1989-01-16 — Page 34

七國民華中

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十年股戊 月

第

半張八紙出 顾胙承就行效

海拔二址地転機

DEAR

:處分類旺

育日九初月二十年辰戊曆夏

想感車候

不是如此之間也就失

干涉,情况更心發不可收拾。 「移」上龍隊之前,俟機上串。若連長不加

唉!人性離道武的是道樣?《安谦 )

有一點空隙,不守秩序者屃裝镄扮檐地

一做,但一樣有不守秩序的情形出現。只要

肘在我钵悧稀搡巴玉的京站有清欄裝

·中華民國僑務委員會頒發登記證台敎新字第一一0號

痛經辉新化舞台

詳裸市北白

[ORKPERAT

1989 中學會考預習專欄

明潇出版社

MILL & DALE PRESS

Physics (16)

C. Y. Mak

Exercise. 16 Pressure, Archimedes

Principle

1. Ir figure 1, block A hangs by a cord from spring balance D and is

submerged in a liquid contained. in beaker By The weight of the Beaker is 2N. the weight of the liquid is 3N and the volume of black A is 1.25x10

(a) As shown in figure 5, if block Y is completely immersed in water, block has to move to the position 0.35 m from the knife edge in order to keep the whole system in equilibrium. Determine the density of the metal block Y.

(b) if the black Yis now increased completely into an unknown liquid 7, then block X has to move to the position 0.32 m From the knife edge to restore equilibrium (see figure 6). Find the density of the unknown liquid Z.

Figure &

報日僑華

期星 日六十月一(九八九一)年八十七國民華中(34)

1.

(a) Let d = density of liquid C

(v) BgV where

T = tension of the cord.

weight of block A

=

+

weight of the beaker

weight of the liquid

8 bouyant force on A

Rreaction of the balance

E on the beaker

A

元朗怡英文書院與柏得中學合辦英詩集

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用铖會收î英語的聽說 講,只號克服了心理障時結項。 可利用的方法。學習外 洪水橋校 1-AVENCER ,能多石英文報章钴础。中西朝冠軍康樂路交

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||日比賽於下午一

density of Tiquid-

-7.5= 10) (1x103). § = 750kgm3 (Ans.)

Let the volume of the lump of

metal be V

Total upthrust

(upthrust on

block A

water+lubthrust on the Flamp

of meta, in water)

(1000)(15)(1x1073)

-4.{1000}(10)V!

Total weight:

Iweight of block A) +

weight of the imp of metal)

8 (8000)(20)V*

(1000)(19)(1×103).

+(1000)(10)V!

8H(8000)(10)(V!)

Mass of metal attached

= (8000)× (4 × 10′′

- 3.2286 kg

(Ans.)

進行

平假元朗上修堂県行,上、下帷禎雨中學。中 校長叡萬對於致辭 棅冠軍康樂路棱、亞軍 作日本冖一月七日)上康樂路荽、烁在大消极 詩集譒反英鼒廣播劇比際集誦比賽中一級冠軍 與柏雨中學聯合舉辦英一 比賽結果如下: 无湖妇及英文沓院礙,就沒有在接。

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合辦英詩集誦比賽

The slope of the graph

10.25 = (3)(2×10^^).

9 - 1250 400

0.05

3.2.

Ans: The density of liquid x is.

3250 kg

(b)(i) Total mass

0.015+0.025.

F. 0.04 kg

From (a)

M: dha

0.04 (d)th](2x10)

M÷CHA

0.04 (d)(h)(2x10

hd = 200

(ii) If h = 0.25 m,

(Arts.)

(0.25)(d) 200

d: 800 kym

the density of the liquid is. 800, x.gm

(Ans.).

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Figure 1

(a) If the spring balance. D reads

SN and balance treads 15N

whats the weight of block A

in air?

(ii) what is the upthrust on A?

(iii) what is the density of liquid

(b) IP block A is now pulled out of

liquid Cy

-(4)

what will balance D reed?

(1) what will balance 5 read?

A wooden block A is weighed in air, in water and in liquid X as shown in figure 2(a), figure 2(b) and figure 2(c) respectively. The readings of the spring balance in each case are ON, 2N and respectively.

(Density of water = 1000, kg:

50

Figure 2(a). Figure 2(b) Figure 2()

(a) Calculate

(1) the upthrust on the block in water.

(11) the volume of the block A. (iii) the relative density of block A..

(iv) the upthrust of the block A in liquid X.

v) the density of liquid X.

(b) If a Jump of metal is attached to the bottom of the block and. causes the wooden block to float in water with its upper. surface just covered by water es shown in Figure 3.

Figure 3.

Find the mass of metal attached. if the density of the metal s 8000 kgm

A metal block is suspended at one of the ends of a uniform red. The

red is supported by a knife edge at

the middle and the whole system is

in equilibrium by hanging a block X

at a distance, 0.4 m from the knife

edge as shown in figure 4..

jerema

·Figure. 4'

--

Figure 5

(a) A flat bottomed tube of mass 0.025 kg and cross sectional araa 2x10 has read shot placed in it. It feats uprighže. in a liquid Xs shown figure 7.

Figure 7

The following results are obtained,

Mass of lead shot

placed Inside the 0.005 0.010 0.015j0.020|0,025 tube (in ky

Depth that,

lanetsed. in!

| Iquid X 61 m)

0.12 0.14 0.16 0.18 0.25

(i) Plot the total mass of the tube. and lead shot against the depth h immersed in liquid X.

(ii)What conclusion on the relationship between pressure, in a liquid and the depth of the liquid cap you draw from the graph.. f) From the graph, determine the. density of liquid X.

(b) 0.015kg of lead shot is placed in the tube. It floats upright in a liquid of density dikgm3: and the depth immersed in the liquid is h meters.

(1) Find the relation between and h.

(ii).If h=0.25m, find the density of the liquid.

5. A block and tackle consisting of three fixed and two movable pulleys is used to raise an aluminium cube "which" completely immersed in water" (density 1000kymas shown in figure 8. If the volume of the cube is 10% the, effort, applied to the machine

is 5x101 and efficiency of the machine is 80%. (Relative density of aluminiuni 2.5)

Figure 8

11000

The system is in equilibrium

(0)

(2)

reading of spring balme 0:5

Rreading of the balance E

= 15N

Substitute into (1) and (2)

3. (a) Let the distance between the knife edge to the end hanging the metal block Y be x.

In figure 4. by taking moment

about the knife edge

Mg(0.4)

P = W/ S

(1)

Where M mass of block X

W

weight of block in

In figure by by taking Toment about the knife edge

(2)

Where is the weight of block

ater

0.35

ྴ་༦ ྃ + R་

(3)

W2+2+3 = 5+15.

(4)

From (4).

મ =15N

(Ans.)

(ii) Substitute the value of

into (3)

. B 1ON.

{Aus

(iii) Since

bouyant force.

weight of equal volume

liquid displaced,

(density of liquid)x(volume

of the block)x(g)

10=(d)x(1.25x10")x(10)

+800kym

(Ans:)

(b) (1) If block A is pulled up out of the liquid, then BO

reading of the spring balance 0

T

0

- 15N (Ans.)

(ii) From (2)

{1000) {g} {V}

B is the. Upthrust)

(5){g (v) {wllectsta

3= 8000kgm

density of the

-3 is

(Ans.) block

(b) In Figure 5, by

Pthe

about the knife

Mg (0.32) Doment

(3).

Where is the weight of the metal block in the liquid Z. (3) (4

0.8W

- upthrust in liquic. 7 ÷WW"

W+2+3 = T+R 15+2+3 15+R

- 0.29

reading of s 5N

:.

Bupthrust on block A in

4. (a)

= 0.25

= 0.2{3}{9}(V).

(where 3 is.

¿

density of ZI

(0:2)(8000)

-3

1600 ky m (Ars.)

(1) Total Mass M = (mass of the tube)

+(mass of lead shot inside the tube)

Tatal mass M |0.03 0.035 0.04 0.045 0.05)

(in kg)

depth h.116 0.12 0.34 0.16 0.18 0.20

the

(a) Let m

mass of block A

V

volume of block &

3

upthryst on block A

in water

liquid X

D.05-

(i). From figüre 4(a)

0.04

mg BN...

Let the tension in the string

0.03.

in figure 4(b) be T

Treading of the spring

0.02

balance = 2N,

0.01

As shown in the force diagram

8 = Teng

9.2

= 2+8

H(in m

(Ans.)

B = 3

(ii) Cross sectional area of the tube A

- 2 x 10-42

V1x10 33

(Ans.)

Pressure at the flat bottom of the tube

(iii) Reletive density of block A

pressure of the liquid at ‘a depth h

Mg

3.

(1) What is the velocity ratio of the system?

(1) What is the upthrust of water on the aluminium cube?. (111) Neglecting the resistance of water, find the acceleration of the cube while it is moving in waler, (iv) Find the position at which the cube will finally stay..! (v) What additional effort. required to apply to the system in order to push the whole cube out of water?

15

=ON

(11) Apply the formula.

10- (1000) (10) (V)

0.8

( {\n\ 5 . )

(iv) Upthrust of block A in liquid

X

B (weight of block. A in air)

(weight of block A in

fiquid X)

BN 0.5N

7.5 N

(Ans.)

From the graph, increases

Tinearly as a increases, therefore, pressure in a liquid increases. linearly as the depth increases.

(111) 9 = (8)(g)(h) where is the

density of the liquid.

M = SHA

the slope of M. against A

The velocity ratio of the system is 5

(11) The upthrust of water on the

cube D

The weight of water displaced -- (1000) (10)(10)

(iii) Let the pulling force exerted

by the string on the cube be 1..

Since V.R.5 and

MA-

TON

therefore,

5x10

[Ans.)

* 100% - 80%

Let the acceleration of the cube be a.

Apply Newton's 2nd law-

1+8-ng = ma

where mass of the cune

(2.5x1000) (10)

2.5x101kg (2x105÷105)-(2.5x10*1(10):

(2.5x101)

a = 2ms-2

(Ans.)

(iv) The cube will finally stay at

tv)

the water surface with part of its volume immersed.

Let the volume of the cube immersed in water be V ner.ce

..

T+B-mg: 0:

where B the upthrust exerted on the cube.

But

=(2.5x101)(10)-2x105 -5x10*N.

B={1000)(10)(*)=10*y* 104V!=5x104

V5m

Therefore, the cube Finally stay at the water surface with half af it's volume immersed.

In order to push the whole cube out of water, the pulling force exerted na the block must at least equal to the weight of the block in air.

Hence, the pulling force =ing

52.5x103

Let the effort required be E'

2.5x109 MUA.

Again apply

M.A.

x 100% e

V.R.

2-5x105

F1

5

* 100% = 80%

6.25 X 10 N

the additional force. required

EE

5.25x10 N-5x101N

1.25x104N

(208.)