High School Physics Questions, Thank You, High School Physics Questions Please Teach Thank you very

Hello landlord, it may be two, because when the force f in the upper right direction, and then the pull force in the vertical direction = mg, the pressure of the object on the ground is zero, so the object is not supported by the force, only by gravity and tension, if there are any questions, welcome to ask, hope! Thank you!

Centrifugal motion itself is not a strictly physical concept. It only roughly depicts the motion characteristics of an object in a circular motion, when the centripetal force weakens or the centripetal force disappears, the object takes the center of the circle as the reference. Both of the situations you mentioned can be said to be eccentric movements.

Of course, your later inference is true.

It is not that the charge moves into the air, there are gases such as nitrogen and oxygen in the air, and the visible material of the wire or something, the voltage is high to a certain extent, and the air will also be ionized as a conductor, we call this phenomenon "air breakdown" The flame of the alcohol lamp is close to the electroscope, and the temperature of the outer flame of the alcohol lamp is higher, but it cannot reach the level of breaking down the air, so only a part of the charge can be moved. It is not right for the charge to move into the air, the air just transfers to the charge

Do you mean to say, using water as a reference?

Time does not change, if it is really an hour, my understanding is: it took half an hour from dropping the hat to finding the hat, and the time from finding the hat to picking up the hat was also half an hour. But one thing I don't understand is that your title doesn't explain it.

Just said that I picked up the hat 8km under the bridge. You didn't say how small the time it took to pick up the hat.

Congratulations, you got it right, it's going to be charged.

1. At any position in the cavity of the homogeneous spherical layer, the net force of the gravitational force of the spherical shell is zero.

2. At any position outside the cavity of the homogeneous spherical layer, the particle is subjected to the gravitational force of the spherical shell, which is equivalent to the gravitational force of all the substances concentrated in the center of the spherical shell to the mass point.

If the particle is on the red circle (on the cavity wall) and is still in the cavity, the gravitational force experienced by the particle is zero with reference to clause 1.

If the interior is filled with homogeneous substance, with reference to Article 2, it is equivalent to the gravitational pull of a solid sphere as large as a cavity.

The answer C is affirmatively correct. The rod does not do work on p or q, but definitely does work on p and q, and the rod does negative work on q and positive work on p, and their algebraic sum is zero. The rod-to-ball force is also not in the direction of the rod.

As for determining whether the algebraic sum of the work done by the rod for p and q is zero, considering it as a whole, it is naturally not difficult to judge the work done by using conservative force. If you want to think about it carefully, it is more convenient to use the angular momentum theorem. The momentum theorem in high school can be considered to be the linear momentum theorem, and the angular momentum theorem is introduced when considering rotation (I learned it at university).

As an extension, I would like to introduce that when a rigid body rotates around a fixed axis, in a short time of t, the work done by the external force on the rigid body is equal to the moment multiplied by the angular displacement (you can not strictly understand it as the moment multiplied by the angle of rotation). If the mass of the rod is not counted, then its moment of inertia (just as mass is a physical quantity that describes the difficulty of changing the speed of an object, and the moment of inertia of a rigid body is a physical quantity that describes the difficulty of changing its angular velocity) is also 0. Here you can analogize that the moment is equivalent to the force, the angle of rotation is equivalent to the displacement, the force multiplied by the displacement is the work, and the moment multiplied by the angle of rotation is also the work.

Same as light springs and light ropes (their mass is not counted, according to Newton's second law, f = ma, no matter what kind of motion the light spring and light rope are in, the resultant force of their force is always zero, for example, the tension of the light rope is equal everywhere, and the elastic force at both ends of the light spring is always equal in magnitude and opposite directions), the resultant moment of the light rod rotating around the fixed axis (the moment of the force of the object on it, not the moment of the object's gravity on the axis of rotation) is always 0, that is, the moment of the object to the rod at p and q is always equal in magnitude, The direction is opposite, and how many degrees p turns, how many degrees q also turns, so the algebraic sum of the work done by the moment must be zero. The work calculated above is the work done by the object on the rod, and according to the conservation of energy, the work done by the object on the rod is equal to the decrease in the mechanical energy of the object; So conversely, the total work done by the rod on p and q is zero! If there is a heavy object hanging at only one end of the light rod, then the force between the light rod and the object is the elastic force, and the direction of the rod must be extended, because the moment of the light rod by the object acting on it must be zero!

This can only be guaranteed by the direction of the elastic extension rod. At this point, tangential acceleration can only be provided by gravity! If you want to use high school physics knowledge to analyze, it is not easy, or even impossible.

In addition to C, it is also correct to easily judge B.

Next, I will use a relatively simple high school method to find the elasticity of the rod to the ball! Let's take a look!

Finally, to sum it up. For the problem of the rotation of the light rod around the fixed axis, if only one end of the light rod is hung with a heavy object, then the force between the rod and the object must extend the direction of the rod; If there are weights hanging from different positions of the rod, the elastic force does not necessarily extend the direction of the rod. But the total moment of force exerted by all objects on the light rod must be 0!

I hope it will be helpful to you.

Solution: During the swing of the ball B from the horizontal position to the lowest point, it is affected by gravity and the force of the rod, and the direction of the force of the rod is to be determined. The gravitational potential energy decreases during the hem and the kinetic energy increases, but whether the mechanical energy is conserved or not is uncertain.

During the hem of the A ball, the gravitational potential energy increases, the kinetic energy increases, and the mechanical energy increases. Since the A+B system only does the work by gravity, the mechanical energy of the system is conserved, the mechanical energy of ball A increases, and the mechanical energy of ball B decreases. So options b, c are correct.

In some problems, the force applied by the rod is along the direction of the rod, but it cannot be concluded from this that as long as the force is applied to the rod, it will be along the direction of the rod. In this problem, balls A and B rotate around point O, and the rod exerts force in both tangential and normal directions. Among them, the normal force does not do work.

The force exerted by the rod on the B-ball is negative for the B-ball. The work done by the club on the A-ball is positive. The mechanical energy of ball A increases, and the mechanical energy of ball B decreases.

To put it more simply, individually, the rod does work on both balls, but the total work is zero, that is, when it does positive work on one ball, it does negative work on the other.

Select AD analysis "When the magnet is close to the coil, the magnetic field of the induced current generated by the coil should hinder the approach of the magnet, so the magnetic field of the induced current is above the n pole, and the magnet and the coil are repulsed, at this time n mg;

When the magnet is far away from the coil, the magnetic field of the induced current generated by the coil should hinder the distance of the magnet, so the magnetic field of the induced current is above the S pole, and the magnet and the coil attract, at this time n mg, so a is correct.

According to the "come and reject" magnet has been to the left, the coil has been to the right, so the friction force on the coil has been to the left, so d is right.

a, in the process of moving to the center of the coil, the downward magnetic inductance line increases, the induced magnetic field is upward, mutual repulsion, n is greater than mg; In the process of moving away from the center of the coil, the downward magnetic inductance line becomes less, and the induced magnetic field is downward, attracting each other, and n is less than mg.

ad.According to the law of stunned times, when the magnet is close, the coil has a tendency to move away from it, so the magnet will go down to the right so the elastic force is greater than the gravity, when the magnet is far away from the coil, the magnet is subjected to upward and right forces, and the magnet has a tendency to move upward to the right, so the elastic force is less than gravity, and the friction force is to the left.

Action and reaction forces: 60n Acceleration: 60 50 = ; 60/75=;

Speed: v1 = , v2 =

1) At the beginning, the east-west speed of the wind is the same as the speed of a person, 5 kilometers per hour; When traveling at 10 km/h, the direction of the wind is 45 degrees from the direction of travel, so the north-south speed of the wind is 10 km/h. The velocity of the wind under the Pythagorean theorem is 125 in the direction of the root number north-east, and the angle of tan = 1 2

a represents the acceleration of the applied force, and x1 represents the displacement of the first ten seconds 1 2*a*t 2=x1

f-umg=ma is solved to f=15n2)a2 to denote the acceleration after the force is removed.

umg=ma2

v*2=2(a2)x x=25mv=at

First use the kinematics theorem s=1 2at2 to calculate the acceleration, and then use the ox two theorem, that is, f-f friction = ma to know the friction force to calculate f. The second question is to calculate the velocity at the end of 10 seconds, use s = v end 2 - v beginning 2 2a and then f friction = -ma the direction of attention to find the acceleration and then use the kinematic formula to find the final distance.

First of all, f (friction) = fn = g = 50 so f = f - f = ma

Because 1 2 at 2=x=50

So the solution is a=1m s 2

So f=ma+f=5+10=15n

The second question is too long, I'll go to your space.

A plank is angled against the wall if the wall is smooth. Knowing that the mass of the plank is m, we can find the support force of the wall to the plank, the support force of the ground to the plank, and the static friction force of the ground.

If the walls are not smooth. If the walls are smooth. It is known that the mass of the plank is m, and the supporting force of the wall to the plank, the supporting force of the ground to the plank, and the static friction of the ground cannot be obtained.

Whether the wall has static friction against the board is determined by whether the wall surface is smooth or not. It is static friction in nature. The magnitude is equal to the difference between gravity and ground support.

a.My answer above is incorrect. Are you sure this is a high school topic?

It seems that there will be three additional conditions for the high school question (1The walls are smooth, 2The coefficient of friction between the ground and the plank u,3

The plank is just out of the critical state, that is, f=un)In this way, the board is subjected to four forces, gravity, the elastic force of the board with the wall and floor, and the friction force of the ground against it, and it is possible to solve each force when only m, , and you are known.

b.If the walls are not smooth, this is beyond the knowledge of high school. Because if the wall is not smooth, there are a total of five forces, which are gravity, two friction and two elastic forces, and in the case of only knowing m, can only find gravity, and there are four unknown quantities left.

An equation can solve an unknown quantity, then for this plank, we can only list three equations, that is, the equilibrium equation of the force in the x and y directions, and the equilibrium equation of m (moment), the unknown quantity is greater than the amount of the equation, obviously there is no solution, because it is not a superstatically determined structure, if it is a superstatically determined structure, it can be solved by adding a deformation coordination equation. So, I don't know what the problem is, the level is limited.........My classmate said that it should be an equation of insufficient conditions, haha!

c.If the wall is smooth, then you can list the third equation, the M moment equilibrium equation, all three unknown quantities can be found (two elastic forces, one friction force), this equation is mglcos = 1 2*mglcos + flsin, about l (plate length), you can find f = wall elastic force = mg (2tan ).

d.In high school, you don't usually use the equation m, and you are generally told that you are in a critical condition.

Second problem: AC pull is 0. If the AC pull force is not 0, then the AC pull force can be decomposed into vertical and horizontal components, assuming that the vertical component is balanced by gravity and BC tension, what about the horizontal force?

Whether there is no force to balance it is obviously not valid. If BC is not perpendicular to AB, then the AC pull force is not 0.

This is not possible to find the support force of the wall to the plank, the support force of the ground to the plank and the static friction of the ground cannot be found, and it is also uncertain whether the wall has static friction against the plank.

If air resistance is not considered, the solution is as follows:

The time from the ground to the highest point and the highest point to the ground is:

t=v/g=1s

Let a block have a mass of m and its horizontal velocity is v1; The other piece has a mass of 10-m and a speed of v2

Combine the velocity images in one RT, as shown in the image on the right.

According to the triangular projection theorem, v = v1v2 = 10 = 100--- s = 100m that is, (v1+v2) t = 100, v1 + v2 = 100 ---

According to the momentum theorem: mv1=(10-m)v2--- according to , v1v2 is 50+20 6m s, respectively, and 50-20 6m s is brought in: m=5-2 6

The mass of the two pieces is 5-2 6kg and 5+2 6kg, respectively

Let the masses of the two pieces be m1 and m2 respectively, and the velocity after explosion is v1 and v2 respectively, then the momentum of the ** process is conserved

m1*v1-m2*v2=0

It is known that m1 + m2 = m

The landing time of the two blocks is equal, set to t, then.

v1*t+v2*t=s

v1/(gt)=(gt)/v2

The four formulas can be obtained:

m1=2kg

m2=8kg

Solution: Let the horizontal velocity of the first block be v1 when it bursts

The second block bursts with a horizontal velocity of -v2

1) Draw a composite diagram of the velocity according to "the speed of the two is perpendicular to each other when landing". Available:

v1*v2=v*v (proportional relationship of right triangles) 2) According to "the distance between two landing sites s=100m", it can be obtained:

v1 x 1 sec + v2 x 1 sec = 100 m

3) According to the conservation of momentum, it can be obtained:

m1*v1=m2*v2

4) According to the conservation of mass, it can be obtained:

m1+m2=10

Solve the above 4 equations together.

v1 and v2 are about , m1, and m2 are respectively ,