I won t do a physics problem sophomore , thank you

The two ammeters are connected in parallel, and the numbers are 2A and 3A, respectively, indicating R1: R2=3R'：2r'= 3:2, then the total resistance of the two tables in parallel is 6r'/5,u=5r+5*6r'/5

Two ammeters in series, U=4R+4(3R'+2r'From the above two formulas:

r=94r'/5

u=476r'/5

When the circuit is not connected to the ammeter, the current flowing through r is i=u r=

The constant voltage between the two points of A, B is E, the internal resistance of A1 is R1, and the internal resistance of A2 is R2, and when parallel, 2R1=3R2 ...1)

e=r(2+3)+2r1 ..2)

In series, e=4r1+4r2+4r.3)

The solution of the above three formulas gives r1=3r 14 r2=r 7 e=38r 7 e r=38 7a

When the circuit is not connected to the ammeter, the current flowing through r is i=e r=38 7a=

The problem is simple, because the first two representations are 2a and 3a respectively, and their voltages are equal, so the ratio of resistance is 3:2, which is set to 3x and 2x respectively, and the ab voltage is set to u, so, according to the first time, u=5r+2a 3x=5r+6x, and according to the second time u=4r+4 (2x+3x)=4r+20x. The two equations subtract x to get 7U=38R, U=38 7R, so if you don't connect to the ammeter, the current passing is 38 7A.

I don't know if there will be a mistake in the calculation, and there will definitely be no problem with the idea, I hope it can help you. ）

Select A, in the other options, since it is a double-turn coil, the magnetic flux in the coil cancels each other out, and there will be no change in the magnetic flux in the EF, and of course there will be no current or voltage. However, in A, the current in the phase line and the neutral line is not equal, and the magnetic flux in the double-turn coil cannot be completely canceled, so there will be a current and voltage in the EF. To make it easier to understand, try drawing an equivalent circuit diagram yourself.

Classmates, you know at a glance that this is a transformer.

It can be seen that the "trip controller" is a voltage-sensitive switch, which has an effect on voltage changes, so answer (b) should be chosen, and the mistake is that "overcurrent protection" has nothing to do with pressure.

Note: The working principle of the transformer can be found in the textbook.

I choose BCDThis is an electric shock protector in which the transformer principle is used. Under normal circumstances, the phase line and the neutral line current are equal to the opposite, and the two magnetic fluxes are equal to the opposite, and the combined magnetic flux is 0, and the induced voltage will not be generated in the EF.

In option A, only the neutral line has a current, and the phase wire is shorted, so the neutral wire is equivalent to the original coil and will induce a voltage in the secondary coil EF.

BCD is chosen, which involves the problem of magnetic flux variation. When a single line is in contact, the original power line has an additional branch, that is, from the person to the ground, resulting in the phase line and the neutral line current are unequal, and the changing magnetic field is generated in the core, resulting in the leakage switch inducing current and acting. Under normal conditions (regardless of whether the current is large or small) and under the contact of two wires, the phase and neutral currents are equal and reversed, and a magnetic field of equal strength and opposite direction is generated through the coil, and the two magnetic fields cancel each other out, and the leakage switch will not generate induced current and will not operate.

The up-slip acceleration is (v(, and the down-slip acceleration is (v(1)-v(

Assuming that the slope is x, we know that when sliding, the frictional force and the gravitational component are both negative work, so the acceleration of the object is gsinx+mu*gcosx, and when the slide, the gravitational component does positive work and the frictional force does negative work.

The magnitude of the acceleration is gsinx-mu*gcosx, so gsinx+mu*gcosx=8

gsinx-mu*gcosx=4

The solution yields GSINX=6 and mu*GCOSX=2

sinx=6/10=3/5

cosx=4/5

mu=2/(10*4/5)=1/4

4 The velocity at the end of the second is v s1 = 1 2 a t*2 = 8a v = at=4a s2 = vt'=40a s1+s2=Let the acceleration be a, find the velocity 4a after 4 seconds, and then calculate it yourself, let t1=4s, t2=10

Friend, the following is the analysis and solution process, I hope it will help you!

Solution: Analysis: (The spring in your question should be fixed to the right end!) The whole situation is consistent with the conservation of momentum:

That is, mv0=(m+m)v--- v=1m s··· The maximum elastic potential energy of a spring is when the spring is compressed to the greatest extent! According to the inscription, the iron is bounced back and happens to be at the left end of the board, and the whole process is only the friction between the iron block and the board, because the horizontal plane is smooth!

The process of the iron moving forward and returning to the spring, the friction between the iron block and the wooden board is equal and set to w, and the maximum potential energy of the spring is e

The current situation is impermeable:

1 2 mv0 2=w+e+1 2 (m+m)v 2··· Which disadvantages.

When returning to the process:

w=e+1/2 (m+m)v^2···

Since the maximum potential energy is e=4j!

I only did it once, and I only calculated it once, and I forgot to understand it! Hope it helps!

If c, b, c, and d are the three points on the circumference of the circle centered on +q, then the potentials of b, c, and d are equal. The increase in the electric potential energy of the same test charge from point A to points B, C, and D is equal (the potentials of points B, C, and D are equal). It can be seen from the fact that the work done to overcome the electric field force is equal to the increase in the electric potential energy

It is tested that the electric field force does equal work during the movement of the charge from point A to points B, C, and D. So C is correct.

An object moves in a straight line on the axis of displacement as shown in the figure below. The sn is the position of the object at the beginning and the end of the first second, the end of the second second, the end of the (n-1) second, and the end of the nth second, then the correct one of the following statements is ().

is the position in the second and the direction is from o to s1

is the displacement in (n-1) seconds, the direction from o to sn-1 is the displacement in the first 2n seconds, and the direction is from s2 to sn

is the displacement in the nth second, and the direction is sn-1 to sn

Solution: a. The position is represented by the coordinates S1, and os1 represents the displacement, so A is wrong;

b. Yes, c, s2-sn is the difference between the displacement of the first n seconds and the displacement of the previous 2 seconds, ==>c is false; d right.

The solution process basically only requires the formula of uniform acceleration linear motion with zero initial velocity: s=at 2 2 The length of a carriage: d=a*4 2 2=8a

The total length of n carriages is nd=a*20 2 2=200a=8na, which is simplified to n=25

First find the elapsed time of the first 16 carriages (because this period is still a uniform acceleration linear motion with zero initial velocity) 16*8a=at 2 2 launch t = 16

Therefore, the last nine sessions will take 20-16=4