A high school physics problem should be analyzed in detail, 3q

Analysis: Know the position of the metal rod A1 and A2 from the time when the metal rod A1 enters the magnetic field (t=0) to A2 leaves the magnetic field, and find the induced electromotive force and current according to the law of electromagnetic induction Answer: Solution: 0-t1(

The induced electromotive force generated by a1: e=blv=

Resistance R and A2 in parallel: R and = R R

r r = so the voltage at both ends of the resistor r u = r and.

r and r e=

Current through resistor r: i1=ur==

t1-t2（

e=0，i2=0

t2-t3 ( Same reason: i3=

Answer: 0-t1(, the current through the resistor r is, t1-t2(, the current through the resistor r is 0

t2-t3(, the current through the resistor r is a comment: this question examines the combination of the induced electromotive force generated by the cutting and the circuit, and the position of the metal rod at different times should be paid attention to in the analysis

The force of the magnetic field on Cd is f=bil=g=, i=1AAB is also f=g=, f=f+g=, and a is wrong.

AB current is 1A, voltage U = IR = 1A * ohm = , C is wrong.

ab velocity, by u=blv=,v=1m s,b error.

Force f does work, w=fvt=, d pair. Pick D

Take a trip here, hahahaha.

Each line can provide a vertical upward force: f=2940*cos30=2528n, so the number of lines: n=30000*g, 2528=119.

According to the analysis of the motion of the object, it can be seen from the title that the object on the smooth horizontal plane moves uniformly under the action of F A, and first decelerates uniformly and then accelerates uniformly in the opposite direction under the action of F B.

Take the direction of F A as positive, and let the acceleration of the object be A1 when F A is acting, and the acceleration is A2 when F B is acting

When the object is under the action of the F Armor.

v1=a1*t ..1)

s=（1/2）a1*t^2 ..2) The object is under the action of fB -v2=v1-a2t .3)-s=v1t -（1/2）a2t^2 ..

4) Obtained from ).

The solution yields a2=3a1

By Newton's law f=ma

It is obtained that f B = 3 f A.

In the whole process, F A and F B do work on the object, the displacement under the action of F A is S, and the displacement under the action of F B is also S, and the work of the combined external force is equal to the change in the kinetic energy of the object.

w = f A * s + f B * s = 32-0 = 32j

w1=fA*s=8j

w2=fB*s=24j

You take a good look at the book and understand the meaning of smoothness, which can be solved by the momentum theorem. I'm in Huainan 5.

If it is appropriate, give a good review.

The velocity in the direction along the rod is the same because the rod cannot be stretched, so the answer is v0cos30= 3 2v0

H is the height of m.

You've got the decomposition of velocity in the wrong direction.

The velocity of m moving to b is the partial velocity of the velocity of a in the direction of the rod, so the straight line where the rod is located should be made as a perpendicular line to obtain v='

Instead of passing v. Make perpendiculars.

This problem can't be calculated in this way, this should be the same as a person pulling a boat, with a circumferential pendulum, the movement is a and velocity, you have to break it down into a circular motion perpendicular to the rod, and then calculate... Know how to get up three.

According to the principle of independence of motion synthesis and decomposition, both boxes in the vertical direction are in free fall! Due to the interval seconds, it is impossible for boxes A and B to be on the same level;

In the horizontal direction, because the two boxes are subjected to the same wind force, and the wind direction is opposite to the initial velocity direction, the two boxes move in a straight line at a uniform deceleration speed in the horizontal direction, and the acceleration of the deceleration is the same. Box A releases the horizontal initial velocity v first. The horizontal velocity of the rear A box is less than V.

Whereas, the horizontal velocity of box B is V. and box A is at the bottom of box B. After that, AB continues to decelerate with the same acceleration, and the horizontal velocity of A is always less than B, so the distance between ABs is getting bigger and bigger!

The answer is definitely c

I think the answer is more reasonable.

As can be seen from the title, when the object is released, it is affected by four forces, the vertical gravity is downward, and the air resistance is upward; Horizontally, both the wind and air resistance are to the left. Because the topic only discusses the horizontal direction, according to the independence of motion, so only consider the acceleration in the horizontal direction, from the question, it can be seen that the horizontal acceleration must be the opposite of the direction of velocity, so list the equation of motion (displacement - time), and then the subtraction of two shifts is a displacement relative to the other, you can see that the displacement of 2 objects is constantly decreasing relative to 1 object, I don't know if it is reasonable

Pick C, that's right.

Let the acceleration in the horizontal direction be -a and the velocity of the aircraft is v, then the velocity of the first box has become v when the second box is exactly leaving the plane'=, the horizontal movement distance of box two s2 = vt-1 2at 2, ( means square ha), the horizontal movement distance of box one s1 = v't-1 2at 2=(,(Of course, the box has a horizontal displacement at the beginning, but this does not affect the solution.) Then s2-s1= , will increase over time.

Flying against the wind, Box 1 slows down horizontally due to the wind, and Box 2 slows down further back.

The object starts out in a uniform accelerated motion.

From Newton's second law, mg = ma

The object accelerates at a distance of x=v and bends 2a

From x=1m 9m, the source of the blind celery body is then moved in a uniform linear motion at a speed of v=2m s.

Therefore, the total hail completion time t=v a+l-x v=5s

f=ma=4000*(-5)=-20000n (the negative sign indicates that the braking force is received in the opposite direction of motion).

m=f/a=900/

a=f/m=

vt^2-v0^2=2as

s=v0^2/2a=15*15/(2*6)=

1.。It's troublesome to write down the analysis, I'll just say the answer, the first 20000n.

3。This question is the same as the first idea.

1 Because the object accelerates upwards.

So the object is overweight.

Therefore, the tensile force f of the wire rope on the cargo is f=ma+mg=2 because the object is thrown upward.

So the acceleration due to gravity is negative.

During the rise of the stones.

is obtained by v=v0+at.

t=x=vot+

So the object falls from the highest point.

is obtained by x=vot+.

x=so the distance of the pebble from the ground x=15+

3.The time it takes for an object to reach its maximum height.

is obtained by v=v0+at.

t=3s total height ascent.

x=vot+

Elapsed displacement.

The first second x = vot+

The second second x = vot+

The third second x = 30*

Fourth second x = 0+

So the object is overweight.

So f=mg+ma=m(g+a)=4mg35 because the maximum bearing capacity of the rope is 36n

And because of its uniform acceleration and vertical ascension.

So the object is overweight and the force on the rope is f=ma+mg

So a=8m s 2

Therefore, the maximum change in velocity is equal to v=v0+at=8*4=32m s.

1 m(a+g)=f=.0 10 kg*(2 first analyze the ascent process v=gt=10*t=4

t=and the velocity to the take-off point is 4m s when falling, and the total experience is 4m s, because the free fall is a uniform acceleration motion, so 4m s+(so 16 2-4 2=2 g s

s=12 m

So 15 - 12 = 3m from the ground

3 After two seconds the velocity is 10m s

After 3 seconds, the velocity is 0m s

There is v2-(v(rear) 2)=2 g s

Just find the s.

After 4 seconds, the velocity is 10 m s, and the direction is reversed, and the altitude is the same as the height of 2 seconds.

4 f=m（g+ g/3)

5 m a=< f- mg

At this time, the rope is satisfied.

So a can be obtained maximum.

So the speed is the maximum.