But I still want to ask about freshman physics

It depends on your major, and many professional college physics courses are only basic courses, not professional courses. For example, I am a chemical engineering major, and I have passed a final exam in college physics, and I can't use it for graduate school entrance examinations.

1.What kind of course is college physics?

College physics is a basic course of science and engineering in universities, through the study of courses, students are familiar with the structure, properties, interactions and basic laws of motion of substances in nature, and lay the necessary physical foundation for the study of subsequent professional foundations and professional courses and further acquisition of relevant knowledge.

2.There are two books on college physics:

The first volume contains classical Newtonian mechanics, conservation laws, continuous mechanics (angular momentum of rotation on a fixed axis), electrostatic and magnetic fields, mechanical waves, interference of light, diffraction, polarization, and thermodynamics.

The next volume includes special and general relativity, quantum physics, the Schrödinger equation, lasers, crystals, band theory, semiconductors, and nuclear physics.

General relativity.

3.There are also university physics experiments:

Experiment 1 Measurement of Young's modulus of elasticity.

Experiment 2 Determination of the moment of inertia of an object.

Experiment 3 Wheatstone Bridge.

Experiment 4 Use of oscilloscopes.

Experiment 5: Research and Measurement of Newtonian Ring Interferometry Phenomenon.

Experiment 6 Michelson Interference.

Experiment 7 Principle and use of polarimeter.

Experiment 8 Depiction of the current field of different electrodes.

These eight experiments may vary from university to university.

After reading the answers to these three questions, I actually have a certain degree of impression of the content of college physics, for a high school student or a freshman, the first impression of reading this catalog is that some of them are very difficult, and some have studied before, which is really good.

Let's talk about the first volume, the first volume is the basic Newtonian mechanics, classical mechanics, this is very simple for some people with a foundation, it is not too difficult to understand, study carefully and everyone still learns to understand.

It is difficult to understand the second volume of college physics, and those who have a little common sense know that the understanding of special relativity is the basis for learning the next volume of physics, and it will subvert your perception of common sense of reality. Ask the teacher more about the understanding of physics in the next volume, and look up more information on the Internet, and you can understand it well. There are a lot of reasoning calculations in it, and it doesn't matter if you can't understand it.

Here, the next volume is more difficult to understand, but the final exam is quite simple, it is the example questions that the teacher usually talks about, and I am proficient in doing it myself and understand it. There are basic types of questions, but they are actually not difficult.

Master the strength of magnetic induction. Magnetic flux; Gauss's theorem in magnetic fields;

Understand the Biot-Savar law. It can be used to calculate the magnetic induction intensity;

Understand the ampere force and Lorentz force, the magnetic moment of the current-carrying coil, and the moment of action of the magnetic field on the current-carrying coil. Magnetic work, which can carry out relevant calculations.

Learn about the motion of charged particles in an electromagnetic field and learn about the Hall effect.

Hello, college physics is mainly about magnetic fields, as well as kinematics, physics in college is really difficult, you need to have high school physics knowledge, but it is not particularly difficult to study seriously.

Through the study of this course, students can be familiar with the structure, properties, interactions and basic laws of natural material motion, lay the necessary material foundation for the subsequent professional foundation and research of professional courses, and further acquire relevant knowledge. However, engineering majors mainly teach basic mechanics and electromagnetism.

Physics is not an ethereal thing, it is actually closely related to our daily life. For example, the manufacture of automobiles involves cylinder models and so on. Only by combining college physics with the things around you can you have a certain interest in college physics and learn college physics well.

a) p=(-1)*(3)*5=15w, p 0 absorption power rate.

b) p = 1 * 1 * 4 = 4 w, p 0 absorbs power.

c) p=(-1)*2*8=-16,p eliminate the power emitted by Wang Jiantan 0.

There is already a negative sign in the previous expression about self-induced electromotive force, and the actual result is the difference between the magnitude of the two electromotive forces, and the total electromotive force in the loop is equal to the superposition of the electromotive forces.

The content comes from the user: Yuemei Frost River.

1.Two parallel "infinitely long" uniformly charged straight lines in a vacuum are separated by a, and their charge line densities are - and + respectively. Seeking:

e=e1+e2=direction: negative along the x-axis.

4.A charged sphere of radius r has a charge density distribution of .

Make a Gaussian surface of radius r1: Yes.

r1+=extraspherical potential.

r1《r5.Shown is a thin, non-uniformly charged rod of length l along the x-axis with a charge line density of = 0(x-a). Take the infinity as the zero point of the electric potential, and find the electric potential at the origin of the coordinates.

Take the length element dx at any position x, and generate an electric potential with a charge o point on it.

o Point total potential.

6.An "infinite" plane with a round hole of radius r in the middle, uniformly charged on the plane, and the charge surface density is .

The field strength of the disc at that point is.

The electric potential is. Electrostatic field 1The radius is with two spherical conductors, each charged, and the two spheres are very far apart, if.

Connect the two balls with a thin wire.

Let the radii of the two spheres be r1 and r2 respectively, and the charges after the wire connection are q1 and q2 respectively, q1 + q2 = 2q, and the potential of the two balls is respectively.

When the two balls are connected, the electric potential is equal, v1 = v2 has.

2q (r1+r2).

2.As shown in the figure, a metal spherical shell with an inner radius A and an outer radius b has an electric charge Q and a little charge Q at the center of the sphere R in the cavity of the spherical shellLet infinity be the zero point of electric potential.

1) Induced by electrostatic electricity, there is an induced charge -q on the inner surface of the metal spherical shell, and a charge q+q on the outer surface

2) (3) Ball center

Draw first.

1。Understand the total displacement (the magnitude and direction of the straight-line distance from the start point to the end point) with the Pythagorean determination; Add several lengths in the question to get the total distance.

2。The average speed, removed in bits, is divided by the total time, and the average rate is divided by the total distance.

Let the whole BAI system accelerate a1 relative to the bottom to the upper DU zhi

motion, let the mass DAOM2 move downward with acceleration a2 relative to the version pulley; And the right is known to be light rope and light pulley, then the tension in the rope is equal everywhere, and the tension force of the rope to the two mass blocks is F1, and the direction is upward. The ground is selected as the reference frame, and the phase is selected as the positive direction, and the dynamic equation of the mass column is selected.

For m1: f1-m1g=(a1+a2)m1; Note: The left side of the equation is the resultant force of m1, and the right side of the equation is the acceleration and mass product of m1 relative to the surface.

For m2: f1-m2g=(a1-a2)m2; Note: The left side of the equation is still the resultant force, and the right side is the acceleration and mass product relative to the surface.

For the whole system: 300-(m1+m2)g=(m1+m2)a1; Note: The product of the combined external force (not considering the internal force) of the entire system on the left side of the equation and the mass and acceleration of the system on the right.

In the above three equations, the unknown quantity is exactly three, the rope tension f1, the acceleration a1, a2, the solution is very simple. Answer: If a2 is found to be less than zero, it means that the acceleration direction of the mass originally assumed is opposite to the actual one, which should only occur in the case of deliberately assuming the opposite.

It is entirely possible to assume that if the mass of a large mass accelerates upward, the acceleration of the solution will be less than 0, which means that the actual result is in the opposite direction to the assumption.

The first question: Why is the amount of charge induced by the inner and outer spherical shells still as much as the amount of charge of the point charge?

This is determined by the law of conservation of electric charge. The point charge q located in the spherical shell generates an electric field, and the spherical shell is in the electric field, and electrostatic induction occurs, so that the inner surface of the spherical shell induces an induced charge -q that is opposite to the point charge, and according to the conservation of charge, there must be an equal amount of induced charge q outside

The second question: why is the internal distribution uneven and the external distribution uniform?

Because the point charge is not located in the center of the sphere, the charge density on the inner surface of the spherical shell that is closer to the charge is larger, and the density of the inner wall of the spherical shell that is farther away is smaller, so the internal distribution is uneven.

On the outer surface of the spherical shell, because the charges repel each other, the result is that they move away from each other, and the end state of this tendency is that they are evenly distributed on the outer surface. It is important to note that if the outer surface is a general curved surface, the areal density of the charge is positively correlated with the curvature.