Solving... Physics is a compulsory subject in high school.

This question examines the synthesis and decomposition of force, with orthogonal decomposition is more clear, because I can't draw a diagram, I tell you to parallel to the inclined plane as the x-axis, to the action point as the origin, to the straight line perpendicular to the x-axis as the y-axis to establish the coordinate system, and then analyze the force on the object, decompose the force that does not fall on the coordinate axis to the coordinate axis, and then use pure geometric knowledge to solve it, and tell you that your force analysis is not thorough, and there is the support force of the oblique block. The method tells you, so let's figure it out for yourself. Because this kind of problem to draw is difficult to solve on the iPad, please understand.

Taking the object as the research object, the force analysis is carried out, and the gravity g is vertically downward; The elastic force fn, perpendicular to the inclined plane, and the thrust f, the Cartesian coordinate system is established in the direction of the parallel inclined plane and perpendicular to the inclined plane, and the equilibrium equation can be established

f + fcosθ = gsinθ；

fn = fsinθ +gcosθ （

Then fn = g

Hello. If you don't understand, please ask.

The Cartesian coordinate system is established with the x-axis along the inclined plane downward and the y-axis perpendicular to the inclined plane. Decompose the gravitational force g and the horizontal force f on the coordinate axes. Because the object is at rest, the net force on the same line is 0List the equations and you'll be able to solve them.

Question 1: vab = vbc, i.e. v is equal, and the displacement of the two segments is also equal. However, the average velocity of the second stage must be greater than that of the first stage, so the second stage has a short motion time and a large acceleration calendar.

Question 2 The vertical upward throwing motion is symmetrical, so half of the time it takes to pass through A twice is the time when the sail is like a while falling from the highest point to the point A, and the highest point of the same object is the same, so the displacement difference between the two periods is the height difference.

Both should move in the horizontal direction of the inclined plane with an acceleration of a = g*tanz.

At this time, the pressure of m on the inclined plane = mg cosz

The acceleration a=g*tanz moves in the direction of the inclined plane.

Pressure n=mg cosz

The initial velocity is zero, but it is not necessarily zero at the beginning of the first 5 seconds.

Speed in the middle = average speed.

The velocity at the middle moment of the first 5s is 15 5=3m s

After 4s, the velocity at the middle moment is 20 4=5m s

Change by velocity = at

t=then a=(5-3).

Let the acceleration be a0, because the muzzle velocity is 0, so there is (0+5a0)*5 2=15, (5a0+9a0)*4 2=20, and a0=??There's something wrong with the title.

The acceleration of the driver's descent is -2m s2, and the time of going down to the bottom of the slope t=10s is greater than 6 inherent dangers when people walk 20 meters.

The dynamic friction factor between the car and the inclined plane is also the same?