First, multiple-choice questions (2 points for each small question, ***24 points)
The title is123455678911112.
Option D C B D A C A B B A A C
2. Fill in the blanks (No.265438, the third question, 2 points +0, the rest 1 point, * * * 28 points)
13. Conservation of fission mechanical energy of gasoline fossil energy 14. Tone vibration timbre
15. Inertia East 16.600 0.06 17. Liquefied friction roughness
18. Add 10.6 19. Crystal 6 73 fixed 20. When the motion state changes, the work transfer is 2 1. (1) Gravity potential (2) Density (3) After the balloon is heated, the air density in the balloon becomes smaller, and the mass and gravity decrease.
Third, answer the question (there are 10 small questions in this question, with ***48 points, and the score of each question is marked before the corresponding question number)
22. Sketch (6 points)
23.(3 points) 79.9 10 7.99× 103
24.( 1) Voltage across the conductor The voltage across the conductor remains unchanged. (2) Measure the rated voltage for many times.
(3) protection circuit
25.(5 points) (1) Horizontal balance adjusting lever (2) 2 (3) Vertical upward 15 (4) Brief description.
26(5 points) (1)① The candle, convex lens and light screen are not at the same height ② The candle, convex lens and light screen are placed in the wrong order (2) The candle is placed in the focus of the convex lens (3) It is nearly concave.
27.(6 points) (1) Connection circuit omitted (2) Drawing image omitted (3)5 1.25 (4)① Add dry battery ② Change the voltmeter range to 0- 15V.
28.(4 points) Principle or application law: moving the pulley can save half the force (1 min).
Operating steps: Lift the wooden block with a movable pulley composed of thin wires and pulleys, and pull the free end of the rope upward with a spring dynamometer at a uniform speed, and record the pulling force F..(2 points).
Expression: g = 2f- 1 (1 point)
Note: Other schemes can be scored if they are reasonable. For example:
Principles and laws of application: the conditions of ups and downs of objects (1)
Step: Add a proper amount of water into the cup and write down the scale corresponding to the water surface (1);
Gently put the wood block into the measuring cup to float, and when it is stable, write down the scale V2( 1) corresponding to the water at this time.
Expression: G=ρ water g (v2-v 1) (1 min)
29.(5 points) (1)R0 = U2/P = 605ω from P=U2/R (1 min).
(2)W=Pt=0.04 kW？ H. ( 1)
(3) p' = 80w× 25% = 20w u' 2 = p' r0u' =110v (1min).
Therefore, a resistor equal to R0 should be connected in series, that is, R' = 605Ω (1min).
(4) Design circuit diagram omitted (1 min)
30.(9 points)
(1) The forces between objects are interactive.
⑵D
(3) Air resistance
A. When pulling, the spring dynamometer should move horizontally.
B. When pulling, the spring dynamometer should move at a uniform speed.
(4)f 1 = p 1s 1 = 2.0× 105 pa×0.25 m2 = 5× 104n
F2 = p2s 2 = 2.5× 105 pa× 1.2m 2 = 3.0× 105n
F = F 1+F2 = 3.5× 105n G = F = 3.5× 105n m = G/G = 3.5× 104kg
(5)Q total = MQ =1.5×103kg× 8.0×107j/kg =1.2×1kloc-0.
w You =ηQ Total = 2.4× 10 10J
W you = Fs F = W you/s = 2.4× 10 10J/6× 105m = 4× 104n。
In 2009, Jiangsu Province (Wuxi) graduated from the junior high school entrance examination.
Reference answers and grading standards of physical simulation questions
First, multiple-choice questions (2 points for each small question, ***24 points)
1.B 2。 B 3。 D 4。 A 5。 C 6。 D 7。 An eight. D 9。 B 10。 D 1 1。 D 12。 B.
II. Fill in the blanks (65438+ 0 for each blank, ***28)
13. The hardness is equal to 14. Contraction C 15. Connecting device Basketball has inertial gravity 16. The supporting force of a and b to a; Horizontal thrust f and friction between the ground and the table; The gravity of A and B and the supporting force of the ground to B are 17. Generator energy saving 18. The number of small nails attracted by sliding rheostat is 19. For example, thermal insulation materials reduce heat dissipation. The combustion chamber is surrounded by water and the like. (Other reasonable answers can also be given) 20.40 8 1 2 1. Parallel 4:122.0.65 0932123.8.001× 65438+.
Three. Problem solving (8 points for questions 27, 30, 3 1, 4 points for questions 28 and 29, 7 points for questions 32, 9 points for questions 33, ***48 points)
Items 32 and 33: (1) There is no calculation process in the answer, only the final answer is not scored; (2) The method of solving the problem is correct, but it is different from the reference answer, so it can be given points according to the scoring standard;
27. 2 points for each number, * * 8 points; Note: The reflection angle (1) must be marked in the question; In question (2), the sign of force must be marked; Otherwise 1 min)
28.( 1 dot per space) 175.6 70 2.5× 103 is too small.
29.( 1 one space per space) (1) speed, distance pushed by wood blocks (2) steel ball towels and glass plates with different qualities (other reasonable answers can also be given points).
30. (Each blank and question 1)( 1)① Tick "X" on DA; ② Changing DA to DC connection ③ The voltmeter will not be able to measure the voltage across the measured resistor (2) 1.8 0.24 7.5 (3)① Considering the contingency of the measured data in an experiment, it is necessary to change the voltage across RX and conduct many experiments; ② Find the average value of resistance to reduce the experimental error.
3 1.( 1) Add water to the bottle (1) Volume change (1) (2)AD(2) (3) The air tightness of the bottle is not good (1) (4) It becomes smaller (65438+.
32.( 1) is greater than (1). In the process of electric heating, water absorbs heat while continuously dissipating heat (1).
(2) In the cooling process of a given time t = t2-t 1 = 300 s, the heat lost by water is:
Q = CMT = 4.2×103j/(kg℃ )×1kg× (84℃-80℃) =1.68×104j (2 points).
The average heat dissipation power of water is: p dispersion = q/t =1.68×104j/300s = 56w (1min).
Answer: The heat lost by water is 1.68× 104 J, and the average heat dissipation power is 56 W. ..
(3) The power of the automatic heater is 56W.
R1= U2/p = (220v) 2/56w = 864.3ω (2 points).
Answer: The resistance of the electric heater for automatic heat preservation is 864.3 Ω.
33.( 1) Two ships drain water at the same speed at the same time (1) Because the gravity of the bridge is borne by two ships, and the positions of the two ships are symmetrical, the two ships bear the same force on the bridge, so each ship has to drain 140 tons. ( 1)
(2) Left; (1) Take point B as the fulcrum, let the supporting force of the left ship to the bridge be F, and according to the equilibrium condition of lever F, CB = G AB/2 (1).
Available: f = gab/2cb = 280×103kg× 9.8n/kg× 55m/2× 40m = l886500n (2 points).
M = f/g =1886500n/9.8n/kg =192500kg =192.5t (1min).
Answer: the displacement of the left ship is at least l 92.5t
(3) According to W=FS, W = GH = 280×103kg× 9.8n/kg× 0.6m =1.6464×106j (2 points).
Answer: Two ships do work against the gravity of the bridge 1.6464× 106J.