View from Bishan-AMK Park, Singapore
View from Bishan-AMK Park, Singapore
Wishing everyone a Merry Christmas and a Happy New Year 2020!
Thankful to have a chance to share my work at SISTIC 2019 and learn from others.
It’s amazing to watch these strandbeasts walk and solely powered by wind.
Wind energy to mechanical energy 🙂
This is a toy version of the wind walker.
Well, that’s Physics =)
In this post, it shows a free-falling ball from a height of 1.0 m. During the impact, the direction of the force on the ground is downwards and the force on the ground by the ball is greater then the weight.
As the ball is free-falling, the only force acting is its weight downwards. Hence a common misconception is to think that the force on the ground during impact is equal to the weight. This is wrong.
The normal force (force on the ball by the ground = stopping force on the ball by the ground) is greater than the weight.
The force on the ball by the ground is equal and opposite to the force on the ground by the ball. Hence the magnitude of the force on the ground is greater than the weight.
Similar concept can be applied if a man jumps off from a height. But in this case, the man’s leg will exert a stopping force over a short distance. That stopping force, once again, is greater than the weight of the man.
This concept is similar to a 2016 O-Level Pure Physics Question P2 Q2, on why the pressure acting on the ground is greater during the jump, compared to when he is standing stationary on the ground.
During the jump, his leg will exert an upward force. This upward force (equivalent to normal force or force on the man by the ground) is greater than the weight of the man. Hence there is a net (resultant force) upwards, causing him to accelerate upwards.
That force on the man by the ground is equal and opposite to the force on the ground by the man. This is an action-reaction pair. Since the force exerted on the ground by the man is greater (greater than weight), the pressure exerted on the floor is greater.
(NOTE: Normal force and Weight is not an action-reaction pair)
In this this question, the displacement-time graphs are given, which are different from displacement-distance graphs.
In the displacement-time graphs of A and B, they show the displacement of that particular point at different timing. E,g, at t = 0s, the A is at the rest position (0 displacement) and at time 0.2 s it is at the maximum displacement. This means A is going up from t = 0 s to 0.2 s.
(a) Amplitude: 1.5 cm
(b)(i) Frequency is the number of complete waves produced in 1 second.
(ii) period T = 0.8s, f = 1/T = 1/0.8 = 1.25 Hz
(c) Closest possible positions of A and B, (refer to the video), is when the
time taken for the wave to move from A to B is T/4 = 0.8/4 = 0.2 s.
speed = distance/time = 38/0.2 = 190 cm/s approx. 200 cm/s
(ii) There are various possibilities in which B can be 38 cm to the right of A. Besides T/4, it can be 1.25T or 2.25 T etc. Hence the speed can be other values.
Refer to the video explanation below
Take note that the question is looking for specific latent heat of vaporisation during boiling. Hence information of mass m1 before heating is not relevant. There is no change of state of the water, it is merely the heating of the water (which involves specific heat capacity of water).
Refer to the worked solution below.
Solutions: Option D
Answer: Option D
Three processes of thermal transfer: conduction, convection and radiation.
Conduction vs Convection:
Radiation vs Conduction & Convection
When the pump is switched on and the air in the jar is gradually removed, the pressure in the jar decreases. There will be fewer air molecules per unit volume in the far. Hence rate of collision of the air molecules with one another and with the wall and hulk will be reduced. As pressure P = F/A, the force acting per unit area decreases, the pressure decreases.
In the marshmallow, there are pockets of air at normal atmospheric pressure initially. As the pressure in the jar decreases, the pockets of air in the marshmallow expands due to this pressure difference. Hence the hulk expands and its volume increases.
Steel is a hard magnetic material. When magnetized, it is not as strong as soft iron. But steel retains its magnetism, hence it is used to make permanent magnets e.g compass need, speaker, fridge door.
Iron is a soft magnetic material. It can be easily and strongly magnetised. But it loses its magnetism easily. Hence it is used to make temporary magnet e.g. Electromagnet.
Do remember that compass is like a freely suspended magnet. It is made of steel!
There are 3 scenarios with slight variations.
Calculate the force F needed to pull the block up the inclined plane.
View the video below to understand how to solve these types of question.
You can also view the solutions below.
Light and sound are both waves. So both carry energy from one place to another.
Light, which is part of the electromagnetic spectrum, is a transverse wave, It can travel through a vacuum at speed 3.0 x 108 m/s. As the light travels from an optically less dense medium (air) to an optically denser medium (liquid or glass), the light undergoes refraction and bends towards the normal due to a decrease in speed.
Light: Optically less dense medium to denser medium:
– speed decreases
– wavelength shorter
– frequency remains constant
Sound is a longitudinal wave. It requires a medium to pass through and it cannot pass through a vacuum. Opposite to light, as the sound travels from a less dense medium (air) into a denser medium (water or solid), the speed increases.
Sound: Less dense medium to denser medium:
– speed increases
– wavelength longer
– frequency remains constant
Refers to the image below to understand how the waves behave in different mediums.
Click here to revise on the calculation of refractive index for light
This simple circuit involves a resistance wire on the meter rule and the use of jockey tapping at different length L on the resistance wire.
Note that this is not a potential divider. Rather this set up works like a variable resistor (rheostat) in the circuit. When the switch is closed (jockey NOT tapping on resistance wire), there is no current flowing as voltmeter (infinity resistance) is connected in series with the ammeter and the battery. Hence the voltmeter is showing the battery’s electromotive force (emf) of 3.0 V.
When the switch is closed and the jockey is tapped on the resistance wire on the ruler, the longer the L (length of resistance wire), the higher the resistance of the circuit, the higher the potential difference (voltmeter) across the resistance wire and the smaller the current through the ammeter. Hence the tapping of the jockey on the resistance wire is similar to adjusting the resistance on the variable resistor.
(If you are wondering why the voltmeter is not showing the emf of the battery when the jockey touches the resistance wire L, it is because in reality there is internal resistance in the battery or even in the connecting copper wire. For olevel theory we assume no internal battery resistance or resistance in copper wire or ammeter).
To learn how to set up the experiment, refer to the video below.
Both ammeter and voltmeter have two terminals, (positive and negative). The conventional current must flow into the positive terminal (+) of the meters and out of the negative terminal (-). If the connection is the opposite, the needle will deflect below the zero marking. Refer to the video below.
Sometimes, the connection terminals on the voltmeter and ammeter are different. Likewise, different types of wire with different connection heads have to be used. Refer to the video to see how are they connected in general.
In this post, we will be going to the basics of setting up the retort stand and pendulum experiment. Most probably, this will be the first experiment which you will perform in the lab. Let’s start with the setting up the retort stand, boss and the clamp.
Next we will take a look at how you set the up the apparatus for the pendulum experiment.
How is the period affected by length, mass of bob, and angle of release? Click on this post to find out more.
Click here to see how to use a digital stopwatch.
Though slinky coil is commonly used to demonstrate transverse and longitudinal waves, you must not quote it as an example for either of the waves.
Transverse Waves (slinky coil)
Longitudinal Waves (slinky coil)
Convection is a process in which thermal energy is transferred within a fluid (liquid or gas) due to the difference in density which creates a current.
The video below shows the convection current in a heated tube filled with water. Colour dye is added to enable us to see the convection current through our naked eyes.
1. Converging lens (convex lens)
Converging lens, also known as convex lens, is thicker at the centre. Below shows some examples.
In O-level, we learned about symmetrical converging lens. i.e. the curvature of the lens are the same on both sides. As light rays pass through the converging lens, the rays come closer together.
Take note that the bending of light, refraction, takes place on the air-glass boundaries on both sides of the lens (as shown above). But for easy drawing, we draw the bending at the imaginary centre vertical which passes through the optical centre as shown below.
2. The 3 Rays
The following 3 rays are important for us to construct the ray diagram and locate the image. We always draw these 3 rays as they have rules to follow, hence guiding us in our drawing.
Refer to the video below for better understanding of the 3 rays.
3. The 4 Key Scenarios
Depending on the distance of the object to the centre of the lens (object distance u), the kind of image you get varies.
Refer to the video below for the better understanding of how the various images are formed.
3. The Pattern
Besides knowing the 4 key scenarios, it is important to know how the image behaves as the object is moved towards the lens.
In general, as the object (starting from a distance of >2f) moves closer to the lens, the image will move further away from the lens and the size of the image becomes bigger.
But when the object is within a focal length, as it moves closer to the lens, the virtual image moves closer to the lens and it becomes smaller compared to the image previously. But the virtual image is always bigger than the object.
Refer to the video for better visualisation and understanding.
4) Other posts on converting lens: