The seventh composition of Chinese exercises of Jiangsu Education Press The one who wrote the investigation report

The seventh composition of Chinese exercises of Jiangsu Education Press The one who wrote the investigation report

Investigation report on the use of lucky money
1） Background:
New year's money is a tradition in our country. The custom of new year's money has a long history. It represents a kind of good blessing from the elders. It is a talisman given by the elders to the children, and it can protect the children's health and good luck in the new year. With the development of social economy, the red envelope under the pillow is more and more popular. So how do you treat and use the new year's money, XXX and I conducted a cooperative survey on the use of lucky money
The purpose of the survey is to understand the situation of students' getting lucky money, make objective evaluation and put forward pertinent suggestions
2） Investigation process
Task division:
1) Title: XXX
2) Search information: XXX
3) Street survey and Statistics: xxx xxx
4) Summary: XXX
Activity steps: (divided into four stages)
The first stage: setting up the topic: determining the object to be studied
The second stage: search for information, two people search for relevant information on the Internet and search for written information
The third stage: street survey statistics
The fourth stage: statistical data, analysis
3） Investigation and analysis
(1) At present, middle school students' lucky money mainly comes from their relatives, generally from their parents, and the amount is fixed every year. The change from the elders is small, but there are some fluctuations every year
(2) Middle school students are 73% satisfied with their lucky money this year. Generally speaking, they are still satisfied
(3) Senior students' lucky money is significantly higher than junior students, because they have grown up, parents and relatives give more than before, and they can manage their own money
(4) Half of the students' lucky money is handed over to their parents, while 40% of the students' lucky money is used to buy books and school supplies, and 10% of the students' lucky money is saved for other purposes
(5) 60% of the students are willing to donate part of the money, 30% have not considered it, and the other 10% are not willing to donate
(6) Students' use of new year's money is more scientific, which shows that we all have the sense of thrift. And in the use of new year's money, we have become more active, which shows that we have the ability to control our own property
5） Investigation suggestions
1. We have basically got the ability to use our lucky money. Parents should not let us hand in all our lucky money. In our questionnaire survey, this situation is not uncommon
2. We should take the new year's money as an expression of family affection, no matter how much it is. Don't care too much about the amount of new year's money, treat and use it correctly
3. The increasing amount of new year's money reflects the continuous prosperity and development of social economy, but there are also a series of problems, which encourage the trend of climbing. Some students also develop the bad habit of waste, which is worthy of our attention
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This is what I wrote. I changed junior high school students into primary school students,
Just make a table,
That's what I wrote,
I got the highest score,

In the second part of the sixth grade, we should imitate finger to write common things and use things to describe people,

The panel of a household electric energy meter is marked with 3000r / kW. H. If a person's electric water heater is used alone, and the dial is turned 210 revolutions in 300s, what is the electric energy consumed by the water heater? What is the heat energy emitted by the water heater? The temperature of water increases by 10 ℃ [assuming that there is no energy at any time, C water = 4.2 × 103J / (kg. ℃)]

The power consumption is 210 / 3000 = 0.07kw. H
07 = 252000 coke
Q=cm△t
M = q / C △ t = 252000 coke / 4200 / (kg. ℃) × 10 ℃ = 6kg

(- 2008) zero power - 3 negative first power + | π - 4 | - (1 / 2) negative second power

(- 2008) zero power - 3 negative first power + | π - 4 | - (1 / 2) negative second power
=1-1/3+4-π-4
=2/3-π

How much electricity does the boiler with voltage of 220 V, frequency of 50 Hz and power of 300 W consume in one hour

W = Pt = 0.3 * 1 = 0.3kw * H = 0.3 kWh

If a, B, C ＞ 0 and a (a + B + C) + BC = 4-2skr (3), find 2A + B + C, SQR is the second root
A sqr(3)-1 B sqr3+1 C 2sqr(3)+2 D 2sqr(3)-2

Choose D
Solution: (it is not strict, but it can be solved. It needs some experience in solving problems)
a(a+b+c)+bc=(a+b)(a+c)
4-2sqr(3)=(sqr(3)-1)^2
If the two formulas are connected, then (a + b) (a + C) = (SQR (3) - 1) ^ 2
By comparison, we can know a + B = a + C = SQR (3) - 1 or 1-sqr (3)
a. If B, C > 0, the result 2A + B + C = 2skr (3) - 2 is obtained

When the power supply voltage is 200V, what is the actual power of the bulb with the mark of "100W / 220V"?
There is a motorcycle battery, open circuit measured terminal voltage U1 = 12V, when a headlamp on the street, the headlamp is marked with 36W / 12V, connected to the headlamp, measured squatting point U2 = 11.6v Q: the electromotive force of the battery; the internal resistance of the battery
The more detailed the answers, the better

1. The resistance of the bulb is r = u & # 178 / P = 220x220 / 100 = 484 Ω, so when the terminal voltage is 200V, its actual power is p = u & # 178 / r = 200x200 / 484 = 82.64w2. The electromotive force of the battery is 12V, the resistance of the headlamp is r = u & # 178 / P = 12x12 / 36 = 4 Ω, and when the voltage of the two ends of the headlamp is 11.6v, the current

Additional questions: let a, B, C, d be integers, and M = A2 + B2, n = C2 + D2, Mn can also be expressed as the sum of the squares of two integers in the form of______ ．

∵m=a2+b2，n=c2+d2，∴mn=（a2+b2）（c2+d2）=a2c2+b2c2+a2d2+b2d2=a2c2+b2d2+a2d2+b2c2=a2c2+b2d2+2abcd+a2d2+b2c2-2abcd=（ac+bd）2+（ad-bc）2∴mn=（ac+bd）2+（ad-bc）2．

It is worth pondering: 2. How to calculate the output power of each power supply after connecting the electrical appliances in series?
How to calculate the output power of one of the two power supplies connected in series with an electric appliance (bulb)?
Set each quantity as: No.1 Power Supply: E1, R1, No.2 Power Supply E2, R2, electrical resistance R
This calculation is derived from exploring the problem of using new and old batteries in series
I remember the formula P = IU = ie-i2r, but I think if there are two power supplies in series, maybe one of them should not satisfy this formula, because the current is not only generated by this power supply?
Also, can p be negative?
The total power refers to the work done by the non electrostatic force to the unit charge per unit time. It can be deduced that the total power = QE / T is equal to ie, regardless of whether I is provided by the power supply itself.
So as of b7707609, I = (E1 + E2) / (R1 + R2 + R) is right,
According to meizhanhua, P output = I2 (R + R2), it can be deduced that if and only if there is a power supply in the circuit, it can be satisfied.
According to rkp321321, there is no basis for dividing the output power according to the internal resistance of the two. P consumption = I ^ 2R, which is proportional to the internal composition, P total = IE and E. if the output power = P1 total - P1 consumption, it is not proportional to anyone.
Because P1 total = IE1, P1 consumption = I ^ 2r1.
To make P1 total > P1 consumption, we can only make I E2 / (R2 + R) p1 total > P1 consumption
If the meaning of output power is P1 total - P1 consumption, then its positive or negative is completely related to the true or false of E1 / R1 > E2 / (R2 + R). I should be right.

After series connection, the total electromotive force should be E1 + E2, the total resistance should be R1 + R2 + R, the total current in the loop is I = (E1 + R2) / (R1 + R2 + R), the output power of No.1 power supply is I2 (R + R2), and the output power of No.2 Power supply is I2 (R + R1). In fact, the electromotive force and its internal resistance will change after the power supply is used for a long time

It's in the supplement
The elemental composition of organic matter a and C H O 3 is obtained by burning 3 G A and 4.48 l oxygen (standard condition) in a closed container to produce CO 2 co H 2O (g). Assuming that there is no residue in the reactant, the gas generated by the reaction is successively introduced into Nong sulfuric acid and alkali lime, and the weight of concentrated sulfuric acid is increased by 3.6 g, and the weight of alkali lime is increased by 4.4 G
1. Calculate the amount of hydrogen atom and carbon atom in 3Ga
2. If the molecular weight of a is 60, the molecular formula of the organic compound is?
3. If a can react with sodium, write out all the possible structural formulas of A

4.48l oxygen weight is 6.4g, sulfuric acid absorbs water, soda lime absorbs CO2 6.4 + 3-3.6-4.4 = 1.4g, so the amount of CO is 1.4g, the amount of CO is 1.4g28 = 0.05mol, so the amount of water is 3.6 / 18 = 0.2, so the amount of hydrogen atom is 0.4mol, the amount of CO2 is 4.4 / 44 = 0.1, so the amount of C is 0.15A