Question on absolute zero

 

LDR and Thermistor

 

Electrons measure the size of nuclei

 

Gravity homework

 

LDR in moonlight & thermistor 3

Gravity revision answers

 

1.       (a)     arrows (at least one) indicating direction is towards the planet.

B1

          All lines looking as though they would meet at the centre judged by eye

At least 4 drawn and care taken
Some of the lines must be outside the planet.

B1

  (b)   (i)      (mg = GMm/r2 and hence) M = gr2/G

Equation needs to be rearranged as shown for C1 mark

C1 M1

         correct substitution M = 24.9 × (7.14 × 107)2/6.67 × 10–11
                                 = 1.9 × 1027 Kg (i.e about 2 × 1027)

A1

  (ii)   correct substitution into V = (4/3)πr3 = (4/3)π(7.14 × 107)3
{= 1.52 × 1024 m3}

If m= 2 × 1027 kg is used

C1

         density = mass/volume = 1.9 × 1027/1.52 × 1024 = 1250 kg m–3

d = 1312 scores 2 marks

A1

[7]

 

  2.     (a)     forces FS and FG acting inwards, force FE acting outwards - all through
centre of proton;
3 forces 2/2, 2 forces 1/2, marked and labelled (2)                                                2

  (b)   FE = FS + FG;
accept FE + FS + FG = 0 allow ecf from (a) (1)                                                     1

  (c)    (i)      FE = Q2 / (4π ε0 r2) (1)
= (1.6 × 10–19)2 / [4π × 8.85 × 10–12 (2.8 × 10–15)2] = 29 N (1)
use of r = 1.4 × 10–15 m (–1) once only                                                      2

  (ii)   FG = m2 G / r2 (1)
= (1.67 × 10–27)2 × 6.67 × 10–11 / (2.8 × 10–15)2 = 2.4 × 10–35 N (1)             2

(iii)    FS = 29 N / same as FE allow ecf (1)                                                          1

  (d)   FE >> FG so FG negligible / insignificant / can be ignored or AW (1)                    1

  (e)    (i)      FE = 0 (1)                                                                                                 1

(ii)     FG = 2.4 × 10–35 N (approx.) allow ecf (1)                                                  1

(iii)    FS = 2.4 × 10–35 N (approx.) (1)

         comment: FS now repulsive (not attractive) or AW
or indicated by minus sign with FS; (1) any 3                                             1

[12]

 

3.       (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]

 

  4.     (a)     force (of attraction) on unit mass (at that point in space/at the
surface of a planet) (1)                                                                                       1

  (b)   (i)      (mgh =) 1500 × 40 × 1.5 × 105; = 9.0 × 109 (J)                                           2

(ii)     (larger as) g decreases with height                                                             1

(iii)    v = 2πR/T; = 2π × 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]

 

  5.     (a)     arrow towards centre of planet                                                                            1

(b)     (i)      g = GM/R2                                                                                               1

(ii)     gS/gO = R2/25R2; gS = 40/25 (= 1.6 N kg–1)                                                2

(iii)    gC/gO = R2/16R2 giving gS = 40/16 (= 2.5 N kg–1)                                       1

(iv)    average g = (2.5 + 1.6)/2 = 2.(05) (1)
Δp.e. (= mgavR) = 3.0 × 103 × 2.05 × 2.0 × 107; = 1.2 × 1011 (J) (2)             3

  (c)    g = v2/r; = 4π2(5R)/T2 (2)
1.6 = 4 × 9.87 × 1.0 × 108/T2 giving T2 = 24.7 × 108 and T = 5.0 × 104 (s) (2)        4

[12]

 

  6.     (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]

 

  7.     It is the force (of attraction) per unit mass.                                                                  B1

[1]

 

  8.     (a)     (i)      Force per unit mass (placed at that point) (1)                                              1

(ii)     g = GM/R2 (1)                                                                                          1

(iii)    Choosing a correct pair of values from the graph, e.g. 6.4 × 106 & 9.8,
10 × 106 & 4.0; (1)
substitute, 9.8 = 6.67 × 10–11 × M/ (6.4 × 106)2 to show
M = 6.0 × 1024 kg (1)                                                                                2

  (iv)  linear graph through origin/from 0 to R (1)                                                 1

  (v)   64 km; 1/100 of R as linear graph under Earth/AW (2)
64000 km; g  1/r2 so for 1/100 g r = 10R (2)                                           4

  (b)   GMe/R12 = GMm/R22; Me = 81 Mm; (2)
Mm = 6.0 × 1024/81 = 7.4 × 1022 (kg) ecf a(iii) (1)                                                3
or any acceptable alternative method i.e. correct method; correct figures;
processed to correct answer

[12]

 

  9.     (i)      The gravitational field strength g is not constant.                                               B1
The student’s value would be greater than the actual value (because the
average magnitude of g is less than 9.81 m s–1).                                                 B1

  (ii)   KE = 1/2 mv2
v = 2πr / T                                                                                                       C1
v = 2 × π × (6800 + 6400) × 103 / 8.5 × 103 / v = 9.76 × 103 (m s–1)                    C1
KE = 1/2 × 1500 × (9.76 × 103)2
KE = 7.1(4) × 1010 (J)                                                                                     A1

  (iii)  A geostationary satellite stays above the same point on the Earth and as             B1
such can be used for radio communications. (the term communications to          M1
be included and spelled correctly to gain the mark).                                           A1
The satellite is not in geostationary orbit
because its period is less than 1 day / 8.6 × 104 s.

[8]

 

  10.   (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]

 

  11.   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]

 

 

Resistivity Cambridge Entrance question

 

Electric power worksheet and answers

Power and Energy

1. A 3 kW immersion heater is designed for use on 240 V mains. Find (a) the current taken from the mains, (b) the resistance of the heater, and (c) the energy used in 1 min,

2. A 2 kW electric fire is designed for use on 250 V mains. What current does it take, and what is its resistance in use?

3.  Question 2 asks `what is its resistance IN USE', Explain why the last two words have been added.

4.  How much energy is stored in a 12 V car battery of capacity 38 A h ? (ie it is capable of delivering 38A for 1 hour before it is flat)

5.  A current of 2 A flows through a resistor of 10 Ω for 2 min, how much energy is transferred?

6. How many lamps marked 175 W 240 V could be used in a circuit fitted with a 5 A fuse if they are connected in parallel? (Remember all components in parallel have the same voltage)

7. An immersion heater marked 3 kW is designed for use on 240 V. If its resistance remains constant, what is its power if connected to a 200 V supply?

8. A fire of resistance 30 Ω is connected to a 240 V supply by (a) cables of negligible resistance, (b) cables of total resistance 2 Ω. Find the power consumed by the fire in each case.

9. A 60 W heater is to be made from manganin wire with a resistance of 3.75 ohms per metre. What length of wire is needed if the heater is to operate on a 12 V supply?

10. What is the total cost of using four 100W bulbs for 6 hours, if a kWh costs 9p. The equivalent low energy bulb has a rating of 21W. How much would it cost to run 4 of these for 24 hours.

11. A 3kW immersion heater is used for 4 hrs. How much will this cost if 1 kWh costs 9.7p

12. A 2.5Kw kettle boils in 13 minutes. How many kWh does it transfer?

13. A TV rated at 102 W uses 0.6 W on standby. How many kWh will it transfer in a day if it is used for viewing for 4 hours in the evening? How much does this cost at 9.3p per kWh.

14. An LCD TV is rated at 134W. A similar plasma TV is rated at 281W. How much will it cost to watch 3 ½ hours on each set if electricity costs 9.5p per kWh?

Extra

Find the power wasted in the cables when 120 kW is supplied through cables of resistance 0.2 Ω (a) at 240 V, (b) at 200 000 V.

( Hint answer for (a) is 50 kW, and (b) is much smaller !)

Power and Energy

1. A 3 kW immersion heater is designed for use on 240 V mains. Find (a) the current taken from the mains, (b) the resistance of the heater, and (c) the energy used in 1 min,

2. A 2 kW electric fire is designed for use on 250 V mains. What current does it take, and what is its resistance in use?

3.  Question 2 asks `what is its resistance IN USE', Explain why the last two words have been added.

4.  How much energy is stored in a 12 V car battery of capacity 38 A h ? (ie it is capable of delivering 38A for 1 hour before it is flat)

5.  A current of 2 A flows through a resistor of 10 Ω for 2 min, how much energy is transferred?

6. How many lamps marked 175 W 240 V could be used in a circuit fitted with a 5 A fuse if they are connected in parallel? (Remember all components in parallel have the same voltage)

7. An immersion heater marked 3 kW is designed for use on 240 V. If its resistance remains constant, what is its power if connected to a 200 V supply?

8. A fire of resistance 30 Ω is connected to a 240 V supply by (a) cables of negligible resistance, (b) cables of total resistance 2 Ω. Find the power consumed by the fire in each case.

9. A 60 W heater is to be made from manganin wire with a resistance of 3.75 ohms per metre. What length of wire is needed if the heater is to operate on a 12 V supply?

10. What is the total cost of using four 100W bulbs for 6 hours, if a kWh costs 9p. The equivalent low energy bulb has a rating of 21W. How much would it cost to run 4 of these for 24 hours.

11. A 3kW immersion heater is used for 4 hrs. How much will this cost if 1 kWh costs 9.7p

12. A 2.5Kw kettle boils in 13 minutes. How many kWh does it transfer?

13. A TV rated at 102 W uses 0.6 W on standby. How many kWh will it transfer in a day if it is used for viewing for 4 hours in the evening? How much does this cost at 9.3p per kWh.

14. An LCD TV is rated at 134W. A similar plasma TV is rated at 281W. How much will it cost to watch 3 ½ hours on each set if electricity costs 9.5p per kWh?

Extra

Find the power wasted in the cables when 120 kW is supplied through cables of resistance 0.2 Ω (a) at 240 V, (b) at 200 000 V.

( Hint answer for (a) is 50 kW, and (b) is much smaller !)

 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