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The distance that A runs from the front S0 of the relay area to the end of the relay area is S A=20+16=36, and the time spent is t=36 9=4s. There are two situations when A catches up with B, 1, B's velocity has not yet reached 8m s, 2, B's velocity has reached 8m s, obviously the second case B's acceleration is larger, just discuss the second case. When A catches up with B, B's velocity is already 8m s, then B starts with a uniform acceleration A B until the speed reaches 8m s, then the time it takes for B to accelerate to 8m s velocity t1=8 a, and then time t2 passes, A catches up with B, then the equation can be obtained.

1 2AT1 2 + 8T2 = 9 (T1 + T2)-16, substitute T1 to get 32 A = 72 A + T2-16, solve T2 = 16-40 A, and because the baton is in the relay area, so T1 + T2 4, T2 4-8 A, get 16-40 A 4-8 A, solve 8 3m s 2 A, this question can also be asked, to make A and B run the fastest speed, how much should B's acceleration be, the answer is as follows:

Because A and B run a total of 200 meters, and A is faster than B, so to maximize the speed of the whole journey, A needs to run as much distance as possible. Therefore, A and B can only run the full distance at the maximum speed when the baton reaches 8m s at the end of the relay area and B reaches 8m s.

Then, the distance that A runs from the front of the relay area s0 to the end of the relay area is s A = 20 + 16 = 36, and the time taken is t = 36 9 = 4s.

B starts with a constant acceleration until the speed reaches 8m s, then the time taken by B to accelerate to the speed of 8m s t1=8 a can be obtained, and the time t2=4-t1 when B runs to the end of the relay area at a speed of 8m s can be obtained.

1 2at1 2+8t2=20, substitute t1 and t2, 32 a+32-64 a=20, a=8 3m s 2

The answer is the same, that is, when B accelerates the most, A and B run the maximum speed.

I think the title is a bit weird, and it's not very expressed.

The meaning of the question is: A chases B at 9m s, and B starts to run when A is 16m behind B, and when B is asked how much acceleration does not exceed how much, A can catch up with B within 20m.

Solution: Directly analyze the situation where B has the largest acceleration, that is, A just catches up with B at 20m.

The time for A to run is (16+20) 9 = 4s=t, so B runs to the position of 20m in 4 seconds, and 1 2at squared = 20 to solve a=

i.e. the maximum is.

The time to reach the conveyor belt must be greater than the time from P to Q, that is, the pipe is the shortest when the vertical conveyor belt is perpendicular, because the time to slide down along other inclined planes to reach the circumference is equal, and it can be deduced that if P is the center of the circle and Pq is the diameter of the circle, then any C must be outside the circle, and the time must be longer than Q.

This means that the tangent point is at point Q, the distance from point P to point Q is the shortest, and the distance from point P to other points on the conveyor belt other than point Q is longer.

1. The ratio of pressure to force area, PA, becomes quietly ignited to call Daqikai.

2. Increase the stress area and decrease, 90000

4. Duan Chang, the density of B is 1000kg m 3 unknown questioning.

The pressure acting vertically on the unit area, the pana is large.

Defibrillation decreases the large contact area.

When the pressure phase is equal to the number of eggplants, the depth of B is large.

bqv=mω^/r=mv^2/r

r=mv/bq

1) It can fly out of the AB boundary, Rmax(1+CoS30°)=1+CoS30°)MV BQ=D

V mask deficiency = 2BQD (2+ crypt3) m

2) Vertical CD boundary flyout d = rcos30° = MVCOS30° bq = 3mv 2bq

v=2√3bqd/3m

uq=1/2*mv^2

u=2b^2d^2q/3

Let the electric field width l, then a=f m=eq m=uq ml=2b 2q 2d 2 3ml

t=v a=(2 3bqd 3m) (2b 2q 2d 2 3ml)= burnt 3l bqd

3)v=√3v=2bqd/m

r=mv/bq=m(2bqd/m)/bq=2d

s=2√([r^2-(r-d)^2]=2√[4d^2-d^2]=2√3d

First of all, the uniform acceleration, the formula s=vot+1 2at 2(vo=0), then s=1 2at 2, the time 2s brings into the formula: the distance of 2s s=2a, the total distance is 4a,

Total distance 4a brings in the formula: t = 8 (root number 8).

Car B catches up with car A, that is, the distance of car B and car A is the same, the time is the same, and the time is set to t

s A = vt = 14t = s B = 1 2at = 1 2 2 t , deformation, t 14t = 0, solution equation t1 = 0 (rounded), t2 = 14s

v B = at=2 14 = 28m s

The distance between car B and car A is the largest, as long as car B is behind car A, then the distance between the two cars becomes larger, as long as the speed of car B exceeds car A, then the distance between the two cars becomes smaller, when the speed of the two cars is the same, the distance between the two cars is the maximum, v B = v A = 14m s, t = v A = 14 2 = 7s

s difference = s A B = vt 1 2st = 14 7 1 2 2 7 = 49m

Height = Total Height Falling Distance = 100 1 2gt = 100 1 2 10 4 = 20m

V end = V0 + AT, 24 = 20 + 2A, A = 2M s

s=v0t+1/2at²=20×2+1/2×2×2²=44m

This question lacks a condition: the quality of the AB conductor rod is not given.

1) When the conductor rod AB moves to the right at a uniform speed, the magnetic field line is cut to generate an induced electromotive force EE BVL

It can be seen according to Ohm's law i e (r+r) of the whole circuit.

i＝e/（r+r）＝2/（3+1）＝

It can be seen according to Ohm's law i u r for partial circuits.

u IR So the potential difference at the AB end (AB is equivalent to the DC power supply, and the potential difference at the AB end is the voltage at both ends of the power supply) is.

2) The potential difference between the pm is the voltage at both ends of the power supply.

3) Magnetic field force formula f bil

The conductor rod moves at a uniform speed.

So the tensile force on the AB conductor rod is also.

4) If the external force is removed, the metal rod will make a deceleration motion with a gradual decrease in acceleration under the action of ampere force (6) according to the kinetic energy theorem.

e=blv= v,i=e rtotal=2 4= a; uab=ir= v;

ump=ir=

Due to the uniform motion of AB, the force of AB is balanced; (The right-hand rule can be used to determine the direction of the ab current from A to B).

i.e. f outside = f a = bil = n

If the external force is removed, the acceleration gradually decreases.

In my opinion, this question q = w electricity = w kean = kinetic energy change = 1 2 * m * v 2, but there is no m. You see if there is a way to ask for quality, if not, ......

Bounty, what about points? If you can't fight for other buildings, don't choose me, hehe.

4. Option AC pair.

Only when the magnetic flux through the wireframe changes, the wireframe loop will have an induced current, A is right, B is wrong; In the process of inducing current, it is the mechanical energy that is converted into electrical energy, C is right (if you investigate the grammatical problem: there is also the conversion of electrical energy into internal energy and it is late, then you don't need C), D is wrong.

5. D pair. It is better to analyze with Lenz's law. From the smaller loops in the diagram, the magnetic flux increases, and the magnetic field of the induced current should hinder the increase of the magnetic flux, and the ob should move counter-circular and clockwise (the area of the loop should be reduced to hinder the increase of the magnetic flux); It can also be seen in a larger loop, the magnetic flux decreases, the magnetic field of the induced current should hinder the decrease of this magnetic flux, and the ob should move counterclockwise (by increasing the area of the loop to hinder the decrease of magnetic flux).

1. Solution: Let the speed through the upper edge of the window be v0, the time to pass through the window is t=, and the height of the window is h1=2m, then.

The velocity passing through the lower edge of the window is vt=v0+g*t=v0+10*, so h1=(v0+vt)*t 2=(v0+v0+4)*, we get v0=3

The height of the object from the upper edge of the window when it begins to fall h=v0 2 2g=2, solution: let the interval between falling balls be t, then.

1 2g(10t) 2=125 solution, t=let the velocity of the third ball be v3, and the velocity of the fifth ball be v5, then.

v3=v0+g(2t)=0+10*(2*

v5=v0+g(4t)=0+10*(4*

The 3rd ball and the 5th ball are separated by s=(v3+v5)*(2t) 2=15m

"2=2 to obtain v=4m s

Because v=at2 so t2=

So x=1 2at2"2=

2=x substitution 1 2*10*t"2=125 t=5st2=t/1o=

x3=1/2at3"2=1/2*10*"2=x5=1/2at5"2=1/2*10*"2=x=x3-x5=

Count the last few counts yourself! If you don't have pen and paper around, you can't do the math!

2 represents the square.

1: Set the top velocity to v; Then there is vt+1 2*gt 2=4 to get v and v 2=2gs to get s 2: the first ball falls time t; 1 2gt 2=125 to get t, then the time interval is t 10 and the falling distance of the third ball is 1 2g*8 2(|The distance of the fifth ball is 1 2g*6 2(2) The distance between the two balls is the subtraction of the upper two formulas.

1. The upper edge of the window, from the acceleration distance formula, the upper edge of v = 3 meters per second. Second. Again the distance formula 2, x = meters.