A physics and mechanics question in the third year of high school

I understand that 1 2gt 2=x can find t=1s, and because the v of the bucket is, the difference between the first drop and the second drop is.

But n cases, when the speed of the turntable is n (n takes any positive integer) to make the water droplets fall on the same diameter on the disk, these n cases are summarized into two types, one is the water droplets on the same side, the other is the interval (once on this side, one on the other side), and the maximum value is the second one!

The second drop is on one side and the third is on the other.

lz you count the distance between the first drop and the third drop in the first case lz I'm only a sophomore in high school Oh, hehe! It's better to give more points.,It's hard for me to play for so long.,It's really troublesome not to find the picture.。

There is h=1 2gt 2

5=1/2*10*t^2

t=1sThe distance between the bucket and the center of the circle when the second drop of water is low.

s1=vt1=

The distance of the bucket from the center of the circle when the third drop of water falls is.

s2=vt2=

When the distance is maximum when it falls on both sides of the center of the circle, I guess you think so, you are wrong.

Because when the water droplets fall, they are not in free fall, but in a flat throwing motion.

The process of falling without dripping is 1 second, and the water is still s=vt= in the horizontal direction for 1 second.

So the position of the second drop of water falling is from the center of the circle.

The position of the third drop of water falling is from the center of the circle.

Sum equals.

The second drop of water with a diameter is on one side of the center of the circle, and the third drop is on the other side, then 2v+3v=

It must be equal, and the ropes are all equally curved and straight.

Therefore, when the drying rack is hung up, when the drying rack is stable and stationary, the node is balanced by three forces, (the vertical downward tension of the node cluster imitation, the tension of two ropes of equal size) If the three forces are balanced, the horizontal component of the tension force of the two virtual Zheng Hall ropes must be balanced, and because the force is equal, the angle must be equal.

This question is a typical ** law problem. The object moves in a straight line from rest, so the direction of the resultant external force is also the direction of the straight line, that is to say, the direction of the resultant force of gravity mg and the tensile force f is along a straight line, according to the triangle rule of vector addition, you draw a diagram, according to this diagram you can see that when f is perpendicular to the resultant external force, f takes the minimum value f=mgsin, so b is correct. Note that when f=mgsin, f is perpendicular to the direction of motion of the object, that is, f does not do work, so the mechanical energy is conserved, and d is correct.

In the range of 0 to 2, tan > sin, it can be seen that f can still be equal to mgtan, but it has two cases, one is the left oblique upward, then the angle with the direction of motion is greater than 2, then f is negative work, and the mechanical energy is reduced; Another situation is to deviate to the right obliquely, when the angle with the direction of motion is less than 2, then f is positive work, and the mechanical energy increases, so c is not right.

b Do a force analysis, when the direction of f is perpendicular to on, it just forms a right triangle with g, and f is the smallest.

Answer BC,1: When the direction of f is perpendicular to on, it forms a right triangle just with g, and f is the smallest; 2: F=MGTAN West Tower, do positive work, the mechanical energy of the particle must increase.

B&d Reasons for choosing B are the same as above.

D is chosen because the external force f is perpendicular to the displacement, so only gravity does the work, and the mechanical energy is conserved.

A cyclist starts to ride in a straight line from a standstill, and he passes a distance of 1m, 2m, 3m, and 4s in the 1st, 2nd, 3rd, and 4th s, respectively.

The instantaneous velocity of the end is.

The instantaneous rate of the end is .

c.The average speed for the first 4s is.

d.The average velocity within the 4s is.

a is false, because it is not possible to obtain the instantaneous velocity at a certain moment according to the conditions;

B false, ibid.

The average velocity of the first 4s = (1 + 2 + 3 + 4) 4 = 10 4 = so the average velocity within the 4th pair of c pairs = 4 1 = 4, so d pair.

First 4s: The period from the beginning of the first second to the end of the fourth second, which is the length of the line segment from 0 to 4 on the number axis.

Within 4s: The period from the beginning of the 4th second to the end of the 4th second, which is the length of the line segment from 3 to 4 on the number axis.

End of 4s: is the moment when the end of the 4th second, the point represented by 4 on the number line.

First of all, the title doesn't say that this is a uniform acceleration motion. So the options for "instantaneous velocity" are all wrong, because there is no instantaneous velocity at all. So the rate is also not available. But the average velocity v= x t

Therefore, v(first 4 seconds) =(1+2+3+4) 4=;

v (4th second) = 4 1 = 4m s

First 4s: refers to the period from the first second to the fourth second.

4s: refers to the beginning of the fourth second to the end of the fourth second, and the time period of 4s refers to the time of the fourth second.

Within 4s: Same as the previous 4s

Within the 4th s: Same as the 4th s

This question is variable acceleration and is not easy to find.

A is based on the average speed of the first 4s, which is not true.

b is calculated according to the uniform variable speed, so it is not correct.

c Average velocity of the total distance in the first four seconds v=s t=(1+2+3+4) 4=average velocity in d v=4 1=4

The first four seconds refer to the first 4 seconds after the start of the exercise, the 4th second refers to the fourth second, and the fourth second, 4s is the moment when the fourth second has just reached after three seconds, and the 4th second is the same as the 4s.

Draw **A:

Within the first second, within the second second, within the third second, within the fourth second.

The first four seconds refer to the overall time, i.e. 4s (1+2+3+4) 4m s= c correct.

The fourth second is the last one in the picture, just like the fourth second! ( Moment )!

The fourth second is marked in the graph, which is 1s (is the time)!! 4 1m s= d correct.

Options a and b do not tell the motion displacement in the fifth second, so they cannot be solved, and it is not yet a uniform acceleration motion!!

Explain your "first 4s, 4s, 4s, 4s", "4s".

Suppose there is a train, and the train has many sections, the first 4s: refers to a period of time, and in the case of a train, it refers to four trains, from beginning to end.

4s: also refers to a period of time, less than the first 4s, is a quarter of it, in terms of trains, it refers to the fourth train, the first three trains are not counted.

There should be no such vague concept at the end of 4s or the beginning of 4s, the end of 4s refers to that moment, and the end of the train refers to the end of the fourth train, excluding the body, and the same goes for the beginning of 4s.

Within the 4s: It is similar to the 4s, but the words are more accurate.

Now when it comes to the topic, if it is a uniform acceleration motion, and the acceleration is 2m per square second, then the distance passed in the 1s, 2s, 3s, and 4s is 1m, 3m, 5m, and 7m respectively, so it can be judged that it is not a uniform acceleration, and it is impossible to decelerate evenly, because the distance traveled is increasing, and it is also not a uniform velocity, so it cannot be judged what kind of motion it is. Look at the CD two items again, the average speed: the distance in a period of time divided by the time, so the C term: the distance in the first 4s is 1+2+3+4=10m, and then divide by 4 to get the average speed; D term: The distance in the fourth second is known, is 4m, and the time is 1s, so the speed is obtained.

3g/2；√3g/6.

The force situation is briefly drawn.

It's strange that there is no direct contact between the ground and C, where does the elastic force come from, let alone the friction force, only if there is elasticity in direct contact, there can be friction if there is elasticity.

The elastic force and friction force of the ground on C are both zero, because elastic force and friction are both contact forces, and C is not in contact with the ground, so there is no elastic force and friction between them!

The elasticity of the B ball is 3 2g The friction of the B ball is 1 2g I am level, however, I can't send **, please forgive me. Pay attention to the force.

This one uses a similar triangle.

The suspension point is subjected to three forces: AC rope tension, BC rope tension, and the object tension is equal to the gravitational force that is constant.

The three forces correspond to the three forces of the triangle.

G sinedc=fbc sinecd=fac sindecWhen point A moves outward, the tensile force in BC and AC becomes larger through the change of each angle.

This should be the balance of junior high school forces.

If a lever is acted upon by three forces, the product of the magnitude of one of the forces and its arm is equal to the sum of the other two forces and their respective arms, then the lever is balanced.

The distance between the two components of point C to the left and right points A and B is shorter, and more force is needed to maintain balance.

Moving upwards, the forces on the AC bc become greater and easier to break.

First, do a uniform acceleration linear motion mgu=ma1, and get a1=2s1=v0 2 (2a1)=1m l

Therefore, before reaching the right end of the conveyor belt, the block has been slid up the slope at a constant speed of 0.

One mgsin = mA2 to get A2 = -6

The distance gained when the velocity is zero.

s=-v0^2/(2a2)=

Therefore, the highest point of the inclined plane was not reached.

Blocks are on the level of the conveyor belt.

umg=ma1

Acceleration a1 = 2m s 2

v^2=2a1s1

Smooth bevels on blocks.

mgsin37°=ma2

a2=6m/s^2

0-v^2=2(-a2)s2

s2=2/3 m

Blocks can reach the top of an inclined plane.

It's been N years since I left school, and physics has been thrown away, so let's talk about it by impression:

It seems that the initial velocity of the particle should be decomposed into horizontal and vertical directions, the horizontal should have no resistance, and the vertical should consider the deceleration effect of g, that is, the height of the rising speed to 0, if it is reached, it can reach the top, and the specific will not be calculated.