The athlete ran a hundred meters and began to stop. Bolt won the 100m at the third consecutive Olympiad

Details Updated on 31.03.2013 12:40

Conditions for the tasks of the city tour in 2003 for the 7th grade.

First step.

Task 1.

In the manufacture of a hollow copper ball with two small holes, another solid copper ball was placed in it with a thread tied to it, the free end of which was left outside. Determine the mass of a hollow ball, if you have: a cylindrical glass vessel, the diameter of which is slightly larger than the diameter of a large ball, a beaker, a felt-tip pen, a glass, water. The density of copper is assumed to be known.

Task 2.

A thin hole is made in the center of a piston weighing 10 kg and with an area of 500 cm2. It is known that if a piston is fixed in a vertical pipe and water is poured on top of it to a level of 10 cm, then 5 ml of water will flow out through the hole in the piston in 1 s. Water is poured into a cylindrical vessel to a level of 10 cm and a piston is placed on top. The piston fits snugly against the walls of the vessel, but can move without friction. How long does it take for the piston to reach the bottom of the vessel?

Task 3.

Some material can be used to make straight wires of various lengths and thicknesses. If you hang such a wire at one end, it may break under own weight, while the wire practically does not change its length. It is known that the maximum length of a wire that does not break under its own weight does not depend on its cross section and is equal to 2.8 m. There are 8 wires 1 m long and with different cross sections (see table). They begin to sequentially hang to each other, starting with the first. Each next wire is attached to the free end of the previous one, as shown in the figure. The mass of the wire connection is very small. How many wires can be hung until one of them breaks, and what is the number of the broken wire?

Task 4.

The athlete, having run a hundred meters, began to stop at the moment of crossing the finish line and completely stopped at a distance of 5 m from it. Determine how long the athlete ran the distance if his maximum speed during the run was 10 m/s. Consider that the athlete's speed increased during acceleration, and decreased evenly during braking, the acceleration and deceleration times are the same.

Second phase.

Task 5.

A lot of boards have been prepared for the circus number, each of which can rotate around the fulcrum. In this case, the support is located at a distance of 1/3 of the length of the board from its edge. The boards are lined up as shown in the picture; a load of 30 kg is placed on the outermost of them. A large family of acrobatic brothers is trying to balance on the boards, with each brother standing on two boards at the same time. The mass of each brother is 80 kg. How many brothers can keep balance?

Task 6.

From an ocean liner with a length of 150 m, moving at a speed of 36 km / h, a boat with people from the ship in distress was found right on the course. From the middle of the liner, a boat was launched into the water, which headed towards the boat at a speed of 72 km / h. From the nose of the liner to the boat, the boat traveled 3 km. Having stopped at the boat for 1 minute and took the distressed, the boat went back at the same speed and moored in the same place of the liner where it was launched. The speed of the boat during the movement is assumed to be constant. Determine the distance traveled by the liner during the entire time of the boat's movement from the moment of departure to the return of the boat to the liner. Plot the speed of the boat relative to the liner from the time from the moment of departure to the moment of mooring.

Task 7.

In a vertically located vessel with sections S 1 and S 2 (S 1 = 9S 2) there are two weightless pistons. The space between the pistons is filled with water. The ends of the vessel are open to the atmosphere. A spring is attached to the upper piston with stiffness k, a weight of mass is suspended from the bottom m. At the initial moment of time, the spring is not stretched, the pistons are fixed, the distance between the pistons h 0 . Find how much the upper piston will sink if both pistons are released.

SOLUTIONS OF TASKS OF THE REGIONAL STAGE

OLYMPIADS FOR SCHOOLCHILDREN IN PHYSICS GRADE 8

2010 – 2011 ACADEMIC YEAR

TASK 1. Water heating.

In this problem, two limiting solutions are possible after the first heating (depending on the final ice temperature):

1). If the initial temperature of the ice is below -2°C, then the reheating will require the same amount of heat and time as was spent on the first heating, namely

Q = mc 1 ∆t (1),

2). If the initial temperature of the ice is 0°C, then you should first melt it, and then heat the resulting water by 2°C, i.e. expend the amount of heat

Q 1 = mλ + mc 2 ∆t.

Substituting the value from formula (1), we find:

Q 1 = (Q(λ + c 2 ∆t))/ c 1 ∆t = 80.6Q.

Target heating time range

τ 1< τ 2 < 80,6τ 1 .

TASK 2. Swimming ice.

According to the condition of the problem, the ball is half immersed in water. This means that it will touch the bottom. In this case, immediately after the overflow, the volume of water in the left vessel will be V / 2 \u003d 50 cm 3 less than in the right one (see figure). Since the water levels in the vessels were also initially the same, then a volume of water equal to V / 4 \u003d 25 cm 3, with a mass m 1 \u003d ρV / 4 \u003d 25 g, should flow from the left vessel to the right one. When the ice melts, the mass of water is compared from the initial value will increase by ρV. Therefore, ρV / 2 = 45 g of water should flow from the left vessel to the right one, of which 25 g flows at the first stage - immediately after lowering into the left vessel of ice. Therefore, when ice melts, the mass of water m 2 = ρV / 2 - ρV / 4 = 20 g will additionally flow from the left vessel to the right one.

Answer: m 1 \u003d ρV / 4 \u003d 25 g, m 2 \u003d ρV / 2 - ρV / 4 \u003d 20 g.

TASK 3. Meter poles.

IN the condition says that after 2 minutes the train was near the column with the number "2". This means that in a given time the train could travel 100 m, 1100 m, 2100 m, 3100 m, 4100 m, etc. Since the train speed is less than 100 km/h or 100/60 km/min, the train does not can travel in 2 min a distance greater than S = (2 min 100 km)/ 60 min ≈ 3.3 km only the following distances are possible: 100 m, 1100 m, 2100 m, 3100 m. The following speed values correspond to them: 50 m/min, 550 m/min, 1050 m/min, 1550 m/min. Since, according to the condition, the distance from the driver's cab to the nearest column with the number "3" is 100 m, then the possible values of the travel time for this distance

Answer: possible time values

TASK 4. Atmospheric paradoxes.

Air pressure decreases with height. Therefore, as the air rises, it expands. Expanding, it does work, spending part of its internal energy. This is the main reason for cooling the air.

TASK 5. Water heating.

Let N balls be transferred from boiling water to the calorimeter. Let us denote the heat capacity of the ball C, the heat capacity of water C in = 4200 J / kg ° C, the temperature of boiling water t k \u003d 100 ° C, the final temperature t. According to the heat balance equation C in (t - t in) \u003d NС (t to - t).

With N \u003d 1 and t \u003d t 1, we get C in (t 1 - t in) \u003d C (t to - t 1).

Substituting the numerical values of known quantities into the last equation, we obtain C in \u003d 3C.

Therefore, for any N, the equation 3(t - t 1) \u003d N (t to - t) is true.

With N=2 we get t=52°C,

With N=3 we get t=60°C.

At t=90°C we find N=21.

TASK 6. At a hundred meters.

The problem is solved graphically.

The graph of the athlete's speed versus time has the form shown in the figure.

The total distance S = 105 meters covered by the athlete is equal to the area under this graph, and the area can be easily found by moving its shaded piece, as shown in the figure. So S = V t, whence t = S/V.

Answer: 10.5 seconds.

1. Seventh grader

A seventh-grader goes to school from home at a constant speed V ═ 2m/s. The distance from home to school is L ═ 103m, and the boy has time just in time for the beginning of the lesson. One day, a seventh grader decides to come home halfway because he forgot to turn off an electrical appliance. Will he be able to get to school by the beginning of the lesson if from that moment on he runs at speed v═ 14.4km/h?

2.Snow

The tourists filled the pot to the brim with snow and melted V ═ 0.75 liters of water out of this snow. Find the volume of the pot if it is known that the water is four times denser than the snow collected by tourists in the pot.

3.Paper

How to find the density of paper if there is a thick notebook in a cage, a coin weighing m ═ 1g, scissors and a lever balance without weights? The side of the cell in the notebook has a length a ═0.5 cm.

4. Amphora

During archaeological excavations, an ancient transparent bottle was found, the lower part of which has the shape of a parallelepiped and is more than half of the entire bottle in volume. Top part the bottle has an irregular shape (see picture). How, having a ruler, a cork for this bottle and unlimited supplies of water, to determine the volume of the bottle?

5. Sprinter

The athlete, having run a hundred meters, began to stop at the moment of crossing the finish line and completely stopped at a distance of 5 meters behind it. Determine how long the athlete ran the distance if his maximum speed was Vmax = 10m/s. Assume that during acceleration and deceleration, the speed of the athlete changed evenly, the acceleration time and deceleration time are the same.

All-Russian Olympiad for schoolchildren 2016-2017 academic year

School tour of the Physics Olympiad

7 Class

1. seventh grader

A seventh-grader goes to school from home at a constant speed V ═ 2m/s. The distance from home to school is L ═ 103m, and the boy has time just in time for the beginning of the lesson. One day, a seventh grader decides to come home halfway because he forgot to turn off an electrical appliance. Will he be in time for school by the beginning of the lesson if from that moment on he runs at a speed of v ═ 14.4 km/h?

Solution:

Changing units of running speed Vrun = 14.4km/h = 14.4x1000m/3600s = 4 m/s
The total amount of student time: Δt = L/v = 103m/2m/s = 51.5s
Spent time walking from home to the place of the forced stop:
The time it took the student to run home and from home to school: t \u003d (L / 2 + L) / Vrunning \u003d 1.5L / 4m / s \u003d 1.5x103m / (4m / s) \u003d 38.625s ≈38 .6s
Comparison of t and Δt/2 shows that the student will not be in time for the beginning of the lesson.

2. Snow

The tourists filled the pot to the brim with snow and melted V ═ 0.75 liters of water out of this snow.

Find the volume of the pot if it is known that the water is four times denser than the snow collected by tourists in the pot.

Solution:

3. Paper

Solution:

To find the density of the paper, we will carry out a thought experiment using the inventory provided according to the condition of the problem.

2Recalculate the number of cells on the left scale pan N l 1Find the thickness of one sheet of paper, equalizing the side known by condition

cells a \u003d 0.5 cm with the end of notebook sheets attached to it. Recalculating the number of sheets obtained by such an adjustment N l , we find the desired thickness d:

d= a/Nl

3Find the volume of paper that balanced the coin Vb:

V b \u003d a a d N l \u003d a² (a / N l) N l \u003d a³ (N l / N l)

We get the desired paper density: ρ \u003d m / V b \u003d 1g / (0.125 cm³ (N l / N l) \u003d

8 (N l /N l) g/cm³2

4. Amphora

During archaeological excavations, an old transparent bottle was found, the lower part of which has the shape of a parallelepiped and is more than half of the entire bottle in volume. The top of the bottle is irregularly shaped (see picture).

How, having a ruler, a cork for this bottle and unlimited supplies of water, determine the volume of the bottle ?

Solution:

parallelepiped shape.

By measuring the length ( but), width (b) and height (h) of the parallelepiped, we get the volume

parts of a bottle filled with water: V p \u003d but b h

Close the bottle with a cork

Flipping the bottle

We measure the height of the air layer h ‘ and find the volume of air above water:

V in =a b h ‘

We get the desired volume of the bottle: V = V P + V in = a b ( h + h ‘)

5. Sprinter

Solution:

To facilitate the solution of the problem, it makes sense to plot the speed of the runner as a function of time. If you have a graph, you may encounter two solutions.

Method 1 ("on the forehead")


It is obvious that the desired time τ, during which the athlete ran the distance, is

is given from the acceleration time τ p and the time when its speed was maximum

τ max: τ = τ p +τ max

τр can be found if we use the fact that the speed during acceleration changed

uniformly: τ p = S p /v Wed . Here S p\u003d 5m (acceleration length, equal by condition to the length

braking), v Wed-average speed during acceleration, equal to V max/2= 5m/s: τ R=5m/5(m/s) = 1s.

τ max is found according to the uniform motion formula when the athlete moved with

constant maximum speed: τ max= (100m - 5m) / 10m/s= 9.5s

As a result, we find the answer to the question of the problem: τ = τ R +τ max= 1s+9.5s = 10.5s

Method 2

If we take into account that, according to the condition, the acceleration and deceleration triangles on the speed drawing are equal, the answer is obtained immediately, taking into account that the distance traveled is equal to the area under the speed graph: τ = 105m/10m/s = 10.5s. For such a decision, if compared with the first, it is appropriate to add two bonus points.

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