Senior Physics Urgently, Urgent Physics Urgent

Let's answer it briefly.

Solution:1Let the time when the block falls to the ground, the downward velocity v when it reaches B, the velocity v1 at point B, and the horizontal velocity v water, which is the horizontal velocity of leaving point A.

v1 can be obtained according to the string theorem of triangles;

2.Let the centripetal force when reaching point o be fn, and the elastic force of the track at point o to the small block is f;

Take the BC surface as the horizontal plane, and the mechanical energy at point B = 0 + kinetic energy = (1 2) * m * v1 2 =;

mcp=37°；So mp=; po=;fn (centripetal force) = mr2 = mv2 r = (f-mg);

Because the surface is smooth, the horizontal velocity of the object v2 2=33 can be obtained when there is mechanical grace conservation;

fn (centripetal force) = mv 2 r = 33 = (f-mg); f=33+mg=43n;

3.According to the force analysis of the motion of the object on the CD surface, the pressure of the object on the CD surface f2=sin37°*mg=6n; friction f=(1 3)*6n=2n;

At this time, the resultant force f3 of the block is divided into two parts, the resultant force and the frictional force of gravity and elastic force; f3=2n+mg*sin53°=10n;

The acceleration of the object a=f3 m=(10 1)m s =10m s; The velocity of the object at the time v3=v1-at=5m s-8m s=-3m s; So the object reaches its highest point and then slides down. The time it takes for an object to reach the highest point t3=v1 g=; Let the highest point be q, and the distance of CQ can be obtained by 5 4m; The resultant external force of the object when it slides down is f4=mg*sin53°-frictional force f=6n;

then the distance of the object sliding qd=(1 2)at 2=; So cd=cq-qd=;

So the height of cd is sin37°*cd=;

After a hasty approach, the answer is not necessarily correct, but the idea should be correct.

I forgot to insert**,Modified the answerIf it can't be displayed**Add me q 290293773 I'll send it to you.。

Don't fall into the mistake of horizontal in this question, and find the key parameters through vertical. In the cd segment, the acceleration is a (vertical cd) = g+. If the length is cd, then cd*sin(53°)=(2*g*vertical cd)*.

The distance of the cd can be found as the answer to question 3. It can be calculated that A(Horizontal CD)=. cd*cos(53°)=v(level)*horizontal cd)*, then v(level), which is the speed required by problem 1.

After finding problems one and three, you can know the total energy, and by finding the vertical height of Bo through the triangular relationship, you can find the velocity of the object at point O, and the pressure can be found according to the centrifugal force and pressure equal, which can be completely solved. Therefore: sin(37°)=, so, sin(53°)=vertical cd) = 8(ms s).

cd=。A(horizontal cd) = 6 (M s s), v (horizontal ) = 3 (M s); ao(vertical)=,v(o)=(v(horizontal)*v(horizontal)+2*g*; Pressure f=m*v(o)*v(o) r=33(n), which is the answer to the second question. The idea is right, but the result is problematic, because the force analysis is an oral calculation, and the questioner is asked to calculate according to the idea.

Do it with conservation of energy and conservation of momentum.

1 Free fall from a to b 1 2GT 2 = H

t=horizontal displacement is s=h so v=

2. Conservation of energy, calculate the velocity of point O, and then use the centripetal force formula to find it.

Solution: For the system composed of AB, it can be seen from the integral method that the pressure of ball A on the horizontal rod is always equal to (MA+MA)*GTherefore, in the whole process, the friction force of the horizontal rod on the ball A is equal to (MA+MA)*G*, since the total length of the rope is unchanged, there is a geometric relationship, when the B ball rises by meters, the A ball moves horizontally by meters.

Let the angle between the starting rope and the horizontal direction be A1 and the final angle is A2, then the B ball velocity longline and perpendicular to the rope direction are decomposed, and then the A ball velocity longline and perpendicular to the growth direction are decomposed, and the simultaneous can obtain va=tan(a)vbSo there is, at the beginning va1 = vb*tan(a1)=4 3m s, and finally va2 = vb*tan(a2)=3 4m s

The whole process can be known by the functional relationship:

w=(ma+ma)*g*

Note: 1. The speed of the actual movement of the object is the combined velocity, and the combined velocity must be decomposed when decomposing, and it cannot be decomposed at will.

2. Grasp the overall method and the isolation method, which is the top priority of high school force analysis.

It's embarrassing! Can't afford to upload**, are you a mobile phone party?

All right! Because it is not constant force to do work, it can only be conserved with energy.

Let's look at b first: because the velocity of b is certain, just look at the change of gravitational potential energy: 5ja is very complicated, he advances, and the velocity is changed to, then you can find w(a)+w(b)=

Oral calculations may not be accurate.

This question mainly solves two problems:

First: velocity decomposition. That is, the object speed, the rope moving speed, and the rope turning speed. The main point to grasp is that the velocity of the object is the combined velocity, and the turning speed of the rope is to see the rope turn in that direction and then disintegrate.

Second: functional relationships. The work done by the rope is equal to the increase of the kinetic energy of the two, the friction generates heat, and the gravitational potential energy of the object increases.

Once the method is mastered, any problem can be solved.

There is no diagram and it is impossible to determine where the ball A and ball B are, and it is always better to have a diagram to solve it.

Let's have a picture, imagination is not good.

To be honest, I don't have enough heart but not enough power, and I can't understand the topic without a picture.

This question is a mock exam question from last year. There is also a first question, you didn't write it, I will only give you a second question. Mn length is L, OMN is isosceles triangle, so oo'=mo'=no'。

If you look at the diagram, when the radius of motion of the particle is l 2, that is, the diameter is l. In this case, the trajectory of the particle with velocity along the positive direction along the x-axis will be tangent to Mn at the point m, and the trajectory of the particle along the negative direction along the y-axis will be tangent to Mn at O'Point, the trajectory of a particle in the negative direction along the x-axis is tangent with mn at the n-point. Particles with velocity above the x-axis will not hit the mn, but there will already be particles with the direction in the third quadrant that will hit no'on, the direction of the particle in the fourth quadrant hits the mo'Above.

The radius cannot be smaller, and it can be smaller than m, n, and o'At three points, no particles hit. then l is the minimum diameter. Solution:

From r=l 2=mv qb, you get b=2mv ql, you have to draw the following diagram, and finally 1 4

1.Solution: (1) is: Draw the direction of the extension of the rope sleeve of the two clan halls on a white piece of paper.

2) In is: first draw the diagram of the tension force f1 and f2 of the two spring scales on the white paper according to the same knowledge or scale. Then the parallelogram is used to draw a diagram of the same scale as the joint spike force f.

3) Medium is: make the end of the rubber band still elongate to the o point.

2.Solution: (1) Diagram omitted.

2) the original length of the spring; About.

3) About 144

Suppose the water flow is used as a reference. The wooden box is stationary, and the boat moves forward at a certain speed, and after walking for an hour, it finds that the wooden box has fallen, and turns around to find the stationary wooden box. In fact, the speed of the two processes from going to back is the same as that of the flowing water, and it has not changed.

Or you imagine the flowing water as the earth, you go forward, you drop something, you go back at the same speed, the velocity in between, the relative displacement does not change. So the time is still 1 hour. Then the wooden box went down the water for 2 hours.

So the speed is.