# Physics 9702 Doubts | Help Page 227

__Question 1075:__

__[Applications > Communicating information > Digital transmission]__**(a)**Distinguish between an

*analogue*signal and a

*digital*signal.

**(b)**An analogue-to-digital converter (ADC) converts whole decimal numbers between 0 and 23 into digital numbers.

State

(i) the minimum number of bits in
each digital number,

(ii) the digital number representing
decimal 13.

**(c)**An analogue signal is digitised before transmission. It is then converted back to an analogue signal after reception.

State and explain the effect on the
reproduction of the signal when the number of bits in the analogue-to-digital
converter (ADC) and the digital-to-analogue converter (DAC) is increased.

**Reference:**

*Past Exam Paper – November 2014 Paper 43 Q12*

__Solution 1075:__**(a)**

An analogue signal is continuously
variable

A digital signal has two / distinct
levels only

**or**1 s and 0 s**or**highs and lows**(b)**

(i) 5

{The number of levels is
given by 2

^{n}where n is the number of bits. The minimum number of levels should be able to convert the range given (between 0 and 23).
When n = 1, number of
levels = 2

^{1}= 2
When n = 2, number of
levels = 2

^{2}= 4
When n = 3, number of
levels = 2

^{3}= 8
When n = 4, number of
levels = 2

^{4}= 16
When n = 5, number of
levels = 2

^{5}= 32
Thus, the minimum number
of bits (that can convert 23) is 5.}

(ii) 1101

{The order of the digital
number corresponds to 2

^{4}2^{3}2^{2}2^{1}2^{0}
13 =

**0**(2^{4}) +**1**(2^{3}) +**1**(2^{2}) +**0**(2^{1}) +**1**(2^{0})
Digital number: 1101 (the first zero may be omitted)}

(Notes on binary numbers and the reproduction of signals can be found at

**(c)**There is a greater number of voltage / signal levels and smaller step heights in the reproduced signal as smaller voltage / signal changes can be seen

__Question 1076: [Pressure]__
The density of mercury is 13.6 × 10

^{3}kg m^{–3}.
The pressure difference between the bottom and the top of a column of
mercury is 100 kPa.

What is the height of the column?

A 0.75 m B 1.3
m C 7.4 m D 72 m

**Reference:**

*Past Exam Paper – June 2007 Paper 1 Q15*

__Solution 1076:__**Answer: A.**

Pressure difference ΔP = (Δh)ρg

Height, Δh = ΔP / ρg = 100000 / (13.6×10

^{3}× 9.8) = 0.75 m

__Question 1077: [Kinematics > Projectile motion]__
A tennis player is standing 5.5m
from the net. She hits the ball horizontally at a speed of 32ms

^{-1}. How far has the ball dropped when it reaches the net? Take g=10ms^{-2}.**Reference:**

*‘AS and A2 Physics,’ by Graham Booth and David Brodie, PG 34 Q3*

__Solution 1077:__
We need to understand to situation
from the information given and make appropriate conclusions from them… This is
a problem dealing with projectile motion.

“A tennis player is standing
5.5m from the net”: the

__horizontal__distance of the player from the net = 5.5m
“she hits the ball
horizontally at a speed of 32ms

^{-1}”: the__horizontal__speed is 32ms^{-1}. Since the ball is hit horizontally, it has**no vertical component of velocity**.
“How far has the ball
dropped when it reaches the net? Take
g=10ms

^{-2}”: We need to find the distance dropped by the ball. This is a**VERTICAL**distance. Additionally, the acceleration due to gravity is also a vertical acceleration with no horizontal component.
For the vertical motion:

Distance dropped = s

Initial vertical speed = 0

Acceleration a = 10ms

^{-2}
Equation for uniformly acceleration
motion: s = ut + ½ at

^{2}
Time for the ball to drop a vertical
distance s = t

But this time t (as defined above) is
the corresponding time for the ball to travel horizontally from the player to
the net.

For the horizontal motion:

Speed = 32ms

^{-1}. Distance = 5.5m. [Speed = Distance / time]
Time t = 5.5 / 32 = 0.172s

Thus for the vertical motion,

Distance dropped, s = (0×0.172) + ½×10×0.172

^{2}= 0.14792 = 0.15m

__Question 1078: [Kinematics > Projectile motion]__
The diagram shows the path of a golf ball.

Which row describes changes in the horizontal and vertical components of
the golf ball’s velocity, when air resistance forces are ignored?

horizontal vertical

A constant deceleration constant
acceleration downwards

B constant deceleration acceleration
decreases upwards then increases downwards

C constant velocity constant
acceleration downwards

D constant velocity acceleration
decreases upwards then increases downwards

**Reference:**

*Past Exam Paper – June 2009 Paper 1 Q8*

__Solution 1078:__**Answer: C.**

When air resistance is ignored, the
only force affecting the motion of the golf ball is its weight, which is
downwards.

Thus the horizontal component of the
velocity of the ball does not undergo any acceleration – this component of velocity
is constant.

The weight causes a constant
downward acceleration (acceleration of free fall) on the ball. During the
upward motion of the ball, its velocity (vertical component) decreases until
the maximum height is reached. Then the velocity increases downwards. However,
the acceleration remains constant during the motion.

A ladder of length 10 m and weight 200 N leans against a smooth wall such that it is at an angle of 60 degrees to the horizontal. A boy of weight 500 N stands on the ladder 1/4 of the way from its lower end. Calculate the normal reaction at the wall and the magnitude and direction of the resultant force acting on the lower end of the ladder.

ReplyDeleteReference: Pan Pacific Physics A Level Volume 1

See solution 1079 at

Deletehttp://physics-ref.blogspot.com/2015/12/physics-9702-doubts-help-page-228.html

What was the gradient in paper 33 from the year 2015 november session? Negative or Positive?

ReplyDeleteDon't know. Haven't seen the paper yet.

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