physics Paper 2.SPM
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Transcript of physics Paper 2.SPM
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