1.B 2。 A 3。 C 4 explosive D 5。 C
chemistry
6.C 7。 C 8。 D 9。 B 10。 B 1 1。 A 12。 A 13。 D
physics
14.C 15。 D 16。 B 17。 D 18。 C 19。 B 20。 A 2 1。 C
Volume 2 (non-multiple choice questions)
physics
f = 6πρVR
23. Solution: (1) After the steel ball is released from the static state, it is acted by gravity G, buoyancy F and viscous resistance F.
F = 6πρV0r(2 points)
F=4πρ0g r3/3 (2 points)
M = 4 π rho R3/3 (2 points)
According to Newton's second law: mg-F=ma (2 points)
Acceleration of steel ball: (2 points).
(2) when V = VM, the steel ball moves in a straight line at a uniform speed, including
Mg-F- f =0 (3 points)
Solution: (3 points)
Note: If students do not consider the buoyancy of steel balls, the rest are correct, so it is suggested to deduct 4 points.
24. Solution: (1) The field strength in the electric field is: E=U/d (2 points).
The work done by electric field force on two balls is: (2 points)
The reduction of electric potential energy is: (2 points)
(2) The work done by gravity on two balls is: WG=mgl (2 points)
From the kinetic energy theorem: WG+WE= Ek-0 (2 points)
Solution: (2 points)
(3) According to (2), there are: (2 points)
A ball is at the lowest point. According to Newton's second law, there are:
(2 points)
Solution: (2 points)
25. Solution: (1) Let the mass of the small metal block and the wooden board be m, and v 1 and v2 represent the velocity of the small metal block and the wooden board A when the small metal block just slides over the wooden board A ... from the conservation of momentum and functional relationship.
MV0 = MVL+2MV2 (4 points)
(4 points)
Solution: VL = 0m/sec v2 = 1 m/sec (unreasonable, omitted) (1 min).
V 1 = 4/3m/sec v2 =1/3m/sec (1point).
(2) Let the speed of the small metal block moving together on the chessboard B from its left end X be v3, and the whole process can be obtained from the conservation of momentum and the functional relationship.
MV0 = Mv2+2Mv3 (4 points)
(4 points)
Solution: v3 =5/6 m/s x=0.25 m (1).
X=0.25, which shows that the metal block has neither stopped on the A board nor slipped out of the B board, which is reasonable.
chemistry
26.( 15)
(1) O=C=O (2 points each)
(2)O2+2H2O+4e-=4OH- (3 points)
2 (3 points)
(3) NH4++HCO3-+2OH-NH3 =+2H2O+CO32-(3 points)
27.( 14 points)
(1) Yang (2 points) 2c 1-2e-= Cl2 = (3 points)
(2) D, Y, F (or KC 1O3, KOH, KCl, note: failure to answer F or KCl will not be deducted) (3 points)
(3) MnO2+4h+2c1-Mn2+c12 =+2h2o (3 points)
(4) c12+H2O2 = 2hc1+O2 (3 points)
28.( 16)
(1)SO3 (2 points) concentrated sulfuric acid (2 points)
(2) Which catheter in device C releases bubbles faster (2 points)
(3)U-shaped pipe (or drying pipe) (2 points) Alkaline lime (2 points, reasonable points)
(4) (3 points)
With the decrease of temperature, the equilibrium of 2NO2-2NO2-N2O4 moves to the right. (3 points)
29.( 15)
Total iodine in (1) mixed solution
=0.02 15 mol (3 points)
Iodine n1(I2) = 25.0×10-3l× 0.234mol/l× 3 = 0.0176mol (3 points).
According to the above reaction formula, it can be seen that 1mol Cu(IO3)2 can completely react with KI to generate 6.5 mo 1 I2.
Therefore, n [Cu (io3) 2] = (0.0215mol-0.0176mol) ×1÷ 6.5 = 0.000600mo1(3 points).
C [Cu (io3) 2] = 0.000600mo1.200l = 0.00300mo1/l (2 points).
(2) Solubility of Cu (Io3) 2: 0.000600 mo1× 414g/mol×1÷2 = 0.124g (4 points).
biology
30.(20 points, 2 points for each space)
(1) character segregation recessive
(2) Sexual blue female () × green male ()
(3) Recessive (blue) parents cannot produce dominant (green) offspring.
(4)C
(5) 1
(6) Amniotic membrane and amniotic fluid
(7) Prolactin
(8)B
3 1.(22 points, 2 points for each space)
I. (12)
(1) down and up
(2) Infiltration is greatly increased.
(3) Answer
Two. (10)
experimental procedure
(2)37.5
(3) Pancreatic amylase
④ Boiling water bath for 2 minutes.
Experimental expectation and conclusion
(1) A is lighter than B.
(2) Two test tubes are the same color (two test tubes are the same color or A is darker than B)