日六月二十年二七九一遥公年一十六國民華中育教僑華
莫四第張六第20日一初月一十年子壬服夏
WAH KIU YAT PO
邹日橋華
reised;, so applying the law?
三期星R
before impact
1973英文中學會考試題預習專欄
conservation of energy
Sum or momenta
mv + M
work
work
院
done
-done
Work done
work done
after impact
物理科
(七)
by
by
over-
effort machine
coming
raisius moving
mu1 + Mu2
=(1 + M}
The block was at rest before.
Suggested answers for issue (6)
Solution
friction parts
impact. V
TE
Mass of the car,m = 1000 kg It moves with uniform velocity of 20 m/s through out,
sina -
mg sma
Friction is uniform for all atages.
On stage I, the engine doesn't work. As it is moving with uniform velocity, the resultant force acting on the car is zero. (XF = ma = 0,
It is the weight of the car that causes the motion. The frictionopposes it to maintain its uniform velocity.
mi ng ging - friction, 1. On stage II, engine works to overcome the friction and keep- ang the car at a speed of 30m/s
mg sine.
P = Fy. - nvg sine
10 20.10.75 13.3 kw
On the steeper part, the slopi
is 1 in 5. mg sina>
If brake is not applied, the car acelerates down the slope. To keep
brakesed uniform, th
force of £!
pawang sina
u = ng sina -
➡ ng sina - ng gin mg(sina sin 0)
1000x10 (— — 75)
-20000
rate of heat dissipated in
the brake system:
20
x 20
26.7 kw
Answer: Power developed on the
level road is 13.3 kw.
Rate of heat converted in the brake system is 26.7
Solution
120
a) Velocity Ratio - V.R.
46
(b) M.A.-=(V.R.)(efficiency)
40 x 0.60 24
(c) If no friction, M.
Load Effort
Load Effort x M.A.
50 x 40 ≈ 2,000 lbf.
If the efficiency is 60%,
Load
≈ 0.6 x 40
24
= Effort x M.A
50 x 24
1200 lbf.
Solution
Mass of the ball, m - 10 kɛ
Initial kinetic energy
Input work = fs
Output work –
Waste work
Efficiency
Input
Since fs 1s greater than W, the work done on the load is less than the work done by the effort, so the efficiency must always be Less than unity.
Suppose that the actual load is W, and the moving parts to be raised are of weight w; the load it raised is (W + w)..
Friction is proportional to the pressure between surfaces in con- tact. It is proportional tɔ, (W+w) So friction donates an additional work of k(W+w), where k is a cons tant for the friction of the mach ine.
The total resistance to be over- come by the effort is then
(W + w) + k(W + w}
= (1 + x)(WL+ w)
Suppose that the machine, with this computed load, is perfectly efficient. The mechanical advan- tage e equals to the velocity ratio
MAVR.
(1+
OW+W) =(V.R.)
MA
{ 1•K) (W+W )
V.R.
= (1-4) (w) + (1+x) (
is
Divide the equation by
(1+k) (W) (1+1) (w)
V.R.
[(1+k)W VR.
W
where. W
The equation can be
Efficiency
(1+k)w
are constant.
-written as
where c and k are constant
Efficiency → k'
8ܪ
If P (effort) is small, the Load is small and the efficiency considerably less than k', but for larger load, P is larger, As negligible. Efficiency becomes greater
A graph of M. A. and Efficiency. ploited against Load is shown as follows:
M.A. Vs Load
Vs Loat
( 接第六張第一頁)
1978
中文中學會考試題預習專欄
徐有祿。
You should describe from your imagination, a time when you
mu
(x+M)
10 x 300
200
英文科 (七)
15 ms
The block was brought to rest after travelling for 15 m by the rough surface. By applying equation of motion; 2a5 v
0152-15 ms
2x15
ma
The frictional force is F
(190 + 10) x 5 x1
loef. of sliding friction
Pressure.
0.75
HYDROSTATICS
Pressure is defined as the avera ge normal thrust (force) per unit
F. cose
area.
wheere P is the average sure,
is the area of the surface on which the pressure exerto,
is the thrust acting on the surface, and cose is the perpendialar
component of the thrust.
At a given point in a fluid, the pressure can act in any directio Thus, p
pressure is a scalar, not.
a vector,
Pressure of a fluid may be mea- sured as:= _
= h£ 8
where his the depth of the
fluid,
is the density of it, Is the local gravity
1ts Newtons per square metre pound-weight per sq. Dynes per square cm.
Other units of atmospheric pressure:
Atmospheric pressure is measured by barometers and is usually cx- pressed in mmHg.
The average value of the baro- metric height at sea-level is 760 mmHg. This is taken as a unit of atmospheric pressure and is called one standard atmospher "(1 atm.)" (
For Meteorological purposes, unit of bar and its sub-wi
llibar is introduced.
bar
bar
103 millibar
= 1706
dyne om
visited a police station to
take part in an identification parade.
Imaginative Approach:
You may pretend to be somebody who had been ill in a hospital and who almost died there.
Imaginative Approach:
You may describe a time when you spent a night in a cell at a police station or detention centre, having been arrested through a mistake in identity topic: A room with a view, Imaginative Approach:
You may write about an elderly lady, almost completely para- lysed, living in a country overseas. Her son buys a house in the country and puts her by a window so that she can look out across that countryside.
Imaginative
Approach:
You may write about a Form 5 pupil who is eager to leave school and who is bored with the view from his classroom. When he leaves school and can- not get a job, he wishes that he were back again looking ati a familiar view,
the same..
A fluid transmits an external- ly applied pressure equally in all directions inside an enclosed vessel.
Archimedes
Principle
An obeject immersed in a fluid experiences. a resultant- upward force owing to the pressure of fluid in it. This upward force is called the upthrust of the fluid on the object.
Archimedeo Principle states. that the upthrust on an object totally immersed in a fluid is equal to the weight of fluid displaced by the object.
The following diagram explains. the principl
1000 J
E
*(v.)
000
height hits
zm(12)
is.halvei
500
The other half is converter": into potential energy E
pa meh = 500 J
(10) (10)h = 500
15 m
When it reaches the maximu
height, H, the potential energ becomes E3 = 1000 J
Ep3
Men = 1000 J
(10)(10)H
= 1000
1- 10 m
Answer: Its kinetic energy is halved at a height of 5 m.
The maximum height it reaches is 10. m.
h) In all machines there is frie tion to be overcome, and ofter the moving barts have to he
Efficiency
180
-75
50
20
LOAD
res Thanosy
(c) The pulley system has a V.R.
of 3. There are two pulleys in the fixed block and only one pulley in the movable. block. However, the man pulls himself up. The V.R. of the sys- tem becomes 4, as he has to pull 4m of the cord before lifting the load ip for 1 m.
If friction and the weight of the novable block of the pulley is ne- gligible, M.A.= V.R. The effort applied is
own weight.
Solution
of his
Mass of the bullet, m = 10 g Velocity of the bullet before impact,
#300 m/
Mass of the block M-190 § Applying Conservation of Momentum, ,with final velocity of the combination as v, then
Sum of menenta
- mu + Mu2
106 (dyn) l
10
10 Nm
10/10
100 Nm
Nm
(10
Converting 1 atm to (a) Nm bars, (c) millibars; with 0.8 m/s/s4
1 atm 760 mmHg = 0.76 mHg density of mercury!
13,6.
kg
cm
- 13.6 (103) (10°)
atm
0.76 x 13.6 x 107, 300 Nm ..101,300
bar
9.8
107
1.013 bar.
1013 mb
mmHg is also known as Torricelli)
(after
torr
1 torr 1 mmHg 133.3 Nm Properties of fluid pressure
1/ Pressure at a point in a rest fluid is the same in all direct- tion.
2/ Pressure at all points of the same level in a rest fluid is
1111
An object (cube) immersed in a fluid. The pressure on Cis great. er than that on B, since h2 a hy
Presure P = Pressure P Resultant horizontal force is 0. But, pressure Po
ressure P
h2 gA >
Resultant force on
upthrust
A
(h2- H1)A is the volume of the
object
Upthrust = Voë
mg
weight of liquid. placed.
Floating Bodies
If a body is totally immersed in a liquid, or gas, the forces act- ing on it are
(1) its weight vertically down-]
wards, and⠀⠀⠀
(2) a force equal to the weight
of the displaced liquid act- vertically upwards.
If (1) is greater than (2) the body will sink in the liquid. If (2) is greater than (1) the body will rise to the surface e and will float partially immerset- sthe condition being that the weight of the body is equal to the weight of the liquid dis- placed by the immersed portion of the body.
If (1) is equal to (2), the body will be in equilibrium, and will remain suspended in the liquid.
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