The physics questions in the second year of junior high school are expected to be clearly explained.

Select C When the sliding blade P is placed at the midpoint of the rheostat, the indication of the voltmeter is U slip = 4V, the supply voltage is U, and the sliding rheostat resistance is R

i=u rTotal = u (r1+.)

U slip = IR midpoint = u (R1 + (1).

When the sliding blade P is placed at the B end of the varistor, the total resistance of the circuit becomes larger, the current becomes smaller, the voltage of R1 becomes smaller, and the voltage of the sliding rheostat increases, so the indication of the voltmeter changes by 2V, which should increase by 2V and becomes 6V:U slip'=6v

i'=u r total = u (r1+r).

U slip'=i'rslip=u (r1+r)*r=6v (2)1)(2) can be solved: r1=r.

then u=12v

Within 15 seconds, the heat generated by the fixed resistance R1 is 60J

u1'=u-u slip'=12-6=6v

q1=(u 2 r1)t=(6 2 r1)*15=60 to get r1=r=9

When the sliding vane p is placed at the midpoint of the rheostat, the R1 voltage U1=U-U slip=12-4=8V;

P1 = U1 2 R1 = 8 * 8 9 = 64 9W When the slide p is placed at the B terminal of the varistor, R1 voltage U1'=u-u slip'=12-6=6vp1'=u1'2 r1 = 6 * 6 9 = 4w The ratio of the electrical power consumed twice p1:p1'For 16:9

Ou: I'll answer when I come back.

Let me tell you this, let p be r at the midpoint, r1 at the b end, and the voltmeter at p point = * (u r) in the same way as the number = (r1) * at b, i take r1 as the resistance of the rheostat, and at the same time see the voltage increased by 2v, then the first indicator is 4V, THE SECOND IS 6V, AND THEN THE FORMULA IS 3 AND (R1) WITH (R1), U CAN BE EXPRESSED BY R1-R, YOU CAN GET U WITH R1-R, BECAUSE R1 AND R HAVE A COMMON TERM R, The amount of variation in the rheostat is the value of r1-r. As long as you know the total voltage U and divide the resistance by dividing the square of the voltage at both ends of the heating resistance by dividing the power p equals the square of the voltage at both ends of the heating resistor, you can get the resistance value of r.

According to the relationship between electric power and voltage, the current and resistance of the electric furnace can be obtained, and then a resistance r and the resistance of the electric furnace are connected in series, the voltage remains unchanged, the power is halved, there is only one unknown r, and the value of r can be easily obtained.

The volume of 70g of liquid level drop is four-fifths of the volume of the B ball. The density of B is.

dWhen the rheostat slider P moves from A to B, the resistance of the sliding rheostat increases, because I=U (R1+R2), so the current decreases, and because U1 U2=R1 R2, U1+U2=U, the total voltage remains unchanged, the sliding rheostat structure increases, and the U2 voltage increases.

Because the sliding plate P moves from A to B, the resistance becomes larger, and because it is a series circuit, the voltage representation number becomes larger; Because the resistance becomes larger, the power supply voltage does not change, and the current decreases according to Ohm's law, so D is chosen

D A to B, R becomes larger, I becomes smaller, voltmeter measures the sliding rheostat, so you become larger.

d.The current indicates that the number becomes smaller, and the voltage indicates that the number becomes larger.

The power supply voltage does not change, the resistance becomes larger, and the current becomes smaller. The voltage at both ends of the rheostat increases.

Solution: Method: Submerge the same volume of iron and aluminum in water, respectively.

Steps: Use a spring dynamometer to measure the gravity of a solid iron block and a block of aluminum.

Submerge the iron and aluminum blocks in the water, respectively, to weigh their gravity in the water.

Calculate the buoyancy of iron and aluminum.

Object gravity The object is submerged in water by gravity buoyancy of the iron block

Conclusion: If the volume of the drained water is the same, then the buoyancy of the solid iron and the aluminum block are the same.

Conclusion of modification: If the buoyancy of the iron and aluminum blocks is the same, then the buoyancy of the object is independent of the density of the object. Question addendum: When the volume of drained water is the same.

Through the experiment**, the students already know that the buoyancy of an object immersed in a liquid is related to the density of the liquid. However, some students speculate that the amount of buoyancy experienced by the object may also be related to the density of the object. Please design an experiment to test this conjecture.

1) Existing equipment: spring dynamometer, thin wire, beaker, water, solid iron and aluminum blocks of the same volume (2) Design experiment: Xiaoli believes that when the buoyancy is large, the method of controlling variables should be used, because it is known that buoyancy is related to the density of the liquid, so they only use one liquid, how to make the volume of the discharged liquid equal?

Your method is: All of them invade the water 3) Write out the experimental steps: 1. Weigh the gravity of lead and iron balls; 2. The total induction into the water, the reading of the spring indication 3 phase subtraction to find the difference and buoyancy 4 difference are equal.

4) Design a record of data

5) Conclusion: If then the question is added:

1. Use a spring dynamometer to make the indication of each immersion of the metal ball in water equal.

My answer is b, c. Here's why:

The circuit diagram is R1, R2, R3 in series, V1 measures the voltages of R1 and R2, and V2 measures the voltages of R2 and R3. Swap R1 and R2, the value of v1 must not change, while V2 may change, because the values of R1 and R2 may be different, so B is wrong; Since the three resistors are connected in series, the currents flowing through them are all equal, which is 1a, and c is a bit bizarre, and the "sum" (my understanding of the sum of the three) should be 3a, so it is also wrong.

ACD should be selectedThe voltmeter is equivalent to an open circuit, and the disconnection is a series circuit, the current does not change a pair, V1 measures the voltage of R1R2, which also does not change, B is wrong, the trunk current does not change, C pair, R2 = R3, each is 2V, R1 is 1V, a total of 5V, D pair.

That's how I solved it, I used friction, you see if it's right, don't think about it again. I still don't know what you're talking about about pulling without friction.

1. The scene outside the window is a real image of handstand and shrinkage (the outside of the window is generally far away from 2 times the focal length, so it becomes a real image of handstand and shrinkage).

2. Yes, 2, real image, handstand.

3. C (when the object is in the focal length, the hole is destroyed into a virtual image with a large Qing finger, and when the object is within 1 to 2 times the focal length, it becomes a magnified real image).

4、 b.A shrunken image (a fiber outside the window is far away from 2 times the focal length, so it becomes an inverted and shrunken image).

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