How can I solve a physics problem wrong, what is wrong?

= This problem is difficult to calculate the acceleration, and it is directly sprinkled with the law of linear motion at a uniform speed.

The fifth seven seconds are the 28th 35th second, and the eleventh 3rd second is the 30th 33rd second.

Because the ratio of the displacement of a uniform acceleration linear motion with zero initial velocity in one second, two seconds, and three seconds is x1:x2:x3 ......= 1：

4:9……So the displacement in the fifth seven seconds is equivalent to the displacement in the 35th second minus the displacement in the 28th second, which is equal to 35*35 - 28*28 = 441, and the displacement in the eleventh third second is equivalent to the displacement in the 33rd second minus the displacement in the 30th second, 33*33 -30*30=189

The second method is to use the instantaneous velocity at the midpoint time to equal the average velocity, and the midpoint moments of 28 35 and 30 33 are all equal to the velocities at this moment As can be seen from the formula for average velocity, x1:t1=x2:t2

t1=7 t2=3 So I can calculate x1:x2=7:3 in the first round of physics review book in my senior year of high school.

Your formula is not wrong, and I have used your method to calculate whether it is a decimal place, close to 3:1 or 3:2

The original intention of the questioner is not to ask you to calculate the acceleration, so it is not accurate.

For such physics problems, you should use the formula correctly and understand the conditions under which the formula applies

s=1 2 at 2 is the displacement of an object with zero initial velocity and acceleration a in t seconds. The initial velocity of zero is a prerequisite for the application of this formula, and the displacement in the fifth 7 seconds, that is, the displacement of 28 and 35 seconds, cannot be directly found by this formula, because the velocity of 28 seconds is not zero, and the precondition of using the formula should be directly calculated by using the formula s=at+1 2 at 2. It is also possible to use the s=1 2 at 2 formula twice to obtain the displacement of 18 seconds and 35 seconds relative to the starting point, respectively, and to obtain the displacement of 7 seconds.

Solution: Inside parentheses are to complete the solution steps!

1> Find the acceleration: f-umg=ma, (this step is not required)!!

2> Find the difference between the displacements of 35 28 seconds: s=1 2 at 2,( s1 =vt+1 2 a35 2-vt-1 2 a28 2).

3> Find the difference between the displacements of 33 30 seconds: s=1 2 at 2,( s2 =vt+1 2 a33 2-vt-1 2 a30 2).

4> proportional results. （s1/s2 =7:3 ）

This topic is a linear motion with uniform variable speed, the time is known, and the displacement ratio is found, so there is no need to find acceleration!

Find the difference between the displacement of 35 and 28 seconds The displacement of the first 35 seconds should be subtracted from the displacement of the previous 28 seconds.

It's the same way to find the difference between the displacement of 33 and 30 seconds, and find the displacement of the first 30 seconds and the first 33 seconds, and then subtract it.

The comparison of the two displacements from above comes out.

The free fall formula is for vertical motion, l is the distance from the starting point on the slope to the landing, and your first equation should be changed to 2lsin gt.

Reason: This motion should be seen as a synthesis of a uniform motion in the S direction and a free-fall motion in the vertical direction. From the figure, several angles are equal to , s and l are equal. h＝ssinα＋lsinα＝2lsinα。

As for your second company, I don't understand your thinking, my thinking is that the S direction moves at a constant speed, S, and it would be nice to combine this formula with the first formula just now.

There is nothing wrong with this question, the speed of the water refers to the speed of the other side of the water, after the hat falls into the water, because the hat is relatively stationary against the water, so the speed of the drift is also the speed of the water on the other side. On the other hand, the topic clarifies that the speed of the ship relative to the water remains the same, but the speed of the ship to the shore changes when the speed is against the water and the speed of the current, one is the speed of the ship against the water plus the speed of the current, and the speed of the ship against the water is the difference between the speed of the ship against the water and the speed of the current. The boat catches up with the hat downstream at point A, which includes the distance of 30 minutes of drifting with the hat in the current and the distance it takes for the boat to turn around and catch up with the hat, and the distance for the hat to continue drifting.

When all movements are based on the shore as a reference, the column formula is more complicated, so a simple solution is suggested, when the water is stationary, the speed of the boat is the same.

There is something wrong with this topic.

According to this topic, after the straw hat enters the water, it should drift with the current, and the speed should be equal to the speed of the water, that is, it is already known that the straw hat has drifted for kilometers after falling into the water for 30 minutes, and the water velocity is directly calculated.

There's nothing wrong with that

This refers to the fact that the velocity of the boat is constant relative to the still water.

Hello! Which physics book is this? I've learned so far, and I don't understand it, thank you!

"The car continues to point B and enters a circular smooth track fixed in the vertical plane, passes the highest point of the track P, and then enters the horizontal track CD. ”

The course of the movement has already been specified, and it makes no sense for you to say now whether it will not be on the orbital CD.

This one. Isn't it normal for the highest point to have a velocity of zero? That's right.

1) Average velocity = total displacement Total time = 100 16 = average velocity has nothing to do with the velocity at a certain moment, only with the total displacement and total time 2) Let the tilt angle of the board be a (0f=g*sina=mg*sina

So the bigger the a, the bigger the sina and the bigger the f.

If you avoid it, fix the plank.

1. The average speed of the whole process must be the total bit removal to the total time, and the other speeds given to you are just conditions to confuse you.

This question uses 100 16=

2. If the inclination angle is too large, the object will start to move without external force, and if the inclination angle is too small, the friction force cannot be completely balanced.

It should be 100 divided by 16 to equal 25 4, right? Because the average velocity is the ratio of the time when a certain displacement occurs to the time corresponding to the occurrence of this displacement, the displacement of the whole process is 100 meters and the time is 16 seconds, so the average velocity is 25 quarters of a second per second.

As for experiments, I think you can do a force analysis of the car and the object, and then analyze it, and you can do a few more sets of experiments.

The 100-meter race, so the whole 100-meter race, a total of 16s, so it is 100 divided by 16

In the first conceptual question, just put numbers into conceptual terms. In order to avoid the second question, too much or too small will cause errors in the test, so you should slowly change the angle to find out the angle at which the block starts to move.

c Because the two conductors of A and B are connected in parallel, the ratio of passing current is 2:1, so the resistance ratio of the two conductors of A and B is 1:2, so when they are connected in series, the voltage ratio of the two ends is 1:2

You're right, the answer is wrong! It should be c.

How to calculate, you should make it clear.

The voltage of the sliding rheostat.

Total current = p=

You write down the process.

Normal light, the voltage at both ends of the bulb, the voltage at both ends of the sliding, series connection, current, rated p=.

I don't know, right?

The data in the original question is incorrect, and it is proved as follows, it is conceivable that if the acceleration of car B is greater, the longer it will take for car A to catch up with car B. So assuming that the acceleration of car B tends to infinity (the acceleration time can be omitted, and it is thought that it can enter a constant state at 12m s immediately, then it takes A the longest time to catch up with B!). by (va vb)*t up to s0

That is, (20 12)*t is up to 84, and t is the longest second, so it can be seen that the longest time is also seconds, and it is impossible to take 12 seconds.

Correction: The speed of car A in the original question should be 17 m s.

Idea: First determine whether car B is caught up in the acceleration phase or at a constant speed. Using the hypothetical method, it is assumed that car B is caught up during the acceleration phase, and the speed of car B is V at this time (if V is launched by 12m s, the assumption is true, otherwise it is not true).

is given by va*t s0 [(vb0 v) 2]*t (find the average velocity for b).

17*12 84 [(4 v) 2]*12 , get v 16 m s The condition in the problem is 12 m s, it can be seen that car B is only caught up at a constant speed.

Let the acceleration time of car b be t, then va*t s0 [(vb0 vb) 2]*t vb*(t t )

17*12－84＝[（4＋12）/2]*t ＋12*（12－t）

The time of car B in the acceleration phase is t 6 seconds.

I saw it clearly. After a little calculation, it seems that there is something wrong with the question.

A chases B, B accelerates from 4 meters per second to 12 meters per second, and then at a constant speed, A has been driving at a constant speed of 20 meters per second. Since speed A is always greater than speed B, the distance between the two cars should be constantly decreasing. It can be calculated that the distance of AB at this time is (20 12) 12 96 meters is greater than the initial 84 meters, right?

Please, I can't see it at all, and I will send it up clearly.

You're too small, aren't you?

The isochronous nature of the free string motion of the b a ball is 2* root number (r g) and the b ball is the root number (2* r g) according to the free fall

cAccording to the periodic formula of a single pendulum, it is 1 4*2*pai*(r g)=pai 2*root number(r g).

Compare the b-ball to the first landing.

I think it's B, which is kind of an energy conversion problem.

Method 1: It can be solved with the help of the kinetic energy theorem and the conversion relationship of gravitational potential energy.

Method 2: Use momentum to solve.

The isochronism of the free string motion of the ball A is 2* root number (r g) and the ball b ball is the root number (2* r g) according to the free fall

cAccording to the periodic formula of a single pendulum, it is 1 4*2*pai*(r g)=pai 2*root number(r g).

Compare the b-ball to the first landing.

Hey, let's discuss it if you have any questions!