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8/8/2019 3. Forces and Pressure
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Mastery Practice 3.1 pg 81
1. (a) minimum pressure = Force
Area maximum
=2 2 2
64
(20 10 )(10 10 )
N
x x m
= 3200 N m-2
(b) Maximum pressure = ForceArea minimum
= 2 2 264
(10 10 )(8 10 )
N
x x m
= 8000 N m-2
_____________________________________________
2. Pressure = Force
Area
=2
36000
0.08
N
m2 2 2
36000
2(400)(10 )
N
m
=2
36000
0.08
N
m
= 450 000 Pa
The platform can withstand the pressure because it is
less than 700 000 Pa.
Challenge Yourself pg 81
1. Pave the soft ground with wooden planks of large
surface area.
This will increase the area of contact.
When the area of contact increases, pressure
decreases.
Hence pressure on the ground is reduced while the
Mastery Practice 3.2 pg 84
1 (a) The tea level is high above the tap. The tea exert
a high pressure at the tap.
(b) When there is very little tea, the pressure exerted
on the tap by the tea is very small. If the
container is tilted, the tea level above the tap
increases and hence the liquid pressure
increases.
2 (a) Pressure = h g
= 8000 x 1025 x 10
= 8.2 x 107 Pa
(b) Force = Pressure x area
= 8.2 x 107 x 0.16
= 1.312 x 107 N
Challenge Yourself pg 85
1.
(i) Water flows in from the top.
(ii) The water column above the sprinkler exerts
pressure to the sprinkles.
(iii) The pressure forces the water sprinkling
out .
(iv) The higher the water column, the bigger the
pressure.
3.1 UNDERSTANDING PRESSURE
3.1
3.1
3.2
3.2
3.2 UNDERSTANDING PRESSURE
IN LIQUID
Water from tap
sprinkler
3.2
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Oil will flow back to the oil reservoir and
platform will be lowered down.
Mastery Practice 3.5 pg 100
1. (a) Buoyant force = 15 10
= 5 N
(b) Weight of water displaced = buoyant force
= 5 N
(c) Volume of water displaced = m = 0.5
1000= 0.0005 m3
Hence, volume of the block = 0.0005 m3
Density of the block = 1.5
Objective Questions pg 104
3.5 APPLYING
ARCHIMEDES PRINCIPLE
3.5 Chap
3
Chapter 3 PRACTISE YOUR
SKILL
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0.0005
= 3000 kg m3
2. Weight of wood = weight of water displaced
100 = (0.8 V)(1000)(10)
Volume of wood, V = 0.125 m3
Weight of wood + weight of stone = buoyant force100 + W = 0.125 (1000)(10)
= 125
W = 25 N
Mastery Practice 3.5 pg 100
1. Velocity of air flowing through A is faster than that
through B. Hence pressure of air at A is lower than
that at B.
2. (a)
(b) The ball will curve off to one side.
(c)
Structured Questions pg 105
1(a) Because the bicycle in Figure 3.49(b) sinks lessinto the sand.
(b) (i) Comparison of the features :
Tyres of bicycle in Figure 3.49(a) are narrower
than that of Figure 3.49(b).
Wheels of bicycle in Figure 3.49(a) has a
bigger diameter than that of Figure 3.49(b)
Two physical quantities :
Area of contact and the pressure
(ii) When the area of contact increases, pressure
decreases
3(a)(i) The pointer moves upwards
(ii) Upper region of the aerofoil is flatter, hence
air flows slower and has higher pressure than
the lower region.
Aerofoil moves downward and the pointer
moves upward.
This follows Bernoullis Principle
(b) The pointer shows a bigger movement
upwards.
(c) To stabilize the racing car at high speed.
(d)
3.6 UNDERSTANDING
BERNOULLIS PRINCIPLE
3.6
Chap
3
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(c) Let off some air from the tyres . Flatten tyres have
bigger areas of contact .
2. (a) When air is pumped out, a partial vacuum is
created in the space between the cup and the
glass panel.
The surrounding atmospheric pressure forces
the cups tightly against the glass panel.
(b) To ensure the contact is airtight.
(c) Air rushes into the cup. Hence air pressure
inside and outside the cup are equal.
(d) The same method cannot be used because the
air in the moon is very thin. The atmosphericpressure is not high enough to press the cup
against the glass panel.
_____________________________________________
When moving through the air, the air flows
faster over the top and creates a region of low
pressure. The higher pressure below the wing
causes a loft which helps the plane to float in
air.