Year 12 Introduction Booklet Answers
Answers for Chapter 2: Physical Quantities/Units
1. The resultant force on a body of mass 4.0 kg is 20 N. What is the acceleration of the body?
m = 4.0 kg
F = 20 N
a = ? m s-2
2. A particle which is moving in a straight line with a velocity of 15 ms-1 accelerates uniformly
for 3.0s, increasing its velocity to 45 ms-1. What distance does it travel whilst accelerating?
u = 15 m s-1
v = 45 m s-1
t = 3.0 s
s=?m
3. A car moving at 30 ms-1 is brought to rest with a constant retardation of 3.6 ms-2. How far
does it travel whilst coming to rest?
u = 30 m s-1
v = 0 m s-1
a = -3.6 m s-2 (negative because it is a deceleration)
s=?m
4. A man of mass 75 kg climbs 300 m in 30 minutes. At what rate is he working?
m = 75 kg
h = 300 m
t = 30 minutes = 30 × 60 s = 1800 s
P=?W
5. What is the maximum speed at which a car can travel along a level road when its engine is
developing 24kW and there is a resistance to motion of 800 N?
v = ? m s-1
P = 24 kW = 24,000 W
F = 800 N
6. Find the current in a circuit when a charge of 40 C passes in 5.0s.
Q = 40 C
t = 5.0 s
I=?A
7. What is the resistance of a copper cylinder of length 12 cm and cross-sectional area 0.40 cm2
(Resistivity of copper = 1.7 × 10-8 Ω m)?
L = 12 cm = 0.12 m
A = 0.40 cm2 = 0.40 (× 10-2 m)2 = 0.4 × 10-4 m2
ρ = 1.7 × 10-8 Ω m
R=?Ω
8. When a 12 V battery (i.e. a battery of EMF 12 V) is connected across a lamp with a
resistance of 6.8 ohms, the potential difference across the lamp is 10.2 V. Find the current
through the lamp.
ε = 12 V
R = 6.8 Ω
V = 10.2 V
I=?A
9. Calculate the energy of a photon of wavelength 3.0 × 10-7 m.
λ = 3.0 × 10-7 m
E=?J
10. Calculate the de Broglie wavelength of an electron moving at 3.0 × 10 6 ms-1 (Planck’s
constant = 6.63 × 10-34 J s, mass of electron = 9.1 × 10-31 kg).
v = 3.0 × 106 m s-1
h = 6.63 × 10-34 J s
m = 9.1 × 10-31 kg
λ= ?m
Answers for Chapter 3: Standard Form
1. Put these numbers in order of size,
5239824 , 25634897 , 5682147 , 86351473 , 1258964755
142586479 , 648523154
5239824 , 5682147 , 25634897 , 86351473 , 142586479 , 648523154, 1258964755
But it is easier to order large numbers when they are written in standard form.
2. Put these numbers in order of size,
5.239 × 106 , 2.563 × 107, 5.682 × 106, 8.635 × 107, 1.258 × 109, 1.425 × 108 , 6.485 × 108
5.239 × 106, 5.682 × 106, 2.563 × 107, 8.635 × 107, , 1.425 × 108, 6.485 × 108, 1.258 × 109
3. Convert these numbers into normal form.
a)
b)
c)
d)
e)
f)
g)
5.239 × 103 = 5,239
4.543 × 104 = 45,430
9.382 × 102 = 938.2
6.665 × 106 = 6,665,000
1.951 × 102 = 195.1
1.905 × 105 = 190,500
6.005 × 103 = 6,005
4. Convert these numbers into standard form.
a) 65,345
b) 28,748
c) 548,454
d) 486,856
e) 70,241
f) 65,865,758
g) 765
5.
Convert these numbers into normal form.
a)
b)
c)
d)
e)
f)
6.
8.34  10-3
2.541  10-8
1.01  10-5
8.88  10-1
9  10-2
5.05  10-9
= 0.00843
= 0.00000002541
= 0.0000101
= 0.888
= 0.09
= 0.00000000505
Convert these numbers to standard form.
a)
b)
c)
d)
e)
f)
7.
= 6.5345 × 104
= 2.8748 × 104
= 5.48454 × 105
= 4.86856 × 105
= 7.0241 × 104
= 6.5865758 × 107
= 7.65 × 102
0.000567
0.987
0.0052
0.0000605
0.008
0.0040302
= 5.67 × 10-4
= 9.87 × 10-1
= 5.2 × 10-3
= 6.05 × 10-5
= 8 × 10-3
= 4.0302 × 10-3
Calculate, giving answers in standard form,
a) (3.45  10-5 + 9.5  10-6)  0.0024 = 1.833333333× 10-2
b) 2.31  105  3.98  10-3 + 0.0013 = 9.193813× 102
(Significant figures are discussed later…)
Answers for Chapter 4: Converting Units to SI Units
Convert the following quantities:1. What is 13.6 eV expressed in joules?
13.6 eV = 13.6 × 1.60 × 10-19 J = 2.18 × 10-18 J
2. What is a charge of 6e expressed in coulombs?
6 e = 6 × 1.60 × 10-19 C = 9.6× 10-19 C
3. An atom of Lead-208 has a mass of 207.9766521 u, convert this mass into kg.
207.9766521 u = 207.9766521 × 1.67 × 10-27 kg = 3.47× 10-25 kg
4. What is 2.39 × 108 kg in tonnes?
2.39 × 108 kg = 2.39 × 108 / 103 kg = 2.39 × 105 tonnes
5. It has been 44 years since England won the World Cup, how long is this in seconds?
44 years = 44 × 3.16 × 107 s = 1.39 × 109 s
6. An TV program lasts 2,560s, how many hours is this?
2,560 s = 2,560/3,600 hours = 0.71 hours
7. The semi-major axis of Pluto’s orbit around the Sun is 5.91 × 1012 m, what is this distance
in AU?
5.91 × 1012 m = (5.91× 1012)/(1.496 × 1011) AU = 39.5 AU
Exercise – Converting Speeds
1. Convert 0.023 kms-1 into m s-1.
0.023 km s-1 = 0.023 × 103 m s-1 = 23 m s-1
2. Express 3456 m hr-1 into km hr-1
3456 m hr-1 = 3456/103 km hr-1 = 3.456 km hr-1
3. What is 30 miles hr-1 in m s-1?
30 miles hr-1 = 30/3600 miles s-1 = (30× 1,609)/3600 m s-1 = 13.4 m s-1
(might be worth noting that 1 m s-1 ~ 2 miles hr-1)
4. What is 50 m s-1 in miles hr-1?
50 m s-1 = 50 × 3,600 m hr-1 = (50 × 3,600)/1,609 miles hr-1 = 1.11 × 102 miles hr-1
5. Convert 33 km hr-1 into ms-1.
33 km hr-1 = 33/3,600 m s-1 = (33× 103)/3,600 = 9.17 m s-1
6. Express 234 miles hr-1 in km hr-1.
234 miles hr-1 = 234 × 1,609 m hr-1 = (234 × 1,609)/103 km hr-1 = 3.77 × 102 km hr-1
Answers for Chapter 5: Prefixes/Converting Unit Magnitudes
1.4 kW
10 μC
24 cm
340 MW
46 pF
0.03 mA
52 Gbytes
43 kΩ
= 1.4 × 103
= 10 × 10-6
= 24 × 10-2
= 340 × 106
= 46 × 10-12
= 0.03 × 10-3
= 52 × 109
= 43 × 103
= 1.4 × 103
= 1 × 10-5
= 2.4 × 10-1
= 3.40× 108
= 4.6 × 10-11
= 3 × 10-5
= 5.2× 1010
= 4.3 × 104
The second column is good enough to type into your calculator and we recommend that when you
are solving numerical problems to stop at this point.
The third column shows true standard form (i.e. the multiplier is between 1 and 10).
(Note that is a good idea to write 10-5 as 1 × 10-5 even though the 1 is redundant but this will help
you to enter the number into your calculator correctly.)
Converting between unit magnitudes for distances.
1.
2.
3.
4.
5.
6.
5.46 m to cm = 5.46/10-2 m = 5.46 × 102 cm
65 mm to m = 65 × 10-3 m
3 cm to m
= 3 × 10-2 m
0.98 m to mm = 0.98/10-3 m = 0.98 × 103 mm
34 cm to mm = 34 × 10-2 m = (34 × 10-2)/10-3 mm = 3.40 × 102 mm
76 mm to cm = 76 × 10-3 m = (76 × 10-3)/10-2 cm = 7.6 cm
Notice that when converting from m  cm or mm, we divide by the conversion factor, but when
going the other way e.g. m  cm we multiply by the conversion factor.
Note that for 5 & 6 I have taken a rather long-winded approach to show the process – it may be
obvious to you that 76 mm = 7.6 cm and it is fine just to do it the short way.
Exercise – Converting Areas and Volumes
1. What is 5.2 mm3 in m3?
5.2 mm3 = 5.2 (× 10-3 m)3 = 5.2 × 10-9 m3
2. What is 24 cm2 in m2?
24 cm2 = 24 (× 10-2 m)2 = 24 × 10-4 m2
3. What is 34 m3 in μm3?
34 m3 = 34(/10-6 μm)3 = 34 (× 106 μm)3 = 34 × 1018 μm3
4. What is 0.96 × 106 m2 in km2?
0.96 × 106 m2 = 0.96 × 106(/103 km)2 = 0.96 × 106 (× 10-3 km)2 = 0.96 km2
5. Convert 34 Mm3 into pm3.
34 Mm3 = 34( × 106 m)3 = 34 (× 106/10-12 pm)3 = 34 (× 1018 pm)3 = 34 × 1054 pm3
Answers for Chapter 6: Re-arranging Equations
Equation Name
Equation
(power of lens)
P
1
f
(magnification of lens)
m
v
u
n
c
v
(refractive index)
(current)
(electric potential)
(power)
(power)
(conductance)
(resistance)
(resistance)
(power)
(power)
(stress)
(strain)
Q
t
E
V
Q
E
P
t
P  VI
I
I
V
V
R
I
1
R
G
P  I 2R
G
V2
R
F

A
x

l
P
First Rearrangement
1  Pf
v  mu
c  nv
Q  It
E  VQ
E  Pt
P
I
I  GV
V 
V  IR
1  RG
I
P
R
V  RP
F  A
x  l
Second
Rearrangement
1
f 
P
u
v
m
v
c
n
Q
I
E
Q 
V
E
t 
P
P
I
V
I
V 
G
V
I
R
1
G
R
P
R 2
I
t 
V2
R
P
F
A

x
l

Further Rearranging Practice
1. a = bc , b=?
b = a/c
2. a = b/c, b=?,c=?
b = ac, c= b/a
3. a = b – c, c=?
c=b-a
4. a = b + c , b=?
b=a-c
5. a = bc + d, c=?
c = (a - d)/b
6. a = b/c – d, c=?
c = b/(a + d)
7. a = bc/d, d=?, b=?
d = bc/a, b = ad/c
8. a = (b + c)/d, c=?
c = ad - b
9. a = b/c + d/e, e=?
e = d/(a - b/c)
Answers for Chapter 7: Using Your Calculator
Exercise A
1.
(7.5 x 103) x (24)
= 1.8 × 105
2.
(6.2 x 10-5) x (5.0 x 10-3) = 3.1 × 10-7
3.
(1.4 x 105) x (2.0 x 104) = 2.8 × 109
4.
4.5 x 103 / 7.0 x 104
= 6.4 × 10-2
5.
4.3 x 10-6 / 6.0 x 103
= 7.2 × 10-2
Exercise B In each case, find the value of “y”.
y = (7.5 x 103)2
1.
4.
= 5.6 × 107
2.
y = (1.3 x 103) x (1.6 x 10-4)
(6.6 x 106) + (3.27 x 10-3)
= 3.15 × 10-8
3.
y = (5.6 x 10-4)2 x (7.8 x 108)
(6.6 x 10-11) x (9.1 x 10-2)2
= 4.48 × 1014
y=
(4.12 x 103) + (6.5 x 102) = 1.60 × 10-2
(2.3 x 104) x (8.1 x 102)
Answers for Chapter 8: Significant Figures
Exercise I - How many significant figures are the following numbers quoted to?
1. 224.4343
2. 0.00000000003244654
3. 3442.34
4. 200000
5. 43.0002
6. 24540000
7. 543325
8. 23.5454353
9. 4.0000000000
10. 4456001
7 s.f.
7 s.f.
6 s.f.
1 s.f.
6 s.f.
4 s.f.
6 s.f.
9 s.f.
11 s.f.
7 s.f.
Exercise II – For the numbers above that are quoted to more than 3 s.f., convert the number to
standard form and quote to 3 s.f.
1. 224.4343
2. 0.00000000003244654
3. 3442.34
4. 200000
5. 43.0002
6. 24540000
7. 543325
8. 23.5454353
9. 4.0000000000
10. 4456001
= 2.24 × 102
= 3.24 × 10-11
= 3.44 × 103
= 2 × 105 (was already quoted to 1 s.f.)
= 4.30 × 101
= 2.45 × 107
= 5.43 × 105
= 2.35 × 101
= 4.00 × 100 (or 4.00)
= 4.46 × 106