scheme chapter 5

8/17/2019 Scheme Chapter 5

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FIZIKMOZAC 2010

CHAPTER 3: FORCES & PRESSUREANSWER

3.1 PRESSUREQuestion 1

(a) Pascal = Nm

-2

(b) A = 800 x 2 x 10-3 = 1.6 m2

P = 500 = 312.5 Pa1.6

(c) A sharp knife has a small surfacearea. So it produce larger pressure on the bread.

Question 2(a) Pressure = force

area

(b) (i) balloon B(ii) pressure on balloon B ishigher

(iii) the surface of the needle incontact with the balloon issmaller than the finger.

(iv) the smaller the surfacearea, the larger the pressureexerted on the balloon.

(v) Pressure increases

Question 3(a) (i) The contact area between

the wheels in Diagram 6.2 islarger

(ii) weight are equal

(b) Vehicle uses the wheels inDiagram 6.2. Because it hassmaller pressure exerted onthe soft ground and will notsink.

(c) Pressure(d) 10 000 : 500

4A 2A2,500 : 250

10 : 1

(e) When the air pressure insidethe wheel lower, the contactarea is larger so the pressureon the ground is smaller

3.2 LIQUID PRESSUREQuestion 4(a)(i) The wall of a dam in Figure 4.2

is much thicker at the bottomthan at the top and withstandthe higher pressure at thebottom of the lake.

(a)(ii) Pressure at B is higher than atA

(b)(i) Dam in Diagram 5.2(b)(ii) 1- When depth increases,

pressure increases.2- Thicker at the base canwithstand high pressure.

(c)(i) Siphon system(c)(ii) Diffrence in water level will

cause different in pressure(c)(iii)

Question 5(a) Depth / density / acceleration

due to gravity(b)(i) PQ > P p(b)(ii) PQ = h ρ g

= 5 x 1000 x 10= 50000 Pa

(c)(i) Different in pressure(c)(ii) Water level at P is same as the

water level in the house water tank // pressure is the samebetween at P and inside thetank.


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No difference in pressure(d)(i) Place the concrete tank at

higher place // on top of hillHigher difference of pressure.

or

Use water pumpIncrease the difference of pressure.

(d)(ii)

Question 6(a)(i) Magnitude : same magnitude of

atmospheric pressureDirections : atmosphericpressure and mercury are in thesame direction //gas pressure direction againstthe direction of mercury andatmospheric pressure //atmospheric pressure acts

downwards(a)(ii) Phg + Patm ,// P g as (a)(iii) Same / equal(b) Gas pressure = atmospheric

Pressure + mercury pressure(c)(i) Mercury level drops and at

same level in both columns(c)(ii) Same pressure // atmospheric

Pressure

3.4 PASCAL’S PRINCIPLE

Question 7(a) Pascal’s principle(b) Show the correct direction(c) the liquid pressure in the main

brake cylinder and the smallbrake cylinder are thesame/equal

1.4

2

4106105

15

F

2. F2 = 18 N

Question 8(a) Pascal’s principle(b) P = 5/2 = 2.5 Ncm-2

= 2.5 x 104 Nm-2

(c) Same pressure(d) F2 = 2.5 x 5 = 12.5 N(e) Liquid cannot be compressed

easily

3.5 ARCHIMEDES’ PRINCIPLEQuestion 9

(a)(i) Archimedes principle(a)(ii) upward: Buoyant force

Downward: weight of Hydrometer

(b)(i) the length of hydrometer submerged in oil is longer thanin water.

(b)(ii) Density of oil is less than water (c)(i) Buoyant force = Wair - Wwater

= 0.25 – 0.22 = 0.03 N(c)(ii) volume of object = volume of

water displaced0.03 = 1000 x V x 10

V = 3 x 10-6 m3

Question 10(a) Pacal(b) Depth(c) (i) Weight of water displaced =

buoyant force = ρVg= 1010 x 2.5 x 10= 25,250 N

(ii) Tension + buoyant force= weight of object

T = 125,000 – 25,250= 99,750 N


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Question 11(a) Mass per volume(b) (i) Density sphere A less than

B(ii) Weight A less than B

(iii) The weight of water displaced by A less than B(iv) Larger weight of sphere,

displaced bigger weight of water

(v) Weight of water displaced =up thrust //When the weight of water displaced increase, up thrustincrease

(c) Archimedes’ principle

(d) Submarine

Question 12(a)(i) Same Volume

Net force zero(a)(ii) Y < X < Z(b)(i) Box Y floats and immersed

partially / box X immersedfully and floats box Z sink

(b)(ii) Greater weight meansgreater mass and greater density.The higher density objectneeds more volume toincrease the buoyant forceto support the weight .

(c) Archimedes principle//equilibrium of forces

Question 13(a) Density is the mass per volume(b)(i) Level of the boat is higher in

the sea than in the river. (thepart of boat submerged in thesea is less than in the river)

(ii) Water displaced in the sea isless than in the river.

(b)(iii) Density of sea water is higher than river water.

(c)(i) The lower the density of water,the greater /higher the volumeof water displaced.

(c)(ii) Weight of the boat = Weight of the water displaced

(d) Archimedes’ principle(e) Ballast tank filled by sea water Weight of submarine > buoyantforce

Question 14(a)(i) Function – for safety

purpose/To ensure themaximum weight limit

(a)(ii) F = mg= 7500 x 10

= 7.5 x 10

4

N(a)(iii) The mark should be higher than the sea water level

(a)(iv)1. Density of sea water is denser

than the density of river water.2. The volume of water displaced

increased when density of liquiddecrease

(b)(i) Up thrust = Weight(b)(ii) Accelerates upwards or moves

Up wards(b)(iii)1. The weight of the air balloon is

decreased2. Buoyant force /up thrust higher

than weight3. The balloon experiences the

unbalanced force.

Question 15(a)(i) Bernoulli’s principle(a)(ii) Y(b) The air moves with a high speed(c)1. The atmospheric pressure

which is higher pushes theliquid up through the narrowtube.


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2. The jet air will force the liquidto be sprayed as fine sprayliquid

Question 16(a)(i) Student mark at the same level

in tube - X,Y and Z-(a)(ii) Atmospheric pressure(b)(i) Water level in vertical tube P is

higher than in R and higher than in P./hp > hR > hQ

(b)(ii) Bernoulli’s principle.(b)(iii) P = hρg

= 0.15 x 1000 x 10= 1500 Pa

Question 17(a) Distance per time(b) (i) Before: water levels are the

same and the roof stayintact.After : water levels are notthe same and the roof riseup.

(b)(ii) Pressure above the roof ishigher compare to pressurebelow

(b)(iii) Speed increases pressuredecreases or vice versa

(c) Bernoulli(d)(i) Q is slower and R is faster (d)(ii) Q is higher and R is lower

Question 18: Kedah 07 The depth of the water in Diagram

9.1 is higher than in Diagram 9.2 The water spurts out in Diagram

9.1 is at a higher rate than inDiagram 9.2

The water spurts out further inDiagram 9.1 than in Diagram 9.2

The deeper the water, the further the distance of water spurt

The deeper the water, the higher the pressure of the water

Question 19: Kedah 07(a) The pressure of water increases

with the depth of the water The bubble expands upon

reaching the surface of thewater//The volume of air bubbleincreases as the depth of water decreases

(b) Buoyant force increases as the

volume of the bubble increases The air bubble moving with

increasing acceleration (volume of air bubble = volume of

water displaced)

(Buoyant force is larger than theweight of the air bubbles)

Question 20: Trengganu 07 A force is applied when you

squeezed at the bottom end of thetoothpaste tube

Pressure is applied to thetoothpaste (tube)

According to Pascal’s principle The pressure is transmitted

equally to the whole tube

Question 21: Perak 07 High altitude low density of air Less collision of molecules with

surface Low altitude high density of air More collision of molecules with

surface

Question 22:SBP 07

B is denser than A. The weight of water displaced is

the same of the weight of the rod. Weight of B is greater than weight

of A B will displace more volume of

water


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Question 23: Trengganu 07 The shape of the wing is aerofoil. The shape of cross section of the

wing causes the speed of airflow The air move faster than above

the wings than below the wing. When the speed of moving air ishigher ,the pressure is lower

Hence air pressure below thewings is higher compare to abovethe wings

there is difference in pressurewhich produce an upwardresultant force.

Bernoulli’s principle

Question 24: Teknik 07(a) B u o y a n t F o r c e : Force experience

when an object totally or partlyimmersed into the liquid

(b) Density of the gas inside the

balloon less dense then air Air is displaced by the balloon

and produced buoyant force The buoyant force is larger than

the weight of the balloon andload and it rises up.

When the buoyant force is equalto weight of balloon and load, itwill float still.

(c) Quantitative problem:(i) Resultant force = 250 – 5

= 245 N

(ii) Use F=ma245 = 5 a

a = 49 ms-2

(iii) air resistance is zero

Characteristics Reason

Used heliumgas

Its light/less densethen air

Mass of load is

20 kg

Total weight of

balloon and theload equal tobuoyant force

Tension allowedof the rope is300 N

To ensure the ropenot break

(ii) Set C Because its usedhelium gas, massof load is 20 kg and

tension allowed isgreater than 250 N

(iii) A is not suitable because mass of the load causes weight of theload and the balloon less thenbuoyant force. The balloon willrise up

( Accept any other set and thereason)

Question 25: Perak 07

Characteristic Reason

Large tyre better stability

Liquid inhydraulicsystem

liquid cannot becompressed

Large mass big inertia

Large base area better stability

Low centre of gravity

better stability

Choose – M Large tyre, liquid inhydraulic system,large mass, largebase area or lowcentre of gravity l


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Question 26: SBP 07

Characteristic Reason

Material madefrom glass

Glass does notcorrode with acid

Small diameter of capillary tube To increase thesensitivity of thehydrometer

High density of shots

Makes thehydrometer staysupright

Big diameter of bottom bulb

To obtaine abigger upthrust

Choose N N is made fromglass, has smalldiameter of

capillary tube,high density of shots and a bigdiameter of bottom bulb.

Question 27:Trengganu 07


A shape of crosssection which isupper side is

longer than thebottom

To produced thespeed of airflowabove the wings

to be higher thanthe speed of airflow below

Large surfacearea of the wing

Produce larger liftforce

Less density of the wingmaterials

Less weight //produce moreupward resultantforce

Higher differencein speed of air

The higher thedifference in

pressureThe most suitable choice is PBecause it hasA shape of cross section which isupper side is longer than the bottomLarge surface area of the wingLess density of the wing materialsHigh difference in speed of air

Question 28: Kedah 07

(i) Diagram 9.3 The weight of the dam is

supported by the forceexerted by the water

(ii) Water in the dam can befiltered and chlorinated to beuses as public water supply

To drive turbines for thegeneration of hydroelectricity//For irrigation//Recreationcentre

Suggestion Explanation

Thicker wall atthe base

To withstandgreater pressureat the bottom asthe pressureincreases withdepth

The wall isconstructedusing stronger materials /Using reinforce

concrete

To avoid the wallfrom breaking / Toincrease thestrength of thewall / To avoid

leakingEquipped withthe water overflowsystem

To avoid flooding / To channel awaythe overflowwater


modification explanation

piston of bigger cross-sectionalarea

Can supportgreater force(weight)

Low densitymaterial

Lightweight //easy to carry

Non-compressibleliquid

Piston can belifted up


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Longer handle Less effortneeded to pressthe small piston

Apply releasedvalve between

small and mainreservoir

Liquid can flowsinto small

reservoir

Question 30: Perak 07

(i) hρg = 0.76 x 13 600 x 10=103360 Pa

(ii) hρg = 0.1 x 13 600 x 10= 13600 Pa

(iii) 0 Pa

Question 31: SBP 08

(a)(i)

Mass devide by volume

(a)(ii)

Density of air in Diagram 9.1is higher than in Diagram9.2.// vice versa

The number of load inDiagram 9.1 is greater thanin Diagram 9.2//vice versa

The height of the balloons inboth Diagram 9.1 andDiagram 9.2 are equal

When the density of the air increase, the buoyant forceincrease

As the density of the air increase, the weight of theload carried increased// .As the density of the air

decrease, the weight of theload carried also decreased

(b) Density of the iron nail is highdensity of water// Average den

cargo ship is lower than the dewater

Volume/ weight of water displaced by the iron nail issmaller

For the cargo ship, thebuoyant force is equal to itsweight .

For iron nail , its buoyantforce is smaller than theweight

modification explanationStreamline shape Decrease/

reduce the water drag/resistance

thick and strongmaterial

To withstandhigh pressure / /pressureincrease withdepth

A d d i t i o n a l

c o m p o n e n t

- ballast tank- periscope

To float or sinkthe submarine

To observeobject outsidethe water surface

Safety feature

Oxygen tank /generator

For respiration


(a)(i)

Gravitational force

(a)

(ii)

Weight lost in Diagram 9.1(b)

> Diagram 9.1(c) // vise versa Apparent weight in Diagram

9.1(c) > Diagram 9.1(b) // viseversa

Density of water > density of oil

The greater the density of liquid, the greater the weightlost / less apparent weight

(iii) Up thrust /buoyant force

(b) Name two correct force(buoyant force and weight)

Buoyant force small becausesmall volume // vise versa

Block sink because weight >buoyant force

Sheet float because weight =buoyant force


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modification explaination

Strong material Can withstandgreat force

Low densitymaterial

Lightweight

Two stageplimsoll line

Save in fresh andsalt water

Big size Can place moregoods

Aerodynamicshape

Reduce water friction

Question 33: Kelantan 08

(a)

(i)

Weight is the gravitational force

acts on an object(a)(ii)

Buoyant force = weight of theboat

(a)(iii)

Sea water is denser

Boat displaced less sea water and gain the same buoyantforce. Therefore boat sinksless in sea water

(b)(i)

Buoyant force = weight of seawater

Displaced

= mg = ρVg= 250 x 1080 x 10= 2.7 x 106 N

(b)(ii)

2.7 x 10 = V x 1000 x 10V = 270 m3

(c)(i)

Specification Reasons

Small stemand long

Increase thesensitivitywhere the scaledivisions are

far apart sothat smallchanges indensity can bedetected.

Glass wall Do not erode

Largediameter of

High upthrusts,displaced more

bulb liquid to beable to floateasily

Lead shoots Hydrometer can stay

upright.P is chosen It has small and

long stem,glass wall,large diameter of bulb andlead shootsused.

Question 34: N9 08

(a) Pressure is defined as the forceacting normally per unit area/Pressure = Force

Area

(b) 1. When the small piston ispulled up, the hydraulic oil isdrawn from the reservoir intothe small piston

2. When the small piston ispushed down , the hydraulicoil is exerted with force and

experienced a pressure3. The pressure is transmitteduniformly from the smallpiston to the big piston.

4. The forced produced raisedthe big piston / The systemcan convert a small inputforce into a bigger outputforce.


Has higher boiling point

So that liquid noteasily boiling/

Has higher specific heatcapacity

So that it can’t beeasily become hot


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Has lower density

So the hydraulic jack is not heavy

Has lower rateof vaporisation

Volume of liquidwill not easily

vaporiseLiquid L ischosen

Reasons: L hashigher boilingpoint, higher specific heatcapacity, lower density and lower rate of vaporisation

Question 35: Kedah 08

(a)

(i)

Archimedes’ principle states

that the buoyant force on anobject immersed in a fluid isequal to the weight of fluiddisplaced by the object.

(a)(ii)

The balloon acted by twoforces: buoyant force and theweight of the balloon.

The density of helium gas isless than the density of surrounding air.

Buoyant force equals to the

weight of the air displaced bythe balloon.

Buoyant force is higher thanthe weight of the balloon.

(c) Large balloon

To produce bigger buoyantforce// increase the volume of air displaced

Use 2 burners

To produce bigger flame //heat up the gas in the balloon

faster Synthetic nylon

Light-weight, strong and air-proof material.

High temperature of the air inthe balloon

Reduce density / weight of theair in the balloon.

Hot air balloon Q is chosen

Because it is large balloon,uses 2 burners / manyburners, uses synthetic nylonand has high temperature of

the air in the balloon.(d)(i)

Weight = buoyant force= weight of water

displacedm x 10 = (10 x 2 x 10-6) x 1000x 10m = 0.02 kg

(d)(ii)

mg = ρVg(0.02) (10) = (0.12 x 2 x 10-4) ρ x10

ρ = 833.33 kg m-3

scheme chapter 5

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