One physics question 1, one physics contest question

The buoyancy of an object is only related to the weight of the discharged liquid. It is believed that the wood floats on the surface of the water and the iron sinks to the bottom. So only the part of the wood below the water level drains the water.

And the whole volume of iron is draining. So in the water it's flam< Iron. However, the density of alcohol is less than that of water, so it is impossible to tell whether the wood is floating on the surface or sinking to the bottom.

Because V is equal, according to F=PVG, so according to the density of the liquid, the higher the density of the liquid, the greater the buoyancy, but because the density of P wood is smaller than the density of water, the wood must float on the water, at this time the iron ball sinks as low as a cup, and I don't know the volume they are discarded, so I can only choose D

I share the opinion of the 11th floor that the density of the wooden block is the density of alcohol.

From Archimedes' theorem, we can get f wood, so choose b

The density and volume of the wood block are unknown and cannot be compared with alcohol. Pick D

The buoyancy of an object is only related to the volume, the iron sinks to a low level, and it is close to the low, and it is not buoyant, while the wood floats and is buoyant, so it is a

The answer is definitely d, because the density of the wooden block is unknown.

The wooden block is placed at rest in the water, so the buoyancy is equal to the gravitational force.

Because the density of iron is greater than that of alcohol, the block of iron sinks in alcohol and the buoyancy is less than the force of gravity.

Because the quality of the wood and iron blocks is not known, it is impossible to determine the wood and iron. Pick D

d unless you know their density.

Option A Because the density of iron is greater than that of water and alcohol, the iron sinks at the bottom, and some of it coincides with the bottom of the cup, resulting in a decrease in the volume of the actual water benefit. Buoyancy is only related to the density of the liquid or gas and not to the density of the object being floated (be careful).

Swimming upstream, the velocity of the boat against the rock is: (3-2)=2 (meters and seconds).

One minute later the boat is upstream of the bridge: 2*60=120 (m).

After the barrel falls into the water, it floats downstream and after one minute is downstream of the bridge: 1*60=60 (m).

Immediately after the discovery, turn the bow of the ship, at this time the barrel is 120 + 60 = 180 (m) When the ship goes down the water, the rate of the opposite shore is: 3 + 1 = 4 (m seconds) Set the barrel and drift x m to catch up, then:

180+x) 4 = x 1 to get 3x=180; x=60 (m) at this time, the barrel distance from the bridge: 60 + 60 = 120 (m).

Solution: This problem needs to consider the relative motion: the speed of the boat on the river = the speed of the boat on the still water The velocity of the water.

Let the velocity of boat A on the still water surface be v (A), the river water speed is v (water), and after the glass bottle is put into the water, the boat A will advance again, and the forward distance is [v (a) + v (water)] * during this period the forward distance of the glass bottle is v (water)*; The distance between the two is [v(a)+v(water)]*water)*a)*;

After the ship turns around, the boat and the glass bottle move in opposite directions, and the rate of approach between the two is: [v (a) - v (water)] + v (water) = v (a), let the time when the boat A meets it and puts the glass bottle in the water after returning to the sea is t A, then:

v (a) * a) * t a.

t A = the same can be inferred.

t B = so: t A = t B.

Let the river velocity be v

T A = A A.

t B = B B.

t A - t B = A A B B.

Equally, if you take the speed of water as a reference system, it is equivalent to two bottles that do not move, and the two boats do round trips, and they are used both when they go and when they come.

Therefore the time is equal.

Equal, both, you can assume that you and I (2 boats) are both on a train (river) at a constant speed, we both go forward, we meet and drop a dollar coin at the same time, and then we continue to walk forward for an hour, and then we 2 stupidly come back again looking for money, how long do you say it will take to find it? Of course, it's hours, but only if no one picks it up.

As shown in the figure, let the red part have a virtual displacement δ and by the principle of virtual work, t( )rδ -mg δ r sin =0, then t( )=mg sin

The maximum tension is mg

When B is jetted, A makes the gas in it eject to the left at high speed, and the gas produces left momentum, which is known by the conservation of momentum, and the object A needs to produce rightward momentum, that is, the right velocity.

When inhaling, B inhales the gas, and after the gas enters, it stops moving by acting with B, and from the outside, the gas is stationary at first, and then it is also stationary, and the momentum does not change, so B does not produce motion.

A will experience a force to the right, so it moves to the right.

B will experience a force to the left, but it is small, it may not move, or it may be left.

Usually when measuring the parallax of a star, the average distance between the Earth and the Sun is taken as the baseline, the distance d is the bottom edge of the triangle, and the distance r from the two ends of the baseline to the line of the star is the two waists of the isosceles triangle, then the apex angle is the parallax, and it can be seen that d = r, d is constant, then it is inversely proportional to r. If the apex angle is 1'', then the sidereal distance is 1 parsec, and now the parallax is'', let the distance be x parsecs, then there is 1''1 parsec =''x parsecs, the distance available is parsec.

The distance between the sun and the earth d=, so the distance of Sirius r=d, where should be converted to the radian system, i.e., =, and substituting the data can obtain r=

I'm sorry, I won't give it to me, I love you hahaha.

No matter how you wiring it, there are only two types of resistance: 2r 3 and 4r 3 (also the maximum resistance).

Even the 12, 13, 23, 24, 34, 16, 15, 65, 64, 45 post resistors are 2r 3;

Even the 25, 26, 35, 36 post resistors are 4r 3.

This kind of question is actually a test of your ability to simplify the circuit, once the circuit is simplified (like my diagram), then the rest of the calculation is very simple. Therefore, to do this kind of problem, the first thing is to simplify the circuit diagram and draw the equivalent circuit diagram.

If you take two vertices, such as 1 and 6, then the resistor has a resistance value, and the total number of different resistance values (excluding 0 ohms) that can be obtained with this resistor is 2

The maximum resistance values are 4r 3 respectively

These 2 connections are 1 (4) to the other point where the resistance is 2r or between is 4r 3

To explain hi me in detail.

The total number of different resistance values (excluding 0 ohms) that can be obtained with this resistor is 2, and the maximum resistance value is 4r 3, respectively

These 2 connections are 1 (4) to the other point where the resistance is 2r or between is 4r 3

If you find two vertices such as 1 and 6, then the resistor has a resistance value, and the two different points correspond to a value, which may be the same or different, and it is calculated by connecting them in series and parallel. Then count them one by one.

Classification discussion, the binding post is the boundary line, the same side (including no matter how the resistance value is one, there are 1 and 2 binding posts on different sides, and in each case, the circuit diagram can be simplified to calculate the resistance value.