the world in a different light
This deck of slides are the ones I will be using for the Symposium on “Leveraging Technology for Engaging and Effective Learning” at the Singapore International Science Teachers’ Conference (SISTC) 2024 on Day 2 of the Conference (20 November). Feel free to download for your reference.
When applying the principle of moments to calculate the magnitude of a force creating a turning effect, where the force is not perpendicular to the length of the object, there are two approaches.
Take the following problem:
A uniform rectangular beam has negligible thickness and weight 850 N. Its length is 5.0 m and it is in contact with the top of a support at point P. P is 0.80 m from one end of the beam.
The beam is held stationary, at an angle of 30° to the horizontal, by a rope that is attached to the bottom corner of the other end of the beam.
Calculate the magnitude of the force T on the beam due to the tension in the rope.
Approach 1: Identifying the perpendicular distance between line of action of the force and pivot
In the applet above, check the box that says “Show perp dist” to see the lines representing the perpendicular distances between each line of action of the force and the pivot P.
Taking moments about P,
Clockwise moment due to T = Anti-clockwise moment due to W
$$T \times (5.0\text{ m} – 0.8\text{ m}) \sin 30^{\circ} = W \times (\dfrac{5.0\text{ m}}{2} – 0.8\text{ m}) \cos 30^{\circ}$$
$$T \times 2.10 \text{ m} = 850 \text{ N} \times 1.472 \text{ m}$$
$$T = 596 \approx 600 \text{ N}$$
Approach 2: Resolving the forces to obtain components of forces perpendicular to the beam
In the same applet, check the box that says “Resolve forces”
Taking moments about P,
Clockwise moment due to $T_{\perp}$ = Anti-clockwise moment due to $W_{\perp}$
$$T \sin 30^{\circ} \times (5.0\text{ m} – 0.8\text{ m}) = 850 \text{ N} \cos 30^{\circ} \times (\dfrac{5.0\text{ m}}{2} – 0.8\text{ m})$$
$$T \times 0.500 \times 4.2 \text{ m} = 850 \text{ N} \times 0.866 \times 1.7 \text{ m}$$
$$T = 596 \approx 600 \text{ N}$$
The simulation below allows students to practise calculating potential differences and currents of a slightly complex circuit, involving three different modes that can be toggled by clicking on the switch.
Link: https://www.geogebra.org/m/jkckp9pr
When resistors \( R_1 \) and \( R_2 \) are connected in series, the total resistance is simply the sum of the individual resistances:
\[ R_{\text{total}} = R_1 + R_2 \]
The current \( I \) through the circuit is given by Ohm’s Law:
\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} = \frac{V_{\text{total}}}{R_1 + R_2} \]
where \( V_{\text{total}} \) is the total potential difference supplied by the source.
The potential difference across each resistor can be calculated using:
\[ V_1 = I \cdot R_1, \quad V_2 = I \cdot R_2 \]
In this mode, resistors \( R_1 \) and \( R_3 \) are in parallel, and \( R_2 \) is in series with the combination. First, calculate the equivalent resistance of the parallel combination:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_1} + \frac{1}{R_3} \]
Thus, the total resistance is:
\[ R_{\text{total}} = R_{\text{parallel}} + R_2 \]
The current through the circuit is:
\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} \]
The potential difference across \( R_2 \) is:
\[ V_2 = I \cdot R_2 \]
Since \( R_1 \) and \( R_3 \) are in parallel, they share the same potential difference:
\[ V_1 = V_3 = V_{\text{total}} – V_2 \]
The current through each parallel resistor can be found using Ohm’s Law:
\[ I_1 = \frac{V_1}{R_1}, \quad I_3 = \frac{V_3}{R_3} \]
Here, resistors \( R_1 \) and \( R_2 \) are connected in series, and the combination is in parallel with \( R_3 \). First, calculate the resistance of the series combination:
\[ R_{\text{series}} = R_1 + R_2 \]
Then, find the total resistance of the parallel combination:
\[ \frac{1}{R_{\text{total}}} = \frac{1}{R_{\text{series}}} + \frac{1}{R_3} \]
The total current is:
\[ I = \frac{V_{\text{total}}}{R_{\text{total}}} \]
The voltage across the parallel combination is the same for both branches:
\[ V_1 + V_2 = V_3 = V_{\text{total}} \]
The current through \( R_3 \) is:
\[ I_3 = \frac{V_3}{R_3} \]
The current through \( R_1 \) and \( R_2 \), which are in series, is the same:
\[ I_{\text{series}} = \frac{V_{\text{total}}}{R_1 + R_2} \]
The voltage across each series resistor is:
\[ V_1 = I_{\text{series}} \cdot R_1, \quad V_2 = I_{\text{series}} \cdot R_2 \]
I modified an existing simulation to demonstrate how the displacement of particles along a longitudinal wave can be represented in graphical form.
Essentially, one would have to determine for each particle, its displacement from the equilibrium position and its corresponding position along the wave’s direction. On the graph, positive displacement indicates movement in one direction (e.g., to the right), while negative displacement indicates movement in the opposite direction (e.g., to the left).
For the full view, go to https://www.geogebra.org/m/kq3e2qjk
Heating and cooling curves are graphical representations that show how the temperature of a substance changes as heat is added or removed over time. They illustrate the behavior of substances as they go through different states—solid, liquid, and gas.
Heating Curve: This curve shows how the temperature of a substance increases as it absorbs heat. The curve typically rises as the substance heats up, with plateaus indicating phase changes, where the substance absorbs energy but its temperature remains constant. Check out the heating curves for water and nitrogen using the drop-down menu.
Cooling Curve: This curve is the opposite of the heating curve. It shows how the temperature decreases as the substance loses heat. Like the heating curve, it also has plateaus where phase changes occur, but this time, the substance releases energy. In addition to water, you can also see the cooling curve for ethanol.
With these ChatGPT-generated interactive graphs, users can change the rate of heat input or released from the substance. They can also read the descriptions that explain the changes in the average PE and KE of the molecules during each process.
A Geiger-Muller (GM) counter is an instrument for detecting and measuring ionizing radiation. It operates by using a Geiger-Muller tube filled with gas, which becomes ionized when radiation passes through it. This ionization produces an electrical pulse that is counted and displayed, allowing users to determine the presence and intensity of radiation.
Svjo-2, CC BY-SA 3.0, via Wikimedia Commons
This simulation (find it at https://physicstjc.github.io/sls/gm-counter) allows students to explore the random nature of radiation and the significance of accounting for background radiation in experiments. Here’s a guide to help students investigate these concepts using the simulation.
Q1: Set the source to “Background” and start the count. Observe the count for a few minutes. What do you notice about the counts recorded?
Q2: Why is it important to measure background radiation before testing other sources?
Q3: Change the source to “Banana” and reset the data. Start the count and observe the readings. How do the counts from the banana compare to the background radiation?
Q4: How do the counts per minute (CPM) for the banana vary over time? Is there a pattern or do the counts appear random?
Q5: Set the source to “Cesium-137” and reset the data. Start the count and observe the readings. How do the counts from Cesium-137 compare to both the background radiation and the banana?
Q6: What do the counts per minute (CPM) tell you about the intensity of the Cesium-137 source compared to the other sources?
Q7: By looking at the sample counts, can you predict the next count value? Why or why not?
Q8: How can you use the background radiation measurement to correct the readings from the banana and Cesium-137 sources?
