Diameter//measure length

Outside jaws

To measure external diameter or the length of an object

0.01 cm

V1 V2

Principle of conservation of momentum

Streamline body

Collision of car// firing a gun

AND gate and Not gate

PP QQ

00 11

11 00

X Y Z Q

0 0 1 0

0 1 1 0

1 0 1 0

1 1 0 1

NOT Gate

AND Gate

Increase the number of coil in the secondary coil

Decrease the number of turns in the primary coil

Because iron can be magnetized and demagnetized easily

P = VI

36 = 241

I = 1.5 A

VsIs = VpIp

(36)= (240)Ip

Ip = 0.15 A

The quantity of heat energy required to increase the tempereture of 1 kg of the substance by 1oC or 1K

The temperature of the soup in the metal pot is higher than the

temperature of the soup in the clay pot.

The specific heat capacity of the metal pot is lower than the specific heat capacity of the clay pot

1. An object with the lower value of specific heat capacity will increase

temperature faster when heated by same amount of energy.

2. The pot that has a lower specific heat capacity requires less heat to increase the temperature

The increase of the temperature is inversely proportional to the specific heat capacity// The lower the specific heat capacity , the higher the increase in temperature

The temperature of the soup in the metal pot is lower than the temperature of the soup in the clay pot after 5 minutes move away from the burner

An object with the lower value of specific heat capacity cools faster due to its lower amount of heat stored.

Contact area diagram 6.1 smaller than diagram 6.2

Weight in diagram 6.1 = weight in diagram 6.2

Vehicles 6.2

Contact area is larger

Pressure

1000 : 500

4X A 2 XA

10 : 1

When the air pressure is the wheel lower, the contact area is larger so the pressure on the ground is smaller

Reflection

Less energy

The building reflects sound

The interval time is shorter

5 cm x 1 ms/cm

5 ms

Show on the graph

From graph, T = 0.005 s, d = 1.7 m

It release 1000 J of energy per second when 240V is supplied to it.

I = P/V

I = 1000/240

= 4.167A

E = Pt

= 1000 X 30 X 30

1000 60

Cost = 15 x 0.23// RM 3.45

Calculate the energy by using E = VIt

P : 240 X 6 X 90// 129 600J

Q : 240 X 5 X 150// 180 000J

R : 240 X 4 X 120// 115 200J

R

Uses the least energy//save energy//save cost