The Scope and Scale of Physics Find the order of magnitude of the following physical quantities. (a) The mass of Earth’s atmosphere: 5.1 × 10 1 8 kg : (b) The mass of the Moon’s atmosphere: 25,000 kg ; (c) The mass of Earth’s hydrosphere: 1.4 × 10 21 kg : (d) The mass of Earth: 5.97 × 10 24 kg : (e) The mass of the Moon: 7.34 × 10 22 kg : (f) The Earth-Moon distance (semimajor axis): 3.84 × 10 8 m : (g) The mean Earth-Sun distance: 1.5 × 10 11 m : (h) The equatorial radius of Earth: 6.38 × 10 6 m : (i) The mass of an electron: 9.11 × 10 − 31 kg : (j) The mass of a proton: 1.67 × 10 − 27 kg : (k) The mass of the Sun: 1.99 × 10 30 kg .
The Scope and Scale of Physics Find the order of magnitude of the following physical quantities. (a) The mass of Earth’s atmosphere: 5.1 × 10 1 8 kg : (b) The mass of the Moon’s atmosphere: 25,000 kg ; (c) The mass of Earth’s hydrosphere: 1.4 × 10 21 kg : (d) The mass of Earth: 5.97 × 10 24 kg : (e) The mass of the Moon: 7.34 × 10 22 kg : (f) The Earth-Moon distance (semimajor axis): 3.84 × 10 8 m : (g) The mean Earth-Sun distance: 1.5 × 10 11 m : (h) The equatorial radius of Earth: 6.38 × 10 6 m : (i) The mass of an electron: 9.11 × 10 − 31 kg : (j) The mass of a proton: 1.67 × 10 − 27 kg : (k) The mass of the Sun: 1.99 × 10 30 kg .
Find the order of magnitude of the following physical quantities. (a) The mass of Earth’s atmosphere:
5.1
×
10
1
8
kg
: (b) The mass of the Moon’s atmosphere:
25,000 kg
; (c) The mass of Earth’s hydrosphere:
1.4
×
10
21
kg
: (d) The mass of Earth:
5.97
×
10
24
kg
: (e) The mass of the Moon:
7.34
×
10
22
kg
: (f) The Earth-Moon distance (semimajor axis):
3.84
×
10
8
m
: (g) The mean Earth-Sun distance:
1.5
×
10
11
m
: (h) The equatorial radius of Earth:
6.38
×
10
6
m
: (i) The mass of an electron:
9.11
×
10
−
31
kg
: (j) The mass of a proton:
1.67
×
10
−
27
kg
: (k) The mass of the Sun:
1.99
×
10
30
kg
.
4.) The diagram shows the electric field lines of a positively charged conducting sphere of
radius R and charge Q.
A
B
Points A and B are located on the same field line.
A proton is placed at A and released from rest. The magnitude of the work done by the electric field in
moving the proton from A to B is 1.7×10-16 J. Point A is at a distance of 5.0×10-2m from the centre of
the sphere. Point B is at a distance of 1.0×10-1 m from the centre of the sphere.
(a) Explain why the electric potential decreases from A to B. [2]
(b) Draw, on the axes, the variation of electric potential V with distance r from the centre of the
sphere.
R
[2]
(c(i)) Calculate the electric potential difference between points A and B. [1]
(c(ii)) Determine the charge Q of the sphere. [2]
(d) The concept of potential is also used in the context of gravitational fields. Suggest why scientists
developed a common terminology to describe different types of fields. [1]
3.) The graph shows how current I varies with potential difference V across a component X.
904
80-
70-
60-
50-
I/MA
40-
30-
20-
10-
0+
0
0.5
1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0
VIV
Component X and a cell of negligible internal resistance are placed in a circuit.
A variable resistor R is connected in series with component X. The ammeter reads 20mA.
4.0V
4.0V
Component X and the cell are now placed in a potential divider circuit.
(a) Outline why component X is considered non-ohmic. [1]
(b(i)) Determine the resistance of the variable resistor. [3]
(b(ii)) Calculate the power dissipated in the circuit. [1]
(c(i)) State the range of current that the ammeter can measure as the slider S of the potential divider
is moved from Q to P. [1]
(c(ii)) Describe, by reference to your answer for (c)(i), the advantage of the potential divider
arrangement over the arrangement in (b).
1.) Two long parallel current-carrying wires P and Q are separated by 0.10 m. The current in wire P is 5.0 A.
The magnetic force on a length of 0.50 m of wire P due to the current in wire Q is 2.0 × 10-s N.
(a) State and explain the magnitude of the force on a length of 0.50 m of wire Q due to the current in P. [2]
(b) Calculate the current in wire Q. [2]
(c) Another current-carrying wire R is placed parallel to wires P and Q and halfway between them as shown.
wire P
wire R
wire Q
0.05 m
0.05 m
The net magnetic force on wire Q is now zero.
(c.i) State the direction of the current in R, relative to the current in P.[1]
(c.ii) Deduce the current in R. [2]
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