Factorization: the square B of 3A + the square C of 9ab

Factorization: the square B of 3A + the square C of 9ab

The square B of 3A + the square C of 9ab
=3ab（a+3bc）

Factorization factor 4x & sup2; + 4xy-3y & sup2; - 4x + 10y-3

(2x+3y-1)(2x-y+3)

If the area of a rectangle is √ 3cm square and one side of it is (3 - √ 3) cm long, how many cm is the other side of it

Length = √ 3 ÷ (3 - √ 3)
=√3(3+√3)÷[（3-√3）(3+√3)]
=(3√3+3)÷（9-3）
=（√3+1）/2

It is known that an asteroid revolves around the sun, its orbit coincides with the earth's orbit plane and has the same direction
It is known that an asteroid revolves around the sun, its orbit coincides with the earth's orbit plane, and its direction is the same, outside the earth's orbit. The time interval between the asteroid and the earth is t, and the nearest distance between the asteroid and the earth is calculated

It is known that GMM / R ^ 2 = m (2pi / T) ^ 2 * r
GMm'/R'^2=m'(2pi/T')^2*R'
And 2pi = (2pi / t-2pi / T ') t
R '= R * [T / (T-T)] ^ (2 / 3)
So the closest distance between asteroid and earth is d = R '- r = R * {[T / (T-T)] ^ (2 / 3) - 1}

It's almost the final exam, who can give a good final simulation paper! (people's Education Press)
The more questions, the better. The more difficult problems, the better. It's better to have more extra-curricular knowledge

1、 Multiple choice questions (4 points for each sub question, 40 points in total)
1. With regard to current, voltage and resistance, the following statements are correct
A. As long as the conductor is connected into the circuit, there is current in the circuit
B. The greater the current passing through a conductor, the lower its resistance
C. There is a current through the small bulb, its two ends do not necessarily have voltage
D. The greater the voltage at both ends of a conductor, the greater the current through the conductor
2. The resistance of an aluminum wire is R. to increase the resistance of the wire connected to the circuit, the method that can be adopted is []
A. Lengthen the aluminum wire and connect it to the circuit
B. Fold the aluminum wire in half and connect it to the circuit
C. It can be seen from the formula r = that increasing the voltage at both ends of the wire or reducing the current through the wire
D. The copper wire with the same length and cross-sectional area is used to replace the aluminum wire in the circuit
3. As shown in Figure 1, when the switch is closed, it is found that bulb a is brighter than bulb B. in this circuit, it is correct to judge the current, voltage, resistance and consumed electric power
A. Ia = IB
B. U a ＜ u b
C. R a = R B
D. P a < P B
4. When the old incandescent lamp is working, it is easy to fuse at the thinnest part of the filament, because compared with the thicker part of the filament with the same length, the thinnest part of the filament is []
A. Maximum power B. maximum current
C. Minimum resistance D. minimum voltage
5. In the circuit shown in Figure 2, the power supply voltage remains unchanged, and after closing the switch s [
A. The voltmeter shows more
B. The ammeter shows more
C. The total resistance in the circuit increases
D. The total power consumed by the circuit is reduced
6. The relationship between the current and the voltage in the two constant resistors A and B is shown in Fig. 3. Now connect a and B in parallel and then connect them at both ends of the power supply with a voltage of 3 V. The following analysis is correct
A. The resistance of a is greater than that of B
B. The voltage of a is greater than that of B
C. The power consumed by a is greater than that consumed by B
D. A consumes more electricity than B
7. As shown in Figure 4, the resistance R1 = 10 Ω, close the switch s, the indication of ammeter A1 is 0.3 A, and that of ammeter A2 is 0.5 a
A. The current through resistor R2 is 0.5 a
B. The resistance R2 is 15 Ω
C. The supply voltage is 4.5 v
D. Resistance R1 consumes 3 W of electrical power
8. With regard to lamp L1 of "8 V 4 W" and lamp L2 of "4 V 2 W", the following statement is wrong []
A. The resistance of L1 is greater than that of L2
B. When both lights work normally, L1 is on
C. When the two lamps are connected in series at both ends of the 12 V power supply, they can light normally
D. When two lamps are connected in parallel, a maximum voltage of 8 V can be applied at both ends
9. When doing the experiment of "exploring the law of voltage in series circuit", Xiao Hong connected the circuit as shown in Figure 5. After closing the switch, she found that L1 was normal and L2 was only weak. You think it is reasonable to analyze the cause of this phenomenon
A. There is an open circuit in bulb L2
B. The bulb L2 is short circuited
C. Due to the high resistance of bulb L2, its actual power is small
D. Due to the small resistance of bulb L2, its actual power is small
10. A dimming lamp circuit is shown in Figure 6. When slide P of sliding rheostat slides to a end, the power of bulb L is 36 W; when slide P slides to B end, the power of bulb L is 9 W; when slide P slides to the midpoint of AB, the power of bulb L is []
A． 16 W
B． 18 W
C． 22.5 W
D． 25 W
2、 Fill in the blanks (2 points for each blank, 28 points in total)
11. A bulb is connected to the lighting circuit, and the current through the filament is 100 mA. During the 20 min power on process, there are some problems__________ J's electricity is converted into heat and light
12. As shown in Figure 7, the power supply voltage is 5 V and does not change. After closing s, the voltage expression is 1 V, then the voltage at both ends of lamp L1 is 1 v_______ 5. The voltage at both ends of lamp L2 is_______ 5. The resistance of lamp L1 and L2 is relatively large_________ The current through the lamp L1 and L2 is the same_________ Large (choose "L1", "L2" or "same")
13. Lamp a is marked with "220 V 60 W" and lamp B is marked with "220 V 100 W". If the length of tungsten filament of the two bulbs is the same, the thicker one is_________ If two lamps are connected in series with a 220 V power supply and used for the same time, what is the ratio of power consumption between lamps a and B__________ .
14. Xiaoming's family has an electric furnace marked with "220 V 880 W". When it works normally, its electric power is_________ W. The current through the wire is________ A. When it is used alone, the electric energy meter marked with "1200 R / kW? H" of Xiaoming's house turns 30 turns in 2 minutes, then the electric power of the electric furnace is 0__________ W.
15. At the peak of power consumption, the voltage of a student's home is 10% lower than that of 220 v. at this time, the actual power of the working "220 V 100 W" incandescent lamp is_______ W. If the "220 V 20 W" energy-saving lamp is used to replace the "220 V 100 W" incandescent lamp, the electric energy can be saved if it works normally for 5 hours every day_______ J.
16. Connect the bulb marked "220 V 40 W" to the circuit of 200 V, the actual power consumed by the bulb_______ (fill in "greater than", "equal to" or "less than") rated power; if the bulb lights normally for 5 hours, it will consume_______ KW? H
3、 Experimental inquiry questions (21 points in total)
17. (10 points) Liu Yang wants to use the experimental equipment given in Figure 8 to explore the relationship between current, voltage and resistance
(1) Before the voltmeter is connected to the circuit, there are two obvious connection errors
①______________________________ ___ ；
②_________________________________________ .
(2) If the error has been corrected and the original current direction does not change, please mark the wire with a pen in Figure 8 and connect the voltmeter to both ends of R1
(3) In this experiment, if we explore the relationship between current and voltage, we should control the current__________ This method of studying problems is called__________________ .
(4) Table 1 is the experimental data recorded by Liu Yang when he explored the relationship between current and voltage___________________________________ .
(5) After the operation of this experiment, if the resistance R1 is changed into a small bulb, the power of the small bulb can also be measured_______ Or______ .
18. (11 points) when Xiaochao was carrying out the experiment of "measuring the electric power of small light bulb", the equipment provided for him on the experimental platform were: 6 V battery, ammeter (0-0.6 a, 0-3 a), voltmeter (0-3 V, 0-15 V), sliding rheostat (50 Ω 1.5 a), small light bulb with rated voltage of 2.5 V (the resistance is about 7 Ω in normal lighting), one switch and several wires
(1) The ammeter should be chosen in the experiment_________ The range of the voltmeter should be selected_________ The measurement range of the system is 0
(2) The physical circuit connected by him is shown in Fig. 9. Before closing the switch, the slide of the sliding rheostat should be adjusted to the right position_______ (select "C" or "d") end
(3) After closing the switch, Xiaochao found that the bulb did not light all the time, the ammeter showed zero, and the pointer of the voltmeter quickly crossed the right most scale line. The cause of this fault may be one of the following situations_______ Or________ (insert serial number)
A. Short circuit of small bulb
B. The filament of the small light bulb is broken
C. The socket of the small bulb has poor contact
D. The positive and negative terminals of voltmeter are connected reversely
E. The rheostat slide is placed at the minimum resistance
(4) After troubleshooting, adjust the sliding rheostat to measure three groups of data as shown in Table 2. Xiaochao found that the resistance of the filament is not equal when the brightness of the small bulb is different. What do you think is the reason for this result__________________________________________________________________ .
4、 Calculation questions (11 points)
19. Using the circuit shown in Figure 10a, Xiao Ming studied the relationship between the current passing through the conductor and the resistance of the conductor. The power supply voltage was constant at 6 v. during the experiment, he did six experiments. After changing the resistance value of the resistance box R 'each time, he kept the voltage at both ends of R' unchanged by adjusting the sliding rheostat, and recorded the corresponding current value, He drew the relationship between the current I and the resistance R 'of the resistance box on the square paper, as shown in Fig. 10B
(1) When recording the first group of data, the resistance value of the resistance box R 'is 4 Ω. What is the current indication at this time?
(2) During the experiment, what is the voltage at both ends of the control resistor box R '?
(3) When recording the second group of data, Xiao Ming records that the ammeter reading is 0.4 A. at this time, what is the power consumed by the resistance box? What is the resistance in the sliding rheostat access circuit?
(4) When recording the fourth set of data, what is the power consumption ratio of the resistance box and the sliding rheostat?
Mid term review test questions
1、 1. D 2. A 3. A 4. A 5. B 6. C 7. B 8. D 9. D 10. A
2、 11.2.64 × 104
12.4; 1; L1; same
13. B; 5 ∶ 3
14. 880；4；750
15. 81；1.44×106
16. Less than; 0.2
3、 (1) ① the positive and negative terminals of the ammeter are connected reversely; ② the conductor is connected to the two terminals above the sliding rheostat at the same time
(2) As shown in the figure:
(3) Resistance; control variable method
(4) When the resistance is constant, the current in the conductor is proportional to the voltage at both ends of the conductor
(5) Electric power
18. （1）0～0.6 A；0～3 V
（2）C （3）B；C
(4) When the brightness of the small bulb is different and the temperature is different, the resistance of the filament increases with the increase of temperature
4、 19. (1) it can be seen from the diagram that when the resistance value of resistance box R 'is 4 Ω, the current through R' is 0.5 a
(2) According to the formula u = IR, the voltage at both ends of R 'is 2 V, such as u' = I1, R '1 = 0.5 a × 4 Ω = 2 v
(3) When the indication of ammeter is 0.4 a, the resistance value of R ′ is R ′ 2 = 5 Ω, then the power consumed by resistance box R ′ is p = i22r ′ 2
＝（0.4 A）2×5 Ω＝0.8 W
R total = = 15 Ω
R slip = R total - R2 ′
=15 Ω-5 Ω=10 Ω
(4) Because when recording each group of data, u ′ = 2 V, u ′ = 4 V, I ′ = I ′
So = = =, i.e