20 points for high school physics questions! Extra points for answering the physics questions in the

This problem can be done as follows: let the radius of the earth be r, the rotation of the earth is t, and the mass is m, then the period of the near-earth satellite is t n, so t 2 = 4 ( n) 2r 3 gm, let the radius of the geostationary satellite be r then t 2 = 4 2r 3 gm, the two are connected together, you can get, r = n 2r, then the height above the ground is (n 2-1) times the radius of the earth.

Represents the period of the near-Earth satellite of the PIE is t=24 nSo there is gmm r = mr4 t .i.e. 4 n r gm = 24 (1).

Geostationary satellites: gmm (R+H) = m(R+H)4 24i.e. 4 (r+h) gm = 24 (1).

(r+h) = n rSo, h=[3rd root number(n)-1]r

Let the radius of the earth be r, the rotation of the earth is t, and the mass is m, then the period of the near-earth satellite is t n, so t 2 = 4 ( n) 2r 3 gm, let the radius of the synchronous satellite be r then t 2 = 4 2r 3 gm, the two are connected, you can get, r = n 2r, then the height above the earth is (n 2-1) times the radius of the earth.

There is no balancing friction;

When the mass of the object is constant, the acceleration is proportional to the resultant external force.

The image is not the origin, indicating that there is no equilibrium friction.

Reason for analysis: When a=0, f ≠0, it means that there needs to be a force to balance with f, that is, there is no equilibrium friction.

Conclusion, when the mass of the object is constant, the acceleration of the object is proportional to the combined external force it is subjected to Reason: The image is a straight line.

1: There is no balance friction. 2: When the mass is the same, the acceleration is proportional to the magnitude of the external force.

Hello, first question: The gravitational force of the keg (g=mg) is converted into the pulling force of the trolley, even if the acceleration is zero at the beginning, but f(g) is already there. Second question:

As can be seen from the image, the larger f is, the greater a, and the mass of the object does not change, so we can get f=ma (Newton's second law). This is a small opinion.

This is because the object moves to the maximum static friction.

A and B two adjacent road signs use 2s, then the intermediate moment velocity: v1 = 15 2 = m s through b and c with 3s, then the intermediate moment velocity: v2 = 15 3 = 5m s The time difference between the two speeds is t=(2+3) 2 = s, then the acceleration a = (v2-v1) t = =-1 m s 2a point velocity:

va = v1-a*1 = +1 = point b velocity: vb= v1 + a*1 = -1 = point c velocity: vc = v2 + a* = 5 =

(va+vb) 2=15 2 (average velocity equal to initial velocity plus final velocity divided by 2) Similarly: (vb+vc) 2=15 3

va+vc)/2=30/5

The answer is the same as above.

The effect of these two forces is the same as that of a force with a magnitude equal to g in a vertical upward direction. That is to say, the magnitude of the resultant force of these two forces is equal to the g direction upward, and the parallelogram rule is followed between the resultant force and the component force.

With these two forces as adjacent sides and an angle of 120 degrees to make a parallelogram, the length of the diagonal is gFrom this, we can find that f= is correct.

When these two forces are in the same direction, 2f=g, f=g is correct. B false.

From the parallelogram rule, it can be seen that the larger the f, the greater the d, and the wrong d.

Select the ACA option: When it is 120, the angle between the two Fs and the vertical direction is 60, F is the hypotenuse, and the vertical component (t) is t=fsin30= Since 2t=g then f=g.

Option C: The vertical component (t) of f at =0 is t=f, 2t=g, so f=g 2 and b are incorrect.

Option d: When it is very large, the vertical component of f (t) is t=fsin(90-) will be small, so f needs to be large to provide enough t so that 2t=g

f=(g/2)*con(θ/2)

When a, f=g f=(g 2) con( 2)=g is correct.

b Regardless of the value of , f=g 2 error f is related to g, .

cWhen =0°, f=g2 f=(g 2)*con( 2)=g 2 is correct.

The larger the d, the smaller the f, the larger the error, the smaller the con(2), the smaller the f, the larger the f.

Then the following statement is correct (a c).

When =0°, f+f=g f=g2

When is 120°, 2 is 60°, and the resultant force of f and f is combined with f to form an equilateral triangle, f=g

The number of electrons moving in the entire ring is n

v = period t = s v

Frequency f=1 t=v s

Current i=q t=qf=nef=n(ev s)n=is (ev).

i=nesv

i is the current, s is the length, v is the velocity, and e is the amount of the meta-charge.

Bringing in can calculate the number of electrons n

I forgot how much, let's do the math myself.

1. While the spacecraft is moving, the earth is still rotating, and the direction of rotation is from west to east, so the projection point is west.

2. According to the figure, it can be seen that the spacecraft has to pass through the equator twice around the earth, just like the periodic motion of the direct point of the sun The longitude difference between the curves is the period of the spacecraft, (

3. The geosynchronous satellite is located above the equator and is relatively unchanged from the Earth's position, so it must pass the initial position again after a period below the subsatellite,24

I don't know if when the spacecraft reaches the other side of the planet is under the satellite, if it is 8 hours, you think about it yourself)

4. According to Kepler's law, r 3 (to the third power of r) is proportional to t 2 (the square of t), and the ratio of the radius is calculated to be 256