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Frage 1 Bericht
(b)) Define the e. m.f. of a battery
ii. A cell X e.m.f. 1.00 V is balanced by a length of 40.0 cm on a potentiometer wire. Another cell Y is balanced by a length of 60.0 cm on the same wire. Calculate the e.m.f. of Y.
With the jockey off the potentiometer wire, the ammeter reading is:
I = 0.20 A
The readings obtained when the jockey makes contact at the stated lengths are tabulated below.
| S/N | Length, L (cm) | Ammeter reading, Ii (A) | L-1 (cm-1) |
|---|---|---|---|
| 1 | 20 | 1.00 | 0.050 |
| 2 | 35 | 0.90 | 0.029 |
| 3 | 50 | 0.85 | 0.020 |
| 4 | 65 | 0.80 | 0.015 |
| 5 | 80 | 0.75 | 0.013 |
The graph of Ii against L-1 is shown below.
From the intercept on the vertical axis, when L-1 = 0,
Io = 0.75 A.
Hence,
\[ \frac{I_o}{I}=\frac{0.75}{0.20}=3.75. \]
Precautions
The e.m.f. of a battery is the work done, or energy supplied, by the battery in driving one coulomb of charge round the complete circuit, including its internal resistance.
For the same potentiometer wire, e.m.f. is proportional to balancing length:
\[ \frac{E_Y}{E_X}=\frac{l_Y}{l_X} \] \[ E_Y=\frac{60.0}{40.0}\times1.00=1.50\ \text{V}. \]
E.m.f. of cell Y = 1.50 V.
Antwortdetails
With the jockey off the potentiometer wire, the ammeter reading is:
I = 0.20 A
The readings obtained when the jockey makes contact at the stated lengths are tabulated below.
| S/N | Length, L (cm) | Ammeter reading, Ii (A) | L-1 (cm-1) |
|---|---|---|---|
| 1 | 20 | 1.00 | 0.050 |
| 2 | 35 | 0.90 | 0.029 |
| 3 | 50 | 0.85 | 0.020 |
| 4 | 65 | 0.80 | 0.015 |
| 5 | 80 | 0.75 | 0.013 |
The graph of Ii against L-1 is shown below.
From the intercept on the vertical axis, when L-1 = 0,
Io = 0.75 A.
Hence,
\[ \frac{I_o}{I}=\frac{0.75}{0.20}=3.75. \]
Precautions
The e.m.f. of a battery is the work done, or energy supplied, by the battery in driving one coulomb of charge round the complete circuit, including its internal resistance.
For the same potentiometer wire, e.m.f. is proportional to balancing length:
\[ \frac{E_Y}{E_X}=\frac{l_Y}{l_X} \] \[ E_Y=\frac{60.0}{40.0}\times1.00=1.50\ \text{V}. \]
E.m.f. of cell Y = 1.50 V.
Frage 2 Bericht
(b)i. Define the term couple as it relates to rotational or oscillatory systems.
ii. Give two practical application of a couple in everyday life.
The metre rule is balanced on the knife edge and its centre of gravity is located at the 49.5 cm mark, which is marked with chalk. The mass printed on the reverse of the rule is MR = 135 g. The 100 g mass is fixed at C, the rule is suspended by two parallel threads of length \(h\) attached at the 10 cm marks, and the separation of the threads is kept constant at d = 80 cm = 0.80 m. The rule is twisted through a small angle about the vertical axis through C and released; the time \(t\) for 20 complete oscillations is taken with a stopwatch. The period is \(T=\dfrac{t}{20}\) and \(T^{2}\) is evaluated. The procedure is repeated for \(h = 40, 50, 60, 70\) and \(80\) cm.
Table of readings
| \(h\) (cm) | \(d\) (m) | \(t\) (s) | \(T=\dfrac{t}{20}\) (s) | \(T^{2}\) (s\(^2\)) |
|---|---|---|---|---|
| 40 | 0.80 | 28.0 | 1.40 | 1.9600 |
| 50 | 0.80 | 29.0 | 1.45 | 2.1025 |
| 60 | 0.80 | 31.0 | 1.55 | 2.4025 |
| 70 | 0.80 | 34.0 | 1.70 | 2.8900 |
| 80 | 0.80 | 37.0 | 1.85 | 3.4225 |
Graph of \(T^{2}\) against \(h\)
Slope of the graph
Two points are taken on the line of best fit: \((h_1, T^2_1) = (40\ \text{cm}, 1.81\ \text{s}^2)\) and \((h_2, T^2_2) = (80\ \text{cm}, 3.30\ \text{s}^2)\).
\[ S=\frac{\Delta T^{2}}{\Delta h}=\frac{(3.30-1.81)\ \text{s}^2}{(80-40)\ \text{cm}}=\frac{1.49\ \text{s}^2}{40\ \text{cm}} \] \[ S = 0.0371\ \text{s}^2\,\text{cm}^{-1} \]Evaluation of \(k\)
Given \(k=\dfrac{S}{Q}\) where \(Q=\dfrac{2}{25\,d^{2}}\), then
\[ k=\frac{S}{Q}=\frac{S\times 25\,d^{2}}{2} \]With \(S = 0.0371\ \text{s}^2\,\text{cm}^{-1}\) and \(d = 0.80\ \text{m}\) so that \(d^{2}=0.64\ \text{m}^2\):
\[ k=\frac{0.0371\times 25\times 0.64}{2} \] \[ k=\frac{0.5936}{2}=0.297\ \text{s}^2\,\text{cm}^{-1}\,\text{m}^{2} \]Two precautions
A couple is a pair of two forces that are equal in magnitude, parallel and opposite in direction, but whose lines of action do not pass through the same point. A couple produces a turning (rotational) effect only, with no resultant translational force. Its moment (torque) is:
\[ \tau = F\times d \]where \(F\) is the magnitude of one of the forces and \(d\) is the perpendicular distance between their lines of action.
Antwortdetails
The metre rule is balanced on the knife edge and its centre of gravity is located at the 49.5 cm mark, which is marked with chalk. The mass printed on the reverse of the rule is MR = 135 g. The 100 g mass is fixed at C, the rule is suspended by two parallel threads of length \(h\) attached at the 10 cm marks, and the separation of the threads is kept constant at d = 80 cm = 0.80 m. The rule is twisted through a small angle about the vertical axis through C and released; the time \(t\) for 20 complete oscillations is taken with a stopwatch. The period is \(T=\dfrac{t}{20}\) and \(T^{2}\) is evaluated. The procedure is repeated for \(h = 40, 50, 60, 70\) and \(80\) cm.
Table of readings
| \(h\) (cm) | \(d\) (m) | \(t\) (s) | \(T=\dfrac{t}{20}\) (s) | \(T^{2}\) (s\(^2\)) |
|---|---|---|---|---|
| 40 | 0.80 | 28.0 | 1.40 | 1.9600 |
| 50 | 0.80 | 29.0 | 1.45 | 2.1025 |
| 60 | 0.80 | 31.0 | 1.55 | 2.4025 |
| 70 | 0.80 | 34.0 | 1.70 | 2.8900 |
| 80 | 0.80 | 37.0 | 1.85 | 3.4225 |
Graph of \(T^{2}\) against \(h\)
Slope of the graph
Two points are taken on the line of best fit: \((h_1, T^2_1) = (40\ \text{cm}, 1.81\ \text{s}^2)\) and \((h_2, T^2_2) = (80\ \text{cm}, 3.30\ \text{s}^2)\).
\[ S=\frac{\Delta T^{2}}{\Delta h}=\frac{(3.30-1.81)\ \text{s}^2}{(80-40)\ \text{cm}}=\frac{1.49\ \text{s}^2}{40\ \text{cm}} \] \[ S = 0.0371\ \text{s}^2\,\text{cm}^{-1} \]Evaluation of \(k\)
Given \(k=\dfrac{S}{Q}\) where \(Q=\dfrac{2}{25\,d^{2}}\), then
\[ k=\frac{S}{Q}=\frac{S\times 25\,d^{2}}{2} \]With \(S = 0.0371\ \text{s}^2\,\text{cm}^{-1}\) and \(d = 0.80\ \text{m}\) so that \(d^{2}=0.64\ \text{m}^2\):
\[ k=\frac{0.0371\times 25\times 0.64}{2} \] \[ k=\frac{0.5936}{2}=0.297\ \text{s}^2\,\text{cm}^{-1}\,\text{m}^{2} \]Two precautions
A couple is a pair of two forces that are equal in magnitude, parallel and opposite in direction, but whose lines of action do not pass through the same point. A couple produces a turning (rotational) effect only, with no resultant translational force. Its moment (torque) is:
\[ \tau = F\times d \]where \(F\) is the magnitude of one of the forces and \(d\) is the perpendicular distance between their lines of action.
Frage 3 Bericht
(b)i. Explain how heat losses by radiation and convection are minimized in a vacuum flask.
ii. State four factors that affect the rate of evaporation of a liquid in an open container.
(a) Cooling curves for tins C and D
100 cm3 of water was used in each tin. The temperature was recorded at one-minute intervals as the water cooled.
| Time, t (min) | Temperature in tin C, T (°C) | Temperature in tin D, T (°C) |
|---|---|---|
| 0 | 85 | 85 |
| 1 | 83 | 84 |
| 2 | 81 | 83 |
| 3 | 79 | 81 |
| 4 | 78 | 79 |
| 5 | 75 | 78 |
| 6 | 73 | 76 |
| 7 | 71 | 75 |
| 8 | 70 | 73 |
| 9 | 68 | 71 |
| 10 | 66 | 69.5 |
| 11 | 65 | 68 |
| 12 | 63 | 66 |
| 13 | 62 | 64 |
| 14 | 60 | 62 |
| 15 | – | 61 |
| 16 | – | 60 |
The two cooling curves, plotted on the same axes, are:
From the graph:
Precautions
(b)(i) Vacuum flask
The silvered surfaces of the double walls reflect thermal radiation and are poor emitters and absorbers; hence heat loss by radiation is minimized. The vacuum between the walls contains virtually no particles, so convection currents cannot occur. The insulating stopper also prevents air circulation at the neck of the flask.
(b)(ii) Factors affecting evaporation
Antwortdetails
(a) Cooling curves for tins C and D
100 cm3 of water was used in each tin. The temperature was recorded at one-minute intervals as the water cooled.
| Time, t (min) | Temperature in tin C, T (°C) | Temperature in tin D, T (°C) |
|---|---|---|
| 0 | 85 | 85 |
| 1 | 83 | 84 |
| 2 | 81 | 83 |
| 3 | 79 | 81 |
| 4 | 78 | 79 |
| 5 | 75 | 78 |
| 6 | 73 | 76 |
| 7 | 71 | 75 |
| 8 | 70 | 73 |
| 9 | 68 | 71 |
| 10 | 66 | 69.5 |
| 11 | 65 | 68 |
| 12 | 63 | 66 |
| 13 | 62 | 64 |
| 14 | 60 | 62 |
| 15 | – | 61 |
| 16 | – | 60 |
The two cooling curves, plotted on the same axes, are:
From the graph:
Precautions
(b)(i) Vacuum flask
The silvered surfaces of the double walls reflect thermal radiation and are poor emitters and absorbers; hence heat loss by radiation is minimized. The vacuum between the walls contains virtually no particles, so convection currents cannot occur. The insulating stopper also prevents air circulation at the neck of the flask.
(b)(ii) Factors affecting evaporation
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