In the physics problem of the first year of high school, I will help analyze it and write out the so

1 From the mathematical knowledge, it is known that the maximum value occurs when the speed of both sides is equal, and this is a rule, because at the beginning, the speed of the police car is small, and the gap is widening, and when the speed of the police car is as big as it, and it is still accelerating, the gap is narrowing.

At this time, t=5+2=7s, the first two seconds difference, s1=10*2=20m, and the rear 5s: police car, truck, s3=10*5=50m

So the maximum distance is s=45m

2 When the maximum speed is reached, it starts for 6s, and the difference is 20+ t=s (v-v1)=22s, and there are 6 seconds left in front of it, so it starts for a total of 28s

1.The distance traveled by the truck is 10t+20 police cars.

The distance between the two cars is the former minus the latter: s=-t 2+10t+20=-2(t-5) 2+45

So, when t=5, s is up to 45m

2.Set it to catch up in t seconds after launching.

The truck had already walked 20m when the police car started, and the total distance when catching up was 10t+20 police car reached 12m s:after 6 seconds

When the two are equal, then 10t+20=12t-36 t=28s

1. At the beginning, the speed of the police car was relatively small, less than the speed of the truck by 10m s, and the distance between the two cars continued to increase (the police car was thrown away); When the speed of the police car is greater than 10 m s, the distance between the two cars begins to decrease (the police car starts to catch up).

So when the speed of the police car reaches 10m s, the distance between the two cars is the largest.

In this case, the time t=10 2=5s, the displacement of the police car s1=1 2*at 2=25m, the displacement of the train s2=10*(2+5)=70m

s=s2-s1=45m

2. Set t seconds after starting to catch up.

The truck had already walked 20m when the police car started, and the total distance when catching up was 10t+20 police car reached 12m s:after 6 seconds

When the two are equal, then 10t+20=12t-36 t=28s

1。First of all, when the police car starts to chase, the distance between the two cars is 2x10=20m, and the driving distance of the police car can be obtained if the time is t, and the truck is 20+10t

Subtraction yields the quadratic function equation: -t 2+10t+20=0, and the maximum distance is 45m when t=5s is obvious.

Then it takes 6s to reach the maximum speed time, and the distance between the two cars at this time is 44m with the first equation, and the speed difference between the two cars after reaching the maximum speed is v-v=2m s, then it is necessary to drive 44 2=22s, plus the acceleration time, a total of 28s

1 ab is definitely not correct There is no weightlessness and overweight in the process of constant velocity Option c Since the speed of the passenger at the beginning is pointing to the right and then to the left down, the passenger's motion state has changed, and it must be subjected to the friction force to the left According to Newton's third law, option c is correct The change in the acceleration of the passenger is not clear in option d So the change in the magnitude of friction is unknown So it is false The answer to this question is c

2 There is the formula p=fv to start (f is the resultant force of the kitten's pressure and friction on the board) In this problem, since the kitten does not shift its position, its forces are balanced, so its net force is zero, because it is subjected to three forces (gravity and the friction and support force of the plank to the kitten), because the gravity is unchanged, so f is constant, and notice that the friction of the kitten on the board is downward to the left, the pressure is lower to the right, and its own gravity is downward, and its velocity is downward, so it has a downward acceleration So its velocity keeps getting bigger, so p=fv shows that p is increasing.

1. There is no answer. But when the elevator is reversed, look at the horizontal direction, the elevator slows down, and the person wants to slow down, so the friction is to the left, and then accelerate to the left after seeing zero, at this time the friction is still to the left, and finally there is no friction in the uniform movement.

2. It should be enlarged all the time and know to leave the plank.

1.Select C Uniform speed movement, force balance, acceleration is 0, neither overweight nor weightless, so AB is wrong; Before and after changing the direction, there is no acceleration in the horizontal direction of the person, the person is not subject to friction, and there is no friction on the escalator, in the process of changing, there is acceleration, and the horizontal acceleration component of the person is to the left (because the horizontal speed of the person changes from right to left), the person is subjected to friction, and the friction force is also generated on the escalator, so the friction force should first increase and then decrease, d is wrong; In the reverse process of the escalator, the person accelerates to the left, and the person receives the friction force of the escalator to the left, and the friction force of the person to the escalator is to the right.

1. Choose C, move at a constant speed, and the acceleration is 0, so it is not overweight or weightless, AB is wrong; The acceleration of the reverse process of the escalator is to the left, so the friction force to the right received by the person is the direction of the friction force of the person to the escalator to the right, c is correct; The change of friction in the reverse process of the escalator is a complex process, there is weightlessness in this process, the pressure decreases, but it generally does not cause skidding, so the friction is still static friction, and the acceleration process time is short, and the speed has a maximum value, so the acceleration is first large and then small, so the friction should be large first and then small.

2. For kittens, the force is balanced, and the resultant force is 0; In the case of planks, the resultant force is constant and the descent accelerates uniformly, so the velocity is proportional to time.

According to p=fv, the power of the kitten increases uniformly over time, and the image should be an oblique straight line past the origin.

First of all, taking B as the object of analysis, walking 5m in 2s, according to V0 2+1 2at 2=S, V0=0, the acceleration A=, then the resultant force of B is: F = Ma=75N, and the friction between B and the ground is ignored, so the friction between B and A is: F=F-F = 45N, the resultant force of A is the friction force, the magnitude is 45N, so the acceleration of A is:

45 20=, then the velocity of a at the end of 2s is: , the displacement of a in 2s: 1 2*, then the displacement relative to b is.

So the answer to this question is: ABC

For b: s=1 2(at 2) a= f-f1 =ma 120-f1=30* f1=45n

For a: f1=ma1 45=20*a1 a1=2s at the end of a velocity: v=at= sliding distance: s=(1 2)at 2 s=

Correct option: ABC

Answer: Let the mass of the last carriage be m, decoupling in the middle, the resistance f1=-km of the decoupling of the last carriage, and the acceleration of motion a1=f1 m=-k(k>0) The displacement s1 passed after decoupling, then v 2-vo 2=2as , s1=v 2 2k After decoupling, the train travels a distance of l with velocity v2, and the resultant force f=km, acceleration a2 a2=km (m-m), v2 2-v 2=2[km (m-m)]l v2 2=2[km (m-m)]l+v 2 acceleration a3=a1=-k displacement s2 s2=v2 2 2k= 2k=m (m-m)l+v 2 2k The distance between them δs=l+s2-21=l+m (m-m)l=ml (m-m).

Isn't the initial velocity and drag force given a proportional relationship to gravity? I just calculated, those two can't be eliminated... Here are my steps to do the question, I hope it will be helpful to you.

Solution: Let the relationship between drag force and gravity be: f=kmg The train starts to move at a constant speed v1.

For the whole train: traction force f=kmg

For the last carriage m: its force is only subject to one resistance f1 = -kmg (force analysis draw it yourself).

Its acceleration a1 = f1 m = -kg

Therefore, the distance it travels after leaving the train is s1 = v12 2a = v12 2kg

For trains, before the traction force is removed: a traction force and a drag force f2=-k(m-m)g

Its acceleration a2 = kmg (m-m).

l=v1t+at2/2 （1）

After the traction force is removed: the instantaneous velocity v2=v1+a1t that has just been removed is subject to only one resistance f3=k(m-m)g

Its acceleration a3 = kg

When the speed of this part of the train is zero, the distance it travels again is s2=2a3 squared of v2

Finally, the distance between them is s=s2+l-s1

In the end, we typed out the answer... It's too complicated... One step is to multiply the left and right sides of equation (1) by 2a1 at the same time

Hope it helps.

Wait a minute, I'll give you a V-T image solution.

From the frequency of 40Hz, it can be seen that the time of each flash is 1 40 = you can know that the initial velocity is meters per second.

g can be obtained from the formula δs=gt

The speed of a is easy to find.

We have just done this problem, the only vertical direction of this problem happens to be one to three to five, then the first point is its throwing point, according to t=1 f, to find the time interval is, and according to delta v=gt, to find g=,,, and then according to the initial velocity = x g delta x,,, to find the initial velocity is, but through the point a is the velocity for the time being, I still won't ask. That's all there is to help you.

First calculate the time t of the flash according to the frequency, calculate t=1 40= s, and then let the first point velocity be v, the system of equations: vt + 1 2gt = v(2t) +1 2g(2t) = substitute t, g and v can be solved, v=0m s g = m s The initial velocity of the ball is vo=( (

The speed of crossing a is:

First, analyze the vertical direction v=v(2t)+1 2g(2t) = because the horizontal velocity is constant, the velocity of point a is calculated using the Pythagorean theorem.

va= (v +vo )=Because the vertical velocity is small, the calculated velocity is approximate, not true.

Solution: The period t1 of the satellite provides the centripetal force of the satellite's circular motion around the moon according to the gravitational force.

gmm r = rw = r(4 )t1, orbital radius r = h+r

And because mg=gmm r, that is, g=gm r

The period of the Earth's rotation is t, the satellite rotates once, the angle of the Earth's rotation is 2 t1 t, and the arc length that should be photographed is s=2 r (t t1)=4 (h+r)( h+r) g) t

Extend 2 ropes at the point o through the o point to do o point to do pq perpendicular line and pq to c to know the angle pqo = b = 30 degrees angle qpo = a = 60 degrees angle poq = 90 degrees.

Then the calculation shows that pc:cq=1:3 gets pc=1 from pq=4, so choose a, do you think right?

Quick Answer:

First of all, d is excluded, and the center of gravity must be between pq;

If the center of gravity is in the middle, the angle of the two ropes will be symmetrical;

Remaining A and C, the center of gravity should be biased to the left according to the rope angle;

So, the answer is a.

If you want to solve the problem in detail, you can take the point p as the rotation point of the bar pq and do a moment balance analysis.

I haven't studied physics in a long time. The first thing to know is that the slope (rate of change, derivative, differential) of the potential curve at a point indicates that the field strength at that point is small. However, it should be noted that the field strength is not directly related to the electric potential, where the electric potential is 0, the field strength is not necessarily 0, and where the field strength is 0, the electric potential is not necessarily 0

kb>0, kc<0, ko=0, ec=false.

B is steeper than C, i.e. EBX is larger than ECX, A is correct.

ubo<0,uoc>0,know wbo=-qubo>0,woc=-quoc

Since the electric field lines always point from the high equipotential surface to the low equipotential surface, the qualitative --- error by o.

The electric field is the gradient of the electric potential, and here ex is the slope of the curve, you should understand.