Junior 3 Physics Answers. There are similarities and differences in the answers.

Because it is in series, the currents through the resistors R1 and R2 are equal; Because it is a constant power supply, the voltage can be considered constant and set to u; The resistance r1 is a fixed resistance, and the resistance is always constant; R2 is a thermistor, so its resistance is variable.

1. Heating R2, R2 increases, which is recorded as R large, R string = R1 + R2 large.

i=u rstring=u (r1+r2 large), you can use p1 = i square r2 large = [u (r1+r2 large)] square * r2 large, and then compare the size, the numerator and denominator are divided by r2 large.

The change of the electric power of R2 is that the denominator increases greatly and the fractional value decreases. The power is reduced.

2. Cooling R2, R2 decreases, which is recorded as R2 small, R-string=R1+R2 small.

i=u r string=u (r1+r2 small), you can use p2=i square r2 small = [u (r1+r2 small)] square * r2 small, and then compare the size, the numerator and denominator are divided by r2 small.

The change in the electrical power of R2 is that the denominator decreases by a large margin and the fractional value becomes larger.

The change in the electrical power of R2 is an increase.

When the temperature of R2 increases, you can think of R2 as a sliding rheostat, which is when the total resistance increases and the total current decreases! The power of the P1 is less than the original power! But according to p=i(square)*r, we get: p2>p1

column formula, and end up with only you and r1, r2.

u=u1+u2

p2=i^2 * r2

p2 = [u/(r1+r2)]^2 * r2u^2 * r2/(r1+r2)^2

The numerator and denominator are divided by r2 ==>

u^2 / (r1/r2+1)^2

If R2 increases, (R1 R2+1) 2 decreases, P2 increases.

If r2 decreases, then (r1 r2+1) 2 becomes larger, so p2 also becomes smaller.

OK.

The R1 resistance is constant, and you understand it as equivalent to a resistor in the battery.

Then if the temperature increases, the resistance of R2 becomes larger, and the voltage U on both sides of R2 does not change, according to P=U2 R, so the electrical power of R2 decreases.

I remember as if the power was maxed out when r2=r1.

The process is not clear, and it seems that junior high school students have a lot of trouble doing it.

32. (1) Why can I only see the presence of birds in room B, but not hear the birds' calls?

2) Because there is a vacuum between the double glazing in room B, and the vacuum cannot transmit sound.

33. Find the speed of the car first, pay attention to the length of the car when passing through the cave, and find the speed of the car to be 10m s. It takes 24s to cross a bridge at this speed, and the distance traveled by the car can be found to be 240m, and the length of the body should also be considered, so the bridge length is 200m.

34. First find the time taken by Xiaogang t=s v=2400m 2m s=1200s. Then find the time of Xiao Ming's movement on the bike, which is 1200s-600s=600s according to the title. Finally, the velocity of Xiao Ming can be found to be v=s, t=2400m, 600s=4ms.

This question can also compare the time taken by the two, Xiao Ming is half the time taken by Xiao Gang, and the distance is equal, so Xiao Ming's speed is twice the speed of Xiao Gang, so it is 4m s.

1. What is the difference between sound and light propagation?

2. Because the vacuum does not transmit sound.

In this circuit, when only the resistance value of the sliding rheostat is changed, the power supply voltage remains unchanged (junior high school physics), and the resistance of the bulb remains unchanged, because the sliding rheostat is connected in series with the bulb, after changing the resistance value of the sliding rheostat, the total resistance of the circuit changes, according to Ohm's law, the total current of the circuit (the number of current representations) will change, and the number of the voltmeter (the voltage at both ends of the bulb) will also change, so that the resistance (the resistance value of the bulb) does not change, and the relationship between current and voltage can be studied (the first empty).

It is also possible to measure the actual power (power below the rated voltage) and the rated power of the small bulb.

1. According to the experimental circuit diagram, the ammeter measures the current of the lamp, the voltmeter measures the voltage of the lamp, and the resistance of the lamp remains constant.

2. Since the current and voltage of the small lamp have been measured, the resistance of the small lamp can also be measured according to RL=U i.

(1) Let the density of the cardboard be , the thickness is H, the side length of the small cardboard is A, the side length of the large cardboard is 2A, the area of the small cardboard is A 2, the volume is Ha 2, the mass is Ha 2, the weight is Gha 2, and the pressure on the plastic bag is Gha 2 A 2= Gh;

The area of the large cardboard is 4a 2, the volume is 4 ha 2, the mass is 4 ha 2, the weight is 4 gha 2, and the pressure on the plastic bag is 4 a 2 4a 2 = gh;

Both have the same pressure on the plastic bag, so they are lifted at the same time.

2) Let the weight of a coin be g, because the weight of cardboard is not counted, the pressure of the large cardboard on the plastic bag is 2g, and the pressure of the small cardboard on the plastic bag is g;

Since the side length of the large cardboard is twice the length of the side of the small cardboard, the area of the large cardboard is 4 times that of the small cardboard. If the area of the small cardboard is s, the area of the large cardboard is 4s;

The pressure of large cardboard on plastic bags is 2g 4s g 2s, and the pressure of small cardboard on plastic bags is g s.

It is considerable that the pressure of the large cardboard on the plastic bag is small, so it is lifted first.

It's a matter of pressure.

The first question is that the thickness of the large and small cardboard is the same, so the pressure on the plastic bag is the same, so it is lifted at the same time.

In the second question, although the thickness of the large and small cardboard is the same, because a coin is placed, the pressure of the small cardboard is relatively large, so the larger one is lifted first.

The answer to the first question is that it will be lifted at the same time, because the pressure of the plastic bag on the 2 cardboards is the same, and the pressure on the plastic bags is equal to the pressure on the plastic bags downward; The second is that the big one is lifted up first because the downward pressure of the large cardboard on the plastic bag is smaller than that of the small cardboard, let the gravity of the small coin be g, the side length of the small cardboard is a, the small pressure p=g a2, the downward pressure of the cardboard p=2g 4a2=g 2a2, and the plastic bag on 2 sheets.

The pressure on the board upwards is equal, so the large ones should be lifted first, and they should be lifted together, and the ratio of the number of coins on the large and small cardboard should be 4:1

Solution: It can be seen from the circuit that the resistance on the left and right sides of the slide is connected in parallel, and the resistance of the two parts is measured by the two ammeters respectively, and the resistance of the pencil lead per centimeter can be set to r by the resistance proportional to the length, and the range of the ammeter is 0, change the position of the slider P on the pencil lead, the maximum ratio of the two current representations is 2:1, the resistance of the AC segment and the resistance of the DB segment are equal, when the slide is located at point C, rac=RDB=LACR, RCD=LCDR=10cm R, The voltages at both ends of each branch in a parallel circuit are equal, and the maximum ratio of the two current representations is 2:

1, i.e. i1=,i2=,

i1/i2 =(u/rac)/(u/rcb)=rcb/rac

rcd+rdb) /rac

rcd+rac) /rac

2 1, solution: rac=rdb=rcd, lab=3rcd rcd 10cm=30cm, u rac =u lacr =, that is, u 10cm r =, u r = 10cm, when the range of the ammeter is 0 3a range, when the indication of the ammeter a1 is 3a, let p be the distance from point a is l, then.

u lr = 3a, ie.

u/r ×1/l =10cm×

3a, solution: l = 2cm, the length range of the slide p is 30cm-2 2cm=26cm, so choose c

It's a bit hard to explain why DD is right.

a.It's supposed to be floating.

b.It's floating in the water, too.

c。The buoyancy of both is equal. Because both are floating, buoyancy = gravity, and gravity is the same, so buoyancy is the same.