2 Force and Motion
4
Force and Motion (II)
Chapter 4 Force and Motion (II)
(d)
Practice 4.1 (p.157)
1
B
2
A
3
D
4
A
Resultants magnitude is 50 N and the
angle between the resultant and the
(a)
horizontal is 37.
6
Let be the angle between the resultant and
2 Force and Motion
3
Chapter 3 Force and Motion (I)
Force and Motion (I)
Practice 3.1 (p.99)
Practice 3.2 (p.107)
1
C
1
C
2
D
2
A
3
C
(a)
Since the ferry moves at a constant velocity,
the net force acting on it is zero.
Take forwards as positive.
Tf =0
T
1 Heat and Gases
2
Chapter 2 Heat and Internal Energy
Heat and Internal Energy
Practice 2.1 (p.28)
1
D
2
C
3
B
4
A
5
(a)
Apply c =
(b) Energy transferred
= Pt = 5 1000 30 60 = 9 106 J
6
Power of the heater
Q 600 1000
=
=
= 1000 W
t
10 60
Time needed =
Q 1
2 Force and Motion
2
Chapter 2 Motion (II)
Motion (II)
(b)
cat as positive.
Practice 2.1 (p.57)
1
A
2
A
3
B
4
D
5
C
6
Car A
Take the initial moving direction of the
s/m
5
0
1
6
26
t/
s
5
(c)
Take the direction towards the left as
positive.
s/m
30
0
Car B
1 Heat and Gases
1
Chapter 1 Temperature and Thermometers
Temperature and Thermometers
Practice 1.1 (p.6)
1
A
2
(a)
5
Let T be the required temperature.
T 0
100 0
Lower fixed point: 0 C
fixed points by using pure melting ice
6
(a)
4
(a)
Infra-red thermome
1 Heat and Gases
4
Transfer Processes
Chapter 4 Transfer Processes
(a)
Practice 4.1 (p.109)
1
C
2
C
3
C
4
(a)
An insulating handle allows us to grasp
the handle without getting burnt.
(b)
(b) Wood
5
Goose-down in the jacket traps air. Since air is
a poor
1 Heat and Gases
5
Chapter 5 Gases
Gases
=
Practice 5.1 (p.167)
1
B
2
A
3
C
= 134 kPa
The final pressure inside the can is 134 kPa.
11
By pressure law,
(a)
The marshmallow becomes larger.
(b)
Since the gas pressure decreases and the
temperature remains un
1 Heat and Gases
3
Chapter 3 Change of State
Change of State
9
Practice 3.1 (p.77)
1
C
2
A
3
A
Let m be the amount of steam needed.
Amount of water needed = 1 m
Energy lost by steam
= energy taken up by water
mslv = mwcw Tw
Energy provided by the heater
m
2 Force and Motion
5
Chapter 5 Moment of a Force
Moment of a Force
8
(a)
normal reaction N
by plank
Practice 5.1 (p.185)
1
P
A
Moment about X
weight W
= Fd = 30 (0.7) sin 15 = 5.44 N
2
D
normal
reaction from
pivot
Maximum magnitude of moment
= FD = 5 (0.2