Friday, December 11, 2015

Physics 9702 Doubts | Help Page 227

  • 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 2n 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 = 21 = 2
When n = 2, number of levels = 22 = 4
When n = 3, number of levels = 23 = 8
When n = 4, number of levels = 24 = 16
When n = 5, number of levels = 25 = 32
Thus, the minimum number of bits (that can convert 23) is 5.}

(ii) 1101
{The order of the digital number corresponds to 24   23         22         21         20
13 = 0(24) + 1(23) + 1(22) + 0(21) + 1(20)
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 × 103 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 / (13.6×103 × 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 at2
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) + ½×10×0.1722 = 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.



4 comments:

  1. 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.
    Reference: Pan Pacific Physics A Level Volume 1

    ReplyDelete
    Replies
    1. See solution 1079 at
      http://physics-ref.blogspot.com/2015/12/physics-9702-doubts-help-page-228.html

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

    ReplyDelete
    Replies
    1. Don't know. Haven't seen the paper yet.

      Delete

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