Physics concept problem! Hurry, hurry!

You have a problem with your understanding of the centripetal force.

The centripetal force is a force named after the effect, and it is not a special force, and the ** of the centripetal force is a common force or the resultant force of several forces. For example, when a small block placed on a record when it is rotated and is stationary relative to the record, when it moves in a uniform circular motion with the record, the centripetal force is provided by static friction. When the train turns, at the right speed, the centripetal force is provided by the gravity of the train and the supporting force it is subjected to**The inner and outer height of the track at the bend); When the speed is too high and the combined force of gravity and support is insufficient to provide centripetal force, the extrusion of the outer wheel rim and the rail provides insufficient centripetal force to the inward elastic force of the wheel rim.

That is, the centripetal force is provided by the resultant force of a certain force or several forces, and there is no centripetal force named by nature.

When the ball moves to the apex of the circle, the gravitational force is provided as a centripetal force, so mg=m*(v2 r).

1. If it is a uniform motion, it doesn't matter whether the vertex is or not 2 You really don't have a clear concept and don't know what the centripetal force is 3 The problem you said is the circular motion of the ball in the vertical plane by gravity and the tension of the string (or the elastic force of the track). The critical condition at the highest point is that gravity provides the centripetal force while there is no tension or pressure at this point. The centripetal force is equal to the gravitational force...

The centripetal force is the resultant force pointing in the direction of the center of the circle...

Agree with the RMM, the whole process does not need to consider the mass change during ascent because of the sentence "After ignition, it is considered to be a uniform acceleration, and after 4 seconds to reach a height of 40 meters above the ground, the fuel is used up".

That formula is s=, knowing that s=40m, t=4s; So v=a t=20m s. At this time, v=20m s, a=-g=-10m s 2, so the speed is reduced to 0 after 2s, and the highest point of the road is reached, so the distance of the deceleration phase is s=20m. The highest point is 20+40=6-m.

1. (1) At the highest point, let the minimum velocity be v1, then gravity completely provides the centripetal force, mg=m*v1 2 l, v1 root number (gl) root number (10* m s

2) When the velocity at the highest point is V2 3M s, it is determined by mg+f m*v2 2 l

The pressure sought is f (m*v2 2 l) mg ( Newton.

2. l v* t is obtained from the knowledge of flat throwing motion, and h g*t 2 2

The gravitational acceleration on the surface of the planet is g 2*h*v 2 l 2

It is known from the substitution relation gm g*r 2 , and the volume of the planet is u sphere (4 3)*r 3

The average density of the planet is p m u sphere (g* r 2 g) [ 4 3)* r 3 ].

(2*h*v^2 / l ^2)* r ^2 / g] / [ 4π/ 3)*r ^3 ]＝3*h*v^2) / (2*π*g*l^2* r )

3. It is obtained by f million g*m*m (r+h) 2.

The gravitational force sought is f million *1000 (6400*10 3 6400*10 3) 2 Newtons.

gr=2m s,2 under the root number. The pressure of water on the bottom of the barrel is f=f-g=mv2 r-mg=2.

A has 4 L sticking out at the right end, then B has 4 L sticking out from the left end, and A and B are considered as a whole and have a symmetrical shape. The length of the right side of A exposing B does not exceed L 2 and will not tip over from the right end of B.

The length of the coincident part is l-l 4=3 4l, and the center of gravity is in the center of the coincident part.

Therefore, the center of gravity of the two is at the distance from the right end of b (3 4L) 2=3 8L, in order to ensure that the two wooden blocks do not tip over, the center of gravity of the two should coincide with the edge of the table. That is, the length of the wooden block b protruding from the edge of the table should not exceed 3 8l

According to the inscription, it is necessary to keep A from tipping over from the right end of B, and not from tipping over from the edge of the table.

In order to prevent A from toppling over from the right end of B, the length of the right end of A protruding from B does not exceed L 2, and similarly it follows: the center of the whole is at a distance from the right end of B (L-L 2) 2=L 4, the length of the right end of B extending from the edge of the table shall not exceed L 4, and the maximum distance of the right end of Aa extending from the edge of the table L 4 + L 2 = 3 4L

It is necessary to find out the center of gravity of the two wooden blocks, so that the center of gravity is just to the edge of the table.

The object is balanced by four forces, and in the horizontal direction, the pressure is equal to the component of the thrust, i.e., n=fsin

In the vertical direction, the components of friction f, gravity g and thrust f are in equilibrium, i.e., f+fcos = g, so f=g-fcos

The table top is smooth if the vertical pencil is slightly disturbed and falls on the table, and the horizontal position of its center of mass is not.

Specialized, falling in the process, by gravity and desktop support force, the following section is the axis, the moment m=i (i is the moment of inertia, is the acceleration), set is the angle between the pencil and the vertical line, m=mg*1 2*l*sin =i =i*(d dt)=i*(d d)*d dt) = i* (d d)* * is the angle between the pencil and the horizontal line).

1 2*l*mg*d *sin =1 2*l*mg*d *cos =i* *d , the integral of both sides finds the relationship between angular velocity and .

The angular momentum theorem mdt=id ,dt=i m(d) is integrated on both sides to give time t

This one... Can it be solved? I think the explanation is too vague, standing upright on the table, if at first specialized.

If it's vertical, because of the balance, there will be no external interference after the genus will always be set up on the table. If you think about it from another angle, even if it is inverted, it has something to do with the shape of the bottom, because different shapes form different moments on the table, and the process of falling down is equivalent to turning, similar to turning 90 degrees, and the time taken will not be the same.

The above is purely personal analysis, I don't know if it's right or wrong. If there are other masters, learn with an open mind.

If the bai pen is regarded as of uniform quality, then the question can be abstracted into a lightweight rod and plate pen with half the length of the dao pen.

When the weight point is the center of mass of the ball, the ball is attached to the rod and can swing freely in the vertical plane, and the trajectory is a circle with the radius of the rod length.

But if you use the knowledge points of high school to deal with it, it is obviously difficult to understand. I use the knowledge points of college physics to involve integrals, which is beyond the curriculum for high school students. In addition, you did not give the quality of the pen, and here you can only make a general statement.

Isn't it right First it slides along the ground (the center of mass is always in the same straight line), then there is a critical angle (at this angle, just off the ground, gravity acts as a centripetal force), and from this angle it starts to move off the ground. So the calculation should be divided into two parts.

Some of the data in this question is foolish, when "I saw the railway maintenance worker in the distance only hit the track with a hammer", this is the last knock, that is, the last time is the time when the last knock is passed to the student's ears after seeing the last knock, so the distance is.

Sound propagation velocity) = 102m

Buoyancy 9*10 -3 n

The drainage volume is 9*10 -7 cubic meters.

CM3 does not change.