A physics problem very urgent, urgent!! Please .

When S is closed, there is only one resistor R1 in the circuit (the current passes through the positive pole of the power supply, S, R1, Table A back to the negative pole), and when the switch S is disconnected, the circuit is connected in series for R1 and R2.

Let the power supply voltage be U, and when the switch S is closed, there is: U= .1) When the switch S is disconnected, there is: u=2)

and p2=i 2r2, i.e., the solution r2=20

From the equations (1) and (2) are equal, and the r2 value is substituted, we get: , and the solution is r1=10 so the supply voltage is: u=

When the switch S is disconnected, the resistors R1 and R2 are connected in series, and the voltages at both ends of the resistors R2 are:

u1=ir1=

s R1 R2 in series when disconnected. Trunk current i= p = i square r2 so r2 = p i square = = 20 ohms.

Supply voltage u= i r1+r2) = = + r1).

When S is closed, the resistor R2 is short-circuited, and R2 can be removed. Supply voltage u= i1*r1 =

In both cases, the supply voltage u is equal, so + r1)= solution r1 = 10 ohms.

So the supply voltage = 10 = 12v

S is disconnected, and according to the partial pressure ratio, the partial pressure of R1 is 10 (10+20) = 1 3 12 1 3 = 4VThe voltage indication number is 4V

This question is not difficult, and the above two have already explained. I didn't want to talk about it more, but I still remind you, don't just say that you're stupid, it's not good to be too modest, you all have strengths, and your drawings are not very good.

It should be a uniform deceleration and linear motion.

Let the magnitude of acceleration be a, the direction is opposite to the direction of motion, and the velocity after 2s is v2, which is defined by the title:

10+v2) 2=8m s, the solution is v2=6m s, so a=(10-6) 2=2m s2

It can be seen that the car stops after (10-0) 2=5s, so the displacement of the first eight seconds is the distance from the start of the deceleration to the stop position:

s=2*5*5/2=25m。

How can brakes move in a straight line at a uniform speed?

It can be seen that the moment of tension is equal to the moment of gravity, m(f)=m(g), that is, tension force * tension arm = gravity * gravity force arm.

The center of gravity is at the midpoint of the rod, so the gravitational force arm is.

So the graph isn't very clear, but if you're not mistaken, x=, f=25n.

So g = is about equal to 10 choose b

The center of gravity can be seen as being in the center of the rod, and it is enough to find the midpoint position.

m A=m B, A=4 B, according to v=m, we can know that the mass must be, and the volume is inversely proportional to the density. Because A: B=4:

1, it can be known that v A: v B = 1:4, because V A: in the question

v B = 1:5, so it can be seen that the volume of B is greater than 4v A, that is, the solid volume, so the hollow object is B.

The volume of the hollow part, the solid volume is 4v, and the hollow object volume is 5v, so the volume of the hollow part is: v

If the volume of A is v, then the volume of B is 5v. B density, then A density 4.

Let's use the hypothetical method.

Assuming that both objects are solid, then mass A is 4v and mass B is 5v. A and B should be of equal quality, and only B has a partial hollow to be satisfied.

A is solid, the mass is 4V, and B is 4V. B solid is 5v, solid B removes 1v to become hollow B, let's take the solid B to dig out 1v, the part that is dug is of course hollow. B density, then the excavated part is 1V.

It is to dig out one-fifth of B. The hollow part is 1 5V B.

It's a simple question.

1. The buoyancy is 10n, because the wooden block is floating, so the buoyancy is equal to the gravity (the wooden block is stationary, it must be subject to the balance force).

2. According to Archimedes' law, the buoyancy of a wooden block is equal to the gravitational force of its water discharge, that is, the buoyancy should be equal to the gravity of water with the same volume as the wooden block 3 4. Let the volume of the block be v, then there is water·3v 4·g=f floating, and v=1 750m is obtained, so the density of the block =g (vg)=

Buoyancy is 10 Newtons.

The density is a quarter of the density of kilograms per cubic meter.

Option B

Pressure p=f s

The pressure is the same, the pressure area of A and B is smaller than the pressure area of B on the ground, so the pressure of A on B is strong but because the mass is the same, so the gravity of A B is the same, so the pressure is the same, I hope it will be useful to you Thank you.

Give you ideas for solving problems.

The first step is to analyze why water molecules escape the moon. It must be too fast, its speed gets rid of the gravitational pull of the moon and thus becomes the "artificial satellite" of the moon, the second step, specific calculations. If the speed is too fast, the centripetal force required is too great, and the formula f direction = mv 2 r, and this force exceeds the combined external force, which is the gravitational pull of the moon.

That is to say, when the F direction is gmm r 2, the water molecules escape from the moon;

But the value of gm is not known, this can be passed by gm=g'r 2; where g' = 1 6g = 10 6 can be found to find the value of the gm product.

TeacherWS wishes you progress in your studies.

Let the resistance wire on one side be R0 and the supply voltage be U

When A and B are terminated on the same power supply, it is equivalent to AC+CB resistor 2R0 and AB resistor R0 in parallel.

The current through ab at this point is i1=u r0

When A and C are terminated on the same power supply, it is equivalent to AB+BC resistor 2R0 and AC resistor R0 in parallel.

The current through AB at this point is i2=u 2r0

Then the ratio of the magnitude of the current passing through AB twice i1 i2 = (u r0) (u 2r0) = 2 1 choose b

b.The first series is connected in parallel with AB, and the second series is connected in parallel with AC.