1990-01-15 — Page 19

一期星 日五十月一(〇九九一)年九十七國民華中

育教日九十月二十年巴己遭夏

郭日僑

position 0.32 m from the knife edge

system is in equilibrium

to restore equilibrium (see figure 6)

Find the density of the unknown liquid Z.

8. + (0000}{10}\? -

(1000) (10)(1x10

(2)

(1000)(10)

0:32

1990 中學會考預習專欄

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MILL & DALE PRESS

Physics (16)

C. Y. Mak

Exercise 16. Pressure, Archimedes.

Principle

In figure 1, block A hangs by cord from spring bajance D and is submerged in a liquid C contained in beaker B. The weight of the beaker is 2N, the weight of the quid is 3N and the volume of block A is 1.25 x 10-303-

4:

(a) A flat bottomed tube of mass. 0.025 kg and cross sectional area"

2 x 10m2 has lead shot placed in it. It floats upright in a liquid: Xas shown in figure 7.

Substitute into (1) and (2)

Since

B

= 21, 4c

3N

8 + {8000}{10}(V)

7x10

reading of spring balance

.D

אל

reading of the balance E 15%

Mass of metal attached (5000) x ( x 104)

0.2286 kg

Figure 6

(ANS.)

WA

+8

-5 + 15

.(3)

(4)

(4)

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

= 15N

(Ans.)

In figure 4, by taking moment about the knife edge

Mg(0.4)=

.. (1)

Substitute the value of WE into (3)

B # 10N

Elquid. X-

iii. Since

Figure 7

The following results are obtained.

Mass of land strat

placed inside the 0.005 0.010 0.015/0.020/0.025

tube (in ky?

Depth (h)

Inmersed in Iquid X-C1n-m

0.12 0.34 0.16 0.18

0.20

bouyant force

(Ans-)

keight of equal. volume liquid displaced

(density of liquid)x(volume

of the block)x(g)

10 = {d)x(1.25x10^3)x(10)

0800 kgm

{Ans:)

If block A is pulled up out of the liquid, then B = 0

reading of the spring balance

Where M Mass of block X

Weight of block Y in air

In figure 5, by taking moment about the knife edge

Mg(0.35) W x .............{2}

Where is the weight of block Y in water

(2):9 (1)

0.35

(where Bis the Upthrust)

Figure 1

(a) If the spring balance 0 reads 5N

and balance E reads 15%,

i.

what is the weight of block A 10 air?

what is the upthrust on A?

iii. what is the density of liquid.

C

(b) If block A is now pulled out of

liquid C,

what will balance, read?

it what will balance E read?

The

A wooden block A is wetched in air, in water and in liquid X as shown in figure 2(a), figure 2(b) and figure 2(c) respectively. readings of the spring balance in each case are 8N, 2N and 0.5N respectively.

(Density of water 1000 kgm

Figure 2) Figure 2(5)

Calculate

Fiquid x

riigure 2(c).

the upthrust on the block in water'.

ii. the volume of the block Ai

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

conclusion ii. What

on the relationship between pressure. in a liquid and the depth of. the liquid. can you draw From the graph.

ii. From the graph, determine the

density of liquid X.

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

Find the elation betw and h.

ii. 1 h 0.250 find the

density of the liquid.

or and tackle consisting of three fixed and two movable pulleys is used torase an aluminium Cube: which completely immersed in water (density 1000 kgm) as shown. "ins figure 8. If the volume of the 10 m3, the effort applied cube

to the machine is 5 x 10 N and the efficiency of the machine is 30%. Relative density of aluminium :=

YONG

(b) 1.

45N

110-Erda (2)

+ 2.

15+2+3 = 15 R = 5N

R

reading of E-is.5M

(a) Let me mass of block A

V

volume of block A

Ans

(Ans.)

upthrust on black A-in water:

B' upthrust on block

liquid X

From figure 4a).

mg - 8N

(1000){g}{V)

(where y

is

density of the

-3

metal and V.15

8000 kgm

volume of the [Ans.) block)

刋畫 商教 日九十月二十年日

(b).

Total fass 0.015 0.025.

-0.04 kg

From (a)

M = dhA

0.04 = (d)(h){2x10^4)

Mdha.

0.04 = (d)(n)(2x10^^)

hd = 200

ii. If 0.25%

(0,25)(d) =

=200

800 kgm

(Ans.)

the density of the liquid is 800 kom 3.

Effort

(Ans.)

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B. mg

The velocity ratio of the system is 5.

11. The upthrust of water on the

cube B

The weight of water displaced

1990 中學會考預

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Economic & Public Af

(1000)(10)(10)

10%..

II. Public Affairs

(Ans.)

(b) In. Figure 6, by taking moment about

the knife edge

Mg(0.32) Wx

Where W" is the block Y in the (3)(1)

0.32

0,0%

fthe metal

upthrust to liquid **

0.2W

=-0.2(g)(g)(V)

(where density of 2).

0.2 8

(0.2)(8000)

1600 kgm

(Aris.

Total Mass M = (mass of tho

tube) + (mass

iii. Let the pulling force exerted by the string on the cube be

Since V.R. 5 and

M.A.

therefore,

5×108

100% 80%

2x105

Let the acceleration of the cuba be a

Apply Newton's 2nd Law.

T4 & mg mathema where m mass of the cube.

(2.5×1000}(10)

- 2.5x10 kg

(2.5x10-|(10)

(2x10+10)

(2:5x107

(1) The Hong Kong Government.

The Governor's Role

Internal Role of The Hong Kong

1. Executive Roler

He is the head o Kong Government chairman of the Council.

Legislative Role:- He is the chairm Legislative Counci laws making - 'th Council are in th approved (enacted Governor.

Judicial Role: He has role of justices:: "and Judges.

jff, the relative density of block

Ag

Cube (10.

Figure 8-

of lead shot.

insider

the

Let the tension in the string in figure 4[b]; bé T.

tube)

kgm)

Total mass M

10.05 02:035 004 0.045 0.05 (3 kg)

120-140446 0.18 0.20 death h (in n}

Total mass Hin kg.

10N

0-05-

0.04

0.03

0.021

(Ans)

0.01

Tv. the upthrust of the block A ih.

Liquid X

the density of liquid X.

(b) If a lump 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 wate: as shown in figure 3.

A

etmetal

wator

Figure 3

Find the mass of metal attached if the density of the wetal is BOO0 kgm 3

A metal block Y 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 From the knife edge as shown in figure 4.

.3.1 m

What is the velocity ratio of the system?

i. What is the upthrust of water

on the aluminium cube?

iii. Neglecting the resistance of water, find the acceleration of the cube while it is moving in water.

iv. find the position at which the

cube will finally stay.

Solutions

system in order to push the whole cube out of water?

(a) Let d density of liquid C.

tension of the cord

weight of block A

weight of the beaker

Treading of the spring balance = 2N

As shown in the force diagram

11. Apply the formula

iii. Relative density of block A

1.

What additional effort is the to required to apply.

B

10 (1030) (10)[V).

1 10:33

ΤΟ

0.8

(Ans.)

Upthrust of black A in liquid

(weight of block. A in air) -(weight of block, A in liquid *)

weight of the liquid

8 bouyant force on A

-SN - 0.5N

R

reaction of the balance E

Figure 4

on the beaker

7.5N

(Ans.)

gV where density of

liquid X

75g1 (101(1x1073

750 kg

(Ans.)

Figure 5

(a) As shown in figura 5, 1 block Y is completely immersed in water, block

X has to move to the position 0.35 in from the knife edge. in order to keep the whole system in equilbrium. Determine the density of the metal block Y..

(b) If the block is now increased

Completely into an unknown liquid

then block X has to move to the

(b). Let the volume of the lump of metal.

be V1..

Total upthrust .

(upthrust on block A in water)

upthrust on the lump of metal in water}

(1000)(10)(1x1073)

+ (1000)(10)

Total weight:

(weight of Block A) +

(weight of the lump of metal).

Cross sectional area of the tube A

10-42 2x

Pressure at the flat bottom of the tube

pressure of the liquid at a depth h

My

2ms

[Ans.)

The Cube will finally stay at the water surface with part of its volume immersed.

Let the volume of the cube

in immersed

water be

hence

where B the upthrust exerted

on the cube.

But

(2.5x10 1110) - 2×105

5x10 N

ax (1000) (10)(V!)

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

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

= 139

2

From the graph, M increase linearly as h increases, therefore, pressure in liquid increases linearly as the depth increases.

*(3)(g)(h) where. g is the density of the 11quid.

[MphA

the slope of M against h = A The slope of the graph

0.0

0.2

= 0.25 kgm

0.25 = (91(2×10′′*)

1250 kg 3

Ans: The density of liquid X

is 1250 kgm3

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2.5810 N

the effort required be E'

Let

M.A, = 2-5x10°

Again apply

M.A.

x 100% e

2.5x105

× 100% = 80%

E6.25 x 10 M

the additional force required.

EE

.25x10′N - 5x10aN

1.25x10′′N

基本面積

(Ans

不折)

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