Ask a physics problem,,I really don t want to sleep if I can t do it.。。

The upper solution is the image method, and the lower solution is the integral.

According to the title, the second depth is x, and the resistance f1=k. Then the second resistance is f2=k(x+1).

Work: w1=f1*s1=k*1=k

w2=f2*s2=k（x+1）*x

and w1=w2

then k=k(x+1)*x

x=(√5-1)/2=

After the second time, is it the third time?

The third resistance: f3=k(x+

then k=k(x+x=.

Let the depth of the second hammer be l

The resistance at 1cm is n, the average resistance of the first hammer is, the workmanship is (1+1+l)*n 2, and the work is (1+1+l)*n 2*l

l = root number 2-1

Let the average resistance of the first time be f, and the second time is f', then there is f*1=w, and in the same way, w=f'*h, and the resistance is proportional to the depth, so that f can be known'=2f, so there is h=, unit cm

It's easy to calculate with points.

Because the resistance is proportional to the depth of the nail into the block, let ff=kx and let the hammer do work as e each time

Then the first hit: e=k 2 times x1 times x1 --x1=1 The formula is w=f times s) The second hit: e=k 2 times x2 times x2

Solution: x2 = root number 2 =

A small piece of ice floe works well.

When p is at c :

r1:rac=u1:uac can also be written as u1 r1= uac rac

rac=20 ω，u1=30v

When p is in b:

r1：rab=u1':uab can also be written as u1'/r1= uab/rab

rab=40 ω，u1'=20v

u=u1+uac=u1'+uab

Substituting the numerical value, i.e., the answer r1=20 , u=60v ;

This is the most intuitive way to come up with simple and straightforward.

Physics problems can be done intuitively, but it's not the easiest way.

According to the principle of resistor voltage division;

u*(r1/(r1+20))=30

u*(r1/(r1+40))=20

obtained: u=60v r1=20 (ohms).

At the beginning, the slide vane is in the middle, indicating that only the general resistance of the sliding rheostat is used. In this case, R1 and Rab each have different voltages.

The voltage occupied by R1 is U*(R1 (R1+20)), which is the indication of the voltmeter at this time.

In the same way, when the rab tract is the largest.

The voltage occupied by r1 is u*(r1 (r1+40)), which is the indication of the voltmeter at this time.

The supply voltage U=30+I1 (40 2) ,U=20+I2 40 , while I1=30 R1 and I2=20 R1 have.

U=30+30 R1 (40 2) ,U=20+20 R1 40 The solution is R1=20 , U=60V

Answer: c d

There are multiple ways to interpret it. If ob=2oa, released from the horizontal, you will find that the increase of kinetic energy of row b is 4 times the increase of kinetic energy of a, and the decrease of potential energy is twice as much, so there are other forces to do positive work on b, that is, the pulling force of ab rope to do positive work on b. According to the law of action and reaction, the force does negative work on a.

Zena. OA rope does not do work, the following technology learners can explain.

oa is perpendicular to the direction of movement of the ball, so don't do work, this, the question is incomplete, where is o, is it swinging downward around o, and what is the rope ab.

Known quantities: l, v, u1, u2, e

Speed up the process. p：1/2mv²=qu1

q：1/2m(2v)²=qu2

u1:u2=1:4

Deflection process. p: q m can be obtained

q：m(2v) l = q· (2v)·b Substituting q m obtained by the above equation can find b

by t(electric field) = l v

t = 2πl/v

t(magnetic field) = t = l 2v

t(p) ：t(q) =1:4

Some of the calculation processes are not written here, but the ideas should be all there, I hope it will help you, and I wish you a happy study

I don't know anything but experience.

A uniform acceleration with zero initial velocity moves in a straight line, and the velocity at intermediate moments is equal to half of the final velocity.

So b is the midpoint of time for AC.

So ab:bc=1:3, so ab=5m

If the velocity of point B is V and the velocity of point C is 2V, then AB segment: V2-0=2GL, BC segment: 4V2-V2=2G*15, and the two equations are divided to obtain L=5M

Let the total distance be s, the time for 15m s is t1, and the time for 20m s is t2, according to the title, there is.

3s 4=15·t1,s 4=20·t2, so t1=s 20, t2=s 80, so the average velocity v=s (t1+t2), substitute the above calculation results, and approximate s, it's OK.

Since the average speed v=s t is set to be s according to the problem, then the total time taken for the whole journey is t=, and the answer can be obtained by bringing them into the formula v=s t.