Physics masters, physics masters, come on

1) The small bulb is labeled "4V 2W'Words.

Rated current i=p u=2 4= rlamp=u2 p=16 2=8 ohms.

2) The lamp ll emits light normally, so the circuit i=

r total = u i = 24 ohm r change 1 2 = 48-8 = 40 ohm r change maximum = 80 ohms.

3) When S1 and S2 are closed, the lamp is short-circuited, and R changes and R is connected in parallel.

When r is maximum, the total work consumed is the smallest, r = 80 ohms.

r total = 80 * 240 (80 + 240) = 60 euros.

p min = u2 r total =

There is no picture. **Know where S1 and S2 are?

Let's elaborate on whether it is in parallel or in series.

R lamp = U2 P = 4 * 4 2 = 16 ohms.

i=p/u=2/4=

r half = (u source-u) i = (24-4) ohm.

r slip = 80 ohms.

p min = p slip + p definite = u source 2 r slip + u source 2 r definite = 24 * 24 80 + 24 * 24 240 =

Answer: The resistance of the bulb L is 16 ohms;

The maximum resistance of the sliding rheostat is 80 ohms;

When S1 and S2 are both closed, adjust the slide p, and the minimum value of the total power consumed by the circuit is.

1) r=u square r 4*4 2 8 ohms.

2) Because L is connected in series with P and the bulb emits light normally, the voltage of L is 4V, so the voltage at the P end is 20V, and the current through the circuit is iP(lamp} U(lamp), so the R U in the middle of P(total) I 20V ohm, so the total resistance of P is 80 ohms.

3) The minimum value of the total power consumed by the circuit is:

Why not understand? Falling freely together, why isn't the acceleration all g? Is the speed of free fall related to mass? Where does the friction come from? Can you give directions?

umg(u>1) can only tell that the ring cannot slide on its own when it is on the rod. Only when the rod is released, it falls with the rod. Otherwise, when the rod is not released, the ring will already fall.

This question is a friction and gravity, there is no need to analyze, don't think too troublesome, it is very simple to mention.

From the moment when the rod touches the ground and moves upward, because it is opposite to the direction of movement of the rod, it is subjected to the upward friction UMG, and u>1, UMG>mg, the friction force is greater than the gravity of the ring, so the ring first decelerates downward motion, then accelerates upward movement evenly, then stands still, and then falls freely ......If the ring doesn't slide out of the rod, the cycle continues.

It is related to UMG(u>1), u>1, the friction force is greater than the gravitational force of the ring, and the ring can move upward.

If u<1, after the rod touches the ground and moves upward, the ring cannot overcome the gravity and moves upward, it has been moving downward with uniform acceleration, knowing that the acceleration is g-umg.

There are two conditions for the occurrence of friction, one is that there is pressure on the contact surface, and the other is that there is relative sliding between the contact surfaces.

Here, whether the rod and the ring are in close contact or not, the contact surface between them always has no tendency to move relative to each other in free fall, that is, if the two objects are separated, the relative velocity of the two objects is always zero, which can be proved by the Galilean falling experiment.

Later, when the rod touched the ground, the speed of the rod changed drastically, and the two objects had relative velocity.

If there is no pressure between the two, then the ring continues to do a freefall motion.

If there is pressure between the two, there is sliding friction between the two, and the rod is acted by gravity and sliding friction to make uniform deceleration work. The ring is subjected to gravity and sliding friction.

When >1, the ring decelerates evenly until it is the same as the speed of the rod.

When >1, the ring accelerates until it is the same as the speed of the rod.

It has nothing to do with friction, because the conditions for friction to occur are that the two objects must have pressure, and in the problem, the ring and the rod are not extruded, that is, there is no pressure, so they have nothing to do with friction.

According to the characteristics of the series circuit, U (R1+R2)=I=5U 6 R1, and the ratio of R2 to R1 resistance is 1 5

Because uab=u-5 6u=1 6u

So r1:rab (i.e., r2)=5 6u:1 6u=5:1 (because the ratio of the voltages across the resistors in a series circuit is equal to the ratio of the resistance values).

Oh, dhdhju123 is in the second year of junior high school. . . If you have any questions in the future, continue to ask me.

It should be the back cross style, then the center of gravity of the student's spine search key should rise to, from which wild to jump to the body over the crossbar, which can be regarded as a uniform deceleration movement with a speed of world dexterity g. (The resultant force is mg).

Then according to the formula v 2 = 2ax, then the velocity is about 4m s

I don't agree with the answer upstairs, if your classmate is using the back-crossing style, then his center of gravity is not raised in meters.

Thinking that half a circumference (maybe the body can't be in harmony with the book), then it is estimated that his center of gravity needs to be raised by meters.

If you follow the form of direct stepping, that is, the center of gravity is raised by meters, this classmate's yard posture should be very strong, that is, jumping up to a meter on the spot.

You must know that the NBA's most bullish bounce is sluggish. . .

At the end of the day, all algorithms are the same: mgh=1, 2mv, 2 v, 2=2gh, the key is the value of h. Hope.

Hello! I can't see exactly what the first line of resistance is in yours! But the answer to this question is 1 4 torus resistance.

Because the current will choose the route with the least resistance, and here the toroids are equal in resistance! So the shorter the length, the smaller the resistance, so the shortest length is the distance between two points ab on the transverse ring! ab The distance between the two points is 1 4 the length of the transverse ring!

So the resistance value is 1 4 toroidal resistance.

Then do the math yourself.

This question is against the sky.