The master is in!! Sophomore Physics!! What a blow to my motivation!

Physics is actually not too difficult, first of all, the concepts and formulas are proficient, and half of it is completed. In fact, physics is mainly highly applied. As long as you understand it, it's equivalent to a simple math problem.

You list the formulas, and the solution is very simple. I think math is much harder to learn than physics, and English is crazy......

Supporting upstairs, physics is actually not difficult.

It's rare that mathematics and English are going to graduate school this year (physics, physics, and quantum mechanics) and I'm probably still hanging on to English, strongly condemning all evil English.

Physics actually feels like it's mostly about pushing down formulas.

You know how all the formulas come to be, how to extrapolate them into other formulas, and basically there aren't too many problems.

I remember when I was in high school, I used a lot of kinetic energy and momentum, and I used it a lot, and it was nothing more than mechanical analysis.

I really can't remember anything else about high school physics.

Physics is mostly about understanding.

Understand one aspect, and all the problems in this area are solved.

You should consciously cultivate physical thinking, think about why you encounter physical problems, know why they are, and think about which knowledge points this physics knowledge is examined after completing a physics problem, and believe that you will become a physics master after 1-2 months.

Pressure on the desktop of the opener: bend early and bury Wang g total s=??

Pressure on the bottom of the pot: 1000*10*

Pressure on the bottom of the pot: ps=1230*

1.The gravity of the kettle after filling the water, except for the weight of the teapot after the water is filled with the contact surface of the mountain call to tease Kaiji p=g s.

Pressure 2The formula for the pressure of the liquid is p= gh

3.The liquid generates pressure with f=ps

What is the quality of water? Otherwise, the pressure of the teapot on the table cannot be calculated.

In the X-T image, the slope represents the speed of the object's motion, the greater the slope, the greater the velocity, and the average velocity is equal to the total bit removal time over that time period.

So in 0-2s, the velocity is 2 2 = 1m s

The velocity in 2s 4s is (6-2) (4-2)=2m s0 The average velocity in 4s is 6 4=

The mass of the fighter should be as small as possible, and the drop fuel tank should be thrown away before the battle in order to increase flexibility; Machine tools are often fixed to the foundation in order to be stable and precise.

The car driving at a constant speed on a straight highway does a uniform deceleration motion after braking until it comes to a standstill, and its displacement changes with time The law of its displacement is s=24t-6t2, then the speed of the car traveling at a constant speed is m s; The magnitude of acceleration while braking is m s2.

An object is free-falling from a height, with an instantaneous velocity of m s at the end of the 3rd second and an average velocity of m s for the first 3 seconds.

When an object of mass m is placed on the floor of the lift, when the lift accelerates vertically upwards with an acceleration of magnitude a, the gravitational force on the object is equal to g=m(g+a); The amount of support force of the floor on the object is .

If a weight of 50 n is placed on the horizontal ground, the combined external force of the object is (80) n under the action of 30 n of vertical upward tension If the upward tension is removed, the pressure of the weight on the ground is (50) n.

1. Reduce inertia and prevent resonance.

2. According to s=v0t+at 2 2, we can know that v0=24m s, a=-12m s 2

3. According to v=gt, it can be seen that v3=30m s; According to h=gt 2 2, the height of the fall in 3 seconds is 45m, so the average speed in 3 seconds is s t=15m s

4. The gravity is still mg, but the force analysis of the object is subject to the downward gravity g and the upward support force f, according to Newton's second law: f-mg=ma can be pushed f=mg+ma

5. The resultant force is zero, because there is no ability to lift the object, and the object is stationary. The force is removed, and the object is subjected to gravity and upward elastic force, and the two forces are balanced, so the pressure of the weight on the ground becomes 50n

Reducing the mass makes it easy to change the state of motion by reducing the pressure.

mg （m+a）g

The magnetic field can do no work, so the velocity of the ions in the circle is unchanged, and it is an ideal collision, so the velocity is always v. Since it is from point A in and point A out, the trajectory of the ion is as follows:

1: Vertically collides with the highest point of the cylinder, and the velocity changes to vertically downward.

2: Vertically collide with the rightmost point of the cylinder, and the velocity becomes horizontal to the left, that is, v3: Vertically touches the lowest point of the cylinder, and the velocity becomes vertically upward.

Now that we know the physical motion process, let's look for specific numbers:

The time of the first injection of ions, the speed of centripetal motion is perpendicular to the connection of the center of the circle, and the time to reach the top is also perpendicular to the line of the center of the circle

The distance of the motion of the object is 4 1 4 circles, and the speed is the same, so t 2 r v landlord do not doubt, q, m'B and other physical quantities are not used, in fact, the most important thing in physics is to analyze the process of motion, and such questions are a little confusing.

Because the velocity direction of the particle when it comes in is directly opposite the center of the circle, it is still shot out from point A after multiple lap infiltration collisions, then the velocity direction of the particle must be perpendicular to the cylinder wall when it collides with the inner wall of the cylinder each time (that is, the velocity direction is reversed and extended through the center of the circle), according to the symmetry, the particle should collide with the cylinder three times, and then it can move to point A to shoot out, that is, the four sides of the cylinder are equally divided, and the total time of the particles moving in it is t period t 2 m (qb), of course, the wheel guessing ridge can also be calculated in this way t 2 r v

Touch it at least 3 times... Just according to the geometry... The key depends on your drawing ability.

This kind of topic is not difficult.

During acceleration, the static friction force is indeed not equal to the centripetal force, it should be that the radius direction component of the static friction force is equal to the centripetal force.

But in the moment before the start of sliding, i.e., in the critical state, the tangential velocity reaches its maximum and no longer accelerates. At this point, all the frictional forces are in the radius direction, so at this critical state, the static friction force is equal to the centripetal force.

This can not consider the process, because the slider only receives the action of static friction in the whole project, so the kinetic energy of the slider is completely provided by the static friction, so only the critical speed of the slider sliding money is required and then the kinetic energy of the slider can be calculated at this time, and this kinetic energy is the work done by friction on the object

The resultant external force provides the centripetal force of the object to move in a circular motion, and the object is only subjected to gravity, support, and static friction. In other words, only static friction can provide centripetal force, and there is no second force that can make it move in a circular motion, so the maximum static friction = centripetal force.

In a circumferential animal with variable velocity, the net force experienced by the object does not point to the center of the circle, but the resultant force can be decomposed into components in the direction of the radius and the tangent direction of motion, where the component force in the radius direction must meet f=mv rThe f(static) in the title should refer to the component of the resultant force in the radius direction, so f(static)max=mv r is also true.

First of all, the centripetal force is not real. Analyzing the Force The centripetal force before sliding is provided by the static friction force. After sliding, there is also sliding friction, but the radius direction component of static friction is equal to the centripetal force, so it can be solved in this way.

You have a point, but the subject is not rigorous.

But if you do what you say, you can't solve it without giving angular acceleration.

The title should be missing one sentence: slow changes in rotational speed.

If it's just a college entrance examination question, you can't think about it so much.

I hope it helps you, if you have any questions, please ask o( o haha

The block goes from stationary to rotating at v speed, gaining kinetic energy. This kinetic energy is the result of the work done by static friction on it, i.e.: w=.

The acceleration of the turntable should also be taken into account, as the block does not necessarily slide in the direction of the radius.

As long as you calculate the change in distance from a standstill to when you slide on the turntable.

Landlord, your considerations are reasonable.

But to see the problem, at the beginning there is no relative sliding between the object and the disc (whether the disc is uniform or accelerating, or curved, polygonal motion, as long as there is no relative sliding, there is static friction).

So it is true that the object is always subject to static friction in the first place (static or dynamic friction is based on the relative tendency of motion between two objects, here it is relatively resting).

Then with the increase in rotational speed.

The centripetal force that the static friction needs to provide increases until it reaches a critical state.

Maximum static friction = centripetal force.

Slow down the block and continue to remain relatively stationary, still subject to static friction.

If you turn a little faster, the maximum static friction will be exceeded, and the object will slide, which is the description of this problem at this moment.

In the process between the time the block moves from rest to before it starts to slide", i.e., the critical state, when it will slide but not slide.

So the column f(static)max=mv r is no problem at all. (The condition that no angular acceleration is provided is considered to be a slow acceleration, i.e., the frictional force along the perpendicular radial component is negligible).

You can continue to ask.

In junior high school physics, the internal resistance of the power supply is not considered, so the voltage at both ends of the default power supply is always the same, if it is more true, you draw the power supply as an ideal power supply plus a resistor, so that the analysis of the voltage of the two sections of the power supply is actually to rise, the specific distribution, according to your equivalent circuit diagram is good.

You write an expression of the total resistance of two resistors in parallel, and then set several sets of data into it yourself, and observe the changes, you will find that it will decrease, so the total current will increase.

1. No return, the voltage size remains unchanged (regardless of internal resistance), and the voltage is still distributed according to the resistance size. When the total resistance increases, the current decreases, and when the total resistance decreases, the current increases.

2。It will increase, like the sum of several numbers, and the sum will increase.

The total voltage does not change with resistance (regardless of internal resistance).

In general, the total voltage is constant, but in the circuit, if there is a power supply, the total voltage is constant. Tandem: the resistance increases, the voltage does not change, the resistance increases, the total current decreases, and the current on each resistor decreases.

Parallel: Similarly, the total voltage remains the same, the voltage of each branch remains the same, the resistance increases, the current decreases, and the total current decreases.