It is known that only one of the three people a, B and C can drive. A says, "I can drive." B says, "I can't drive." C says, "a can't drive." if only one of the three people is telling the truth, who can drive?

Suppose that a can drive, then what a and B say is true, which is contradictory to the known, so a can't drive, if B can drive, then what a and B say is false, what C says is true, which is in line with the meaning of the topic, if C can drive, then what B and C say is true, which is also contradictory to the meaning of the topic, so B can drive, a: it's B who can drive

In the following electrical appliances, the principle of magnetic field acting on current is used to work
A. Electric fan B. electric blanket C. incandescent lamp D. electric cooker

A. The electric fan makes the fan blade rotate by the movement of the electrified wire in the magnetic field. A, B, C and d all make use of the thermal effect of the current

Zhao said: A is No. 2, already No. 3; Qian said: C is No. 2, B is No. 4; Sun said: D is No. 2, C is No. 3; Li said: D is No. 1, B is No. 3. And Qian, sun and Li were only half right. Q: C is No

No.3
Assuming that Qian said that C is No. 2 is right, then sun's "C is No. 3" is wrong, and his "D is No. 2" is in conflict with the previous hypothesis. Therefore, Qian said that "B is No. 4" is correct. Push to Li's view, Ding is No. 1, and then push to sun's view, C is No. 3

Xiao Ming observed the incandescent lamp, energy-saving lamp and electric fan in his home. The TV plate says 220v40w. Which one gives the most heat when it works normally at the same time?

Incandescent lamp, because the incandescent lamp is a pure resistance electrical appliance, it is calculated according to the square of I multiplied by R, and the so-called energy saving is also a pure resistance circuit, and the electric energy consumed by friction when the electric fan rotates is relatively small. Use p = UI to calculate the power of non pure resistance electrical appliances, and use the square of I multiplied by R to calculate the heating power of pure resistance circuit. TV is between them. Therefore, electric lamp has the largest heating, followed by TV, Electric fan is the smallest. Yes, it's also a classic problem of high school electricity

How to segment the passage of Camellia

One is a paragraph, three is a paragraph, and four is a paragraph

Marked with 220 V, 40 W incandescent lamp, energy saving lamp, fan, TV, which produces more heat? Why?

Incandescent lamp!
In theory, it converts all electric energy into internal energy (heat)
In addition to others, there are also some electrical energy into other types of energy

logical reasoning
School a and school B hold a chess match. Five players from each school compete five times a day for five days. The five players from school a are Ding Yi, hu er, Zhang San, Li Si and Wang Wu
1. Ding Yi's opponent on the first day meets hu er the next day
2. The player who was defeated by Li Si on the third day won Wang Wu on the fourth day
On the fifth day, Wang Wu's opponent made a draw with hu er
4. The player who won Zhang San on the fifth day lost to hu er on the third day
Q: who did the players who played against ding on the third day compete with on the fourth and fifth day?

This question is not simple, there are many cases, and they are all correct (please see the summary for details)
Day 4: hu er
Day 5: Li Si
Or
Day 4: Li Si
Day 5: Wang Wu
Or
Day 4: Zhang San
Day 5: Li Si
Or
Day 4: hu er
Day 5: Wang Wu
All right
Here I will briefly introduce various situations and the detailed inference method of one of them, and let you infer the others by yourself: (the inference of the first situation is divided into the following steps)
This question should make a hypothesis about the competition. Just as the second word of each person's name in school a is arranged in an arithmetic sequence, we assume that "Ding Yi is 1, hu er is 2 (and so on) ". Suppose that the students in school B are a, B, C, D and E
Then, according to the meaning of the question, we write the following expression (hypothesis) on the draft paper:
Day one: a -, B -, C -, D -, e——
The next day: a -, B -, C -, D -, e——
Day three: a -, B -, C -, D -, e——
Day 4: a -, B -, C -, D -, e——
Day 5: a -, B -, C -, D -, e——
Then pair in 1, 2, 3, 4 and 5 one by one
1. Suppose that Ding Yi's opponent on the first day is a, then hu er's opponent on the second day is a (Ding Yi's opponent on the first day meets hu er on the second day.)
On the first day, Ding Yi's opponent is a (the same below)
Day one: A-1, B -, C -, D -, e——
The next day: A-2, B -, C -, D -, e——
Day three: a -, B -, C -, D -, e——
Day 4: a -, B -, C -, D -, e——
Day 5: a -, B -, C -, D -, e——
2. Similarly, suppose that Li Si's opponent on the third day is B, then Wang Wu's opponent on the fourth day is B (the player defeated by Li Si on the third day wins Wang Wu on the fourth day)
Write on draft paper:
Day one: A-1, B -, C -, D -, e——
The next day: A-2, B -, C -, D -, e——
The third day: a -, B-4, C -, D -, e——
Day 4: a -, B-5, C -, D -, e——
Day 5: a -, B -, C -, D -, e——
3. Because Wang Wu's opponent on the fourth day made a draw with hu er on the fifth day (Wang Wu's opponent on the fourth day was B), so hu er's opponent on the fifth day was B
On the draft paper:
Day one: A-1, B -, C -, D -, e——
The next day: A-2, B -, C -, D -, e——
The third day: a -, B-4, C -, D -, e——
Day 4: a -, B-5, C -, D -, e——
Day 5: a -, B-2, C -, D -, e——
4. According to: the player who wins Zhang San on the fifth day loses to hu er on the third day. Suppose Zhang San's opponent on the fifth day is C, then hu er's opponent on the third day is C
Write on the draft paper:
Day one: A-1, B -, C -, D -, e——
The next day: A-2, B -, C -, D -, e——
The third day: a -, B-4, C-2, D -, e——
Day 4: a -, B-5, C -, D -, e——
Day 5: a -, B-2, C-3, D -, e——
Let's look at the line against B, which lacks 1, 3
On the first day, A-1, then the first day against B can only be 3, and the second day against B can only be 1 (each school has five players for the round robin, five games a day, five days in total)
Write on the draft paper:
Day one: A-1, B-3, C -, D -, e——
The next day: A-2, B-1, C -, D -, e——
The third day: a -, B-4, C-2, D -, e——
Day 4: a -, B-5, C -, D -, e——
Day 5: a -, B-2, C-3, D -, e——
Then let's look at line C
Now on our draft paper, C has not competed with 1, 4, 5, and because 1 has competed with others on the first day and the second day, the column of C on the first day and the second day can not be 1
So we write on the draft paper:
Day one: A-1, B-3, C -, D -, e——
The next day: A-2, B-1, C -, D -, e——
The third day: a -, B-4, C-2, D -, e——
The fourth day: a -, B-5, C-1, D -, e——
Day 5: a -, B-2, C-3, D -, e——
Then there are the difficulties
This is because there is more than one arrangement in the match. Because the arrangement on the draft paper is all in line with the meaning of the question now, and we can no longer accurately present the opponent of the next match. That is to say, we can only rely on the topic assumption now
First of all, let's confirm the position where 1 can appear. It can be seen from the draft paper that 1 can't appear in the first, second or fourth day. It can't appear in column a, that is to say, the position where 1 can be written is D, E on the third day, or D, E on the fifth day
There are four positions
Because the question of the title is who the players who play against one on the third day compete with on the fourth and fifth day
Now we just need to make a hypothesis, choose any one of D and E on the third day and write 1
(1) let's assume that it is D and 1 on the third day, then according to the meaning of the title, we can know that 1 on the fifth day can only be written on E
Complete on the draft paper
Day one: A-1, B-3, C -, D -, e——
The next day: A-2, B-1, C -, D -, e——
The third day: a -- B -- 4, C -- 2, D -- 1, e——
The fourth day: a -, B-5, C-1, D -, e——
The fifth day: a -, B-2, C-3, D -, E-1
Now for the title is not so clear, still can not effectively exit all the competition
We can only make a hypothesis. We know that a on the third day can't be 1,2,4; a on the fourth day can't be 1,2,5; a on the fifth day can't be 1,2,3
Let's assume that on the third day a will play against 3, then on the fourth day a will only play against 4, and on the fifth day a will only play against 5
Show it on draft paper
Day one: A-1, B-3, C -, D -, e——
The next day: A-2, B-1, C -, D -, e——
The third day: A-3, B-4, C-2, D-1, e——
Day 4: A-4, B-5, C-1, D-E——
Day 5: a-5, B-2, C-3, D-1, E-1
At this time, it's easy to launch D opponent on the fifth day is 4, and e opponent on the third day is 5
It is shown on the draft paper as follows:
Day one: A-1, B-3, C -, D -, e——
The next day: A-2, B-1, C -, D -, e——
The third day: A-3, B-4, C-2, D-1, E-5
Day 4: A-4, B-5, C-1, D-E——
The fifth day: a-5, B-2, C-3, D-4, E-1
At this time, still can not continue to push down, only in a hypothesis
We can see from the draft paper that E on the first day can be used for 2,4; E on the fourth day can only be used for 2,3
[1] let's assume that E on the first day is 2, then we can deduce that E on the fourth day can only be 3, and E on the second day can only be 4
On the draft paper:
Day one: A-1, B-3, C -, D -, E-2
The next day: A-2, B-1, C -, D -, E-4
The third day: A-3, B-4, C-2, D-1, E-5
The fourth day: A-4, B-5, C-1, D-3
The fifth day: a-5, B-2, C-3, D-4, E-1
At this point, we can push D on the fourth day, and the right one is 2,
Because the opponent of C in the second day can only be 4 or 5, and the opponent of 4 in the second day is e, the opponent of C in the second day is 5
So the first day of the C opponent is 4
It's easy to quit the first day of D is the fifth

It is known that only one of the three people a, B and C can drive. A says, "I can drive." B says, "I can't drive." C says, "a can't drive." if only one of the three people is telling the truth, who can drive?