Help solve the third problem of Urgent Physics

Solution: The answer is A. A selection arrives.

The analysis is as follows: let the length of the whole journey be l.

For car A: Set the required time to t1

Then there is: 1 2*t1*v1+1 2*t1*v2=lt1=2l (v1+v2).

For car B: The time it takes.

t2=1/2*l/v1+1/2*l/v2

l(v1+v2)/2v1v2

Because v1 ≠ v2

v1-v2)^2>0

v1+v2)^2-4v1v2>0

v1+v2)^2>4v1v2

l*(v1+v2)^2>4v1v2*l

Compare t1 and t2 and know that t1 therefore: A arrives first.

If you have any questions about this question, please ask!

Let the distance between the two places be s, and the time of A is t

v1*t/2+v2*t/2=s

t=2s/(v1+v2)

For B, let the first s 2 take time t1 and the last s 2 take time t2

t1=s/2v1

t2=s/2v2

t'=t1+t2=s/2(1/v1+1/v2)=s(v1+v2)/2v1v2

Let's find t-t'

t-t'=2s/(v1+v2)-s(v1+v2)/2v1v2-s/2*(v1^2-2v1v2+v2^2)-s/2*(v1-v2)^2<0

So T arrives first.

A correct.

I saw it clearly, and I chose C to arrive at the same time. This problem can be reversed, that is, the 2 cars are driven opposite, then the operation mode is the same.

If a is selected and the total time of A is t1, the total time of B is t2, and the total distance is s, then A has t1 2(v1+v2)=s, and B has 2t2v1v2 (v1+v2)=s.

1. According to the kinetic energy theorem, fs=1 2mv 2

So the displacement ratio = 1:1

Because s=(1 2mv 2) Zhongling leaky potato (2 mg) is about m2, according to the kinetic energy theorem.

fs=1/2mv^2

So s=200m f

There are two directions to this question f, but the work can be sold to add up.

3. Let the B speed be 3V, so the C speed is 4V

4v)^2-(3v)^2=2gs

The solution yields v=2 5m s

So the c velocity is 8 5m s

The displacement is ((8 5) 2-0) (2*10)=16m

1) F=UMG MA, Tremor High V0=AT, Eggplant Ruler S=AT2 2 Because U is the same, So A is the same, and because V0 is the same, so T is the same, so S is the same, so S1:S2=1:1

2)f×s=mv0^2/2,s=(mv0^2)/(2f)=200m/f

3) Let the time from A to B be 3t, and the displacement is S1

Then the time from A to C is 4t, and the displacement is S2

So s1=[g(3t) 2] 2=9gt; 2 quietly2,s2=[g(4t) 2] 2=16gt; 2 2s2-s1=7gt; 2 2=7m

So s2=16m

1.Because the ratio of mass is equal to 1:2, the ratio of positive pressure is 1:2, so the ratio of the posture number of the friction force is 1:2, so the ratio of acceleration is 1:4, so the ratio of displacement is 1:4

(3) Solution:

As can be seen from the table, the acceleration a1=2m s 2 at a-b, the motion time of a-b is t1 and the velocity of a-b satisfies: v1=a1*t1

Looking at the process of b-c in reverse, it is a uniformly accelerated linear motion with an initial velocity of 0. The acceleration of c-b a2 = 5m s 2 and the motion time is t2

The velocity of c-b satisfies: v2=a2*t2

There is v1=v2 at point b

And because t1+t2=

From the above equations, t1=2s can be obtained

That is, the time taken to transport to point B is 2s(It's easier later.) The velocity of t= is the velocity of looking forward from the middle second of **. )

v=3-5*.01=

(1) Acceleration a=; ，f=ma=2n

2) acceleration on a smooth slope a=1;

That is, GSIN (angle) = 5, bevel angle = degree;

3）v=3-5*

Accelerated decline when weighing "weight" in factIt is in this case that the elevator supports people

By Newton's second law, there is:

mg-n=m·g/5

Solution: n=4mg 5=4 500 5=400n, choose b correctly.

So you don't really understand n0

The answer skips a step, and the complete one should be like this:

The impulse of a single particle f0δt=mv1+

Here δt should be the collision time, because of the instantaneous effect at the time of collision, δt is very short.

But what we're asking for is the total impulse of a large number of particles, how many particles are here?

It should be n0*t, because n0 is the number of particles per unit time, and only multiplied by time is the total number.

Take a closer look:

At this time, the impulse is the total impulse, and f is the total effect produced by all the particles, so what is the collision time at this time?

Because the impulse of all particles has to be considered, and the particles hit the baffle almost uniformly and continuously (after a period of stability), so how much time is taken is the same (we can take a certain period of time in between, because the total impulse is the same in the same time interval), it turns out that the collision time of a single particle is no longer meaningless, or the total impulse is no longer an instantaneous effect, but the duration of all particles in the time interval taken by it. For example, take t time in the answer.

So in the answer, f in ft is the total force that is continuously exerted, t is the duration of the force n0t where n0 is the number of particles per unit time, and n0t is the total number of particles that lasts t time.

I think the n0t here is what you understand as n0.

The charged ball is electrically unknown, and is subject to the vertical downward gravity, the vertical and rod oblique upward support force, the vertical direction of the electric field force, and the Lorentz force and friction force after movement.

It is easy to know that the speed of the BC segment is greater than that of the AB segment, and the Lorentz force QVB is also larger than the BC segment, and the electric field force and gravity remain unchanged.

by Newton's second law.

The support force is large in the BC segment, and the friction force is also large in the BC segment. aa-b

Paragraphs b-c. Gravity.

The electric field force is equal to the work respectively.

Supportiveness. Lorenz didn't do anything.

The friction force is a negative pin and the BC section is large.

The average acceleration in the a-b segment is greater than that in the b-c segment.

The velocity of the ball passing through point b is greater than 2m s, so a is wrong.

B friction does negative work BC section is large, so B is wrong.

c The electrical properties of the charged ball are unknown, and the positive or negative work of the electric field force is unknown, so it cannot be said that the potential energy of the ball must increase, so C is wrong.

There is a possibility that the ball has reached before point b.

Gravity. Supportiveness.

Electric field force. Lorentz force.

The combined spike force with friction is zero, and the uniform linear motion is done, by the kinetic energy theorem, (support force.

Lorentz force deficit does not do work) It is easy to obtain that the work done by the gravitational force and the electric field force of the ball from b to c is equal to the work done by overcoming the frictional force, so d is right.

To sum up, choose D