Masters please enter junior high school physics questions, junior high school upper physics questi

…Have you forgotten about buoyancy?

With the same mass, the volume of A must be 3 times that of B. It can be assumed that the volume of A is 3 cubic meters and that of B is 1 cubic meter.

n=g-f float.

Then nA = 6 10 4-1 10 3*10*3 n=3 10 4nnB = 6 10 4-1 10 3*10*1 n=5 10 4n That's it!

Because both objects A and B are denser than water, they both sink to the bottom. Since both objects have equal mass, the ratio of their volumes should be 3:1.

We might as well assume that the volume of A is 3m, then the volume of B is 1m. The pressure of A and B on the bottom of the container is equal to the support force of the bottom of the container on A and B. So now let's find the support of A and B for the bottom of the container.

f A = g A - f floating armor From the formula of gravity and buoyancy, it can be known: g A = 2 * 10 cubic * 10 N per kilogram * 3m i.e. g A = 60,000 N f Floating A = 1 * 10 cubic N * 10 N per kilogram * 3m i.e. f Floating A = 30,000 N then the pressure of A on the bottom of the container F A = 30,000 N

In the same way, we can find : f B = 50,000 N.

Therefore, the pressure of the two balls of A and B on the bottom of the container is 3:5

That is, the ratio of the support force of the bottom of the container to the two balls of A and B is 3:5

Is it good to give a few points as a reward and is missing points!

Option C because A density is 2 times 10 cubic kilograms per cubic meter, B density = 6 * 10 cubic kilograms per cubic meter, and the mass is equal. Therefore v a:v has = 3:1 and c is chosen because the gravitational force on the object is equal to the amount of liquid it dispels

Option b takes into account the buoyancy experienced by the ball. Buoyancy is equal to the gravitational force of water that is the same as the volume of the sphere. Set the ball weight 6, then A buoyancy 3, B buoyancy 1Therefore, the ratio of A and B is (6-3):(6-1)=3:5

According to Archimedes' principle,

In the same solution) sunk in water, the buoyancy ratio is equal to the volume ratio, let the volume of A be 3, f float 1=, fn =, then the volume of B is 1, f float 2=, fn =.

1.If the flywheel speed of a single-cylinder four-stroke diesel engine is 960r min, the diesel engine completes (480) work cycles per second, does (480) times of external work, and completes (1920) strokes. 2.

A single-cylinder four-stroke diesel engine, its speed is 20r s, in 1h it completes (144000) strokes, each work needs ( ) s, if the gas pushes the piston in the power stroke is 2205j, then the power of the diesel engine is (22050) in the power stroke, the gasoline gas pressure can reach 5 10 5pa (note, read as 5 times 10 to the 5 power, the same is true), the pressure of diesel engine gas can reach, If the piston area and piston movement distance of the gasoline engine and the diesel engine are the same, the ratio of the work done by the diesel engine to the gasoline engine in a power stroke is (c) a 1:2 b.

1:1 c. 2:

1 :14.Answer the questions.

A single-bar four-stroke gasoline engine, the speed of the flywheel is 600r min, the average pressure of the gas to the piston in the power stroke is 5 10 5pa (the pressure remains unchanged in this stroke), the piston stroke is, the cross-sectional area of the piston is 100 square centimeters, and the power of the gasoline engine is found. (Work done in minutes: 600 2 = 300 times, total work:

W=FL=PSL=5*10 5PA*100 10000*, power 300000 60=5000W=5KW.

Fill-in-the-blank questions. 5.The compression stroke of a gasoline engine is the process of converting mechanical energy into internal energy.

If the efficiency of the gasoline engine is 25%, and the gasoline consumed for 1 hour is 480g, the gasoline engine does external work (5520000) j(q gasoline = calculation question: The parameters of Huayang 1870 series solar water heater are as follows: vacuum tube diameter length (mm) 70 1800 specification (branch pipe) 15 Appearance size (mm) 1840 1597 1765 insulation layer (mm) 60 water tank capacity (l) 150 lighting area (m2) 2 1 (1) It is known that the average solar radiation energy obtained by each square meter of the area facing the sun in our city within 1h is 3 106j, if the time of receiving solar energy every day is calculated as 8h, and the water heater can convert 50% of the received solar energy into the internal energy of water, then under normal circumstances, how many degrees Celsius can the water heater heat a tank of water with an initial temperature of 20 in a day?

It is known that C water = 4 2 10 3 J (kg

Answer: The energy received by the water heater is:

3*10^6 * = j

q suck = j * 50% = j

m = v = 1 * 10 3 kg m 3 * m 3 = 150 kg δt = q suction cm = j ( j kg· 150 kg).

It can be raised by 40 degrees, and it can be raised from 20 to 60.

According to the title. Known.

The volume of a small stele va = 180-100 = 80 cm3 = mass m = 140 g =

Stone stele m = 30kg

Untie. =m/v=

m= *m=1750*30=52500kg. The weight of the stele is 52500kg

According to the title. Know.

Aluminum = kg m3m aluminum = 27g =

v1=v2=

Water = kg m3 (everyone knows).

Solution (1) v Al = V2-V1 =

Aluminum ball = m v =

Aluminum balls Aluminum.

Aluminum balls are hollow.

2) V empty = v aluminum ball - v aluminum = v aluminum ball - m aluminum aluminum.

The volume of the hollow part of the ball is.

3) m ball total = m aluminum + m water = m aluminum + water * v empty.

If the water balloon is filled, the gross mass row is coarse.

I don't know the format of the answer, it should be slow.

Choose my Saw. 3q

The correct answer is D, and the analysis is as follows:

A correct. Because at -40, the mercury has solidified, and the mercury thermometer cannot be used.

b is also correct. The normal luminous temperature of an incandescent lamp is more than 2,000 degrees Celsius. Tungsten for filament is to use tungsten to have a high melting point.

c correct. Aluminum and magnesium have close melting points, melt at high temperatures, cool down, and solidify almost simultaneously.

d error. The freezing point of aluminum and copper is very different, and when cooling, copper solidifies into copper lumps, while aluminum does not.

Therefore, the correct answer is d

In this problem, a mercury thermometer does not correctly display -40 because it has become a solid state below -39. bBecause tungsten filament is used in incandescent bulbs, it is correct. In the description of C and D, there is a bit of a problem, because for C, because the magnesium metal will burn when it is heated to about 500 when exposed to air, it is impossible to simply put it in a container to heat it to melt to obtain magnesium aluminum alloy, and it is possible to make an alloy only when the container is in the protective gas, so it should be incorrect.

However, this characteristic has nothing to do with "melting point and boiling point", and it seems that the answer to d where the melting point is too far away is "incorrect", but copper and aluminum can be eutectic. It seems that it is up to the specific teacher to determine it.

A is incorrect, BCD's statements are all correct, the method of elimination.

r= u i=(9-6)v (2-1)a=3 ohms.

u=u1+u2=ur'+r*ir=9v+3 ohms*1a=12v, this is what you write when you write your homework, if you want to explain.

q 862128329

Answer: I don't know if there is a problem with this question!

a. Error. Because in Figure B, its pressure decreases and the friction decreases, so F A > F B;

b. Error. Because this method does control the roughness of the contact surface and changes the magnitude of the pressure, but the indication of the dynamometer is no longer equal to the magnitude of the sliding friction, it cannot be studied experimentally;

c. If b is not equal, its indication is equal to a component of friction + gravity along the inclined plane;

d. This experiment does control the equality of its contact area, but according to the ** element of this question, it is not necessary to control its area to be equal. And other factors that cannot be controlled remain unchanged. Even the amount of pressure has changed.

To sum up, I think there is a problem.

1.When you hear the sound, see that the train is 200 meters away from the mouth of the right tunnel.

If the distance between people is x meters from the right crossing, then 100-x meters from the left crossing.

The speed of the person is a and the speed of the vehicle is b

100-x)/a=(100+200)/b x/a=200/bx=40a/b=1/5

2.Suppose the average speed of his car in the second half of the journey is V, the total distance is 2s, and the half distance is S.

Then the first half of the journey takes S 20, and the second half of the journey takes S v then (S 20 + S V) * 40 = 2s (total time multiplied by average speed = total distance) and then calculated, S can be eliminated.

In the end, it turns out that s (2v)=0, so it is impossible.

This kind of question is set up first, write out the time used for each period, or the distance traveled in each period, and then list the formula, and the list is the simplest formula. Hope.

Solution: A. A moves on the horizontal plane and B moves on the inclined plane, then the pressure of A is greater than the pressure of B, so the friction force received in A is greater than the friction force received by B So A is wrong

b. The roughness of the contact surface in the two experiments is the same, and the pressure is different, so the relationship between the sliding friction and the pressure is studied, so B is correct

c. In B, the tensile force must not only overcome the magnitude of the frictional force, but also overcome the gravitational force, so the magnitude of the tensile force is greater than the friction force, so c is wrong

d. In these two experiments, the roughness of the contact surface was controlled, so D was wrong

Therefore, to solve this problem, it is necessary to grasp the factors that affect the magnitude of sliding friction: the magnitude of pressure and the roughness of the contact surface In this problem, the roughness of the contact surface remains unchanged, and when the inclined plane changes, the magnitude of the pressure changes, so the relationship between the magnitude of friction and the magnitude of pressure should be studied At the same time, it is necessary to master the application of the control variable method

A is wrong should be the other way around B is correct I just don't know how he got it f B c wrong is not equal to should be greater than d wrong and how smooth the contact surface is.

Chose B because the experiment was problematic, but his starting point was to study the relationship between sliding friction and pressure, although the results were wrong! There are no other answers!