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Physic kertas 2

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Mercury

Expansion / increase in volume

thermometer X

The smallest division is smaller // able to detect the smallerchange

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Rate of change of displacement// Velocity = Displacement Time

Between D and E

Total displacement= [ ( x 4 x 10) + (6 x 10) + ( x 2 x 10)] [1/2 x 2 x 10]

= [20 + 60 + 10] [10]= 80 cm

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increases linearly

As the temperature increase, the speed of the molecules increase. Therate of collision between molecules and container wall increase.Therefore, pressure increase

- 273 oC

0oC = 273K

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P = P T T

200 = P ___273 + 27 273 + 80

P = 235.3 kPa

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Force is an action that can change the type of motion of the object

which is in a straight line// Pushing or pulling action on an object

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MEASURE of the ability of the conductor to resist the flow ofcurrent through it

The length of wire in Diagram 6.1 is longer

The potential difference in Diagram 6.1 is bigger

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The incidence angle in denser medium produce a 90 refracted angle.

n = 1/ sin 42n = 1/0.669n = 1.49 or 1.50

Total internal Reflection

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refractive index of the inner core is greater than the outercladding.

total internal reflection can occur

A bundle of optical fibres can transmit more information (An o pt icalf ibre is very sm al l in d iameter )

Optical fibre has high flexibility.

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Concave reflector

converge wave// focus the signal

At focal point

Waves converge at focal point// Signal focus at focal point

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Big

can collect more wave/ signal

Type X is most suitable

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SECTION B

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Force per unit surface area// Pressure = Force Surface area

- Mass of load is equal- Depth of sinking in D9.2 is less than D9.1 // Vice versa

- Area in contact with the soft ground in D9.2 is bigger than D9.1

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When depth of sinking increases, the pressure increases

When area in contact increases, pressure decreases

1. When the piston is pushed in, the air flows out of the nozzle with high speed2. Creating a region of low pressure above the narrow tube3. The higher pressure of the atmospheric air acts on the surface of the liquid

causing it to rise up the narrow tube4. The liquid leaves the top of the narrow tube through the nozzle as a fine

spray.

http://localhost/var/www/apps/conversion/Video%20f5Phy/Bernoulli's%20Theorem%20And%20Its%20Application.mp4

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Tak perlu fikirkansangat ini!

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Criteria Explanation / ModificationStructure of thedam

The design to

ensure safety

The material usedfor the dam

Additionalcomponent toproduceelectricity

1. Build a dam that has thicker wall at the base,to withstand higher pressure at the bottom

1. Made of concrete , not easily break

1. Equipped with water overflow system , avoid

overflow/ flooding2. Build high wall , store more water / avoidoverflow

1. Build turbine at the bottom of the wall, highwater pressure will turn the turbine toproduce electricity

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Speed of wave is uniform // Wavelength is uniform

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Aspect Reasons

Location of resort ATTHE BAY

Features to reduce theerosion of shoreBUILD BARRIERS

WITH SMALL GAP &HIGH WALL BARRIER

Features to enablechildren to swimsafely TO BUILDSANDBANK

Features to protecthotel from strong windTO GROW TALL TREE

1. Calmer water and lower amplitude waves

because energy diverged// spread out.

1. Build barriers with small opening, water wavesare reflected and diffracted so less energy wavereach the shore.

2. Build high wall barriers, to protect beach fromhigh waves.

1. To create shallow area, speed and wavelength ofwave decreases.

1. Grow tall trees between the beach and hotel, thetrees spreads out the wind energy

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SECTION C

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Springs constant ,k is the ratio of force to the extension // k = F / x

Spring X

Spring X is thicker Spring X is stiffer Spring X extend shorter than spring Y for the same force k = F / x , extension of spring is shorter, the value of k is higher

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Characteristic Explanation

Small diameter enough space for the spring to be installed //more stiffer

High elastic limit can support heavy loadhigher spring constant small compression of the spring

small natural frequency reduce bumping

S is chosen small diameter , highest elastic limit, highestspring constant and small natural frequency

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The distance between focal point and optical centre

Virtual, upright and magnified

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Aspects Explanation

f o > f e Higher magnification of image

Eyepiece is thickerthan objective lens

Shorter focal length and higher power

Diameter of theobjective lens is

greater than diameterof eyepiece.

To ensure more light can enter the telescope to

produce brighter image

Distance between twolensesS = f o + f e

f o + f e is the distance of normal adjustment whichwill produce a sharp image.

Chosen arrangement :K

Because f o > f e , eyepiece is thicker than objectivelens, diameter of the objective lens is greater thandiameter of eyepiece and distance of two lenses isf o + f e

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so ,

1 = 1 + 1f u v1 = 1 - 1v f u1 = 1 - 1v 20 30v = 60 cm

P = 1/ff = 1/5 = 0.2 m = 20 cm

u = 30 cm, v = ?

m = vu

m = 60/30= 2

Real, magnified and inverted