Wird geladen....
|
Drücken und Halten zum Ziehen |
|||
|
Hier klicken, um zu schließen |
|||
Frage 1 Bericht
Using the above diagram as a guide, carry out the following experiment:
(b)i. State four characteristics of the image of an object formed by a plane mirror.
ii. State two Conditions necessary for total internal reflection to occur in a medium.
The prism is placed on its traced outline ABC and light is aimed at the point O on face AB at an angle of incidence \(i\). The plane mirror is held against face AC. For each value of \(i\) the emergent direction is fixed by the no-parallax pin method: pins \(R_1\) and \(R_2\) are set on the incident line NO, and pins \(R_3\) and \(R_4\) are set while viewing through face BC. Line \(R_4R_3\) is produced to meet NO produced at T, where the angle \(\theta\) is measured, and the angle \(e\) is measured at D. The construction is shown below.
| S/N | i / ° | θ / ° | e / ° |
|---|---|---|---|
| 1 | 25 | 62 | 28 |
| 2 | 30 | 54 | 32 |
| 3 | 35 | 45 | 35 |
| 4 | 40 | 40 | 37 |
| 5 | 45 | 30 | 41 |
A graph of \(e\) (vertical axis) against \(\theta\) (horizontal axis) is plotted, and a straight line of best fit is drawn through the points.
Two widely separated points are read from the line of best fit: \((\theta_1,e_1)=(10,49)\) and \((\theta_2,e_2)=(83,20)\).
\[ s=\frac{\Delta e}{\Delta\theta}=\frac{20-49}{83-10}=\frac{-29}{73}=-0.397 \]The magnitude of the slope is \(|s|=0.397\) and the intercept on the \(e\)-axis is \(c=53\). Therefore
\[ k=s^{-1}=\frac{1}{0.397}=2.52 \]Antwortdetails
The prism is placed on its traced outline ABC and light is aimed at the point O on face AB at an angle of incidence \(i\). The plane mirror is held against face AC. For each value of \(i\) the emergent direction is fixed by the no-parallax pin method: pins \(R_1\) and \(R_2\) are set on the incident line NO, and pins \(R_3\) and \(R_4\) are set while viewing through face BC. Line \(R_4R_3\) is produced to meet NO produced at T, where the angle \(\theta\) is measured, and the angle \(e\) is measured at D. The construction is shown below.
| S/N | i / ° | θ / ° | e / ° |
|---|---|---|---|
| 1 | 25 | 62 | 28 |
| 2 | 30 | 54 | 32 |
| 3 | 35 | 45 | 35 |
| 4 | 40 | 40 | 37 |
| 5 | 45 | 30 | 41 |
A graph of \(e\) (vertical axis) against \(\theta\) (horizontal axis) is plotted, and a straight line of best fit is drawn through the points.
Two widely separated points are read from the line of best fit: \((\theta_1,e_1)=(10,49)\) and \((\theta_2,e_2)=(83,20)\).
\[ s=\frac{\Delta e}{\Delta\theta}=\frac{20-49}{83-10}=\frac{-29}{73}=-0.397 \]The magnitude of the slope is \(|s|=0.397\) and the intercept on the \(e\)-axis is \(c=53\). Therefore
\[ k=s^{-1}=\frac{1}{0.397}=2.52 \]Frage 2 Bericht
You have been provided with an accumulator E, a standard resistor Rx, two resistance boxes RB\(_{1}\) and RB\(_{2}\), two keys K\(_{1}\) and K\(_{2}\) and other necessary apparatus.
(b)i. Explain what is meant by the potential difference between two points in an electric circuit
ii. A cell has an e.m.f. of 3 V. When it is connected across a resistor of resistance \(4\Omega\), a current 0.5A passes through the circuit. Calculate the internal resistance of the cell.
E.m.f. of the accumulator, \(E = 2.0\ \text{V}\).
For each setting the resistance in \(RB_1\) was made equal to the resistance in \(RB_2\) (\(R\) in \(RB_1 = R\) in \(RB_2 = R\)). With \(K_1\) and \(K_2\) closed the potential difference \(V_1\) across the standard resistor \(R_x\) was read, and \(V_1^{-1}\) evaluated.
| \(R\ (\Omega)\) | \(V_1\ (\text{V})\) | \(V_1^{-1}\ (\text{V}^{-1})\) |
|---|---|---|
| 1 | 1.50 | 0.667 |
| 2 | 1.30 | 0.769 |
| 3 | 1.15 | 0.870 |
| 4 | 1.10 | 0.909 |
| 5 | 0.90 | 1.111 |
The graph of \(V_1^{-1}\) (vertical axis) against \(R\) (horizontal axis), with both axes starting from the origin, is a rising straight line.
Taking two widely spaced points on the line of best fit, \((0.3,\ 0.60)\) and \((6.5,\ 1.22)\):
\[ s = \frac{\Delta (V_1^{-1})}{\Delta R} = \frac{1.22 - 0.60}{6.5 - 0.3} = \frac{0.62}{6.2} = 0.10\ \text{V}^{-1}\,\Omega^{-1} \]The intercept on the vertical axis (value of \(V_1^{-1}\) where the line cuts \(R = 0\)) is
\[ c = 0.57\ \text{V}^{-1}. \]The potential difference between two points in an electric circuit is the work done (in joules) in moving one coulomb of positive charge from one point to the other. It is measured in volts.
Given \(E = 3\ \text{V}\), \(R = 4\ \Omega\) and current \(I = 0.5\ \text{A}\):
\[ E = IR + Ir \] \[ 3 = (0.5 \times 4) + 0.5r \] \[ 3 = 2 + 0.5r \] \[ 0.5r = 1 \] \[ r = \frac{1}{0.5} = 2\ \Omega. \]Antwortdetails
E.m.f. of the accumulator, \(E = 2.0\ \text{V}\).
For each setting the resistance in \(RB_1\) was made equal to the resistance in \(RB_2\) (\(R\) in \(RB_1 = R\) in \(RB_2 = R\)). With \(K_1\) and \(K_2\) closed the potential difference \(V_1\) across the standard resistor \(R_x\) was read, and \(V_1^{-1}\) evaluated.
| \(R\ (\Omega)\) | \(V_1\ (\text{V})\) | \(V_1^{-1}\ (\text{V}^{-1})\) |
|---|---|---|
| 1 | 1.50 | 0.667 |
| 2 | 1.30 | 0.769 |
| 3 | 1.15 | 0.870 |
| 4 | 1.10 | 0.909 |
| 5 | 0.90 | 1.111 |
The graph of \(V_1^{-1}\) (vertical axis) against \(R\) (horizontal axis), with both axes starting from the origin, is a rising straight line.
Taking two widely spaced points on the line of best fit, \((0.3,\ 0.60)\) and \((6.5,\ 1.22)\):
\[ s = \frac{\Delta (V_1^{-1})}{\Delta R} = \frac{1.22 - 0.60}{6.5 - 0.3} = \frac{0.62}{6.2} = 0.10\ \text{V}^{-1}\,\Omega^{-1} \]The intercept on the vertical axis (value of \(V_1^{-1}\) where the line cuts \(R = 0\)) is
\[ c = 0.57\ \text{V}^{-1}. \]The potential difference between two points in an electric circuit is the work done (in joules) in moving one coulomb of positive charge from one point to the other. It is measured in volts.
Given \(E = 3\ \text{V}\), \(R = 4\ \Omega\) and current \(I = 0.5\ \text{A}\):
\[ E = IR + Ir \] \[ 3 = (0.5 \times 4) + 0.5r \] \[ 3 = 2 + 0.5r \] \[ 0.5r = 1 \] \[ r = \frac{1}{0.5} = 2\ \Omega. \]Frage 3 Bericht
You are provided with a retort stand, clamp and boss, a pendulum bob, a piece of thread, and other necessary apparatus. Carry out the fo lowing experiment:
(b)i. What is meant by the period of oscillation of an oscillating body?
i. Explain the acceleration of free fall due to gravity.
The pendulum obeys \(T = 2\pi\sqrt{\dfrac{l}{g}}\), so that \(T^{2} = \dfrac{4\pi^{2}}{g}\,l\). Here the length used for the horizontal axis is \(L = l - 30\ \text{cm}\), and the time \(t\) is measured for 20 complete oscillations, giving \(T = \dfrac{t}{20}\).
| S/N | l (cm) | t (s) for 20 osc. | T (s) | T² (s²) | L = l − 30 (cm) |
|---|---|---|---|---|---|
| 1 | 130 | 45.70 | 2.29 | 5.22 | 100 |
| 2 | 110 | 42.00 | 2.10 | 4.41 | 80 |
| 3 | 90 | 38.00 | 1.90 | 3.61 | 60 |
| 4 | 70 | 33.10 | 1.66 | 2.74 | 40 |
| 5 | 50 | 28.70 | 1.44 | 2.06 | 20 |
Reading two widely separated points on the line of best fit, \((L_{1}, T_{1}^{2}) = (100,\ 5.22)\) and \((L_{2}, T_{2}^{2}) = (20,\ 2.06)\):
\[ s = \frac{\Delta T^{2}}{\Delta L} = \frac{5.22 - 2.06}{100 - 20} = \frac{3.16}{80} = 0.039\ \text{s}^{2}\,\text{cm}^{-1} \]The line cuts the \(T^{2}\) axis at
\[ C = 1.3\ \text{s}^{2}. \]Taking \(\pi = \dfrac{22}{7}\):
i. \[ k_{1} = \frac{4\pi^{2}}{s} = \frac{4\left(\frac{22}{7}\right)^{2}}{0.039} = \frac{4 \times 9.878}{0.039} = \frac{39.51}{0.039} = 1013\ \text{cm s}^{-2} = 10.13\ \text{m s}^{-2}. \] This is the acceleration of free fall due to gravity, \(g\).
ii. \[ k_{2} = \frac{C}{s} = \frac{1.3}{0.039} = 33.33. \]
i. The period of oscillation of a body is the time taken for the body to make one complete to-and-fro movement (one full oscillation).
ii. When a body falls freely under its weight alone, the rate of increase of its velocity with time, caused by the Earth's gravitational pull, is the acceleration of free fall due to gravity. It is directed vertically downwards and has a value of about \(10\ \text{m s}^{-2}\) near the Earth's surface.
Antwortdetails
The pendulum obeys \(T = 2\pi\sqrt{\dfrac{l}{g}}\), so that \(T^{2} = \dfrac{4\pi^{2}}{g}\,l\). Here the length used for the horizontal axis is \(L = l - 30\ \text{cm}\), and the time \(t\) is measured for 20 complete oscillations, giving \(T = \dfrac{t}{20}\).
| S/N | l (cm) | t (s) for 20 osc. | T (s) | T² (s²) | L = l − 30 (cm) |
|---|---|---|---|---|---|
| 1 | 130 | 45.70 | 2.29 | 5.22 | 100 |
| 2 | 110 | 42.00 | 2.10 | 4.41 | 80 |
| 3 | 90 | 38.00 | 1.90 | 3.61 | 60 |
| 4 | 70 | 33.10 | 1.66 | 2.74 | 40 |
| 5 | 50 | 28.70 | 1.44 | 2.06 | 20 |
Reading two widely separated points on the line of best fit, \((L_{1}, T_{1}^{2}) = (100,\ 5.22)\) and \((L_{2}, T_{2}^{2}) = (20,\ 2.06)\):
\[ s = \frac{\Delta T^{2}}{\Delta L} = \frac{5.22 - 2.06}{100 - 20} = \frac{3.16}{80} = 0.039\ \text{s}^{2}\,\text{cm}^{-1} \]The line cuts the \(T^{2}\) axis at
\[ C = 1.3\ \text{s}^{2}. \]Taking \(\pi = \dfrac{22}{7}\):
i. \[ k_{1} = \frac{4\pi^{2}}{s} = \frac{4\left(\frac{22}{7}\right)^{2}}{0.039} = \frac{4 \times 9.878}{0.039} = \frac{39.51}{0.039} = 1013\ \text{cm s}^{-2} = 10.13\ \text{m s}^{-2}. \] This is the acceleration of free fall due to gravity, \(g\).
ii. \[ k_{2} = \frac{C}{s} = \frac{1.3}{0.039} = 33.33. \]
i. The period of oscillation of a body is the time taken for the body to make one complete to-and-fro movement (one full oscillation).
ii. When a body falls freely under its weight alone, the rate of increase of its velocity with time, caused by the Earth's gravitational pull, is the acceleration of free fall due to gravity. It is directed vertically downwards and has a value of about \(10\ \text{m s}^{-2}\) near the Earth's surface.
Möchten Sie mit dieser Aktion fortfahren?