Senior 1 physics, ask for a detailed explanation, thank you

Well, you are misled by stereotyped thinking, b does a linear motion with a constant acceleration, its velocity does not necessarily increase, if the direction of acceleration is opposite to its direction of motion, then it will decelerate to zero and then accelerate.

Therefore, the velocity of points A and B can still be the same again at some point.

And when the speed is the same again, the speed of both is -v0

I don't want to count it at all、、There are multiple answers to ......A does a uniform circular motion with period t=2 r v0

Let a and b have the same velocities after time t, then t=(n+1 2)t(n=0,1,2,3,4...

For b there is -v0=at+v0

The solution is a=......(Bring it in yourself = =||.)）

The acceleration of b is negative.

After passing through one room, the direction of velocity changes to the opposite.

Let t be, b becomes -v0

t=-2v0/a

At the same time, the velocity of a is -v0, i.e., a has passed (n-1 2) cycles:

t=(n-1/2)t=(n-1/2)*2πr/v02v0/a=(n-1/2)*2πr/v0

a=-v0²/[(n-1/2)πr]

n>0 and is a natural number.

b do deceleration. The v0 direction is the positive direction.

t=2πr/v0

t=nt+t/2 n
..

v0-v0=at

a=-2v0^2/（2n+1）πr

Let the total time for the object to fall be t seconds, then the displacement of the last second can be expressed as .

s1=1 2*gt; 2-1 2g(t-1) 2, i.e., 35=1 2*10*t 2-1 2*10*(t-1) 2, and t=4v=gt=40m s

The acceleration of both is a, and the tensile force of the rope is f:

m1g-f=m1a

f-m2g=m2a

Solution: f=2m1m2g (m1+m2).

The spring dynamometer reads = 2f = 4m1m2g (m1 + m2).

The second method: m1g-m2g=(m1+m2)a to find the acceleration of the whole link.

m1g+m2g-t=m1a-m2a Newton's second law of the whole.

t is the number of spring scales.

The formula p=fv

When P represents the rated power and V represents the maximum speed, the traction force F is equal to the resistance of the car, that is, the car has no possibility of acceleration.

Therefore, if the resistance of the straight pavement is f, then p total = p1 + p2 = fv1 + kfv2 = kfv3

The solution is v3=(v1+kv2) k

The friction force has static friction force and dynamic friction The static friction force can vary between 0 and the maximum static friction, and the dynamic friction force is determined by the roughness of the contact surface and the friction factor determined by the material and the positive pressure on the contact surface f=un The contact surface is determined The object is determined Then the dynamic friction force is unchanged Therefore, the object does a uniform linear motion on the table under the action of 10n horizontal tensile force Description The dynamic friction force is 10n No matter how much force (greater than 10N) is used to pull the object, the dynamic friction force is 10N

For other objects, assuming that the tensile force of 20N makes the block move in a straight line at a uniform speed, it is explained that the dynamic friction is 20N, and the combined kinetic friction will remain constant at 20N.

f= g=, direction horizontally to the left (frictional force is opposite to the relative direction of motion.) When the object is subjected to a force f1 = 15n to the left horizontally, the net force to the object is 15 + 10 = 25 n in the direction of the horizontal to the left.

f= g, the factor is constant, the positive pressure is constant, and the relative direction of motion is constant.

f unchanged.

If you don't think of it this way, the maximum static friction can be regarded as = sliding friction, then it is 10N.

f = 10 N, direction to the left.

f=10N The direction is to the left, because the object is still moving, and it is still experiencing sliding friction, which is equal to the product of the positive pressure and the friction factor, and has nothing to do with the horizontal force.

The first empty 10n second empty left The third empty 10n The fourth empty right The reason why the third empty is 10n and not 5n is because the maximum dynamic friction of the object is 10n!! FYI.

At constant velocity, f=100*, horizontally to the left; Same as above, because the instantaneous velocity v is not 0 at this point, the object is still moving to the right

The maximum static friction is constant.

This question is not complete, first of all, it should be stated that the acceleration of the car is constant when changing gears, and it should also be assumed that the acceleration of the car and the acceleration of the deceleration a are the same.

Let the deceleration time of the car be t, then the acceleration time is (12-t) according to the title:

Deceleration process: v slowest = v0-at

Acceleration process: v0=v slowest +a(12-t).

Solution: a=4 3 m s 2 So the velocity at the end of 2 seconds v1=10-4 3*2=22 3m s also gives the velocity at the end of 10 seconds.

If you don't understand something, ask me again, the teacher is willing to help you!

Is the acceleration the same before and after?

The conditions are not enough to solve.

Because the wooden block 1 2 moves at a constant speed, the acceleration direction is zero and scattered, which can be regarded as a whole, and the elastic force is the internal force, so.

f-u(Shen's m1+m2)g=0---

For m2 there is f-um2g-kx=0---

The joint solution yields x=um1g k

So the length d=x+l=um1g destroys k+l

Remember to take oooooooo

b The desktop is smooth, if the early impulse of the object is m, and the object of the two phase beams is 2m, then when the land slag is pulled and the thin rope is broken, a=f1 2m t=ma gets t=f1 2

If the table top is rough, when the string is broken, the tensile force of the wire is t

f+ma=t t+ma+f=f2 gives f1=f2

Select C if the desktop is smooth, F1=T

If the table is rough, f2=t+f

The speed of the car is 48km h=(40 3)m good no s

The distance traveled when the car catches up with the tractor s=vt=(40 3)*30=400m

Then the distance traveled by the tractor is 400-200=200m, and the distance traveled by the tractor is half of the car, and because the time is the same, the speed is also half of the speed of the car, that is, 24km h

48km liter nucleus h=

Set the tractor speed to be v

Compared to the towing machine, the speed of the car is.

Then 30=200 (, the solution is v=20 3 m s

48km h=40 3m pure Tsai Ming s

The speed of the car has always been 48km h, and the cousin Xunming car does a uniform speed straight line movement.

v0t=200+vt

v=20/3=24km/h

48km h to Lippi (40 3) m s tractor speed which orange difference v=((40 Wu 3)*30-200) 30=(20 3)m s=24km h