F= BIL

Emf and internal resistance 2 (y12)

y12 Thermistor 3

Electric Field Exam Questions (all of them in no particular order)

Alpha Particle Charged plates

E Field calc

E Field Bunsen Flame

Electron in E Field 3

Electron in E Field

Electron in E Field 2

Experiment with scales and rods

Force in Field

Internal Resistance of a cell

Internal Resistance

Q 1,2 & 3 Look up the answers in your text books.

Q4       Total R = 2 + 5.5

            I =  1.5 V/Total R

            Pd across cell = pd across external R = I(5.5) = 1.1V

Q5       emf = 1.5V

            Lost volts = (1.5 – 1.2) V

Internal r = lost volts/0.30A = 1.0 ohm

External R = 1.2V/0.30A = 4.0 ohm

Q6       lost volts = 100A (0.04 ohm) = 4V

            Only 8V across external circuit so bulbs are dim

Q7       I=V/R= 1.25V/25 ohm = 0.05A

            Internal r = lost volts/I = 0.25V/0.05A = 5 ohms

            Total R with 10 ohm = 10 + 5 =15

            I = E/R = 1.5/15 = 0.1A

            V across terminals= V across 10 ohm R = 0.1A(10 ohm) = 1V

Coulomb's Law Practical

Electric Power and Cost

1 (a)      P = VI               P power in W

                                    V p.d. in V       

                                    I current in A

3000 = 240 x I

I=3000/240 =12.5A

(b)      V = IR

240=12.5R

R = 240/12.5 = 19.2 Ω

(c)      3 kW = 3000 W = 3000 joules per second

Energy used in 1 min = 3000 x 60 = 180 000 J

2        8A       31.25Ω

3        Resistance depends upon temperature

4        1 641 600 J

5        4800J

6        6

7        2.083 kW

8        1920W 1687.5 W

9        .64m

10.     21.6p  19.1p

11      £1.16

12      0.54kWh

13      .42 kWh        .112p + 3.79p = 3.9p

14      4.46p            9.34p

Extra  50kW  0.072W

Nitrogen Atom

1.       (a)     Positive as E-field is downwards/top plate is positive/like charges repel/AW (1)  1

  (b)   (i)      k.e. = QV; = 300 × 1.6 × 10–19 = (4.8 × 10–17 J) (2)                                2

  (ii)   1/2mv2 = 4.8 × 10–17; = 0.5 × 2.3 × 10–26 × v2 so v2 = 4.17 × 109;
(giving v = 6.46 × 104 m s–1) (2)                                                             2

  

Ke of Satellite

Slight mistake here left an m out of the 1/2 GMm/r

oops - forgot to multiply by the 10kg for Ke so that should be 2.01 x 10^8 J
 and GPE is for 1km = 1000m so GPE = 0.402 x1000 = 402J

Gravity Apollo parking orbit

Gravity QUestions

 1.      (i)      r has been increased by a factor of 3 from the centre of planet.                    C1
g = (40/32 =) 4.4(4) (N kg–1)                                                                          A1

  (ii)   M = gr2 / G
M = (40 × [2.0 × 107]2) / 6.67 × 10–11                                                            C1
M = 2.4 × 1026 (kg)                                                                                        A1

  (iii)  M = ρV = 4/3 πr3 ρ                                                                                         M1
g = GM / r2  r3 / r2 (Hence g  r)                                                              A1

[6]

2        The astronaut is accelerating / has centripetal acceleration (1)
and the space station has the same acceleration (1)
a person does not feel gravity (1)
only feels forces applied by contact with the walls of the space station (1)
no support force from the space station (as they have the same acceleration) (1)      4

MAXIMUM (4)

[4]

 3.      Period 24 hours (1)
Satellite must stay locked into Earth’s period of rotation (or wtte) (1)

          Plane Equatorial (1)
Centre of orbit must be centre of Earth because axis of orbit must be spin axis of Earth (1)
(gravitational force above equator is only force available to provide
centripetal force in a synchronised orbit, otherwise an engine is required)

          Direction Same as Earth’s rotation (1)
(otherwise satellite and Earth would counter rotate)

[5]

  4.     (a)     i. F = GMm/r2 or F α Mm/r2 with labels (1)                                                       1
ii. finite universe contracts/ resultant force on stars (1)                                    1

  (b)   Any 2 from
i. (satellite B) has larger circumference/smaller velocity
(satellite B) Gravitational field strength is less
(satellite B) Centripetal force is less                                                                  2

          ii.(accept calculation from either satellite)
r13/ T12 = r23/ T22 (1)
satellite A                                satellite B
r23 = 70003 × 57.22 / 1.632      r23 = 671003 × 57.22 / 1.632 (1)
r2 = 75,030 km                        r2 = 75, 320 km (1)                                           3
(= 75,000 km)                         (= 75,000 km)

  (c)   Land-based are (any 3) 1 mark for each
more light can be collected/ made larger
more stable
more manoeuvrable
cheaper to build/repair
longer lifetime/ not exposed to high velocity particles
greater access                                                                                                   3

[10]

  8.     (i)      4.5 (N kg–1)                                                                                                      1

(ii)     g = (–)GM/r2                                                                                                     1

(iii)    g ∞ 1/r2;so value is 40/9 = 4.4(4) (N kg–1) ecf c(i)                                            2

[4]

× 2.0 × 107/(4.5 × 103) = 2.8 × 104 m s–1 (2)
½ mv2; = 0.5 × 1500 × (2.8 × 104)2 = 5.9 × 1011 (J) (2)                                             4
aliter: F = mv2/R; = mg; so ½ mv2 = ½ mgR;= 6.0 × 1011 (J)

[8]

y12 Nov Ass

Part A

1. D
2. E
3. C
4. A
5. D
6. A
7. A
8. B
9. D

Part B

SHM

Divided Circuits

1. Two resistors of 3Ω  1.4A and 7Ω 0.6A are connected in parallel. If the total current through them is 2 A find the current in each resistor. 

2. Two resistors of 2 Ω 2A and 4 Ω 1A are connected in parallel. If the current through them is 3 A find the current in each resistor.

3. Three resistors of 2 Ω 4 Ω 0.5A and 8 Ω 0.25A are connected in parallel in a circuit. The current in the 2 Ω resistor is 1 A. What is the current in the other two resistors? What is the total current in the circuit? 1.75A

4. Three resistors of 4Ω 0.5A, 5Ω 0.4A and 20Ω 0.1A are connected in parallel. If the total current through them is 1 A find the current through each resistor.

5. Ammeter X reads 2 A. What are the readings of am­meters A, 4A B1A, C 1A and D 4A?

6. A resistor of 2 Ω and another of 7Ω are connected in series, and a 3Ω resistor is connected in parallel across the other pair. If the total current through the network is 2 A, find the p,d. across the 7Ω resistor. 3.5V

The Volt

Volt

                                V = E/Q                 1 Volt = 1 Joule / 1 Coulomb

1.       Calculate the potential difference across a bulb if 20 C transfers 4 J of electrical energy.

V = 4 J /20 C = 0.2 V

2.       Calculate the emf across a dynamo if 600 J of energy is transferred by 25 C of charge.

V = 600 J / 25 C = 24 V

3.       A bulb has a potential difference of 6V across it. Calculate the energy is transferred by 90 C.

E = V x Q = 6 V x 90 C = 540 J

4.       A train motor operates on a potential difference of 25 kV .Calculate the charge that transfers  800 kJ of electrical energy.

Q = E / V = 800000 J / 25000 V = 32 C

5.       A torch bulb runs off a cell of emf 1.5V Calculate the energy transferred by a charge of 0.27 C

E = V x Q = 1.5 V x 0.27 C = 0.41 J

A bulb takes a current of 0.25 A. How many Coulombs per second? 0.25 C per second It transfers 60W of powerHow many Joules per second? . 60 J per second

Calculate the emf of the supply. V = E ÷ Q = 60 / 0.25 = 240 V

Circular Motion _London Eye

(a)     (i)      speed v = 2π r / t
v = 2 × π × 122/2 /(30 × 60) (1)
v = 0.21 m s–1 (1) allow 0.2 m s–1                                                           2

  (ii)   F = 12.5 kN × 16 = 200 kN (1)                                                                1

(iii)    W = F × s or
= 200 k × 2 × π × 122 / 2 (1) ecf (ii) allow ecf for distance from (i)
= 7.7 × 107 J (1) allow 8 × 107                                                                2

  (iv)  P = W / t, energy / time or F × v or
= 7.67 × 107 / (30 × 60) (1) or ecf (iii) / (30 × 60)
= 42.6 kW (1) allow 43 kW, only allow 40 kW if working shown             2

(v)     •        Friction force at bearing opposes motion so not useful (1)
•        Friction force of tyres on rim drives wheel, so is useful (1)
•        Electrical energy supplies power to drive wheels /
          useful implied (1)
•        Input energy (electrical or energy supplied to motor)
          is converted into heat (1)

          Last point to do with the idea that once moving with constant speed e.g.
•        All work is done against friction
•        No input energy is converted into Ek
•        All input energy ends up as heat
•        Any other relevant point relating to energy (1)                              5

  (b)   (i)      k = F / x
= 1.8 × 106 / 0.90 (1)
= 2.0 × 106 Nm–1 (1)                                                                               2

the pendulum bob is travelling in a circle (1)
so it is accelerating towards the centre (1)
(it has a constant speed in the time interval just before vertical to just
after vertical)

          bob is not in equilibrium (1)
so the tension must be (slightly) larger than the weight of the bob (1)                        3

MAXIMUM 3

Electric Current

Electric current

1. Calculate the average electric current in a wire when a charge of 150 C passes in 30s? I = Q/t = 150/30 = 5A
2. The current in a small torch bulb is 0.20 A. What total electric charge which passes through a point circuit in 12 minutes? 1.4 x10²How many electrons pass through this point in this time? 1.4 x10² /1.8 x 10^-19 = 9 x 10²⁰
3. In the circuit opposite the three bulbs are identical and reach hill brightness when the current is 0.20A. Bulb B is observed to be at full brightness. Which (if any) of the other bulbs will be at full brightness, and what is the current in each of them? B,C

B = 0.2A, C = 0.2A, D = 0.05A

4. A new electric cell was joined in series with a bulb and an ammeter. The initial current was 0.30 A. At subsequent intervals of 1 hour the readings on the ammeter were: 0.27A, 0.27A, 0.26A, 0.25 A, 0.23A, 0. 19A, 0.09A, 0.03 A, and at 9 hours the ammeter reading had become negligibly small.
   (a) Plot a graph of current against time.
   (b) What does the area under the graph represent? Total charge
   (c) How much electric charge passes through the circuit when a current of 0.10 A passes for 1 hour? 360C
   (d) What is the total electric charge which passes through the cell in the 9 hours? 6.3kC
5. In a gas discharge tube containing hydrogen the current is carried partly by (positive) hydrogen ions and partly by electrons. An ammeter in series with the tube indicates a current of 1.5 mA. If the rate of passage of electrons past a particular point in the tube is 6.0 x 10¹⁵ s^-1, find the number of hydrogen ions passing the same point per second. 3.4 x10¹⁵ s^-1

Bobsleigh

Mark scheme for conical Pendulum

1 Explanation

There is a resultant (or net or unbalanced) force

􀀹

Plus any 3 of following:-

Direction of motion is changing 􀀹

Velocity is changing 􀀹

Velocity change implies acceleration 􀀹

Force produces acceleration by F = ma (or N2) 􀀹

Force (or acceleration) is towards centre / there is a centripetal

force (or acceleration) / no force (or acceleration) parallel to

motion

􀀹

No work done, so speed is constant 􀀹 Max

3

Mark Scheme for Washing Machine circular motion 6

Stone on String

Child on a Swing

Test on Momentum

Circular Motion

Power of ten error on Q1 should be 1.57

internal resistance questions