 # Solutions to SQA examination

### Paper I Solutions

```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
OR
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.

END OF QUESTION PAPER