so on. D. In turn.The player who
was unable to make the next move.Who wins the game with the right -
the first or the second?(Option: the winner is the one who could
not make the next move).Home1. 25 matches.Take turns, each
Statement 1, 2 or 3 matches (light version 1 or 2).The winner is
the one who at the end of the game will be an even number of
matches.Who wins: beginner or the second?2.A- natural.The two take
turns playing smash all of the available range (the first time
during the whole set - the number ofA) to natural or long-terms 1
natural factors.In one move, one number is divided by 2 terms or 2
factor.The player who can not make the next move.Who wins the game
if properly?a)A= 10b)A= 11Theoretical Physics6-7 class1. Boy Loew
boating and water photographed beach.When he published the photos,
I could not understand where the image bank, and where his
reflection in the water.Specify the number of possible ways to do
this.
2. Flatland - a flat country.It has only two dimensions.How can
I attach the wheel to the car in Flatland?(Make-axis, as in the
three-dimensional world, it is impossible - she has nowhere to hang
around there).3. In carrying out accurate weighing scales placed in
a hood, from under which evacuated.Why do it?4. Mikhail Lomonosov
issedoval movement of air in mines complex profile.In what
direction will show through the winter and summer in the mine, the
illustrations?
8-9 class1. Boy Serge watched TV and decided to photograph it
interesting image.When he published the photos, I saw them on the
light and dark bands.How Come?What can we say about the device
gate?[task involves, but has been withdrawn.]2. Towards each other
fly two identical dumbbells.They elastically collide.How will fly
dumbbells after the collision is over?Consider the case
ofv1=v2andv1= / =v2.
3. The load falls on a vertical spring.At what height it will
have a top speed?
4. In the foam float hidden two pieces of lead.What sluagh float
will:1) sink:2) protrude from the water;3) to tip over the surface
of the water?foam 15, 2, 2, 1,15 -> 10, 2, 2, 1,10 -> 3, 3,
2, 21 After the exchange of rubles.25 kopecks.Copper Copper
determine what was silver.(Note: in 1983 in the Soviet Union were
in the course of the coin 1 rub. 50 kopecks., 20 kopecks., 15
kopecks., and 10 kopecks. white, conventionally called the silver
and coins 5 kopecks., 3 cop., 2 kopeks. 1 cop. yellow,
conventionally called copper).3. (8 cells.) In 1983 the plane point
and a circle of radius 1. Prove that there is a point on the
circumference, the sum of distances between all of these points is
not less than 1,983 in 1983.4. (9 cl.) Two circles intersect
line.Prove that the angleABC= angleDEM.
Theoretical Physics5-7 cl.1. When will emerge from the bath
faster one and the same volume of water - when it is a man or when
it is not?2. On the popular lecture said that the height of the
Eiffel Tower 302 meters 15 centimeters. "Well, no," thought one of
those present.Why he thought so?3. Archimedes had to lift a heavy
load, and had two arms:He thought - "for us, weightlifters, it's
just right."Which lever he chose?8-9 cl.1. Task 1 for 5-7 cells.2.
Task 2 for 5-7 cells.3. How to determine the weight of the boat,
having a coil of rope?It takes place during the boat ride.4. When
you turn on the speakers in the sequential translational network
open one disables all in parallel - one short izamykanie disables
all.How to include?Experimental Physics (5-9 cells).1. (5-7 kl.)
How to measure the diameter of the thread usual line?2. How to
weigh a book that weighs about 2 times larger than the limit
dynamometer?3. On the inclined plane rolled 3 bubble from the glue
- empty, half-full with water and.Why half-stops before?(8-9 kl.)
Which part of the energy is spent on friction bubble?4. Those
bubbles are.With a weak push the base falls empty, with an average
- complete.Why such a sequence?5. (8-9 kl.) If narisovanna star *
look through the slit, cut with a razor in the paper, the visible
beam 1-2.If you rotate the gap, the rays begin to spin.How
Come?Chemistry1. The ideal syringe needle is lowered into the water
and pull the piston.How does the rate of filling the syringe on how
much pushed the plunger and how quickly he pushed?Explain your
answer.2. Can be a simple substance molecule polar?If not, why not,
if so, give examples.3. As a fairly well-equipped laboratory light
the alcohol lamp, if you forgot your home matches?Come up with more
examples.4. In the last century there was much debate about the
structure of the benzene molecule, which has a composition of
C6H6.Try to draw all of the structural formula, which can describe
the composition of such a molecule.Note that the C atoms can be
connected double and triple bonds.5. Why concentrated hydrochloric
acid fumes in air, and concentrated ammonia solution - no?6. It is
known that solutions of iodine in water and alcohol have a brown
color, and iodine vapor and carbon tetrachloride solution -
purple.What could you explain this fact?7. When bubbling xenon,
methane under a pressure of 1 atm.through water at 2OC precipitate
colorless crystals of Xe * 5.75H2O and CH4* 5.75H2O, which easily
decompose with the release of the gas to under heating.What do you
think, what is the structure of these compounds?8. Can one get two
acid salts of the two acids - salt, two bases - salt, the salts of
the two - base.If the process is possible - Write the reaction - if
not, try to justify their opinions.9. How does the reaction rate on
the concentration and temperature / ceteris paribus /?How could you
explain this relationship?10. Some waste production is oil
containing water.It turns out that this oil burns even better
dry.How can you explain this and how to do it?11. What, in your
opinion, different substances corresponds to the following
formulas: FHH | | | Cl-CH Cl-CC-Br | | | Br FFPlease note that due
to carbon directed corners of a tetrahedron.Draw the molecule
thereof.12. What are the inorganic substances in living organisms
and why are they needed?13. 1.3 liters made of 130 g of a clear
crystalline substance, which is in the process of dissolution
zhidgosti level has not changed, but its temperature is
lowered.After complete dissolution of the substance liquid is
evaporated, with remaining 0.13 g of solid residue.Determine what
kind of fluid and a substance dissolved in it.Competition in
MathematicsTasks1. On the sides of the hexagon was recorded six
numbers, and each vertex - a number equal to the sum of the two
numbers on the allied side.Then, all the numbers on the sides and
top of a single number erased.Is it possible to restore the number
on the top?2. Vertices A, B, C of a triangle are connected to
points A1, B1, C1, which lie on opposite sides (not top).Can the
midpoints of AA1, BB1, CC1lie on the same line?3. Three chess
players A, B and C played a match tournament (everyone with
everyone played the same number of games).Could it be that by the
number of points won first place A, C - last, but by the number of
victories, on the contrary, A finished last, C - the first (for the
victory is awarded one point for a draw - a half-point)?Problem
solving in mathematics competition1. Yes I Am.Assume that a number
of1, a2, a3, a4, a5, a6are recorded on the ribs 6 in order of the
polygon counterclockwise.Then, six-sided polygon vertices will be
recorded (counterclockwise) of the following:A1+ a2, a2+ a3, a3+
a4, a4+ a5, a5+ a6, a6+ a1.We can assume without loss of generality
that the only erased in the top number - is a1+ a2.Then it can be
recovered by the formula:(A2+ a3) - (a3+ a4) + (a4+ a5) - (a5+ a6)
+ (a6+ a1).After removing the parentheses it turns out that all the
numbers except a1and a2, enter the right pane, double, and with
different characters.2. The middle of said segments lie on the
respective average lines of the triangle (as an average parallel to
the base line).They also lie inside the triangle (point A1, B1,
C1can not celebrate the vertices).Since the line can not cross the
three sides of the triangle, the midpoints of the segments and can
not lie on the same line.3. Yes I Am.Place in the match-tournament
determined by the difference between the number of wins and the
number of lesions.Here is a table of the tournament in 8 circles
(as in match-tournament contenders in the 50s and early 60s) that
meets the specifications (in the row corresponding to the player
and the specified number of victories over the other
players):ABCglassesvictory
A128.53
B0484
C237.55
Competition for mathematical gamesTerms Games
1. Honeycomb (1x1 squares) are arranged in a 9x9 square.In the
central box instead of honey - tar.In one move is allowed to make
vertical or horizontal incision (squares on the sides) and eat one
of the cut pieces, in which there is no tar.Play two of the moves
made in turn, loses the one who has to eat tar.Who's winning?(You
can consider the case where the tar is not in the center.)2. On a
strip of size 1x8 in the 4th, 6th and 8th cells are chips.Each move
any of them can be pushed to the left by any number of cells, so as
not to jump over another chip and not get to the occupied
space.Played by two, take turns.The player who can not make a
move.Who's winning?3. The arrow on the clock shows twelve hours.In
one move is allowed to move in a clockwise direction to two or
three hours.The winner is the one who put the arrow on the
eleven.Who will win?4. The cake is in the shape of a regular
pentagon.In one move is allowed to make an incision along the side
or diagonally.The winner is the one whose turn is after the cut a
piece containing cherry, located in the center of the cake.Who will
win?Consider a plywood table on which is painted a regular pentagon
in the center of which stands a vase, and the loser is the one
whose turn after vase falls.5. In a series of written seven
minuses.In one move, you can fix on the plus one or minus two
adjacent.The player who can not make a move.Who will win?Consider
the eight minuses in a row;daisy with eight or nine petals.The
story of what math gamesIn this section, we did not give solutions
of games offered at the competition.Otherwise, you will not be
interesting to play them.But it is useful to discuss what
amathematical game.With conventional games like chess, checkers, or
of them have in common is that there are two players chasing
opposing goals.A difference is that chess would be a fatal blow, if
it turned out that the white, for example, with the right game
always win.A mathematical games just interesting to invent the
correct ways to play.Mathematics, however, are able to prove that
the right way to play there in the ring games.But to find the way
to chess and Go is not yet possible, even using super-computers
(for checkers situation is somewhat different).But those games that
we'll play a lot easier, and so they manage to find out who wins
the game at the right.And how do understand what the right
play?After a sequence of moves in the game alternates.Suppose for
example, I wrote down the right, winning sequence of moves for
me.After all, the enemy is not obliged to follow it!However, if it
is possible to formulate a rule, following which one player always
gets a prize at all possible moves of the second, it is clear that
the outcome of the game is defined (it should be, of course, know
this rule).Mathematics in such cases we say that the first player
hasa winning strategy.It is the search of a winning strategy and
one of the players is a mathematical game.Of course, if the rules
of the game you like, you can it some time to play with his
friends, but after a while you will notice that those wishing to
participate in the game with one of the parties is getting
smaller.And then, when the strategy will be known to all, the game
play will not be none.Let's look at some examples of mathematical
games.Task 1.There are two piles of stones, 8 pieces each.Two
persons play the following game: in one move is allowed to take a
handful of any number of stones.Who can not make a move - lost.Find
a winning strategy for one of the players.Before embarking on the
search strategy, we advise you to play this game with someone else,
or at least one of the two players at once.If you have formed a
definite opinion about who wins in this game, try to formulate a
rule that should guide the player, ie,strategy.After that, read on
and compare their solution with ours.Note that each turn number of
stones in one of the piles decreases.So the game can not last
indefinitely, no matter how the players play.Solution 1.The winner
is the second player.To do this, he must adhere to the following
strategy (called "symmetric strategy"): after each of the first
player moves need to equalize the number of stones in a pile.Why
such a strategy is called symmetric?Not only in this game, but also
in many other positions, some are symmetrical.So, the symmetric
strategy is that each turn a player achieves a symmetrical
position.In this case, the position symmetrical, if the number of
piles of stones in the same way.So check that the symmetric
strategy is the solution of the game and provides a second player
win.The first question is whether the player will always follow
it?To answer this question, think about the position in which the
second player can not equalize the number of stones in a pile?Only
if the stones in a pile ALREADY equally.But this situation to move
the second player can not occur!Indeed, initially have a
symmetrical position.The first player to take a certain number of
stones, and the position becomes asymmetric.Now the second
question: whether the second player wins, sticking symmetric
strategy?As already mentioned, the number of stones in the pile is
constantly decreasing.The course can not be done when there are no
stones.But this position is symmetrical, and there she was after
the move of the second player.Objective 2.There are a handful of
stones 4 to 9, 10, 12 and 18 stones respectively.At its course the
player can take one of a handful of 1 or 2 stones.Who can not make
a move - lost.Determine the outcome of the game (ie, to find a
winning strategy for one of the players).This task is more
difficult to last.However, try to solve it, and then read the
solution given here.Solution 2.The situation in which it is
impossible to make progress in this game is also unique stones ---
all players.It turns out that the first player can make sure that
he took the last stone.A winning strategy for the first player: do
such a course, the number of stones in each pile divisible by 3.Is
it always the first player can follow this rule?Note that on the
first turn possible: take one stone from the heap, in which 10
stones.The second player has to take its course from one of a
handful of 1 or 2 stones.If its course in every pile of stones was
the number divisible by 3, and if the first player follows the
preceding rule, this is true, then before the next move of the
first player in exactly one pile of stones is not the number is
divisible by 3. It is clear that the first player can follow the
preceding rule, taking from this heap remainder of dividing the
number of stones on the 3.So, after the move of the first player
the number of stones in a pile to be divided by 3. But the game
ends when the stones in piles - 0 0 0 0 - iethe first player
wins.Competition in PhysicsAssignmentsLower grades (6-7)1. Why
icicles are usually located only one side of the roof?2. Air and
water are transparent.Why is opaque fog?3. Sergei Archimedes
believes that he is lucky with good weather.More than once he
noticed that when the clouds surrounded the entire horizon, namely
the sun shines on it.Does Serge rights?Upper grades (8-9)1. Sergei
Archimedes believes that he is lucky with good weather.More than
once he noticed that when the clouds surrounded the entire horizon,
namely the sun shines on it.Does Serge rights?2. Wider or narrower
than the moon moonlit path on the water (meaning angular
dimensions)?3. Rats and other rodents some teeth stay sharp all the
time, despite the constant stitching.This is because their teeth
consist of materials with different hardness.Explain how such
self-sharpen your teeth and how they are distributed substances
with different hardness.Competition in AstronomyQuestions1. When
people built the oldest observatory?Why is it needed?2.
Eratosthenes in the 2nd century BC, knowing nothing about other
countries, to measure the size of the Earth.How is it managed?3.
How to change the speed of the satellite, if it starts to slow
down?4. They say: "light and not heat."About what I may say so, and
why?And whether there is other way around?5. Why is the planet in
the sky after the Sun go?6. To the sky wants to make all well
verree great impression, and in fact quite small and
nevzrachnenky?7. What constellations in the sky most ancient?Do you
know why they are so named?8. Why do not we see how stars are
born?Can we see how they die?9. When humanity begins to travel to
other planets that we will be interested in it?10. What are the
similarities between the universe and the cotton candy?Answers to
questions of competition on astronomy1. The answer to this question
depends on what is meant by the observatory.A.If this religious
building, at least to some extent related to astronomy, the oldest
known structure - Stonehenge - the remnants of a giant megalithic
stone structure, built at the turn of the Stone and Bronze Ages
(1900-1600 BC) on the territory of modern England.At Stonehenge
carried out not only ritualistic ceremonies, but also, due to the
special orientation of the individual parts of the structure, was
conducted through calendar days, marks the beginning of the
seasons.B.If you have in mind for the construction of regular
visual observations of stars, it is an observatory of ancient
Babylon (1st half of the 2nd millennium BC).There were widespread
predictions of important events occurring on the basis of the
heavenly signs, and thought: the more precise observations, the
more certain prediction.It appeared in Babylon associated with
astronomical events mathematical time-scale (8th century BC).B.If
we talk about observatories equipped with tools, then we should
talk about ancient Greece, about the age of antiquity.
2. Eratosthenes lived in Alexandria in northern Egypt.And in the
south of Egypt was the city of Siena, which was a wonderful well.He
was vertical and deep, and at noon, the longest day the sun
illuminates its bottom, iewas at its zenith.Eratosthenes measured
figured in one of those days the sun zenith angle (angle to the
vertical, it was 7.2O), imagine the Earth ball, put Siena and
Alexandria on one meridian and ascribe 7.2Omeridian arc distance
between cities.Knowing this distance and the fact that a full
circle is 360O, it is not difficult to find the proportion of the
circumference of the Earth's meridian.Eratosthenes was 46,000
km.Excellent accuracy!3. If the satellite will begin to slow down,
it will spiral closer to the ground.Hence, it has a circular orbit
at an initial vertical velocity component to appear.If braking is
slow enough, then for each new round circular orbit can be
considered, but a smaller radius H. Sincecircular velocity V = (GM
/ (R + H))1/2(R - radius of the Earth, H - height above the earth,
M - mass of the Earth, G - gravitational constant), we see that
Vcrincreases with decreasing H.iewith slow growing and normal
braking, and the tangential component of the velocity, and hence
the full velocity vector.The growth rate is due to the completion
of the kinetic energy of the satellite mv2/ 2 due to the potential
energy mgH.4. "Light, and does not heat" - can be said about any
source that emits in the optical range at low power.On the
contrary, you can say about the powerful sources of electromagnetic
radiation with a peak in the infrared region of the
spectrum.Example - iron.5. The solar system is highly
flattened.Planets very slightly spaced from the central plane in
which they revolve around the sun, all in one direction.This plane
is called the ecliptic.Therefore, the observer on Earth, and it
would seem that the planets and the Sun on the background of
distant stars move in the sky on the same line.6. Solution of this
problem is not given, becausethe collector could not learn what it
had in mind the authors.7. It is difficult to say whether the names
endowed beautiful configuration brightest stars Adam and biblical
patriarchs.Ancient texts mentioning the constellations date from
the second half of the second millennium BCPerhaps the oldest known
- Old Babylonian text (1700 BC).This is the prayer to the starry
gods.Referred to the constellation - participants myths (monthly
myths, according to the lunar cycle).Obviously, the first in this
regard, attracted the attention of circumpolar constellations,
especially the Big Dipper and the North Star in Ursa Minor, because
the rotation of the sky takes about an axis passing near the North
Star.These constellations are the first to receive sustainable
titles (based on mythological), which in most cases are not
associated with modern, inspired later periods (Antiquity, the
Middle Ages, the era of geographical discoveries).8. Stars are born
when compressed gas and dust clouds, with a sufficiently high
degree of compression when warming up in the center is sufficient
to "ignite" thermonuclear reactions.Death of a star occurs when the
star is no longer in a stable condition under normal parameters
(density, size, somehow comparable with solar).In this sense, the
stars in store for several kinds of deaths, according to their
initial weight.When M RCOOH+3R'OH
MR(x) 3M +R '(3Y) +6036x + 453y + 51
7.423.226.72
0.14 mol0.07 mol0.21 mol
We hope that a solution to this proportion did not take long at
any of the aforementioned scholar nor a physicist, and they will
find that MRacid - 46, alcohol - 32, and saponified orthoester of
formic acid - HC (OMe)3.3. Again - the logic does not work
honors.The calculation for ordinary metals A and B gives the
answer, not provided for periodic system:the mass fraction of
hydrogen24 / (a + b + 192 + 216) = 0.0453a + b = 121,8
the mass fraction of oxygen320 / (a + b + 192 + 216) = 0.5176a +
b = 210,3
4. (Atomic mass A and B taken as a and b, respectively).And it
does not happen.Hence, not only the hydrogen contained in water of
crystallisation.5. It is assumed that one of the "metals" - not
metal, but something containing either oxygen or hydrogen (in this
case rectilinear invalid payment).Scrabble recalls that happens
also ammonium, behaves quite "metallic", a physicist calculates it
logically.Therefore, ammonium alum.If a = 18 (ammonium), then b =
210-18 = 192 - is iridium Ir.Formula alum -NH4Ir (SO4)2*
12H2O.Well, those who are afraid of words that are not in the
school curriculum, it would be better not to go to the Olympics
(especially since there is nothing new in the properties of iridium
with respect to aluminum casting is not required).6. We form the
equation for the reaction:CaO + H2O = Ca (OH)2;56 g CaO react with
water to form a 74 g Ca (OH)2,CaO A g react with water to form
(74/56) A g Ca (OH)2Thus, in the solution will contain 1.32 g A Ca
( OH)2Since the weight of the solution (on the basis of the
reaction equation) is equal to the sum of the weight of the
starting materials and (A + B) r, starting material is completely
soluble (no precipitate) have= 132 * A / (A + B) (%).7. We form the
equation for the reaction:CuO + H2= Cu + H2O;1 mol of CuO - 1 mol
Cu,80 g / mol - 64 g / mol,80 g of CuO give 64 g of Cu.
Hence, the mass of the sample taken after passing over it
hydrogen decreases in 80/64 = 1.25 times.8. Methods color celluloid
based on his restoration various nitrogen-containing compounds.On
this and many other interesting things in organic synthesis is well
written in the book Shpaustsusa "Journey into the world of organic
chemistry."
Tasks1. (7-11) on the plane is a square, and invisible ink
applied point.Man in special glasses sees the point.If you draw a
straight line, he said, is whether a point on the line, and if not
lies, he says, on which side of this line is invisible point.What
is the minimum number of lines necessary to carry out, to see
whether the invisible point lies inside a square (not outside and
not on the border)?2. (7-9) Is there a 100 natural numbers such
that their sum is equal to their least common multiple?3. (7-9) The
rectangle ABCD (AB = a, BC = b) It so happened that turned the
pentagon area S (C laid down in A).Prove that S 2NO2NO + O2->
2NO2In the absorption solution CO32-:1) H2O + NO + NO2+ 2CO32-->
2HCO3-+ 2NO2-2) 4NO2+ O2+ 2H2O + 4CO32--> 4HCO3-+ 4NO3-Case 2)
is impossible.Hence, the gas mixture before absorption - NO,
NO2and, possibly, N2.After absorption - N2+ NO.Suppose, in terms of
1 mole of a mixture (0.5 mole of N2O and 0.5 mol2) to get the x mol
NO2.We have:Number N: 2x + y + 2z = 1Number O: 3x + y = 1The volume
ratio of y + z = 5/8Hence, y =11/20, x =3/20, z
=3/40.Reacted1/2-3/40=7/40mol N2of 0.5.Hence, reacted 70% N2.7. You
can use any reasonable method of analytical determination of H2,
Zn2+and Fe2+.For H2- determination of redox-potential of water for
Zn2+sulfide method for Fe2+- [Fe (salt)]3-.The most promising
method is based on detecting the presence of lubricants, for
example, using adsorbents or analyzing a surface layer of water on
the presence of organic impurities.
Competition in Mathematics
TasksGrades 7-91. From the hot tub faucet filled in 23 minutes,
from a cold - 17 minutes.Mary opened the first hot tap.How many
minutes it should open the cold, so that when filling the bath with
hot water nalilos 1.5 times more than the cold?2. The length of the
height of the rectangular trapezoid ABCD AB equal to the sum of the
lengths of the bases AD and BC.In what respect bisector of angle
ABC divides side CD.3. At each kilometer of highway between the
villages of Elkin and Palkino should post a sign on one side of
which it is written, how many kilometers to Elkin, on the other -
to Palkino.Boris noticed that each column is equal to the sum of
all digits of 13. What is the distance from Elkin to
Palkino?10-111. Any more numbers among all the numbers from 100 to
999: those in which the average number of both more extreme, or
those in which the average figure is less than both the extreme?2.
Circle held several (finitely many) different chords so that each
of them passes through the middle of any other of the studies
chords.Prove that all these chords is the diameter of the circle.3.
Is there a convex polyhedron has 12 ribs, which are, respectively,
12 and parallel to the diagonals of the faces of a cube?Additional
tasks (for all classes)1. Find the sum of angles of MAN, MBN, MCN,
MDN and MEN, drawn on graph paper as shown.2. 3. * -ABCDE4. | | | |
| |5. * - * - * - * - * - *6. | | | | | |7. * - * - * - * - * - *8.
| | | | | |9. * - * - * - * - * - *10. | | | | | |11. * - * - * - *
- * - *12. | | | | | |13. MN - * - * - * - *14.
15. Dan paper circle.Can I use the scissors to cut it into
several parts that make up the square of the same area?(Cut
permitted under direct and arcs).16. In the bazaar selling fish,
big and small.Today, three large and one small stand together as
much as three large and one small yesterday.Is it possible to find
out from these data that the more expensive: one large and two
small today, or five little yesterday.17. Peter cut a rectangular
piece of paper in a straight line.Then he cut in a straight line
from one of the resulting pieces.Then he did the same with one of
the three pieces of the resulting etc.Prove that, after a
sufficient number of cuts can be selected among the 100 pieces of
the resulting polygons with the same number of vertices (e.g.,
triangles 100 or quadrangles 100, etc.).Problem solving in
mathematics competitionGrades 7-91.To the hot water in the bath
turned out 1.5 times more than the cold, cold tap must fill 2/5
baths and cold - 3/5 (to find out, it was possible to identify the
volume of the bathroom for 1, the amount of cold water for x; if x
+ 1,5x = 1, where x = 2/5).But if the hot tap must be open all
(2/3) * 23 = 69/5 minutes, and the cold - for (2/5) * 17 = 34/5
minutes.Hence, the need to open the cold tap over (69/5) - (34/5) =
35/5 = 7 minutes.2.We prove that the bisector of angle ABC chord
divides DC half.Let O - midpoint of the side DC.Draw through the
points A and O line, let M - the point of intersection of this line
with the line BC.Then triangles AOD and MOC will be equal to the
second feature (DO = OC by hypothesis,/AOD =/COM as vertical;/=
ADO/OCM, as the lines AD and BC are parallel).This means that AO =
OM, t. E. The triangle ABM - isosceles.But bisector at the vertex
of an isosceles triangle is the median and the height from which
the bisector of angle ABC passes through the middle of AM, that is,
the point O, and then divides DC into equal parts (ie. A. O -
mid-DC).
3.Let the distance from Elkin to Palkino n kilometers (by
hypothesis, n-integer).We number the columns from Elkin to Palkino
order.Consider the 9th column, t. E. A pillar, spaced from Elkin 9
kilometers (clearly, n>10).Then, one side is written 9, and on
the other: n-9.The next column is written on one side 10 and the
other: n-10.If n does not end at 9, then the n-9 would not have
ended with at 0, and hence the amount of digits of n-9 and n-10
would be equal, but then on the 9th column sum of the digits would
be 8 more than on the 10th, which is impossible.Hence, n ends at
9.If n> 49, then the sum of the digits in the 49th column is
greater 13. This means that for n is the only such features: 19,
29, 39, 49.If n = 19, then on the 9th column sum of the digits is
equal to 9 + 1 + 0 = 10 - a contradiction.If n = 29, then on the
9th column sum of the digits is equal to 9 + 2 + 0 = 11 - a
contradiction.If n = 39, then on the 9th column sum of the digits
is equal to 9 + 3 + 0 = 12 - a contradiction.There is only one
possibility: n = 49.It is easy to check that in this case, the sum
of all columns of numbers will be equal to 13 (it is sufficient to
check only for 9, 19, 29, 39, and 49 poles, think why).10-111.Note
that if a number greater than the average number of both extremes,
then the number of 999-a figure less than the average of both
extremes.Therefore, some of the numbers from 100 to 999-100 = 899
the same amount of numbers with the highest average figure and the
lowest average number (which can be divided into pairs).But there
are still a number from 900 to 999. It is clear that none of them
are numbers with the greatest average number, but there is the
lowest, such as 901. Therefore, greater numbers of those who have
less than the average number of both extremes.2.Note that the
smaller the distance from the center O to the circumference of the
chord, the greater the chord length.Since the final number of
chords, among them is the smallest in length, for example, AB.By
hypothesis, it passes through the midpoint of chord K of some
other, say, CD.If the point of intersection of AB and CD is not
well and the middle of AB, the distance from point O to CD will
obviously be greater than the distance from O to AB (t. To. OK will
be greater than the length of the perpendicular from the point O to
AB) therefore, the chord has a smaller CD than AB, length -
contradiction.So, CD goes through the middle of AB, where the
perpendicular from the point O on these chords are the same.This is
possible only if the AB and CD are the same, or if the AB and CD
intersect at the center.The first is impossible, then AB and CD -
diameters.Likewise, it is proved that all the other chords pass
through O.
3.The answer is: there is.Example.Construct a cube whose edges
are twice more than the initial, middle note and its faces.It is
clear that one can construct an octahedron with vertices at the
marked points, with all its edges are equal halves of the diagonals
of the cube faces constructed.Thus, this is the desired
octahedron.Additional tasks (for all classes)1.Note that/=
MAN/RES,/= MBN/QER,/MCN =/PEQ,/MDN =/NEP.Hence we see immediately
that/MAN +/+ MBN/MCN +/+ MDN/MEN =/MES = 45o
2.Let us assume that this can be done.Then the edge of each
piece is made up of lines and arcs of circles.For each arc define
its "angular measure", equal to the ratio of length to the radius
of the arc, with taken with a positive sign if the radius vector
drawn from the center of the circle, sticks "inside" our paper
circles and negative if it sticks in outside.Now we assign each
piece of the amount of angular measures its arcs.It is easy to see
that if we "sostykum" together two pieces A and B, the sum of the
angles measures piece A + B is equal to (the sum of the angular
piece measures A) + (sum of the angles measures piece B).This is
because when two arcs are glued, the amount of angular steps equal
to 0.Hence we see that the sum of the angles measures of the arcs
on the circle should ovpadat with the sum of the angular measures
of the arcs on the boundary of the square.But this is not the case
for the circle is equal to the number 2, and for the square -
0.3.Let 'fish prices ": today the big fish worth bC, and small
mC.Yesterday most cost bv, and the little - mv.Then from the
conditions of the problem we have two equations3bCm +C= 5bv, 2bCm
+C3b =vm +v.Hence, we obtain:5mv= (2bCm +C-3bv) 10b 5 =C+ 5mC-3
(3bCm +C) = bC2m +C.That is five little yesterday cost as much as
one large and two small today.4.Note that when Petya dissections
obtained only convex polygons.In addition, we note that the cutting
convex n-gon in a straight line, we get da new polygon with n1and
n2and n vertices, respectively1+ n23 * 99 + 4 * 99 + 5 ((3 * 99 +
1) -99-99)(Ie. A. Of the numbers n1, ... is at most equal to 99 3
to 99 4 equal, and the remaining number is not less than 5).Then12
+ 4 99 *>99 * 7 + 99 * 5 + 5therefore 4>5 - a
contradiction.So, after 3 * 99 + 1 Petya's sections there are 100
polygons with the same number of vertices.
Competition in PhysicsTasks1. From the tower height H jumping
stunt tied cord.The length and rigidity of the cord are chosen so
that stunt stops near the ground.Made some hesitation, stuntman
hangs above the earth at an altitude of L. Find the maximum speed
that the stunt was during the jump.2. Water is poured into the
bottle to the point where the bottle begins to narrow, and the top
was poured kerosene (kerosene lighter than water and not miscible
with it).The bottle is then shaken several times and put in
place.How to change the pressure on the bottom of the bottle?3.
Ball from pind-pong table falls from the Ostankino Tower and
elastically strikes the ground.Determine the acceleration of the
ball immediately after take-off.Qualitatively draw a graph of
acceleration versus time.4. Narrow plate of length L lies on a
rough horizontal table.Plate starts to rotate at a constant low
speed by applying a force (finger) on the side at one end of the
plate.Determine where the point around which the plate.5. Consider
the following model of a clock.The light beam is successively
reflected by two mirrors arranged parallel at a distance L. Each
time the beam is reflected from the mirror 2, the electronic
circuit for increasing the clock 2L / c, c - velocity of light.Show
that if these clocks give some speed v in a direction parallel to
the plane of the mirror, the speed of the clock is reduced, and
find how many times (this is - the famous "relativistic reduction
of the time").(Reminder: the speed of light in any frame of
reference is the same and equal to c = 300,000 km / sec) .6. As you
know, on a moonlit night on the water forms the light path (where
should be a reflection of the moon).Explain (qualitatively) its
shape.As it depends on the width of the wave heights (or ripples)
on the water?7. Before you caliper.Determine (with his help) what
color - red or green corresponds to the shorter wavelengths of
light.Note:one of the first who vysskazala the idea that the color
is determined by the wavelength was Euler.He believed that the red
color corresponds to a smaller wavelength than the green.Come up
with a simple way to check this.Was Euler rights?Problem solving
competition in physics1.The speed stunt will be a maximum at a time
when the maximum of its kinetic energy.The kinetic energy at any
given time is equal to the work committed over stuntman external
forces at a given moment.External forces that should be considered
in this problem are the force of gravity and the force of the
elastic cord.Let the cord length is l, stiffness k, mass stunt m,
the gravitational acceleration g.If we denote by x the distance
from the top of the tower, which is flying stunt to this point in
time, the work force of gravity at this point is mgx, and the work
of an elastic spring -kx2/ 2.Work elastic force is negative, since
this force is directed against the movement stunt.We obtain the
expression for the kinetic energy stuntmv2/ 2 = mgx-k (xl)2/ 2 = -
(k / 2) (x - ((m / k) g + l))2+ (m2g2/ 2k) + mglThe value of this
expression is maximal and equal to (m2g2/ 2k) + mgl, when (x - ((m
/ k) g + l)) = 0, i.e.m (vMax)2/ 2 = (m2g2/ 2k) + mgl.From
WherevMax= ((m / k) G2+ 2GL)1/2.It remains unknown to us here to
express the magnitude of the problem through data.To do this, we
rewrite the conditions of the problem in the form of equations
containing these unknown quantities.What stunt stops at the ground,
means that the work of the external forces is equal to 0 when the
tensile cord still Hl, ie0 = - (k (Hl)2/ 2) + mgH.That stunt
eventually freezes at a height L from the ground, means that the
elastic force of the pinch adjustment equal to the force of
gravity, i.e.k (HLl) = mg.For computational convenience we denote u
= k / m.Then our terms and conditions can be rewritten as two
equationsu (Hl)2= gH, u (HLl) = gwith respect to two unknown l and
u.Solving this system and substituting the solution into the
expression for the maximum speed, we get the answer:vMax= ((G2/ u)
+ 2GL)1/2= G (H + L-(2H (L (H / 2)))1/2).2.Choose a column of
liquid from the surface to the bottom of the bottle.Shaking up to a
column made up of clean water column and the column of kerosene
over the water column.Pressure on the bottom after shaking changes
so as to change the weight of the selected column, which at the
same size will now consist of water and kerosene.So, find out the
column will be easier or harder.First consider that a very narrow
neck.Then the kerosene fraction in a mixture with water is very
small, and the weight of the column of our practically become equal
to the weight of the column of pure water.This column will be so
severe initial composite column as the column weight of kerosene
less than the weight of the column of water, which replaced it in
the neck after shaking.Hence, the bottom pressure will
increase.Then it remains to note that the column weight of a
mixture of kerosene and water can be equal to the weight of the
composite column to shake only when the width of the neck of the
bottle becomes equal width.3.The ball is very light, so you need to
take into account air resistance to its motion as well as it is
necessary for the fall in the air a feather light.As the air
resistance increases with increasing speed of the object in the
air, then falling with a high tower, the ball eventually attains a
speed at which the air resistance force becomes equal to the
gravity of the earth ball - its weight.After that, the speed of the
ball stops changing, because the sum of forces acting on it is
equal to 0. With this speed and it will reach the ground.The ball
is very elastic, therefore, after hitting the ground, the magnitude
of its velocity will not change much, but the speed will be
directed up now.The force of air resistance, remains the same in
size, that is equal to the weight, will be directed down now.At
this point the value of the sum of the forces applied to the ball,
will be twice the strength of his attraction Earth, and thus the
acceleration of the ball is twice the gravitational acceleration
g.4.We assume that a moving plate portion of a fixed length s of
the friction force acting ks, where k - coefficient of friction
which is independent of the position on the plate portion, that is
the same throughout its length and is independent of the velocity
plot.Then, in order to maintain uniform motion applied finger force
F must uravnoveshivapt total force of friction:F = kL.That the
movement is a uniform rotation around a fixed center means that no
external force with respect to this center is 0. Let l be the
distance from the pin to the center of rotation.Then the value of
the moment of the force F is equal to Fl = kLl.We calculate the
friction torque acting, for example, on a segment from the center
to the end plate, which acts on the finger.Coordinates of a point
on this segment can be regarded as the distance from it to the
center of rotation.We divide the interval into N small plots points
with coordinates x0, x1, ..., xN(in this case it turns out that x0=
0 and xN= L) and add the value of the friction torque acting on
these sites .The magnitude of the frictional force applied to the
portion of xito xi + 1, k is equal to (xi + 1-xi).Distance from the
center of rotation to the point of application of this force can be
approximately regarded as the distance to the middle of the plot
(xi + 1+ xi) / 2.Then the value of the friction force at the moment
of this portion is equal to k (xi + 1-xi) (xi + 1+ xi) / 2.Adding
points in all areas, we obtain((1/2) k (xN)2- (1/2) k (xN-1)2) +
... ((1/2) k (x3)2- (1/2) k (x2)2) + ((1/2) k (x2)2- (1/2) k (x1)2)
+ ((1/2) k (x1)2- (1/2 ) k (x0)2) == (KL2/ 2) - (k02/ 2) = kl2/
2.Likewise we find that the magnitude of the friction torque acting
on the segment on the other side of the center is equal tok (Ll)2/
2.Given the direction of the forces, we write the condition for the
vanishing of the sum of the moments of the external forces0 = Fl -
(kl2/ 2) - (k (Ll)2/ 2) = kLl - (kl2/ 2) - (k (Ll)2/ 2).From this
equation we find that the distance from the finger to the center of
rotationl = (1- (21/2/ 2)) L5.Let us find the timebetween two
successive reflections of the beam from the mirror to the observer
2 K, which sees our model moves with velocity v parallel to the
plane mirrors.To do this, the length of the path traveled by the
beam between the reflections, divided by the speed of its
movement.The speed of the beam in the reference frame K (as in any
other frame of reference) is equal to c = 300,000 km / sec.It
remains to find the path length (see. Figure), which is equal to
A1B + BA2= 2A1B.You can write the equationA1B2= A1C2+ BC2.where BC
= L, A1,C = A1A2/ 2.Express the members of this equation in terms
of the data of the problem and the required time.A1A2- is the
distance at which moves our model in the time between two
reflections, A1,A2= v.On the other hand, A1B + BA2- is the distance
that light passed during the time between two reflections, A1B +
BA2= c, and hence, A1with B =/ 2.Substituting these expressions in
our equation, we obtain the equation for:(C/ 2)2+ (v/ 2)2= L2,from
where= 2L / (C (1- (v / C)2)1/2).Thus, an observer noted K time
between two reflections='/ (1- (v / C)2)1/2.where'= 2L / C - the
time between these two events marked by an electronic circuit, for
which our model is stationary.This means that the speed of the
clock, driven by an electronic circuit, after imparting a velocity
model v, will be less than the clock speed observer ostavshekosya
stand still in/'= 1 / (1- (v / c)2)1/2times.
6.If the water surface was mirror smooth, no track, we would not
have seen.Water would create a clear reflection of the moon,
visible in one specific point on the surface of the mirror.If we
see a track, this means that the surface waves act as small mirrors
which are due to the different tilt send us moon light from points
that are away from the place of its reflection mirror
smooth.Variation in position of the points on which falls to us
moonlight, i.e. the track width, the greater the slope of a strong
possibility of small mirrors, that is, the higher the water
waves.If viewed as a wave of convex reflecting surface is incident
to the beam on the surface of its reflection from all points on the
surface of a cone, which is wider, the more convex surface, or
higher than the wave (in the case where the surface is flat, i.e.
zero wave height This cone is compressed into a reflected
beam).Adjudged whether reflected from the water space of the light
to the observer, i.e. whether it is in place within the lunar
tracks depends on whether the observer is located within the cone
of rays created at that location with a maximum wave height.If we
consider the maximal cones of reflected rays at different locations
of the water within which enters the observer standing on the
shore, it turns out that the path widens when approaching the
observer.7.All that we need in this experiment from the calipers -
a narrow slit that can be made between his wings.To zaintereovatsya
this gap, we must remember that the intensity of the received light
waves passing through the slit depends on the angle distribution
after the gap.For waves of a given frequency, this dependence
appears in the form of alternating-receiver on the screen (or the
retina of the observer) of light and dark bands parallel to the
slit.This phenomenon is called diffraction.The fact that the waves
arriving at a given point from different angles slit may amplify or
attenuate each other depending on the difference of the lengths of
the paths that they passed.If the wavelength is much greater than
the width of the gap, the waves coming from different points of
cracks practically do not differ from each other and reinforce each
other znachittolko regardless of the propagation angle.For an
observer it looks as one broad band light.Dependence of the
intensity on the angle becomes stronger, and therefore the width of
the strips and the distance between them is smaller, the shorter
the wavelength.Thus, if the distance between the strips of one
color is less than the other, the colors have a wavelength of
shorter length.And if you look at the bands of red and green colors
of the light transmitted through the leaf calipers, it becomes
clear that Euler was wrong.
Competition in ChemistryTasks1. Masa piece of chalk and 50 g of
calcined residue was dissolved in 1 liter of water.Through the
solution at room temperature was omitted chloro obtained by
reacting 250 ml 20% HCl solution (= 1.1 g / cm3) with an excess of
potassium permanganate?a) Which substances contained in the
solution?b) Evaluate the content in it the main reaction product.c)
Determine the mass fraction of calcium carbonate in the original
piece of chalk, considering that chlorine has reacted completely.2.
The combustion oxygen unknown substance formed 3.6ml water and 2.24
liters of nitrogen (standard conditions).Relative vapor density of
the substance to hydrogen is 16. Determine the molecular formula of
the substance.What properties it possesses?Where to use?3.
Determine how many grams of a 10% solution of sulfur oxide (6) into
pure sulfuric acid and 60% solution of sulfuric acid is necessary
to prepare 480 g of a 90% solution of acid.4. For a complete
combustion of the organic substances required twice less oxygen
than for complete combustion of the next member of the homologous
series.It can be any compound.5. Four chemical element designated
by the letters A, B, C, D. Pick such reactions, which can be
encrypted as follows:a) 2CA + A2= 2CA2b) D4C3+ 6B2A = 3CB4+ 2D2A3a)
CB2A2+ CB4A = C2B4A2+ B2A6. Metal plate weighing 50 g after staying
in a solution of hydrochloric acid decreased in weight by 1.68%,
with 0.336 liters of gas stood out.From what can be made of metal
plate?7. How to change the rate of reaction between the molecules
of nitric oxide (2), and oxygen, if a sealed vessel filled with a
mixture of these gases in a ratio of 2: 1, the pressure increase is
3 times?Problem solving competition in Chemistry1.Assuming that the
chalk is pure CaCO3, we have 0.5 moles of calcium carbonate.For
equationCaCO3= CaO + CO2amount of CaO and 0.5 mol amount.After the
process ofCaO + H2O = Ca (OH)2obtain 0.5 g of 37 mol or lime.In 250
ml of 20% HCl solution sorderzhitsya 250 * 1.1 * 55 = 0.2 g of
hydrogen chloride, which is 55 / 36.5 = 1.507 mol.For
equation2KMnO4+ 16HCl = 2MnCl2+ 5Cl2+ 2KCl + 18H2Oobtain 1.507 *
5/16 = 0.471 mol or 17.19 g of chlorine.a) reaction of lime with
chlorine at room temperature proceeds according to the equationCa
(OH)2+ Cl2= CaOCl2+ H2OIt is evident that the lack of chlorine, and
may be present in solution in addition to the main product
CaOCl2and Ca (OH) Cl, CaCl2, Ca (OH) OCl, Ca (OCl)2, Ca (OH)2.b)
The maximum content CaOCl2solution is(0.471 mole * 127 (g / mol)) /
(100 g + 37 g + 17.9 g) = 0.0567 or 5.67%c) If reaction (1) passed
quantitatively, the lime, and accordingly, the oxide and calcium
carbonate was 0.471 mole.In this case, the mass CaCO3: m = 0,471 *
100 mole (g / mol) = 47.1 Then, the(CaCO3) = 47.1 / 50 =
0.942.2.The substance may comprise hydrogen, nitrogen,
kislolrod.However, the relative molecular mass of the substance 16
* 2 = 32 limits the number of oxygen atoms to one or eliminates
it.By the condition of nitrogen formed 2.24 liters, which is 0.1
mol.Hydrogen 0.4 mole water;0.1 mol of N2were 0.2 moles of nitrogen
atoms.The ratio of nitrogen and hydrogen in the compound is 0.2:
0.4 = 1: 2.Simple formula - NH2.MR(NH2) = 16;MR(N2H4) = 32 to
coincide with the defined problem statement.From the formula of the
starting material H2N-NH2.This hydrazine at room temperature it is
a colorless liquid that is used as one of the propellant
components.3.10 g of 10% oleum contains 10 g of SO3and 100 g of 90%
sulfuric acid.If the count content of SO3, then 90 g of H2SO4is
contained in all (18/98) * 90 = 16.53 g of water, the rest
100-16,53 = 83.47 g oleum is SO3.Similarly, in the case of 60%
sulfuric acid: 40 + 60 * (18/98) = 11,02 + 51,02 = 40 g of water
and 48.98 g of SO3.In 480 g of 90% sulfuric acid contains 0.9 * 480
= 432 g of P2SO4or (80/98) * 432 = 352.65 g SO3and (480-352,65) =
127.35 g Water .Suppose we must take x g oleum, then a 60% acid -
(480-x) In the oleum contains (x / 100) * 83.47 g SO3, and acid
((480-x) / 100) * 48.98 g of SO3, for a total of 352.65 g of
SO3.Hence:(X / 100) + 83.47 * ((480-x) / 100) * 48.98 =
352.65.Solving the equation, we get x = 340,8 g of oleum, and a 60%
acid, respectively 480-340,8 = 139.2 g4.For the oxidation of the
group -CH2- (homologous difference) need three oxygen
atoms.Therefore, the starting material must also be oxidized by
three oxygen atoms, then the next member of the homologous series
will be oxidized by six oxygen atoms.CxHy+ (x + (y / 2)) O2= xCO2+
(y / 2) H2O,Cx 1 +H2 + y+ (x + (y / 2) +1,5) O2= (x + 1) CO2+ ((y /
2) +1) H2OExamples of substances oxidized by three oxygen atoms
(the first term gamologicheskih series):CH3OH - methanol,HOOCCH
(OH) COOH - gidroksimalonovaya acid,CH2N2- diazomethane,HO-CH2-COOH
- hydroxyacetic acid,HO-C ( COOH)3- trikarboksimetanol.5.A -
oxygen, B - is hydrogen, C - carbon, D - aluminum.6.Reaction
equation: M + nHCl MCl =n+ (n / 2) H2.The weight loss was 50 *
0.0168 = 0.84 g of metal x is isolated (n / 2) * 22.4 L H2, and
0.84 g - 0.336 H L2, x = (0,84 * n * 22.4) / (2 * 0,336) = 28n ofIf
a monovalent metal, m. E. N = 1, it is the relative atomic weight
28 (silicon).However, silicon does not dissolve in hydrochloric
acid and may be monovalent.For n = 2 the relative atomic mass is 56
tons. E. Metal - iron.You can check the options with n = 3 and 4,
and make sure that the correct answer - only iron.7.Reaction
equation: 2NO + O2= 2NO2.Denote the concentration of NO letter a,
O2- b, then to the pressure change speed reuktsii V1= ka2b.With
increasing pressure, the concentration increased by 3 times, skrost
is V2= k (3a)23b = 27kab;increasing the speed V2/ V1= 27
times.Competition in MathematicsTasksGrades 7-9 (or younger).1. By
the banks of the Nile came a company of six people: three Bedouin,
each with his wife.At the shore is a boat with oars, which stands
only two people.Bedouin can not allow his wife was without him in
the company of other men.Can the whole company to cross to the
other side?See comment on the condition of the jury of this
problem.2. In triangle ABC, angle A is 120 , the point D lies on
the bisector of the angle A, and AD = AB + AC.Prove that the
triangle DBC - equilateral.3. Written in a circle 7 natural
numbers.It is known that in each pair of adjacent numbers divided
by one another.Prove that there exists a pair of adjacent numbers
and not with the same property.10-11.1. Number of 1/42 spread in
infinite decimal fraction.Then crossed 1997 th digit after the
decimal point, and all the figures standing to the right of
strikeout numbers have shifted to the left one.How many more: new
or original?2. Is it possible to cut an equilateral triangle into
five distinct isosceles triangles.3. Antique dealer bought 99
identical in form of ancient coins.He was told that exactly one of
the coins - a fake - this is easier (and real weigh the same).How
using pan scales without weights for weighing 7 to identify the
counterfeit coin, if an art dealer does not solve any coin weigh
more than two times?Problem solving in mathematics competition1.
(7-9).The jury's comment on the condition of the problem 1 for
grades 7-9The jury believed that the wife can not "be in the
company of another Bedouin without her husband" even infinitesimal
time.That is the situation when a woman swims in a boat to the
shore, where there is her husband, but there is another Bedouin,
and without leaving the boat sails away, was, according to the
jury, contrary to the statement of the problem.However
inadvertently, this condition explicitly in the final text of the
problem has not got, and when checking for his failure to score was
not reduced.Any other decision of the jury students could rightly
challenge, since the concept of "being with" uniquely defines the
above situation.Back to the condition of the problem.
We introduce the notation: B1, B2 and B3 - Bedouins, x1, x2, x3
- their wives,[}- boat.Bank of the NileNileThe other bank of the
Nile
B1 + x1, x2 + B2, B3 + R3[}There is no one
B1, B2, B3 + R3[X1, x2}There is no one
B1, B2, B3 + R3{X1]R2
B1, B2, B3[X1, R3}R2
B1, B2, B3{X1]R2, R3
B1 + x1[B2, B3}R2, R3
B1 + x1{B2 + x2]B3 + R3
X1, x2[B1, B2}B3 + R3
X1, x2{R3]B1, B2, B3
R2[X1, R3}B1, B2, B3
R2{R3]B1 + x1, B2, B3
There is no one[X2, x3}B1 + x1, B2, B3
There is no one[No one}B1 + x1, x2 + B2, B3 + R3
2. (7-9).Note on the segment AD point K such that AK = AB.Then
by hypothesis KD = AC.ABK is equilateral triangle, since it has two
equal sides and one of the angles of 60 .Therefore, the triangles
ABC and KBD are equal on both sides (AB = KB and KD = BC) and the
angle between them (the angle BAC = angle BKD = 120 ).So the sun =
BD and angle DBK = angle CBA.Adding to both sides of the last
equality angle KBC, get the angle DBC = angle KBA.Thus, triangle
DBC - isosceles 60 at the vertex, so equilateral.13. (7-9).Connect
pairs of adjacent numbers so that the arrow came from the fold
(so-called number that is divisible by the divisor) to the splitter
(if neighboring numbers are equal, then the direction of the arrow
is chosen arbitrarily).The total number of arrows is odd (7), so
they can not alternate directions.Hence, any two adjacent arrows in
the same direction: x -> y -> z.This means that x is divided
by y, and y is divisible by z.That is, x = ay and y = bz (a and b -
integers, a = x: y and b = y: z).Therefore, x = ay = a (bz) =
abz.It follows that x divided by z (potomuchto x: z = (abz): z =
abz: z = ab (z: z) = ab * 1 = ab).1. (10-11).Number 1/42 can be
represented in the form of a periodic decimal fraction.The period
begins with two decimal places (two) and consists of 6 numbers:
238095 (1/42 = 0.0238095238 ...).Set it is possible, for example,
by dividing by 42 one "column".Since 1997, when divided by 6 gives
the remainder 5, 1997, I figure recorded the same number as the
fifth - zero, and the next - a nine.Hence, the new number is
greater.Infinite periodic decimals are indicated as follows: 1/42 =
0.0 (238095).2. (10-11).You can.Let ABC - this equilateral
triangle.Note on the AC side point D, so that AD = DC.Note on side
BC point E, so that BE = CE.Note on the segment BD point F, so that
BF = AF.Note on the segment AF point G, so that AG = GF.DEC
(equilateral), DGF (equilateral), BED (BE = ED), BFA (BF = FA), AGD
(AG = AD) - the required equilateral triangles.The possibility of
constructing the conditions specified are distinct and equilateral
triangles are almost obvious from the figure, a rigorous proof of
these assertions is long and uninteresting and is not included
here.There are other options for cutting.3. (10-11).First put on
the two scales to 13 coins, then (if the balance is in equilibrium)
uberm them and put on 11 more of who took no, then 9,7,5,3 and 1 to
as long as one of the bowls do not outweigh .If this does not
happen, then after the seventh weighing (when the scales will be
only one coin) will be only one coin, which was not involved in
weighing.It is false.If some kind of weighing bowl outweighed, then
the false coin is in the other side.The total number of coins in
this bowl denote 2k + 1 (every time we we place on one side of an
odd number of coins), and we've used 7-k weighings, with each coin
weighed no more than once.Therefore, it remains to find a false
coin in a group of (2k + 1) ones in k weighings, each coin weighing
not more than once.This can be done to break all the coins in the
group, except one, divided into pairs k and the weight of the coins
successively compare each pair.If some weighing balance is
violated, then the lighter coin is counterfeit.Otherwise, false
coin - the remaining single.
Competition in Physics
On a sheet with the task was given this explanation:After the
task number in parentheses indicates how classes, this problem is
recommended.To solve problems not of their class is allowed.Pay
attention to the quality of physical explanation: the work, which
is good, with explanations resolved two or three tasks will be
highly appreciated - higher than the work in which there are fuzzy
reasoning on many tasks.
Assignments1. (7-9) Compares two experiences.1. On the table is
laid wooden ruler thickness of 1-2 mm and a length of 50 cm so that
slightly more than half the line lay on the table, and the rest
hung.If you click on the overhanging part, the line will bend
easily.2. Line also is laid, but the portion of the line that is on
the table is laid a sheet of paper.Now, if you quickly press the
overhanging part, ruler became like heavier.Explain why this is so
and describe the phenomena that occur when pressing on the line.2.
(7-9) It is possible to imagine such a possibility, that in some
countries the duration of the day was defined as the time between
adjacent sunrise.What time of year in this country, "day" was the
longest, and in some - the shortest.3. (7-9) Why eventually radius
bend of the river slowly increases?4. (9-11) a) Try to estimate the
order of magnitude of the coefficient of friction end of your pen
on the desk or paper as follows: first put pen almost vertically,
propped on her finger, and then tilt the handle up until it begins
to slip.b ) Estimate the order of work on writing your work at the
competition in physics.There is a view only the mechanical work on
the printing on the paper;costs of other forms of energy are not
counted.Note: If the task requires something to evaluate in order
of magnitude, the greater accuracy is required;need just an order
of magnitude when the error is 10 times is quite acceptable.5.
(9-11) Contemporary Galileo, professor at the University of Padua
Sanktorius invented the first thermometer, which was a glass ball
filled with air and the glass tube is connected to the
atmosphere.The tube is placed in a drop of water, which overlaps
message cavity vessel with the external environment.With this
device Sanktorius are diagnosed his patients (he was a doctor).The
patient took the ball in his mouth, and a minute later on the
Status of water drops Sanktorius determines what the patient
fever.Depend on whether such a thermometer readings of the weather,
in particular whether they will be different in clear weather and
rain?6. (10-11) Try to invent a device that creates a magnified
image of the object and it does not use a curved surface.7. (10-11)
Try to come up with such an experience, in which the charge is
moved, and the magnetic field does not arise.
In summing up all the tasks are considered equivalent
(paragraphs a) and b) of the problem 4 were considered separate
tasks).Reply to the competition in physics1. The thing is, of
course, a piece of paper.However, the line has become "harder" is
not due to the extra weight of a piece of paper (its mass is very
small), and atmospheric pressure (this is known enough experience
to demonstrate the atmospheric pressure).When pressing the sharp
edge of the line hangs it rotates around the edge of the table by
lifting the middle of the sheet of paper, whereby a vacuum is
created under the sheet (air pressure below atmospheric) and
pressures above the plate - atmospheric (or slightly more).As a
result, a paper sheet, and, consequently, the line force directed
downwards.The resulting space region with reduced pressure is
communicated with the atmosphere only through the "channels" along
the edges of the ruler, in which also the pressure is less than
atmospheric (for two reasons: due to the vacuum existing in the
fact that the air moves through "channels" on the paper a high
speed).If the force due to the pressure difference in the channels
and air would be enough to drag the table pressed him atmospheric
pressure edges of the paper (with a strong impact on the ruler),
channels slam shut, the air inlet under the sheet of paper at a
time to stop and line will seem heavy.This description has been
compiled on the basis of the jury's own observations.In various
literature and the works of the participants of the tournament
there are other explanations, slightly different than shown.All
explanations, correctly reflecting the essence of the phenomenon,
the jury will be counted as correct.2. It is clear that since the
"Earth rotates once on its axis in 24 hours, at any time of the
year" (as written almost everything, but a lot on it and stopped),
the difference between adjacent sunrise will be less than 24 hours
when the day is increasing and more 24 hours increased when the
night.Ie from the date of the winter solstice (21 or December 22)
until the day of the summer solstice (21 or June 22) night in this
country will be less than 24 hours, and at other times - more.It is
known that the highest intensity change of day length reaches
during the spring and fall equinoxes.(You can check this on the
calendar, but you can find a purely geometrical reasons - think
what.) So, the shortest day in this country - the vernal equinox
(March 21 or 20 depending on the year), and the longest - on the
day of autumn equinox (September 23).(Recall that the tournament
was held on September 28.) The exact values of the times of sunrise
and sunset for Moscow can be found, for example, in "Diary of
Moscow schoolchildren."Equinox(spring and autumn) - the moment of
passing the center of the Sun in its apparent motion on the
celestial sphere through the celestial the equator.At this time,
the length of day and night is the same.Solstice(summer and winter)
- moments when the height of the sun above the horizon at noon, the
maximum and minimum.3. This task (question) the jury stuck in
option at the last moment as a problem for younger students (as
problems for the lower grades in the variant lacking), hoping to
get around this answer:"Because of the centrifugal force of the
water interacts strongly with the outer coast than with internal
and stronger blurs it, thereby increasing the turning radius (bend)
".However, immediately after the commencement of the audit work it
became clear that the issue is not as simple as it was supposed to,
and clearly not for the junior.Briefly turning process River is as
follows.On the right section of the river is known, the maximum
speed at the center ("Bistrica") and decreases as it approaches the
shore due to the friction of them.At the turn of water tends to
inertia continue straight for her turn must apply a force, with the
larger, the more its speed (by rotating the ball of mass m on a
string in a circle of radius R with a velocity v thread tension
force is mv2/ R).The result is a sort of water velocity and flow
rate in the outer bank of rotation is greater.We assume the
obvious, that the intensity of erosion of the coast increases with
increasing flow rate.Also, we assume that a straight section of the
river, this rate is an equilibrium, since there is a form of the
shore does not change with time (speed increases the width of the
river increases, which leads to the erosion of beaches, t. E. To
increase the width of the bed. But because of the this rate of flow
is reduced, leading to the shores of namyvaniyu (deposition of
sedimentary rocks). And vice versa.).Then, if a straight section of
the flow velocity at both banks was an equilibrium, then when you
turn after the "sorting" on the outer coast of the speed is greater
than the equilibrium and it will be eroded, and at the inner - less
than the equilibrium, and it will pan out.In addition to "sort" the
water velocity at the turn of the river there is another effect:
after the "slow" the water was near the inner bank, and "fast" -
from the outside, "slow" the water begins to "crush" the "fast",
which leads to an even greater increase in the rate of the
latter.On the other hand, the "fast" water after the interaction
with the external bank becomes turbid (there are eroded from the
shore rock particles) and loses its speed.Then she pushed to the
inner bank of the available water flows faster.At the inner bank,
in turn, loss occurs "haze" in the precipitate, i.e.additional
namyvanie it.The above reasoning is a simplified description of the
process at the level of the school curriculum in physics, which
obviously does not take into account many important circumstances,
so it is certainly not correct in terms of geology.As it turned
out, this task was interested in even the Albert Einstein, in which
"Reasons Education meanders in rivers and so-called Beer law "can
find an explanation for the many features of the flow of rivers,
for example, why the rivers of the northern hemisphere blur mainly
the right bank, and the south - left.And in 1997 on this topic has
been defended his doctoral dissertation in which the problem is not
completely solved (so far!).4. Somehow, this problem has caused
discontent decisive (maybe because we are the most essential part
of their work, not mechanical, not interested): "if so try often,
all the tables in the school will be inscribed."Nevertheless, many
approached correctly to estimate the friction coefficient.(Although
at the last moment, when the tasks have already been handed out,
the jury became worried - because if you specify ktr= 0.1, in a
wide range, from 1 to 0.01, the error does not exceed a given
accuracy. However, students on this temptation did not fall . And
rightly so - because the jury will evaluate the argument, not the
final figure.)In the figure all the forces acting on the pen are
indicated by a bold arrow.Rxand Ry, which are not real forces, and
the projections R, shown for convenience only and therefore
designated by arrows a lower fat content).The vertical component of
the forces of reaction finger and paper should be equal (for
example, from the condition that 0 torques about the center of the
pen - or pen begins to rotate, we believe that the mass m pens
distributed uniformly along its length).Together, they must balance
the force of gravity mg (otherwise the handle will fail under the
table), so they are both equal to mg / 2 (N = mg / 2 and Rx= mg /
2).Pen fall when the friction force F on the papertrbecomes
smaller, than the horizontal pressure force component pen on
paper.But where does this component, because the handle simply
presses down its gravity?No, it still relies on a finger to solve
this problem and the response of this support R and gives us the
horizontal component of Rx.Two angle indicated in the figure symbol
a (alpha) as well as angles with mutually perpendicular sides: the
reaction force of the finger R, acting on pen, fountain pen
perpendicular and the vertical projection of the horizontal force
perpendicular to the paper surface.By definition, the tangent of an
acute angle in a right triangle is equal to the ratio of the leg
opposite the angle to an adjacent.Then TG = A Rx/ (mg / 2) (as
defined by the tangent to the angle a, which "rests \"
finger).Hence, Rx= (mg / 2) tg a.This force (Rx) can be balanced
only by the force of static friction Ftr.p.pen on paper, which, in
turn, can not exceed the force of friction Ftr= kN = k (mg / 2}.
Therefore, at the time, when the pen in a continuous decrease in
the angle of a fall,Rx= kN(mg / 2) tg a = k (mg / 2)ktr= tg aThus,
estimating the angle of a tangent and finding it (and even better,
you can experiment by definition, building a right-angled triangle
containing the angle a, measure its legs of improvised means, such
as the edge of a sheet of graph paper, and calculating their
ratio), we obtain an estimate of the coefficient of friction.Now
just find the mechanical work - it is the product of the frictional
force on the handle passed way. It remains to estimate the weight
of the handle and assess the overall length written by you (for
example, assuming that all of the letters - is the height of
squares corresponding to your handwriting) - and the problem is
solved.
5. Actually Sanktorius are diagnosed "for 10 pulse beats," but
to solve it is not essential, but only adds to the long text of the
problem.Maybe because the problem could not read to the end, it was
making less than we expected.Since the first thermometer was
connected with the atmosphere, it influenced his work.Moreover, the
thermometer affect atmospheric conditions at the time of its
manufacture.Indeed, if the pressure inside the tube is increased
due to increased temperature of the patient, the doctor could see
that the droplet has moved by a greater amount.But if the air
pressure in the day was more than daily calibration of the
thermometer, the droplet could remain in place or does not reach a
normal level of temperature.But of the many messages
Gidromettsenrta all already known link between atmospheric pressure
and clear or rainy weather.So the weather affect the reading of the
thermometer.But on the other hand, do History has brought to us
information about the doctor-charlatan, inaccurately diagnosed
patients with fever?Rather, Sanktorius before measuring the
temperature of the patient, the thermometer calibration is
performed by measuring the temperature at, and only then - in a
patient.Jury these details are not known for certain, but indirect
evidence of them is the fact that the patient briefly ("for 10
pulse beats") held a thermometer at home, because he was already
heated professor!This example shows that it may not be the right
answer monosyllabic (effect-no effect), is important to correctly
reasoning that, depending on the additional assumptions lead to
different (but good!) conclusions.6. A list of devices that satisfy
the conditions of the problem, these students: the camera obscura,
crossed prisms, holography (hologram recorded at a single
wavelength and reproduced at the other), the diffraction grating
with concentric rings, different ways to create optical density
inhomogeneities in the medium (lighter in the air is dissolved in
water, salt crystals).There are other options, but they are not
offered to the students and are very hard to describe, so no more
description.7. Take the spherical surface (e.g., a balloon) at
which electric charge is uniformly distributed.To begin to change
its radius without moving her center.Located on the surface charges
will obviously move, with the process of moving is centrally
symmetric.Consequently, all of the physical phenomena that result
from this process should also have a center of symmetry. *The only
possible (from geometrical considerations) in this situation, the
magnetic field configuration is this: the magnetic field lines
diverge uniformly in all directions from the center of the
sphere.But this is impossible, sinceMagnetic lines of force may not
be endless. *Obviously, the condition of the problem satisfy any
centrally symmetric motion of the charges and the superposition of
such movements.The jury is not known whether such a situation in
the other cases.Note:For an asterisk claims in the scientific world
there is no single opinion.We rely on the school curriculum and are
not going to contribute to the debate on these issues in modern
physicsInformation about the competition jury in PhysicsOn
estimates of the competition in physics tournament them.Lomonosov
1997Competitions in physics and mathematics tournament it.Lomonosov
Moscow Olympiad in mathematics, Tournament cities and some other
similar events are held the same korrellirovanoe a lot of people,
so in all these events is used for many years has developed a
system of ratings:0 - -.- + + / 2 + - +.+!0, the problem is not
recorded in the- In the solution contained only wrong, have no
relation to the correct solution or contained in the problem
statements-.the problem is not solved, but the decision has at
least one nontrivial true statement pertaining to the right
decision and not contained in (all estimates are reported to
participants, so stamped assessment "minus" in this case may give
the student a misconception that in his work incorrectly absolutely
everything, including this very statement)- + Problem is solved
correctly, but the decision has significant progress in the right
direction.For the right decision, but their absence, usually put +
- + / 2 in the problem can not be solved sorting options - the
correct answer without explanation.In the task, where necessary,
for example, to prove the necessity and sufficiency of something -
for the right solution to only one of the parts.In other problems -
if the solution has a sufficiently large number of true moves, but
to build one right solution still can not.+ - The right decision
with a significant failure.+.the problem is solved correctly, but
there are insignificant flaws+ Problem is solved correctly+!the
problem is solved correctly and in the decision contains the
original and / or substantially beyond the school curriculum
(corresponding class) points that the jury considers it necessary
to emphasize.Spelling errors are not taken into account (because it
is a competition in physics, not in Russian, but also because the
primary school children in solving complex problems of the Contest
can not do without grammatical constructions that they have not
learned in school).Although, unfortunately, there are so many
errors ...Arithmetic errors do not affect the meaning of the
solution, not taken into account (is reasonable to assume that
students 7th grade and older are able to perform arithmetic
operations, such errors are not caused by lack of knowledge, and
other factors).If the work has only correct strikeout decision, it
is counted as correct.If the solution of the problem can not be
obtained by brute force of a small number of options (yes / no
increase / no change / decrease), and her work has several options,
including the right, count the correct option (even if it is
crossed out).Available for solving incorrect statements that have
no semantic connections with the rest of the decision, the
assessment is not reduced.The criteria for determining the
winnersGrade 6 and underDiploma if one + / 2 or betterOfficially,
the tournament is held for students grades 7-11, but talented
schoolchildren also successfully participate in the tournament.This
year, the students of class 5 and younger participate in the
competition in physics is not taken.The competition was attended by
two sixth-grader:Gaifullin Sergey school '10 Zhukovsky, Moscow
Region,evaluation - +.+.0 0 0 0 0(This is a very good result for
grade 6)Kolychev Alexander School '58 Moscowevaluation 0 + / 2 +
0/2 0 0 0 0(The discussion of the jury also decided to award him a
diploma, that is the criterion for grade 6 is purely formal)7 and 8
classesAround if one + / 2 or betterAround if the three - + or
betterDiploma if one + - or betterGrade 9 and olderAround if one +
- or betterAround, if three - + or betterdiploma if the two + - or
betterThe phrase "or better" does not mean subjective view "better
or worse", and the presence in the specified number or those
estimates or higher.Any formal justification for the above criteria
for determining the winners do not exist.The decision was made
after checking all the works under discussion based on the
experience of last year's results, the actual situation at work,
personal opinions of the members of the jury.CheckThe audit
involved 7 (NN Konstantinov, undergraduate and graduate students,
physics).Each work independently verified first 2 times, then
recheck again (usually the third person), which, when there is a
discrepancy between the estimates and the estimates and its own
opinion to hold discussions with the other reviewers (sometimes
quite tense), and set the final evaluation.At the closing ceremony
the students could challenge them (this year, wishing there was in
the past - were).When checking the work of the different classes
were presented the same requirements for the physical side
solutions, but takes into account different levels of intellectual
development of students of different ages in the evaluation of
logical correctness and validity of the decision.The difference
between the classes into account in the criteria for determining
the winners.Of course, when checking have been developed certain
formal criteria, but each work still tested individually, in
accordance with common sense, and the criteria themselves are often
formulated in the form of words or phrases such as "prism",
"centrifugal force".Bringing them to the formal compliance with all
the assessments require breaching a lot of work for what the jury
had no opportunity now (see. The last paragraph), and the
publication of the criteria without such a reduction can cause
massive resentment among the tournament participants, their
teachers and parents.Therefore, the criteria are not published.7
checks - to put it mildly, is not enough (in mathematics at about
the same amount of work they were 37).But more than just was not
wishing, most physicists were limited excuses and valuable comments
and innovations.If any of the readers willing to make a personal
contribution to the work tasks Tournament next year, its
implementation and verification work, the jury happy to support
this initiative.Selection problemsAt the competition were invited
to a lot of problems (7, one of them with two points), but to get
high score was enough to solve a small part of them, as can be seen
from the above-stated criteria.For each task, indicate which
classes students recommended to solve it (for each class - at least
3 tasks to provide student the opportunity to choose a favorite
subject for the task and to avoid a situation where the student can
not solve any problem, because I do not know exactly the formulas
and definitions that are needed for this).Students can solve
problems and not the class, summarizing the results of all the
tasks are taken into account equally.This rule came as follows: The
tournament has been held for more than 20 years and its rules have
changed over time.In particular, in 1992/93, and a few years before
the tournament formally held for grades 7-9, and for 10 and 11
classes at the same time and in the same places Tournament held
cities in mathematics.But the ubiquitous high school students still
leaked on Turlov.Combat it was useless and in the 1993/94 school
year, they were invited to the event "officially", and for them was
made a separate option in physics.Due to the fact that it
successfully solved and several (about 15) schoolchildren in HS
1995/96the jury decided to conduct an experiment and offer a
"senior" mladsheklassnikov all, it was found that almost half of
mladsheklassnyh degrees in physics (mainly in part) due to problem
solving "older" version.It became apparent that their division
would be extremely unfair to such students.The complexity of the
task (for the class) approximately corresponds to the level of
regional Olympiads (Juniors - fans challenges - can solve the
problems of high school, and high school students, especially for
fans of a problem is the university level ( 7, about the movement
of the charge and the magnetic field). Relatively low complexity of
most tasks and assessment criteria are due to lack of time (7
competitions for 5 hours) and low (unfortunately) the general level
of school physical education in Moscow (much lower than the
mathematical or humanitarian).The content of the competition
problems in physics this tournament is somewhat different from
those of the school, district, city and university
competitions.Jury tries to avoid formal synthetic problems, a key
element of the decision which is the formal knowledge and
combination formulas and definitions (such problems and so much on
the above activities and entrance examinations), mathematical
problems "in the physical Skin" (in the tournament have a
competition in mathematics) very simple (comforting) problems
(students should not be in vain to spend their decision time and
effort required in other competitions).Preference is given to
quality problems and issues, a simple experimental tasks (which are
shown or in the audience, and each participant can do them in
place), the tasks for which it is necessary to understand the
content of the physical process or phenomenon, evaluation tasks.The
jury also fundamentally waived task, which is difficult to
check.Back to the list of contests
Competition in ChemistryTasks for competition in chemistry
providedby the Moscow Chemical LyceumHouse of scientific and
technical creativity of youth in Moscow.Solving problems of the
above-mentioned organization, unfortunately, are not provided for
that tournament organizing committee apologizes.Participants of the
tournament that the jury expects to receive from them three of the
proposed solutions of problems of choice.For each task to grade on
a scale from 0 to 5. The final evaluation is the sum of scores for
the three tasks otsennnye highest number of points than the other
(points a, b, c, objectives 5 assessed on a 5-point scale
individually and summarizing the results were considered separate
tasks; unfortunately, it has not been announced to participants in
advance).If such problems but three student solved yet sufficient
other, the amount increased by 1.Tasks1. (Grades 9-11) According to
the information preserved in folk tradition, in the first century
of our millennium there was the idea that there is metal as much
and planets.Write the chemical symbols of these metals.2. (Grade
9-10) As, using as the starting materials iron filings, salt,
baking soda, chalk acid battery for refueling, water and air, to
obtain various iron compounds.The equations of the corresponding
reactions, indicating the conditions of their implementation.3.
(9-10 grade) proposed a formula of inorganic substances A and for
which you have the following transformation:A + B -> BDT-> B
+ LE + F -> WA + F -> AndG + H -> R + D+ E B -> DB + D
-> DA + H -> R + EB +