Solutions to SQA examination

1999 Higher Grade Physics

Paper I Solutions

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1. C		11. B			21. A
2. A		12. B			22. A
3. E		13. B			23. D
4. E		14. D			24. A
5. C		15. B			25. D
6. D		16. C			26. D
7. C		17. E			27. C
8. B		18. D			28. B
9. D		19. C			29. D
10.D		20. B			30. E

31. a. V(vertical)=V(resultant)xsin30 V=14xsin30 V=7(m/s)
b. The maximum height reached is calculated using:
v2=u2 + 2as
s=(v2-u2)/2a Note that it is the vertical components of motion that are being considered in this equation and that at the maximum height the vertical component of the velocity is zero, reducing the above equation to: s=-u2/2a With increasing q u(vertical) is increased. Thus the vertical displacement(height), calculated using the above equation, is increased. 32. Use Boyles law P1V1=P2V2 to solve this problem. P1=20x105Pa V2=P1V1/P2 V1=0.01m3 V2=20x105Pax0.01m3/4x105Pa P2=4x105Pa V2=0.05m3 V2=? 33. Energy gained by the electron in the electric field is calculated using E=qV. The kinetic energy of the electron is calculated using E=mv2/2 Equating these two equations gives: mv2/2=qV or v=(2qV/m)1/2 v=(2x1.6x10-19x2500/9.11x10-31)1/2 v=(8x10-16/9.11x10-31)1/2 v=(8.78x1014)1/2 v= 29.63x106m/s 34. a. Use the potential divider voltage formula V(R1)=[R1/(R1+R2)]xV(S)for this problem. V(R1)=Voltage across fixed resistor=1.8kW V(S)=Supply Voltage=6V V(R1)=[1.8kW/(1.8kW+1.2kW)]x6 V(R1)=0.6x6 V(R1)=3.6V b. The Voltage in the circuit is divided according to the equation V(s)=V(R1)+V(variable). When the resistance of the variable resistor is increased the voltage across it increases. This means that the voltage across the fixed resistor must decrease if the two voltages add up to equal the supply voltage. 35. a. Use the equation I1d12=I2d22 I1=0.60W/m2 I2=I1d12/d22 d1=1.5m I2=0.60W/m2x(1.5m2)2/(4.5m2)2 I2= ? W/m2 I2=0.067W/m2 d2=4.5m b. The light from the laser does not spread out in all directions as is the case with a standard filament bulb. This means that the intensity of the beam does not decrease with distance. The intensity at 4.5m and at 1.5m have a value of 400W/m2. 36. a. Two nuclei join in this nuclear reaction, therefore, it is described as a fusion reaction. b. The mass of the product 189F is less than the sum of the masses of the two reactants, 147N and 42H. This difference in mass, called the mass defect(m), is converted into energy (E). The amount of energy is calculated using the equation E=mc2, where c represents the speed of light. 37. a. Photoelectric emission is used to describe the process where electrons absorb quantised energy, of a sufficient amount, from electromagnetic radiation to enable them to escape from the metal in which they are bound. b. The threshold frequency describes the lowest frequency that the electromagnetic radiation can have to stimulate photoelectric emission.


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