Can the law formula of high school physics motion be directly used Shenma 1 3 5 and the like to seek

It depends on the difficulty of the topic, if it is a relatively simple topic, then it is best to deduce the process and then use, if it is a more difficult topic, and using this formula is only one of the steps, you can use it directly, there is no hard and fast rule.

Landlord, you have a question.

Formulas can be used directly.

As long as the conditions are met, it is the law of movement.

That 1:3:5 is just the end result of the derivation.

For example, there is a formula for the difference between two adjacent displacements at the same time.

x1-x2=at²

This formula is the result of simplification, and it was originally derived from several basic formulas.

Chemistry is relatively simple, grasp the most basic principles of the Blapai acid-base trans Heying, precipitate, redox should be able to solve 90% of the reaction, as for physics, you can do three or four of each type of problem to summarize something by yourself every time you slip should be about the same.

I saw you write the principle and thought you were going to ask about the principle of chemical reactions, so this book came in

.Kepler's third law: t2 r3 k( 4 2 gm) {r: orbital radius, t: period, k: constant (independent of the mass of the planet, depending on the mass of the central object)}

2.The law of gravitation: f gm1m2 r2 (g ?m2 kg2, the direction is on their line)

3.Gravity and acceleration due to gravity on celestial bodies: gmm R2 mg; g gm r2 {r: celestial radius (m), m: celestial mass (kg)}

4.Satellite orbit velocity, angular velocity, period: v (gm r)1 2;ω＝gm/r3)1/2；t 2 (r3 gm)1 2{m:central celestial mass}

5.1 (2, 3) cosmic velocity v1 (g r ) 1 2 (gm r ) 1 2 ; v2＝；v3＝

6.Geosynchronous satellites gmm (R+H)2 m4 2(R+h) t2{H 36000km, H: height above the Earth's surface, R: radius of the Earth}

Note: (1) The centripetal force required for the motion of celestial bodies is provided by gravitational force, f to f thousand;

2) the law of gravitation can be applied to estimate the mass density of celestial bodies;

3) Geosynchronous satellites can only operate over the equator, and the period of operation is the same as that of the Earth's rotation;

4) The orbital radius of the satellite becomes smaller, the potential energy becomes smaller, the kinetic energy becomes larger, the velocity becomes larger, and the period becomes smaller (together with three antis);

5) The maximum orbital velocity and the minimum launch velocity of the Earth satellite are both.

gmm/r^2=mg

In general, it is used to represent gm in the form of g,r to solve celestial problems.

gmm r 2 = f direction = mv 2 r

Calculations such as finding the linear velocity of a celestial body in flight.

I can send it to your email if you want.

Kepler's third law: a3 t2=k[a is the semi-major axis, t is the period} Find the mass of the earth: m=gr g2 Find the mass of the central celestial body: m=4 2r3 gt2

The law of gravitation: f=gm1m2 y2 f=gmm r 2 This has a wide range of uses, and you can know the central celestial body and its own velocity, as well as the radius of rotation.

f=w 2mr angular velocity of its own mass and radius of rotation.

f=v 2rm linear velocity self-mass and radius of rotation.

f=ma centripetal acceleration of its own mass.

f=mg (only on the surface of the central object).

1.Kepler's third law: t2 r3 k( 4 2 gm) {r: orbital radius, t: period, k: constant (independent of the mass of the planet, depending on the mass of the central object)}

2.The law of gravitation: f gm1m2 r2 (g ?m2 kg2, the direction is on their line)

3.Gravity and acceleration due to gravity on celestial bodies: gmm R2 mg; g gm r2 {r: celestial radius (m), m: celestial mass (kg)}

4.Satellite orbit velocity, angular velocity, period: v (gm r)1 2;ω＝gm/r3)1/2；t 2 (r3 gm)1 2{m:central celestial mass}

5.1 (2, 3) cosmic velocity v1 (g r ) 1 2 (gm r ) 1 2 ; v2＝；

v3＝6.Geosynchronous satellites gmm (R+H)2 m4 2(R+h) t2{H 36000km, H: height above the Earth's surface, R: radius of the Earth}

Note: (1) The centripetal force required for the motion of celestial bodies is provided by gravitational force, f to f thousand;

2) the law of gravitation can be applied to estimate the mass density of celestial bodies;

3) Geosynchronous satellites can only operate over the equator, and the period of operation is the same as that of the Earth's rotation;

4) The orbital radius of the satellite becomes smaller, the potential energy becomes smaller, the kinetic energy becomes larger, the velocity becomes larger, and the period becomes smaller (together with three antis);

5) The maximum orbital velocity and the minimum launch velocity of the Earth satellite are both.

Newton's second law is a simple application.

The force f of the object is obtained

It can be calculated that according to the principle of parallelograms, f together = 2 * cos30 * 50n = common direction of 50 3n = f at the angle of the corner, maybe 50n bisector.

2。According to Newton's second law:

f together = m * a

Obtain a = f m = m 2 kinematic formula: vt = v0 + a * t br > s = 0 * t + 1 2a * t 2

The initial velocity of the object is 0 uniform acceleration motion, so v0 = 0 such an object at the end of velocity v 3 seconds = a * t = m s = m s displacement within 3s = 1 2 * a * t 2 = 1 2 * 9m = m.

Hmmm......Hello landlord.

Your question involves physics and mathematics, and I will elaborate on it in separate sections.

1. Newton's laws of motion, to be more precise, Newton's second law. (You are absolutely right when you say Newton's laws of motion), f=ma, which means that the resultant external force is proportional to the acceleration, from this equation we can know that a=f m, f is the resultant external force and not a single force.

Since it is Newton's law of motion, it means that kinematics are involved, so the formula of four linear motions with uniform velocity should also be known.

For example, v -v0 =2ax, v=v0+at, x=v0t+1 2at, x=at, you can also find the acceleration separately (if you don't understand, you can ask).

The point is that physics is a very logical subject, so the landlord will study more.

2. The trigonometric function in junior high school mainly involves several that is, in the right triangle, sinna = opposite side hypotenuse, cosa = adjacent edge hypotenuse, tana = opposite side adjacent edge, generally only these three points are examined. The rest belongs to the content of compulsory 4 in the first year of high school.

I don't know how to continue to ask, and the owner of the tower is satisfied.

Newton's second law f=ma, so a=f m

There are two ways to find acceleration, one is Newton's second law, and the other is kinematics.

Playing with physics formulas is like playing with blocks and a Rubik's cube. If someone else builds the blocks, turn the Rubik's Cube and give it to you, it is interesting to have wood.

It's also hard to get on the exam.

IMHO, you can't learn physics if you ask others for formulas!

The output power and input power generally refer to the working parameters of the motor.

Input power generally refers to the electrical power consumed by the motor;

The output power generally refers to the mechanical power that the motor converts electrical energy into mechanical energy, which is the output;

Because the coil of the motor has a certain internal resistance, it will convert a part of the electrical energy into internal energy and lose it in the form of heat, so the input power is generally greater than the output power, the formula: P in = ui, P out = FV, P heat = i * i * r, P in = P out + P heat; Cui = Pu Out + Yi * Yi * La