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### Answer page - L to Z.

posted Oct 16, 2015, 8:44 PM by Upali Salpadoru   [ updated Jul 31, 2019, 3:20 PM ]
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L
• Light - Introduction.
• Q.1.0

•  Phenomenon Can the wave theory explain? Can the particle theory explain? Reflection Yes. Yes. Refraction Yes Yes Interference Yes No. Diffraction Yes No Photo electric effect. No Yes.

 Q. No. Answers Q. No. Answers 2.03.04.0 1 – B.    2- C,   3 – D  4 – B , 5 – A , 1 42˚, 2 – Image.  3 - O˚  4 – Red and blue,  5 – Red.1       Refractive index (1.3)                   =  sine i. / sine r.             Sine r. =  sine i. / 1.3                              =  sne 20˚ / 1.3          =  0.3420 /1.3            =  0.2630      r.  =  15.3˚  2  Refractive index        =  Sine 90˚./ sine r   1.3 =  1/Sin r.     Sine r. = 1/1.3           =  0.7602Critical angle   r. = 50.28˚ 5.06.0 1 Image,   2  Real,  3   Inverted ,   4  Usually smaller,  5.  Same size.1 – A ,  2 – C,  3 – A, 4 - C,  5 – D,

### 1.0 i) and ii) 2x2+1=5 marks

2.0 i)

ii) Let the sun’s rays fall on the lens perpendicular to the plane as shown. The distance between the lens and the converging point will be the focal length.

5x2=10 marks

3.0 Show how the 5 rays will emerge through the lens.10 marks.

4.0 Show how the 5 rays will emerge through the lens.

10 marks. 5.0Red and blue spots are 2 light emitting diodes.

i) Where will be the 2 sharpest images be?

ii) If the lens is moved to F  what changes can be observed in the images?

5x3=15 marks. 6.0

i) Draw a ray diagram to show the image of the object shown.

ii)  Real,Inverted, smaller.

10x2=20 marks. 7.0

The diagram shows an image formed by a double convex lens. Use ray diagrams to find the position of the object.

10 marks.

8.0 Anobject 2cm high is placed 15 cm away from a convex lens whose focal length is 5cm.

i) Position of the image                                        .ii) Size of the image.

 1/v + 1/u = 1/f1/v + 1/15 = 1/ 51/v= 1/5  -  1/151    =   3 -1 v       15v= 15/4 =3 .75 cm. M=     Height of image =   Image distance          Height of object       Object distance.               h/2    =   3.8 / 15                H    =  2x3.8 /15

10 marks.

9.0

What lenses are normally used for the following.

.i)  Camera,   -    Convex lens  ( sometimes convex+ concave)

ii) Flash light, -  Concave to spread the light.

iii) Door viewers (peepholes.)- concave  to see a large area.

iv) Near nearsightedness, concave

v) Slide projector.  convex.

10 marks

M

Machines.

 1.0                                  W x 0.1 = 5 x 0.3          W = 15 N 2.0(W+5) x 0.5 = 16 x (0.5 + 0.5)       W+5    =  16/0.5              W =  32 -5                     = 27 N

3.0
 Taking L as Pivot                          ACW moments = R x 6                          CW moments =  3000x3  +  400 x 4                                      6 R =  9000 + 1600                                         R = 10600 / 6                                             =1767 N Taking R as pivotCW moments = L x 6ACW moments = 3000 x3  +40 x 2        6L =  9000 + 800             =  9800/6              =1633Verification:-Downward forces =  3000 + 400 = 3400Up ward = 1767 + 1633 =  3400

4.0     4.1   A couple.   4.2  Torque from one hand =  100 x 0.3 Nm.

For two hands              = 30x2  = 60 Nm.

Mirrors.

 1.0  Image is virtual, (behind the mirror)  laterally inverted ( not inverted) same size. If you rotate the mirror by 90 deg.  you will know why we give the above answer. 2.0 3.0 Image is virtual behind the mirror same size laterally inverted. 4.0i)  A piece of paper -  diffuse reflection.ii) A water surface - Some rays will go through while some will undergo regular reflection producing a parallel beam. Outside surface Inside surface
 5.0 Rear view mirror of a car  - diverging. 6.0 Virtual small.

7.0

A 2-cm object  is placed at a distance of 25 cm from a concave mirror having a focal length of 15 cm. Calculate (i) the image distance (ii) the image size.

Focal length = 15 cm.,    Object distance = 25 cm.,  Object height = 2cm.

Formula:-       1/ object distance  +  1/ image distance   =  1/focal length.

1/25   +  1/ di                          =  1/ 15

1/di     =  1/15  -  1/25

di      =    75 /  5 - 3

=   37.5

Machines.

# Machines - Inclined plane.

7.1   W=  weight x height
W = 1000 x 30   =  30,000  J
7.2   W = mgh
= 70 x 10 x 30 J
= 21000 J
7.3   Work done on cart  = Force x displacement.
30,000    =   F x 50
F    =  600 N
7.4  Efficiency  =useful work/ total work
=30,000 / 30,000 +2100.
= 30,000 / 32100
=0. 9968  (As a percentage   99.7 %)  (10 x 4 = 40 mark

Machines - Levers.

 1.0 Effort x Effort arm  = Load x Load arm. 10 x  W=  5  x  40     W    =  200/10   = 20 N 2.0 The centre of gravity of the ruler has to be at 50 cm. mark. Therefore the distance from G to pivot is 0.2 m.ACW Moments = CW Moments          W x 30  =  30 x 80                  W  =  80 N
 3.0 3.1      Clockwise moments= anti clockwise moments      R 1 x 6  = 3000 x 3  +  400 x 2      R 1       =  (9000 + 800)  /  6                 = 9800/6   =   1633 N

1

1

3.1

Clockwise moments= anti clockwise moments

R 1 x 6  = 3000 x 3  +  400 x 2

R 1       =  (9000 + 800)  /  6

R 1      =  9800/ 6   =  1633 N                                                                      2

3.2

Total up force = Total down forces

R 1  +   R 2    =  3000  + 400

R 2    =  3400   - R 1

R 2   = 3400 -  1633   = 1767 N                                                                1

4.0

4.1   A couple.                                                                                                                   1
4.2  Torque of a couple =  One of the forces x  diameter.
100 x  0.6     =    60. Nm.                                                                                        1

5.0
5.1      Fulcrum is in the centre of the bolt.
Rotating of the nut forms a couple.
Moment of a couple =  F  x  Diameter
Clockwise moments = Anticlockwise moments
)
(F x 0.07) +   10 x0.1     =  40 x 0.35
0.07 F      =   (14 – 1) / 0.07
F =   186 N                                                                   1

5.2      .Anticlockwise moments = Clockwise moments.
F x 0.07   =   (10 x  0.1)  +  (40 x 0.07)
0.07    F    =    1 +  14
F =  15/0.07      =  214 N                                                                            1

5.3   When lifting the weight of spanner will be acting against you.                                           1

Machines - Pulleys and Wheels.

1.1 ….. 2

1.2    Velocity ratio =   Load / Effort

2  =   L   /   15

L =  2 x 15    =  30  N

Weight of pineapple =  Load -  weight of pulley.

= 30 – 2   = 28 N.

Therefore mass of pineapple =   2.8 Kg.

= 28 / 15    =  1.9

1.4   Efficiency   =  Mechanical advantage  /  Velocity ratio

=  1.9  /  2    =  0.95

E %  =  95 %.

80/100    =     MA / 2

MA  =  0.8 x 2        =   1.6

Using equation  MA  =   Load / Effort

1.6   =  700/ Effort.

Effort   =  700/1.6        =  437.5 N

3.1    ……10  m

Velocity ratio =   Distance  Effort moves / Distance Load moves.

=  20 / 10   =  2.

3.2   Method 1.                     Total Load   =  Weight of log + weight of hanging pulley

=  1200  +  5      =  1205  N

As the load is supported by 2 strands each strand wwill get  Load / 2

= b1205 /  2     =  602.5  N

As the single fixed pulley does not alter the magnitude the man will have to give a minimum  force of  602.5 N.

Method 2.

Velocity ratio = Ideal Mechanical advantage

2       =        1205  /  Effort

E        =       602.5  N

3.3

=  1200 /  602.5     =   1.99

Efficiency =   Mechanical advantage / Velocity ratio.

E     =   1.99/2   =  0.996

=  99.6  %

3.4 ……………Lower.

3.5………….. Lower

4.0

 Question A B 4.1  How many strands are supporting the load? 4 5 4.2  What is the velocity ratio? 4 5 4.3  What will be the effort required to lift a load of 482.0? (482+18)/ 4= 125 N (482+18)/5= 100 N 4.4 What is the mechanical advantage? 482/125= 3.9 482/100= 4.8 4.5 What is the percentage efficiency? 3.9 / 4x100=  97.5% 4.8/5=  96 %

V.R =   rw  / r a     =   18/3   =  6

Considering the Velocity ratio as the Ideal Mechaanical advantage, we get

6 =  750/  E

6 E =  750

E  =  750 /6  = 125 N

7.0
7.1  Cogwheel ratio =    16 : 64
=  1 :  4
As the rare wheel is attached to the rare cog wheel it will make 4 turns.
7.2  For 1 turn of pedal the rare wheel will rotate 4 times.
The circumference of wheel =  2pi r
= 2 x 3.14 x0.4  m
The distance bike will travel  = 2 x 3.14 x 0.4 x 4

= 10.05  m

• ## Momentum

Q. 1.0

1.1  A   =  4000 x 6   = 24,000. Kg.ms-1

1.2  B   =  5500 x4 = -  22000 Kg.ms-1 .

1.3  A + B Combined momentum   24000 + (– 22000)  =  2000

Combined mass  =  9500

As momentum is conserved

2000 =  9500 . V

V   =   2000/9500 = 0.21 ms-1 to Right.

Q. 2.0

2.1       Using Pythagoras theorem;

R2 = 8 2 + 6 2   = 64 + 36  = 100

R =   10 kg,ms-1

2.2       Momentum = Mass x Velocity

6  kg.ms-1    =  m x 13.5

M  = 6  / 13.5   =  0.44 kg.

2.3        Force x time =  Impulse = Change in momentum

F x t =   10 kg,ms-1

F  =  10 /0.06  N

=  166.67 N.   36.9  west of the original direction.

Q. 3.0

3.1 ..0.

3.3 Momentum of Part 3,  =   √ 22 + 12

= 2.24kgms-1

3.4    Momentum = mv

∴ v =2.24÷ o.1

= 22.4ms-

Using trigonometry to get the angle:-

### Tan 0.5 =27°

3.0

i.               90

ii.              20gms-1.

iii.            15gms-1

iv.             As the direction is 90 we can use the Pythagoras theorem to get the resultant.

R2 =  202 + 152

R = 25g.ms-1

v.              37

Motion Linear

Q.1.0

 i. Velocity = distance/ timeV = 50/ 2V= 25 ms-1. ii.Time = 50 + 30 /25          =   80/25           = 3.2 s. iii.  a = Change in velocity / time          = 25 /5            =  - 5 m.s-2
 i. Velocity = distance/ timeV = 50/ 2V= 25 ms-1. ii.Time = 50 + 30 /25          =   80/25           = 3.2 s. iii.  a = Change in velocity / time          = 25 /5            =  - 5 m.s-2

Q.2.0

 i.            V = d  / t.    150 = d x 5  ∴  d   = 150 x 5           =    750 m. ii.  a  =(V f – Vi )  /  t.     a  =(0 - 150) / 10         = -15 m.s-2 iii.   method 1.d = V1t + ½ at2   d = (150 x 10)  + -15x100 d=  1500 -1500/2 =     =  1500 - 750  = 750 mMethod 2.   V.av = (Vi + V.f) / 2            = 75 m/s   d=  V xt     =  750 m.

### i. What is the maximum velocity attained?   ii.     Find the positive acceleration ..        iii.    Find the acceleration during the reducing speed.   iv.    Find the height freely fallen before the chord tightens.

 i.40 ms-1 ii.a =(V 2 – Vi )  /  t.   = (40 - 0) / 4.5   = 8.9 ms-2 iii.a =(V 2 – Vi )  /  t    = (0- 40) /4.5    = - 8.9 ms-2 iv. d =V.av x t V.av = 1/2  (V f +Vi )         = 1/2 (40 + 0)        = 20 ms-1  d =V.av x t      = 20 x 4.5      =4.4 m.

### Motion in a circle

1.0

a.    The distance of the circular path  is  =  2 πr       =  2x 3.14x 4  m

Period of rotation                               =   3 s

Speed            =  d/t     =      (2x 3.14x 4)  /  3

= 25.12 ms-1

b.    Acceleration = v2/r    =   v2 /r     =   25.12 x 25.12 /  4

=157.8 ms-2

c.   Acceleration is towards the centre of the circle.

d.   The tension is due to centripetal force and the centrifugal force.

Force =   mass  x acceleration.

=  0.25 x 157.8  N

=132.8  N

2.0

a,   Acceleration =  v2/r     =    100 x 100/ 10     =   1000 ms-2      to the centre.

b,  F= ma    =   800 x  1000  = 800,000  N

c,  800,000  N

3.0

Radius of circle is the sum of half width of each

Radius =  1.2/2 +  1.4 /2  =  0.6 +  0.7  =1.3 m

Acceleration  =     4π2 r / T          =   4 x (3.14)2 x 1.3 / T

=   4x  9.86 x 1.3 / T

=  51.3 /T

Force =  mass x acceleration = 40 x a

380 =  40 x  51.3 / T

320  =  /T

T  =  2050/ 320  =      6.4  seconds

a.     Accceleration =  51.3 / 6.4   = 8  ms-2

a=  v2/r

V2 =  ar

V =   ar

b.  Velocity                      = 8 x 1.3 =       10.4

=  3 ms-1

c.  Shell moves at a tangent to the circle.

• # Newton's Laws.

 Newton’s Laws   1 and 2 only i. ii. iii. iv. v. Marks 1.0 100m/s2 310 m/s. 3,100kg.m/s 3,000 kg.m/s 1000 kg.m/s2 5 2.0 15 kg.m/s 0 15 kg.m/s 15kg.m/s 75 N 5 3.0 . 0.75m 20m/s 0.0375s 48,000kg.m/s 1,280,000 N 5 4.0 Equal. A. 2000m/s2 500m/s2 1.6 m 5

Asessment

18 - 20  Excellent           13 - 17 Very good                10 -   12    Good         5 -  9  Fai

Nuclear Changes

1.0

1.C  2. A  3. B,  4. E, 5 D.

2.0

1.236.   2. 56.  3. No charge,  4. Inside the nucleus. 5.  C.

3.0

1. H+ + Cl-  HCl   ……………….. Chemical change.

2. Mg24 +   n1 Na24 + H1     ………..Artificial transmutation.

3. 6C14  7N14+ -1e0……(β)……….. .Beta decay. / Beta emission.

4. 86 Rn222  = 2H4  +84 Po218 ………...Alpha decay /Alpha emission.

5. 92U235 + 0n1 + 92U236  ………….Nuclearreaction.

Q.4.0

1. Isotopes.  2. A neutron. 3.  Ba, Kr and three 0n1.

4. Forming U-236 and splitting them forming three Ba, threeKr and emitting six 0n1.

5. A chain reaction.

5.0

1. 124.

2. Excess neutrons over a stable ratio.

3.  3.7 x2 = 7.4 s.

4.  84Y215

5.  Alpha and Gamma.

# Motion - Projectiles.

Q.1.0    1.1  ….At  45°

1.2…..At  90°

Q.2.0     2.1….Yes.

2.2….If you select the stone it would be because it experience less air friction .The stone has    a      high density and therefore a small surface area .

…..If you select the stick it would be because it has a greater surface area and therefore the chance of hitting the target is greater.

Q 3.0

3.1

0 = V1 +  2x -10 x 30

V12 =  600

V =  √ 600   = 24.5 m   upward.

3.2

It will come down at the same speed but in the opposite direction.

24.5 m downward

Question 4.0

4.1.

Using formula:-    a = V2-V1  /t

-10 =  0 – 150 /t

-10 =  -150/t

t.  =   15 s.

4.2.

Distance = velocity x time.

d.=  150 x 15 m   =  2250 m

Question 5.0

(5.1)

Vertical component of velocity =  R sin 45°

=14 x 0.707   = 9.9 ms-1

Using formula  a=  Д V/t

t =  Д V/a

=9.9 / -10   = 0.99 seconds.

Assuming that the horizontal velocity does not change and the total time is double that for the projectile to rise  we get the range from this.

Range =  R cos Θ x0.99x2     =  9.9 x1.98

= 19.6 m

(5.2)   Velocity of the put when coming down will be the same as the starting velocity at the same level. This is R cos Θ

Using the formula            v22 =V1+ 2ad.

V2 2 = 9.92    + (2x 10 x 2 )

V2  =  138

V  =   11.7  ms-1

Average velocity while falling the lat 2m

Vav  =  V1 + V2 / 2

Vav   =   (9.9 + 11.7)  / 2        =   10.8ms-1

Time =  distance / velocity

t.  =  2/ 10.8   =  0.19  seconds

As the horizontal speed remans the same

d  =  R cos Θ  t

= 9.9 x 0.19  =   1.9 m.

The total displacement  =  19.6 +  1.9   =   21.5 meters

Q

R

Rotatory Motion.

1.0

1.1  b.        1.2 a. 1.3    d.   1.4   c      1.5   d.

2.0

2.1    40 rev in   120 s

1 rev.    in 120/40   =  3 s.

2.2   One rev is 2π   radians.

Time taken   =  3 s.

Angular velocity =  2π / 3     =  6.28/3  = 2.1  rad s-1.

3.0

3.1       . τ  = F r.

= 25 x0.16  Nm  = 4.0 Nm.

3.2                τ  = Ft x r

Ft =    τ / r = 4.0 / 0.08  =  50 Nm.

3.3       Angular displacement of large cog wheel  =   2 π

Circumference                    =2 π r 1

Rotations of small wheel =  2 π r 1  /  2 π r 2 = r1/r2  =  0.08/0.04   =  2

Rotations of rare  wheel   = 2

3.4              Distance cycle will travel == 2 x 2 π r 3  =   3.77  m

4.0

4.1 . α  =   (  ω f – ω i )   /  time

=  ( 7  - 0 ) / 4     =  1.75  rad s-2.

4.2   0.  ( horizontal line shows  uniform velocity )

4.3       ω ( av)   =   ( ω f + ω i )   / 2

= 8 / 2 =  4 Rad s =1.

Angular displacement in 5 seconds    5 x 4   = 20  rad.

2 π   radians will be   1 rev.

20 rad   will be       20/   2π =  20/ 6.28   =  3.18  turns.

S

• Simple Harmonic Motion.

Q 1.0

1.1-  a,  1.2- c,   1.3 - a,  1.4-b.   1.5- c .  1.6- b

Q. 2.0

i.  Swinging of a chandelier.

iv. Plucking a string in a guitar.

These cannot be taken as SHM.

ii. A palm can get pushed by the wind and it can come back only the wind speed decrease.

iii. When a child jumps up it accelerates down by the gravitational force. In SHM the the whole time acceleration has to be directed towards the centre of motion.

v.  The same answer as for above.

Q. 3.0.

i.                     10g= 0.01kg.

Potential gravitational energy that can be released = mgh.

=  9.8x0.01 x0.2 J

= 0.0196 J

ii.                  Potential energy changes to kinetic energy.

Kinetic energy=  ½ mv2.

½ mv2.    =  0.0196

v2.=  0.0196/ 0.5x0.01

v   =         3.92  = 1.98 ms-1

iii.                Zero.                              Iv.   Up (push by the water in the other arm)

iv.                Va=  v1 +v2 / 2

=   0.99 ms-1

v. The distance particle has to cover in one cycle.

d  =  vavx t.

Period is      t =  d/va

t =0.4 /0.99

= 0.4 seconds.

0.4s ………..= 1cycle

1.              = 1/0.4 =  2.5 Hz.

Vi  .  Throughout the oscillations the force acts towards the equilibrium position.

• # Speed Velocity and Acceleration.

Q. 1.0

The distance train has to travel = 50 m

Time taken                                = 2 s

There fore speed                     = 50m / 2s    =  25 m.s-i

In passing the platform the train travels the length of the train plus the length pf the platform.

=  59m + 30 m

Using formula                         S  = V .  t

Making t subject                    t  = S / V

t =  80 m  /  25 ms -1

= 3.2 s.                     \

Q. 2.0

i.    The velocity at cruising speed is shown by the blue section of the graph.
The area of this part        =  150 x 15
=   2250     (The area below graph)
The displacement at cruising speed = 2250m.

ii.    Change in velocity               =   V2  -  V 1
Time taken                       =   10 s
Therefore acceleration  =  ( 0 – 150) m.s-1   / 10  s
=   - 15 m.s -2

iii.     The displacement during landing  =  Area below red graph
=  Area of green triangle
= ½ base x height
=   (150  x 10) / 2
=      750
The displacement                 = 750 m

Q. 3.00

i.    Maximum speed     =  35 ms-1
ii.    This is only for three seconds.
Using formula     a =  ( V2 – V1 ) /    t
a  =  ( 30 – 0 )  / 3 m.s -1
a   =    10 m.s -2

lll.       Using the same formula
a =  ( V2 – V1 ) /    t
a  =  ( 0 – 35m )  / 4 m.s -1
=  8.75 m.s -2

Iv,      Area of the green part.     45 m.

Q.4.0

Equation 1.         acceleration   =  change in velocity  /time

a   =  (V2  - V1 ) /

Therefore           t     =  (V2  - V1 ) / a
t   = 30 s-1   / 6 ms-1
=  5 s
l.

l.

 Period First second Second second Third second Fourth second Velocity avg. 35 25 15 5
1

ii.

 Period First second Second second Third second Fourth second Displacement 35x 1 =35 m 25X1=25m 15x1 =15m 5x1 =5

iii.

 Time / s 1 2 3 4 Displacement 35.m 60.m 75.m 80.m

iv.

## T

• TORQUE

1.0

CW Torque = ACW Torque

J X2  =  225 x 2.5

J  =  562.5 / 2 …………………..=  281.25 N.

Total downward force =  281.25 + 225 ……………..=  506.25

Support force = 506.26 N.

2.0

Taking A as Fulcrum

CW Torques=  8 xB

ACW = 2X10) +4.5 X 25

AS the torques cause no rotation  8B = 20 +    112.5

B=      132.5 / 8………….16.6  N

Total downward force  =   10 + 25  = 35

Then A gets     35 - 16.6  …………………………………….18.4 N

3.0

1. T =  Fxd

= 50 x sine30 x 0.25

= 6.25 Nm.

B. At 90 .

C. The weight of spanner will act down through tnhe centre of mass.

4.0

4,0A

1. Weight.

2. Normal reaction

3. Foreward force.

4. Vertical component of upward force.

5. Friction.

B

1 Weight cos 30 = R

R = 40x 0.87

R = 34.8 N

2. Foreward force =  W sine 30

=  40 x0.5

= 20 N

3.  If the friction stops sliding the torque = foreward force x radius.

T = 20 x 0.35

=  7 Nm.

## U

Units of Measurement

1.        The volume of a solid block  is  20 m3 .  The density is 6 kg m3. What is the weight of the mass?

Answer:   Mass  = volume x density.        20 x 6 = 120 kg.

Weight -  mass x gravitational force.           120 x  9.8   =   1176  N.

2.        John getting a standing start runs 100m. in  15 s. gradually increasing velocity . Find the final velocity?

Average velocity =  (Vi + Vf ) /  2

100/15  = 0 + Vf / 2

Vf   =  2x100/ 15   …… V13.33ms-1.

3.        Jane has a mass of 50 kg and her cycle  20 kg. On riding at a speed of  10ms-1 a snake decides to cross her path. She brings the bike to a stop in 2 seconds. What was her braking power.

Kinetic energy of cycle and rider = ½ mv2.

Ke. =   70/2  x  10=  3500 J.

Power = J/t

=  3500/2    ……=  1750 W.

4.        1.0 kW. Steel kettle of 250g. is used to heat 2 litres of water at 20C . Find the minimum time it will take to boil the water.

Heat absorbed by water =  mct ………….2kgx 2400 x (100-20)

=  2x2400 x80

=  432000 J

Heat absorbed by kettle =  mct ………….0.25 x490 x 80

=  9800.  J

Total heat absorbed =  441800 J

Power =  J/t      ……….t=  J / p  ……..=  441800 / 1000

T=  441.8 s……. 7.4 minutes.

=

5.        A car and passengers having a mass of 1500 kg. speeds up from 15m-s     30 m-s   in 3 seconds. Find the force car has used.

Acceleration =   (vf – vi ) / t  …………….(30 – 15) / 3

= 5 ms-2

Force =  mass x acceleration…………..= 1500 x 5  N.

=  7500   N.

• ## 1.01. - d.  2. - b.   3. -  c   4 – a,

2.0  c.       3.0-  a&b.
4.0

.to 8. Obtain the answers from this diagram.

Fig. 5. Free body force diagram.

9.             860 N (Approx)

Trigonometry solution

Sine 60°  =   h/6  =  0.866

h =  6x 0.866    = 5.196

Tan 60°  =  5.196/ x = 1.7321

X = 5.196 /1.7321   = 3

Therefore base  = 4+3= 7

Tan Θ  =  5.196 / 7 =0.7423

=37°

Sine 37° =   5.196/ R =0.6018

R =  5.196/ 0.6018   = 8.6

Actual force is  860 N

10.          A force equal in magnitude to resultant in the opposite direction.

## W

• WAVE PROPERTIES
• Q.1.0   D

Q.2.0

6.5x10-7/  1x10-4  = …..x / 3)

0.02m

3.0

1.

Loud sounds are a result of constructive interference while faint sounds are at lines of destructive interference.

2.

V =fλ…………………………λ= V/f

= 330/264 m.

λ     = 1.25m.

3.   Tan θ =   2λ/3

= 2x 1.25/3…………=2.5/3 ….=0.833

Θ    =  39.8°
• Q.4.0

1. W…………...500Hz.

2. N………….. f a. =  Vw  x f w

Vw-Vs

= 330 x 500 / 330-20

=  165000 / 310………..= 632.26 Hz.

3.  E…………...500 Hz.

4   S ………...f a. =  Vw  x f w

Vw+ Vs

fa    = 330x500 / 330 + 20

= 165000 / 350……...471.43 Hz.

2. As she would be getting closer, she will receive more waves. Apparent frequency would go higher and pitch will increase.

• ## WORK , POWER  and Efficiency

1.1 F = ma,   F= 45x9.8   =  441 N. down.

1.2 Reaction =  441 up.

1.3  w= fd.,      W= 441+ (25x 9.8) x 3000

=  441 + 245

= 686 x3000  =   2058000 J

1.4         P = w/t.     P = 2058000/ 30x60     = 60 W.

 2.1  2500 x 8 = 20000 J. 2.2.  2500 W. 2.3   Power =  Work/ time                2500 = 20000 / t                     t.  =  20000 / 2500 .    t.=  8 s. Q 3.0 Answers 3.1   3 (800 + 600) = 4200 N.     , 3.2   3 ( 800+600)/30 = 140  W.    ,  3.3   800 x 3 = 2400  J. 3.4   (800 x 3) ÷ 5 =480 N.     , 3.5   512 – 480 =  32 N. Q.4.0  Answers. 4.1  Work=  5000 x 9.8 x 25 =  1225000 J  ( 1225 kJ.). 4.2   746 x 2 = 1492  J. 4.3    P =  W/t  ,   t. = W/p  .          t.   = 1225000 / 1492.               = 821 s   =    13.7 minutes. 4.4   According to the given HP of the pump in 20 mts. ( 20 x 600 = 12000 s)  it should perform  1492 x 20x60 J. Actual work done was = 1250000. Efficiency =  1225000 /  1790400  =  0.68 Percentage efficiency = 0.68 x100 = 68. %. Q. 5.0 5.1  400 N.     5.2.  400 x 12 x 0.2  =  480. J.   5.3.    480/ 5  =  96  W     , 5.4  P.E g. = mgh                   = 40 x 10 x 2.4                    =    96 J. 5.5 All the potential energy gets converted to Kinetic energy =  96 J.   ( 4 x 5 = 20 Mark

WAVE PROPERTIES