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1) Kepler discovered: "The orbit of all planets around the Sun is elliptical, and the Sun is at a focal point of all ellipses (Kepler's first law); The ratio of the quadratic of the semi-major axis of the orbits of all planets to the quadratic of the orbital period is equal (Kepler's third law)".

It can also be directly argued: Kepler's first law, Kepler's third law;

2) On the basis of his previous research, Newton proved by his extraordinary mathematical ability that the gravitational force between the planet and the sun is directly proportional to the mass of the planet and inversely proportional to the power of the distance from the planet to the sun, that is, the law of universal gravitation;

3) Cavendish's main contribution was: "By skillfully using the torsion scale device, Zhidao measured the gravitational constant g with relative accuracy and calculated the mass of the earth for the first time in the laboratory." He is known as the first person to weigh the earth. ”

4) Assuming m is the mass of the Sun, m is the mass of a certain planet, R is the distance between them, and T is the period of the planet's revolution, then the mass m of the Sun can be found as follows:

The centripetal force required for the planets to move in a uniform circular motion is.

f=mrω^2=mr(2π/r)^2

Whereas the centripetal force of planetary motion is provided by gravitational force, so.

gm′m/r^2=mr(2π/r)^2

This can be solved.

m′=4π^2r^2/gt^2

3: Increase. It is usually undetectable, because what we call "detection" is measured with corresponding instruments in classical physics, and the change of mass with velocity is non-classical physics, which is derived by the ulnar formula, so it is usually undetectable.

4: No, you can't. The force acts on an object, and the object does have some acceleration, but this is limited to classical physics. In non-classical physics, the law of the ox does not apply, so it cannot exceed the speed of light. Einz Nezstan also pushes out no object that can be faster than the speed of light.

5: Gets bigger. Becomes smaller (weightless Lee).

Of course, my explanation doesn't necessarily apply to test-oriented education, because test-oriented education is often not the same as the truth (haha, off-topic).

(1) Hearing 66 sounds, that is, walking 65 sections, so after walking x=65 25=1625m, 1 and a half minutes, that is, t=90s, so there is v=x t, v=1625 90=18m s.

2) Find the average velocity in the first 2s, there are x=1+3=4m, t=2s, v=4 2=2ms

To find the average velocity in the last 2s, there are x=5+7=12m, t=2s, v=12 2=6ms

To find the average velocity of all motion time, there is x=1+3+5+7=16m, t=4s, v=16 4=4m

With the average speed: because the uniform acceleration is followed by the uniform deceleration v to 0, so the average speed of the uniform acceleration process (or the average speed of the intermediate moment) is consistent with the average speed of the uniform deceleration process, so the whole motion process can be regarded as a uniform linear motion, 15m divided by 10s to get the average speed of the movement, that is, the average speed of the uniform acceleration motion process, then the maximum speed can be found, like this kind of problem, to make good use of the average speed, the uniform speed movement into a uniform speed solution.

Note: If you don't understand the first sentence, draw an image, and a v t image can combine uniform speed motion with uniform speed motion as soon as you draw it.

Let the maximum velocity be v, then the average velocity of the acceleration process is v1=(0+v) 2. The deceleration process can be regarded as a reverse uniform acceleration, so the average velocity of the deceleration process is v2=(0+v) 2

The total forward displacement s=v1t1+v2t2=(t1+t2)v2=vt 2

By question: s=15m, t=10s, so v=3m s

If the maximum speed is 2V, then the average speed of uniform acceleration and uniform deceleration is V.

So vt1+vt2=v(t1+t2)=s,v=s (t1+t2)=,2v=3m s.

Because the first process is a uniform acceleration, the second process is a uniform deceleration, the average speed of the two processes before and after is (1 2) (v at the beginning + v at the end), and the first process is zero at the beginning of the velocity, and the second process is at the end of the velocity, so it is not difficult to know that the average velocity of the whole process is (1 2) v (v is the maximum velocity).

The average speed of the whole process is 3 (m s) at the maximum speed

Let the acceleration of uniform acceleration be a and the acceleration of uniform deceleration is -a, and the maximum velocity is vs=1 2at 2+1 2at 2

t+t=10

at=at (reverse uniform acceleration).

v=3

Solution: The maximum speed is v

Then the acceleration and deceleration times are set to t1 and t2 respectively, because they are all uniform speed movements, so the average speed can be taken.

t1+t2=10s

v/2)*（t1+t2)=15m

v=3m s is obtained

Draw a V-T diagram, the bottom of the triangle is the time 10s, the height of the triangle is the maximum velocity v, the area of the triangle is the displacement s=15, and the v=3m s is obtained according to s=15=1 2*10*v

Kai talks about h=10m s

s=v1*t1+v1*t2+

250(m)

v flat 1 = s t = 250 20 =

v Ping 2 = (S-V1*T1) T2 = 150 10 = 15 (m s) v=a*t1 = 2 * 5 = 10 (m s) branch base.

50 = t3 = 10 (Mengsun Jin S).

t=t1+t2+t3=5+120+10=135(s)