Junior high school physics, the process of asking for help, thank you.

I'll help you :

First question: R1 R2 is in series when both 1 and 2 are disconnected. So now the total resistance of the circuit is rtotal = r1 + r2 = 10 + 10 = 20 ohms.

So the total voltage U total = i1 x r total = x 20 ohms = 8v.

Second question: When 1 and 2 are closed, r1 is shorted, so the lamp and r2 are connected in parallel. So the current through r2 = 8v 20ohm =

And because the trunk current = a at this time, the current through the lamp l = a - = at this time.

So the resistance of the lamp l rl = 8v = 16 ohms .

Then according to P lamp = the square of the amount of the amount of rl = the square of 12v, 16 ohms = 9w. Find out that the rated power of the lamp is 9W

Q3: Because R1 is required to consume the maximum amount of power, the situation of the circuit is that both 1 and 2 are disconnected, and the rheostat is adjusted to the leftmost end, so only R1 works in the circuit at this time. At this time, R1 has the largest electrical power, and the electrical energy consumed at the same time is also the largest.

According to the formula w = square of u r1 x t = square of 8v 10 ohms x 60s = 384 j

The answers to these three questions are all given to you, and it is too tiring to write by hand, and some formulas are not very standardized, so you have to sort it out yourself.

I hope it will be helpful to you, and I wish you progress in your studies!

Analysis: Set the power supply voltage to U

1) When only the switch S is closed, R1 and R2 are connected in series, which can be seen according to the title.

u=i（r1+r2/2）=

Density, the object is 20cm long, immersed in water 20, and the water is only 10cm deep, so it actually sinks to the bottom and cannot float.

Oh, let's first analyze the method of doing these questions.

1) It can be analyzed by the characteristics of the circuit and the characteristics of S2;

2) The temperature of the pot when boiling water can be known from the boiling point of the water, and the temperature of the pot can be judged by the characteristics of the switch;

3) When the rice cooker is in the heating state, the resistance r1 works, when it is in the heat preservation state, the two resistors are connected in series and work together, according to the formula p=u r, the ratio of the electric power consumed by the rice cooker in the case of heat preservation and heating can be found

4) Formulas are available.

5) First find the heat absorbed by the water when the water is heated to boiling according to Q=CMT, use the efficiency formula to find the heat emitted by the rice cooker, and then find the heating time according to Q=W=PT=U RT

Note!!! The efficiency used here is 84%, and the title says 80%, please replace it yourself. (Because I had done 84% of it on my roll before, I copied it directly).

S2 is a temperature control switch, the chef switch is pressed, which is equivalent to S1 closed S2 open, when the temperature of the food in the pot reaches 103°C, the switch S1 will automatically disconnect and disconnect when disconnecting external force will not be automatically turned off. The S2 is an automatic temperature control switch that automatically disconnects when the temperature of the food in the pan reaches 80°C and the temperature drops below 70.

U*U R1 220*220 = 1000 watts of power when cooking.

Insulated power supply U*U (R1 + R2) 220 * 220 (484 + watts.

1) Rice for cooking. Only from the insulation and control S1 of S2 is closed.

2) The temperature control switch is broken.

3) Approximately equal to 1.11 minutes (i.e., 1000. ）

4) Q suction = cm (can start for this t) = pt = u * u * t * 84 r = 4200 * 80 (220 * 220 * 84) = 1000 sec =

t = minutes (number to minute.) ）

In fact, S2 is a temperature control switch, when you press the switch when cooking, it is equivalent to S1, close S2 to disconnect, when the food temperature in the pot reaches 103, the switch S1 will automatically disconnect if there is no external force after disconnection. The S2 is an automatic temperature switch that will automatically disconnect when the food temperature reaches 80 and close when the temperature drops below 70.

Therefore, the power when cooking rice is u*u r1, which is 220*220 watts.

When keeping warm, the power is U*U (R1+R2), that is, 220*220 (484+W (1) can be cooked. S2 only takes the insulation and controls the S1 disconnect.

2) The temperature switch is broken.

3) Approximately equal to 1:11 (i.e.: 1000.)）

4) Q suction = cm (t end -t beginning) = pt = u*u* t * 84% r = 4200 * seconds = minutes.

i.e. t = minutes (bring the number into the approximate score.) ）

I'm sorry, classmate, I don't know.

The answer is 30 degrees.

Because the light energy that wants to return coincides with the original light. If the midpoint of the incident on the pn is x, then the angle of the angle pox can be calculated to be 180-90-60=30 degrees.

According to the principle that the angle of incidence is equal to the angle of reflection. Find the angle moa equal to 30 degrees.

Hope it helps.

The original resistance r can be regarded as a three-stage resistance in series, that is, r = r 4 + r 2 + r 4, after removing the small square, it is equivalent to the middle r 2 because the cross-section becomes the original 1 2, so the resistance becomes the original r 2 2 = r

After the change, it becomes r1=r 4+r+r 4=3r 2 inverse ratio, so the current becomes the original 2 3

1. Draw three lines horizontally and vertically on the original square, evenly divided into 4 * 4 = 16 small squares 2, assuming that the resistance of each small square is r, the large square composed of can be regarded as 4 small r horizontal series and then 4 groups of longitudinal parallel connections. So, the equivalent resistance of a large square is r.

3. After cutting out the middle four small R's, from the horizontal axis, divide the remaining part into upper and lower pieces. The lower piece is in the shape of a "concave", and the top is just turned over. These two pieces are connected in parallel.

4. The concave resistance below is "2 R and + 2 R strings + 2 R and 2", and it is easy to find its equivalent resistance of 3R.

5. The hollow-shaped resistor is two 3R in parallel, that is.

6. Ohm's law: when the voltage is constant, i is inversely proportional to r, so i'=(2 3)i

The law of resistance r= l s

Let the previous resistance be r

The resistor after that is divided into three parts.

The length of the first and third parts is l 4, and the resistance is also r 4, and the length of the second part is l 2 and the width is l 2 (which can be understood as the cross-sectional area is l 2), and the resistance is r

The total resistance is r 4 + r 4 + r = r 3 2, and the current is the original 2 3

First, find the number of roots n of the rope of the pulley block

The number of roots of the rope should not be less than 4, assuming n = 4, the gravity of the pulley block is g'=4*140-500=60nThe assumption is reasonable, n=4

When lifting an object weighing g = 600n:

Tensile force f=(g+g')/4=(600+60)/4=165n

This question needs to be drawn, otherwise it will not be clear.

Beijing South Jinan West 8:00 9:32, 1 hour 32 minutes, 2 minutes of parking is not counted.

Jinan West Nanjing South 9:34 11:46, 2 hours and 12 minutes.

Nanjing South Shanghai Hongqiao 11:48 12:55, 1 hour and 7 minutes.

The distance from Beijing South to Shanghai Hongqiao: S 1318km, train running time: T 4 hours 55 minutes 295 60 hours, average speed: V=S T=1318 (295 60) 268km h

Average speed: v1=s1 t1=406 (92 60).

The distance from Jinan West to Nanjing South S2=1023-406=617km, running time: T2=11:46 9:34 2:12 132 60 hours, average speed: Sensheng V2 S2 T2=617 (132 Feng Chunfeng 60).

The distance from Nanjing South to Shanghai Hongqiao: S3 1318km 1023KM=295km, running time: T3=12:55 11:48 1:77 60h, average speed:

v3＝s3/t 3＝295km÷（67/60）h≈。

Therefore, it is the fastest from Jinan west to Nanjing south, and the slowest from Nanjing south to Shanghai Hongqiao.

1.The time is 2h16min and the distance is 617km2The whole journey is 1318km, the time is 4h55min=295min, and the average speed is 1318 and 295 is about equal.

Buzhou lead 92 is approximately equal to.

v2 = 617 136 is approximately equal to.

v3 = 295 69 is approximately equal to.

So the second segment is the fastest, approximately.

Are you also a sophomore? I've done this question, but I didn't learn physics well The third question should be to find the speed of these 3 paragraphs v=s front cover t, and then compare and compare.

Solution: From the meaning of the question, it can be seen that the power supply voltage is 3V, the rated voltage of the bulb is, and the selected power supply voltage is too small;

As can be seen from the title, the resistance value of the sliding rheostat access circuit when the switch is closed, which is wrong, and the sliding rheostat slider should be placed at the maximum resistance value before closing the switch;

The power supply voltage is 3V, the voltmeter range should be 0 3V, it can be seen from the illustrated voltmeter, its indexing value is, the indication is 1V, the indication is 1V, the indication is from the illustrated ammeter, and the sliding rheostat voltage is 3V-1V=2V, I=U R

The maximum resistance of the sliding rheostat is the maximum slip = U slip i

2v As can be seen from the title, the voltmeter is connected in parallel with the sliding rheostat, and the sliding vane of the sliding rheostat is at the midpoint, then the resistance value of the sliding rheostat to the circuit is r=r slip maximum 2

10. The power supply voltage is an integer multiple, a, if the power supply voltage is, the voltage at both ends of the sliding rheostat when the bulb emits light normally is, and the circuit current is i=

The bulb does not shine properly;

b. If the power supply voltage is 6V, the sliding rheostat voltage is when the bulb emits light normally, and the circuit current is i=

The bulb can emit light normally, and the rated power of the bulb is p=ui=;

c. If the power supply voltage is, the sliding rheostat voltage when the bulb emits light normally is, and the circuit current is i=

i=u r, at this time, the bulb resistance r=

The bulb resistance increases with the increase of temperature, and the bulb resistance cannot be less than 15, so the power supply voltage cannot be.

Therefore, the power supply voltage can be 6V, and the rated power of the bulb can be the same as the plexus.

Therefore, the answer is: Problems in equipment selection: the power supply voltage is too small; Problems in operation: the sliding vane is not placed at the maximum resistance value before closing the switch; 20；②6v；

Xiaomei's problems in the selection of equipment and the operation process are:

Problems in the selection of equipment: the power supply voltage is too small; Problems with operation: The sliding vane before the closing switch is not placed at the maximum resistance, and the maximum resistance of the rheostat she uses is 20 ohms

Xiaomei corrected the above problems, redesigned the circuit for experiments, and was able to accurately judge whether the small bulb was emitting normally by observing the number of the voltmeter When she moved the sliding blade of the rheostat to the midpoint position (that is, the resistance of the rheostat to the circuit was half of the maximum value), she found that the small bulb happened to glow normally, and the voltmeter pointer happened to be on the scale line, then the power supply voltage was 6 volts, and the rated power of the small bulb was watts

p=i^2r

rl=20ω

When p slides to the ab midpoint:

When U=P slides to point A, the voltage stool wheel is still 6V

When p slides to point A.

i=i1+il

i1=r1=u/i1=6/

pl=u 2 rl=36 jujube god letter 20=

Ten: 1. Contact, relative motion, obstruction, friction.

2. Friction and inertia.

3. Force, increase. , 4, horizontally to the left, 4

Eleven: 1, the size of the pressure, the contact area.

2. Pressure, 1 square meter.

3. Equal, the larger, the density of the liquid, the density.

4. Equal to, less than.

5、mg/3a2，1:1，2:3

Twelve: 1, pressure, 760, 1 10 5, increase, decrease.

2. The greater the fluid, the greater the depth.

3. Airplane, large.

4. Increase and decrease. 1、a