Physics Elective 3 2 How to Prepare for Winter Vacation

First of all, in fact, 90 is not very bad for physics, you have to be confident. For this kind of science, the knowledge in the textbook is actually very empty, and there is not a lot of substantive content. If you are a strong self-directed learner and don't want to go to an after-school tutorial class, you can refer to the following methods (which I did at the time):

If you want to prepare, the first thing to do is to buy a tutorial book with more comprehensive knowledge content, more systematic content, and optional 3-2 textbook analysis with after-class exercises. Then make your holiday plan, how much content to watch in a day, pay attention to the preview must be done after class practice (preferably the same day or the next day), it doesn't matter if you can't finish it, you can write selectively (for example, don't do big questions and more difficult and troublesome questions first, after all, your mastery of knowledge after your preview is definitely not 100%, and the rest of the questions can be done after the teacher finishes the class after the school starts). In addition, the winter vacation is not long, and you have to juggle other subjects and other things, so you don't have to finish all the preview, and even if you finish all the preview, you probably will forget about the rest by the time the teacher talks about it.

You can preview the content of the 2 chapters or the content before the midterm exam according to the actual situation, because even if you only have a part of the preparation, if you do it carefully, you will obviously feel that you have gained a lot after the start of school! Everyone is slowly getting through this period, come on, do it with your heart, you will definitely be able to do a good job!

If you feel that you are poor, then you will never be able to learn well, compare the determination to build a cantilever thorn stock, and then relearn the laws and theorems in physics, that is the foundation, you must lay a good foundation!

Commonly used formulas in electromagnetism.

Electric field strength: e=f q

Point charge electric field strength: e=kq r

Uniform electric field: e=u d

Electric potential energy: e

q Potential difference: u

Work done by electrostatic force: w = qu

Capacitance definition: c=q u

Capacitance: c = s 4 kd

The motion of charged particles in a uniform electric field.

Acceleration uniform electric field: 1 2*mv quv

2 qu m deflection uniform electric field:

Vertical acceleration: a=qu md

Vertical displacement: y=1 2*at

1/2*(qu/md)*(x/v₀)²

Deflection angle: =v v =qux md(v) microscopic electrosheltered reed current: i=nesv

Non-electrostatic work of power supply: w= q

Ohm's law: i = u r

Series circuits. Current: i i i

Voltage: uu u u

Parallel rattling circuit.

Voltage: u = u = u =

Current: ii i i

Resistance in series: rr r

r resistors in parallel: 1 r

1/r₁1/r₂1/r₃

Joule's law: q=i

rtp=i²rp=u²

r electric power: w = uit

Electric work: p=ui

Law of resistance: r = l s

Ohm's law for all circuits: =i(r

r) = u.

u. Ampere: f=ilbsin

Magnetic flux: macrobeam = bs

Electromagnetic induction. Induced electromotive force: e=nδ t

Wire cutting magnetic inductance line: δs=lvδt

e=blv*sinθ

Induced electromotive force: e=lδi δt

Commonly used formulas in electromagnetism.

Electric field strength: e=f q

Point charge electric field strength: e=kq r

Uniform electric field: e=u d

Electric potential energy: e

q Potential difference Yinhuiyan: U??

Work done by electrostatic force: w???qu???

Capacitance definition: c=q u

Capacitance: c = s 4 kd

The motion of charged blue mass particles in a uniform electric field.

Acceleration uniform electric field: 1 2*mv??quv??

2 qu m deflection uniform electric field:

Vertical acceleration: a=qu md

Vertical displacement: y=1 2*at

1/2*(qu/md)*(x/v??)

Deflection angle: =v v??=qux/md(v??Micro current: i=nesv

Non-electrostatic work of power supply: w= q

Ohm's law: i = u r

Series circuits. Current: i??

i??i??

Voltage: uu??

u??u??

Parallel circuits. Voltage: u??=u??=u??=

Current: II??

i??i??

Resistance in series: r

r??r??r??

Resistors in parallel: 1 r

1/r??1/r??1/r??

Joule's law: q=i

rtp=i??

rp=u??

r electric power: w = uit

Electric work: p=ui

Law of resistance: r = l s

Ohm's law for all circuits: =i(r

r) = u.

u. Ampere: f=ilbsin

Magnetic flux: =bs

Electromagnetic induction. Induced electromotive force: e=nδ t

Wire cutting magnetic inductance line: δs=lvδt

e=blv*sinθ

Induced electric front momentum: e=lδi δt

People's Education Edition High School Physics (Elective 3-2) Formula.

1.φ=bssinθ

is the magnetic flux (wb) b is the magnetic induction intensity (t) s is the area (m) sin is the sinusoidal value of the angle between the direction of the magnetic field and the surface of the conductor;

e is the induced electromotive force (v) and n is the number of turns (turns).

is the amount of change in magnetic flux (wb) δt is the time of change in magnetic flux (s);

The derivation formula is e=n=ns=nb=blvsin b is the magnetic induction intensity (t) and s is the area (m).

s is the area of change (m), b is the intensity of magnetic induction (t), l is the effective length (m), and v is the velocity (m s).

sin is the sine value of the angle between the direction of the magnetic field and the direction of motion;

Derive the formula: f amp = q = n p amp = p electricity =

f ampere force (n) vm is the maximum speed (m s) r is the total external resistance ( ) r is the total internal resistance ( ).

r conductance is the resistance of the conductor itself ( ) p ampere is the power of the ampere force (w) p electricity is the electrical power (w) v is the speed (m s);

Self=le, self-induced electromotive force (v) l is self-inductance coefficient (h), δi is the change self-inductance current, (a), δt is the change time (s);

e is the electromotive force (voltage) (v) em is the peak value of the electromotive force (voltage) (v) is the angular velocity of the coil rotation (rad s) t is the time (s);

em is the peak value of the electromotive force (voltage) (v), n is the number of turns (turns), b is the magnetic induction intensity (t), and s is the area (m).

is the angular velocity of the coil rotation (rad s);

t is the period (s) f is the frequency (hz);

um==i is the RMS value of the current (a) IM is the peak value of the current (A) U is the RMS value of the voltage (V) um is the peak value of the voltage (V);

is the voltage at both ends of the primary coil (V) U2 is the voltage at both ends of the secondary coil (V) N1 is the number of turns (turns) of the primary coil N2 is the number of turns (turns) of the secondary coil;

Derive the formula: n1i1=n2i2

i1 is the current in the primary coil (a) i2 is the current in the secondary coil (a) n1 is the number of turns (turns) of the primary coil n2 is the number of turns (turns) of the secondary coil;

Temporary cramming, just look at the concept, definition or something Physics questions are not necessarily the exams that are usually done, they usually don't, and they still won't be before the exam. Let's try harder.

It's really hard to rely on the usual accumulation to want to assault, so I will look at the formula before the exam, and if you can set it, you can set it.

The answer is "c". As long as it is noted that the direction of the magnetic flux inside the bar magnet is opposite to the direction of the magnetic flux outside the magnet, and the non-uniformity of the distribution of magnetic field lines outside the magnet, it is not difficult to draw conclusions when calculating the magnetic flux through a certain cross-sectional area.

C, the calculation of magnetic flux is required to calculate the number of net magnetic flux, and the internal magnetic flux of the bar magnet is inside the magnet, the external magnetic flux is dispersed, if the total magnetic flux inside the magnet is 100, then the return inside B is set to 40, then the magnetic flux of B is 60, a radius is large, so there must be 50 returned in A, then the magnetic flux of A is 100-50=50, so C