Section B
5ai) force per unit nass
ii)The def is force per unit mass , so F/m = GMm/R2 divide by m gives g = GM/R2.
b) The density = 2.52 x 107 kgm-3
bii) The density is greater in the centre as the outer layers compresses the inner layers.
ci) g = 1.2 x 1012 Nkg-1 cii) acc = 1.59 x 107 ms-2
iii) comparing this values , a is less than g , the normal force is greater than 0 , particle will not leave.
d) The accelerating protons has ke and when it approaches the star and undergo decceleration and emit x rays. The decrease in KE is converted to Xray photon.
6. The gradient of the graph is not constant hence the net force not constant . Since Weight is constant , the air resistance veries with speed.
b) gradient of the graph at velocity = 0 gives the magnitude of free fall as no drag force act at v = 0.
c) count the area under the graph. height = 19.2 m
cii) Energy lost = KEi - mgh)/KEi = 0.4
d) i) acc of the ball is gradient of the graph at v = 10 ms-1.
ii) Fd = ma - mg
Fd = 0.35(13-9.81) = 1.12N
e) The area below the x axis is the same above the x axis. Height of rise and fall is the same. Since the gradient will be less slopping, the area is found to extend for a longer time. Gradient of graph decrease when the object is moving downwards.
7.b) Charge = It = 72C
ii) E = 108 J iii) E = 104 J iv) R= 6 ohm
ci) new R = 6.8 ohm by taking the emf to be 3V and internal resistance = 0.5 ohm cii) The current increase leads to higher temp, hence resistance increase.
d The resistance of thrmistor and fixed resistor is too large compared to the internal resistance.
ii)1. p.d = (3/4000 + 2000) 4000= 2.0 V
2 pd = 3/2000 + 1800)x 1800 = 1.4 V
iii) for the p.d to be 1.2 V , R = 2700ohm
The same 2700 ohm will give a pd = 1.8 V and not 2.4 V as desired.hence no fixed resistor can acheived the desired range of 1.2 V to 2.4 V .Note that part iii) the pd is across the fixed resistor but part ii) the pd is across the thermistor.
