Đề bài và lời giải kì thi HSG quốc tế IPhO 2011 - Thái Lan môn vật lý

... Fm-1 Error analysis is not required Part Resolution of digital calipers [1.0 point] As the relative position of the parallel plates is varied, the capacitance changes with a pattern This set-up ... values Code 33J 68 82J 151 Capacitance value (pF) 34 68 84 150 Experimental Competition: 14 July 2011 Problem Page of Part Calibration Perform the measurement of f using the given capacitors of ....

Ngày tải lên: 27/07/2015, 17:17

... Competition: Problem 14 July 2011 Page of Cylinder with holes plus ball inside Adhesive tape Pin cap Stop watch Thread (for balancing) Base plate Ruler Caution: The thin pin is sharp When it is ... on the central axis of the cylinder Assume that the cylinder is uniform and the mass of the end-caps is negligible

Ngày tải lên: 27/07/2015, 17:17

... graph around this range Experimental Competition: Question 14 July 2011 Page of C vs x graph 250.0 200.0 y = 17.455x - 504.54 C (pF) 150.0 100.0 50.0 0.0 10 20 30 40 50 60 70 x (mm) From this graph, ... These maximum C are plotted (on the Y-axis) vs nodes (on the X-axis.) 200 180 C_max vs Node graph 160 y = 19.924x + 82.04 140 120 100 80 60 40 20 Node 0 This graph is linear of which the...

Ngày tải lên: 27/07/2015, 17:17

... MODIFIED Q2_EXPERIMENT_SOLUTION_14JULY.DOCX Experimental Competition: Question 14 July 2011 Page of II For small-amplitude oscillation about any point O the period T is given by considering the equation: ... Competition: Question 14 July 2011 Page of Method (a) T R(s2cm) 16 14 12 10 R (cm2 ) 0 50 100 150 200 250 300 Calculation from straight line graph: slope   0.04108  0.0007 s2/cm, y...

Ngày tải lên: 27/07/2015, 17:17

... Antenna”, Publications of the Astronomical Society of Australia, 2009, 26, pp.12 8-1 32.) Theoretical Competition: 12 July 2011 Question Page of B C Earth A FIGURE Enlarged view of the three spacecrafts ... The sides (or „arms‟) are about 5.0 million kilometres long The LISA constellation is in an earth-like orbit around the Sun trailing the Earth by 20 Each of them moves on a slightly i...

Ngày tải lên: 27/07/2015, 17:17

... Theoretical Competition: 12 July 2011 Question Page of 2.6 If this spherical bubble is now electrified uniformly with a total charge q , find an ... increase in its radius, find Given that (1 q2 R4 0 R0 R [0.7 point] 2.8 What must be the magnitude of this charge q in terms of t, a , s , 0, R0, Pa in order that the bubble will float motionlessly

Ngày tải lên: 27/07/2015, 17:17

... Theoretical Competition: 12 July 2011 Question Page of 3.2 Find the expression for the force f acting on the ion due to the polarised atom Show that this force is attractive regardless of ... neutralized, and the atom is, in turn, charged This process is known as the “charge exchange” interaction What is the cross sectional area A b0 of this “charge exchange” collision of the atom as ... si...

Ngày tải lên: 27/07/2015, 17:18

... 1 and  are related by two Sine Rule equations sin   sin 1 R ……………………… (7) sin sin 1     r2 Rr Substitute (7) into (6)  M  m R  r2  R  r  m ……………………… (10) m R  ,Eq (10) gives ... figure) m r2  M r1 ……………………… (6) r1  r2 sin 1 m  sin  M 1   Equations (5) and (6): ……………………… (7)  The equation (4) then becomes: Equations (8) and (10): sin 1     ……………...

Ngày tải lên: 27/07/2015, 17:18

...  … (15) In the last equation, the surface tension term has been ignored From balancing the forces on the half-sphere projected area, we have (again ignoring the surface tension term) q P ... 0.8% when the temperature decreases from 307.1 K to 300 K The film itself also becomes slightly thicker.) The drag force from Stokes’ Law is F  6 R0u … (9) If the bubble floats in the updraught, .....

Ngày tải lên: 27/07/2015, 17:18

... Competition: Question Solution Page of  The potential energy of the ion-atom is given by U   f dr ……….………………………(6) r  Using this, U   f dr   r Q ……………………………………………………………(7) 32 2 r [Remark: ... according to the Principle of Conservation of Energy:  Q 2 mvmax   mv0 2 32  r …………… (8) …………… (9) Eqs.(12) & (13): 2  b   Q mv0  b       1 32 2 b  rmin   rmin  Q ...

Ngày tải lên: 27/07/2015, 17:18

... n +1) 1 (1 − x) n( n 1) 1 (1 − nx) (1 + x) n( n 1) (1 − x) n( n +1) = n( n 1) n( n 1) n( n +1) n( n +1) n( n − 1) n( n + 1) (1 + x) 1 (1 − x) − (1 + x) (1 − x) 1 2 n( n 1) n( n +1) n 1 1 = (1 ... n( n 1) = n( n +1) n( n +1) n( n 1) n( n 1) n( n + 1) n( n − 1) (1 + x) 1 (1 − x) − (1 + x) (1 − x) 1 2 n( n +1) n( n 1) n 1 1 = (1 + x) (1 − x) ( n + 1) (1 − x) − ( n...

Ngày tải lên: 12/07/2014, 15:20

... Newton, ta có 1005 u 20 12 = 2k C 20 11 320 11 2k 14k = 320 11 + 1005 k=0 2k C 20 11 320 11 2k 14k k=1 Do < 2k < 20 11 với k 1005 20 11 số nguyên tố nên 2k C 20 11 = 20 11 2k−1 C 20 10 20 11 2k Mặt khác, theo ... 1005 2k−1 C 20 11 320 12 2k 14k−1 k=1 Đến đây, cách sử dụng tính nguyên tố 20 11, ta thấy 2k−1 C 20 11 2k 2 C 20 10 20 11 = 20 11 2k −...

Ngày tải lên: 12/07/2014, 15:20

... = Hay x + + = x + = a) Từ x + x +1 x x x 2 x4 + x2 + 1 25 21 = x2 + + = x + ữ = = x x x 4 x = (2 điểm) x + x + 21 x2 b) P(x) = x + x +1 P(x) > với x 0, P(0) = 1 = x + + ( x + 2) Với ... có (1) (2 x 5) + x + + x 2 x + = ( x + 3) + ( x 1) = 2x + + 2x = (2) *) Với x (2) có dạng: x + + x = 2x = x = x =...

Ngày tải lên: 02/12/2013, 02:11

... = Hay x + + = x + = a) Từ x + x +1 x x x 2 x4 + x2 + 1 25 21 = x2 + + = x + ữ = = x x x 4 x = (2 điểm) x + x + 21 x2 b) P(x) = x + x +1 P(x) > với x 0, P(0) = 1 = x + + ( x + 2) Với ... có (1) (2 x 5) + x + + x 2 x + = ( x + 3) + ( x 1) = 2x + + 2x = (2) *) Với x (2) có dạng: x + + x = 2x = x = x =...

Ngày tải lên: 02/12/2013, 02:11