If you want to find notes for each section of your course click on the individual sections from the drop down menu. Happy hunting! It is all there!
If you want to find notes for each section of your course click on the individual sections from the drop down menu. Happy hunting! It is all there!
As Mr Clydesdale says “A very good band”
The date is really 2020, but I need this post under the main Electricity notes section.
Mr Sharkey demanded I take screenshots of the traces for his OneNote ClassNotebook as he was made to leave against his will! So he asked, and I did!
We’ve plugged in a 1.5 V cell to the picoscope, put a voltmeter in parallel and noted the reading on the voltmeter and the looked at the value on the picoscope.
The picoscope was picking up some of the electrical signals from the computers and power around the room.
Notice on these images the reading on the picoscope and the voltmeter are the same. The cell is a source of D.C, direct current. In direct current the current /charges only flows in one direction. The free electrons in the ciruit are always drifting around the circuit in one direction.
When the polarity is reversed (swapping the positive and negative connections to the cell) the trace moves below the zero line showing that the current is now in the opposite direction. The voltmeter reads -1.5 V (the negative indicating that the current is in the opposite direction).
When we connect up to the A.C supply of the usual school power supplies we can see that the trace indicates the current flows in both directions. We can tell this as the trace of the voltage goes above and below the 0 V line on the picoscope. The trace shows a wave indicating the voltage and hence current is changing direction and magnitude many times per second. In the case of the mains voltage the frequency of the supply is 50 Hz.
Notice that the reading on the voltmeter reads 6.69 V. The power supply is set to 6V, but the peak of the trace is greater than this, about 9.5 V. The peak voltage of an A.C. trace is always greater than the quoted voltage of the supply. This is because we want to be able to compare A.C and D.C traces and so the quoted value is 1.414 times smaller than the peak voltage, try this.
When the polarity is reversed it makes no difference to the trace.
Another power supply in the Department is the 5.0 V regulated power supply. We can see this is a D.C trace and that the value of the voltage and hence the current is steady.
We can see when the polarity is changed (the connections to the power supply are swapped over) We can see the the trace of the voltage goes below the zero line, indicating the current is moving in the opposite direction. The voltmeter reading is the same as the value on the picoscope.
However, when we connect the picoscope to the usual Lockmaster power supply on the D.C. setting we get a rather unusual trace. The trace is D.C, remember direct current tells us that the current remains in one direction. However, the voltage and hence current isn’t constant. This is an unsmoothed D.C trace, and is common in cheaper power supplies. The trace never goes below the zero value on the screen.
Reversing the polarity shows us that the voltage is opposite, we get a negative value on the power supply but the trace never goes above the line. The current remains in one direction.
So in summary
In DIRECT CURRENT the current always moves in one direction.
In ALTERNATING CURRENT the current changes direction, usually many times per second. The current also usually changes magnitude (size).
With cells or regulated power supplies the D.C trace gives a constant value. In an unregulated trace the current also changes magnitude, but never direction.
(It is measured in units of time, e.g seconds, minutes, days, years, millions of years!)
From the Yellow Chemcord Book- this is how to answer the questions HALF LIFE QUESTIONS
Chemcord have kindly giving permission to upload these questions here. If you thought they were useful you can buy the National 5 Revision books soon:
half life Questions A print out for those who would like a copy of the National 5 Chemcord revision questions on half life. Here are the questions written out: HALF LIFE QUESTIONS
Time
in minutes |
Corrected
Count Rate in c.p.m. |
0
1 2 3 4 5 |
100
58 32 18 10 5.6 |
a) Plot a graph to show these results.
b) Estimate the half life of the source from these results.
22. In an experiment with a source, carried out in an area where there is a high background radiation, the following results were obtained.
Time (s) | Count Rate
(c.p.m.) |
0
30 60 90 120 150 180 210 240 270 300 |
88
72 60 52 44 39 36 34 32 29 30 |
a) Plot a graph to show these results.
b) Estimate the background count rate.
c) Estimate the half life of the source from these results.
ANSWERS
For Questions 2-6 (to find t ½ when Ao and A known)
Step
For Questions 7-10 (to find the final activity when t and t ½ are known) Step
For Questions 11-14 (to find Ao when A, t ½ and time are known)
Steps
For Questions 15-17
Step
The activity of a radioactive source decreases time. However the rate of decrease slows with time. Because of this, and because the decay of individual atoms is random and unpredictable, theoretically a radioactive source will never completely lose all of its activity. The time taken for half of the atoms in a radioactive sample to decay is a constant for that source called the half-life of the source. So the half-life of a radioactive source is the time period during which the activity of the source falls to half of its original value. The half-life of some sources is as low as a fraction of a second; for others it is many thousands of years.
Finding the half-life of a radioactive source
Apparatus: Geiger-Muller tube, Scaler counter or ratemeter, Source (eg.sealed protactinium-234 radioactive source and drip tray).
Instructions:
Half life and safety
To measure the half-life of a radioactive source, the level of the background radiation is first measured. Then the count rate with the radioactive source present is measured over a suitable period of time using a suitable detector such as a Geiger-Muller tube connected to a scaler. A graph of the count rate (with the source present), corrected for background radiation, is plotted.A suitable count rate value is chosen, say 80 counts per minute, and the time at which the source had this count rate, t1, is marked as above. In a similar way the time t2 at which the count rate is half the previous value, 40 counts per minute, is found. The half-life of the source is the time period t2 -t1. Any starting value can be chosen, the time period for the count rate to halve in value will always be the same.
EXAMPLE
In six years, the activity of a radioactive isotope drops from 200 kBq to 25kBq. Calculate the half-life of the isotope.
SOLUTION: original activity = 200 kBq
⇓
Activity after 1 half-life = ½ ×200 kBq = 100 kBq
⇓
Activity after 2 half-lives = ½ × 100 kBq = 50 kBq
⇓
Activity after 3 half-lives = ½ × 50kBq = 25 kBq
So 6 years represents 3 half-lives, thus one half-life is 2 years.
There are several safety precautions that must be taken when handling radioactive substances.
In addition there are several safety precautions relating to the storage and monitoring of radioactive substances.
The equivalent dose received by people can be reduced by three methods:
Stay safe and keep under your annual dose of 2.2 mSv!
Here are the results from the Protactinium Generator Experiment. Your task is to correct for background (take the background count per second away from the count rate) and then plot a graph of count rate (cps) against time (s). Remember the count rate was taken every 10 s but shows the value of the count rate (for one second)
Background count rate (c.p.m.) 48.0, 46.0, 42.0
Average background count rate (c.p.m.) 45.3
Average background per second (c.p.s.) work it out!
Time | Count rate |
---|---|
(s) | (cps) |
0 | 80.3 |
10 | 73.9 |
20 | 67.3 |
30 | 60.5 |
40 | 55.2 |
50 | 49.6 |
60 | 45.7 |
70 | 41.5 |
80 | 37.4 |
90 | 34.1 |
100 | 31.3 |
110 | 28.5 |
120 | 25.9 |
130 | 23.9 |
140 | 21.7 |
150 | 19.4 |
160 | 17.6 |
170 | 16 |
180 | 14.9 |
190 | 13.4 |
200 | 12.3 |
210 | 11.2 |
220 | 10.2 |
230 | 9.2 |
240 | 8.4 |
250 | 7.5 |
260 | 6.9 |
270 | 6.4 |
280 | 5.7 |
290 | 5.3 |
300 | 4.7 |
Here are the results for the Indium-116 half life experiment. Warning, do not plot a graph in 15 minute intervals or you will have more difficulty finding the half life. Make the scale ten minute intervals.
You can track the experiment yourself through the link below
Background count = 31 c.p.m.
Time | Time from start | Count rate |
---|---|---|
(hours) | (mins) | (c.p.s.) |
9:00 | 0 | 675 |
9:15 | 15 | 570 |
9:30 | 30 | 452 |
9:45 | 45 | 375 |
10:00 | 60 | 328 |
10:15 | 75 | 275 |
10:30 | 90 | 219 |
10:45 | 105 | 181 |
11:00 | 120 | 164 |
11:15 | 135 | 149 |
11:30 | 150 | 126 |
11:45 | 165 | 106 |
12:00 | 180 | 90 |
Here are three more examples for you to practice producing a half life graph and for finding the half life of Protactinium
Background Count (cps): 5, 5, 3, 5, 4, 5, 5.
Time (s) | Count in 10s |
---|---|
0 | 308 |
10 | 279 |
20 | 240 |
30 | 217 |
40 | 197 |
50 | 182 |
60 | 165 |
70 | 158 |
80 | 145 |
90 | 129 |
100 | 116 |
110 | 109 |
120 | 98 |
130 | 89 |
140 | 80 |
150 | 75 |
160 | 66 |
170 | 60 |
180 | 55 |
190 | 50 |
200 | 46 |
210 | 43 |
220 | 35 |
Background Count (cps): 4, 3, 5.
Time (s) | Count in 10s |
---|---|
0 | 841 |
10 | 752 |
20 | 693 |
30 | 622 |
40 | 593 |
50 | 544 |
60 | 481 |
70 | 443 |
80 | 392 |
90 | 364 |
100 | 341 |
110 | 301 |
120 | 272 |
130 | 251 |
140 | 243 |
150 | 211 |
160 | 204 |
170 | 183 |
180 | 172 |
time (s) | Count in 10s | corrected count in 10s | Corrected Count Rate (cps) | Background count (cp10s) |
---|---|---|---|---|
0 | ||||
5 | 105 | 100.5 | 10.05 | 5 |
20 | 95 | 90.5 | 9.05 | 5 |
35 | 81 | 76.5 | 7.65 | 3 |
50 | 90 | 85.5 | 8.55 | 5 |
65 | 86 | 81.5 | 8.15 | 4.5 |
80 | 73 | 68.5 | 6.85 | |
95 | 76 | 71.5 | 7.15 | |
110 | 79 | 74.5 | 7.45 | |
125 | 53 | 48.5 | 4.85 | |
140 | 53 | 48.5 | 4.85 | |
155 | 58 | 53.5 | 5.35 | |
170 | 61 | 56.5 | 5.65 | |
185 | 60 | 55.5 | 5.55 | |
200 | 43 | 38.5 | 3.85 | |
215 | 60 | 55.5 | 5.55 | |
230 | 48 | 43.5 | 4.35 | |
245 | 45 | 40.5 | 4.05 | |
260 | 44 | 39.5 | 3.95 | |
275 | 42 | 37.5 | 3.75 | |
290 | 47 | 42.5 | 4.25 | |
305 | 28 | 23.5 | 2.35 | |
320 | 31 | 26.5 | 2.65 | |
335 | 36 | 31.5 | 3.15 | |
350 | 32 | 27.5 | 2.75 | |
365 | 24 | 19.5 | 1.95 |
You should now have had plenty of practice at finding the half life graphically, nothing should phase you now.
Here are the booklets for the D&G Question Booklets. I’ll upload them in word and pdf format.
word pdf Waves & Radiation waves-rad-dg waves-rad-dg Energy & Electricity nrg-elect-dg2 nrg-elect-dg Dynamics & Space dynam-space-dg-ps dynam-space-dg
ds-n45-question-book-2-forces-sept-2012 Some questions for you to practise from Anderson High School.
nat-5-open-ended-questions-booklet
ds-n45-question-book-2-forces-sept-20121
Here are some definitions to learn for the waves topic. Remember you must be able to spell:
Term | Definition |
Amplitude (A) | the distance from the middle of the wave to the top (or the bottom) measured in metres. Maximum displacement from the mean position! |
amplitude, (A) | maximum disturbance of the particles in a wave. (or distance from middle to top of wave) (m) |
Angle of incidence | the angle between an incident ray and the normal (a line perpendicular to the reflecting surface at the point of incidence) (°) |
Angle of reflection | the angle between a reflected ray and the normal (°) |
Angle of refraction | the angle between the light ray in the more optically dense material and the normal. (°) |
Critical angle | The critical angle is the angle of incidence above which total internal reflection occurs. (°) |
Diffraction | occurs when wave meet a barrier, the waves bend around an obstacle. Long waves diffract more thank short waves. |
Energy and waves | Waves transmit energy. The greater the amplitude the more energy is transferred. |
Frequency, (f) | the number of waves per second. Frequency is measured in hertz (Hz). |
Frequency, (f) | number of waves produced or passing a point per second. (Hertz or Hz) |
Law of reflection | The angle of incidence = the angle of reflection |
Longitudinal wave | In a longitudinal wave the particles move along the line of the direction of travel of energy. |
Normal | a line at 90° to the surface at the point of incidence, (from which all angles are measured.) |
Period. (T) | Time for one wave to pass a point or time for one wave to be produced. (s) |
Principle of reversibility of light | The principle of reversibility of light states that a ray of light which travels along any particular path from some point A to another point B travels by the same path when going from B to A. |
Reflection | when a wave “bounces off” a surface we say it is reflected. (Particles can also reflect) |
Refraction | when light waves travel from one material to another the waves slow down and there is a reduction in wavelength in the optical thicker material. Unless the waves enter along the normal there is also a change of direction. |
Speed, (v) | rate of covering a distance. Number of metres travelled per second. (ms-1) The speed of the waves is represented by the formula |
Total Internal Reflection | When a wave hits a boundary at an angle larger than the critial angle the wave is entirely reflected if the material on the other side of the boundary is less optically dense. |
Transverse wave | In a transverse wave the particles move at 90 degrees to the direction of the flow of energy. |
Wave | a way of transferring energy. |
Wave speed | the speed at which the wave travels |
Wavelength | the distance between the same point on successive waves. |
Wavelength, (λ) | the distance between two successive points on a wave. (metre or m) |
Wavelength, (λ) | The wavelength of a wave is the horizontal distance between two adjacent troughs or crests or any two corresponding points on the wave |
The pressure, volume and temperature of a gas all affect one another. This would make the results of an experiment to investigate changes in all three at once complicated to understand. This problem is overcome by making one of them stay constant, whilst the relationship between the other two is investigated.
This Law is about the variation of (a fixed mass of gas) volume with the pressure of a gas at steady temperature. The apparatus shown in the diagrams below may be used to find how the volume of a fixed mass of gas varies with pressure at a constant room temperature.
A syringe can be compressed to increase the pressure which is measured using a pressure sensor. The fixed mass of gas, usually air is trapped in the syringe, and its volume read from the scale.
Alternatively a foot or bicycle pupil is used to increase the pressure which is measured on the Bourdon Gauge. The fixed mass of air is trapped in the capillary tube by a bead of mercury, and its volume measured on the scale.
How Pressure is related to volume for a constant mass and temperature of gas.
Pressure (kPa) | 100 | 111 | 125 | 143 | 167 | 250 |
---|---|---|---|---|---|---|
volume of air column(cm 3 ) | 50 | 45 | 40 | 35 | 30 | 20 |
Use the results to show a relationship |
It is found that when the pressure in increased the volume of the gas decreased so that:
pressure × volume = constant
p × V = k
The pressure of a gas is caused by the molecules hitting the walls of the container. Reducing the volume results in a shorter distance between the walls and so the number of molecules hitting the walls increases- resulting in increased pressure.
This law looks at the variation of pressure with temperature at constant volume. The apparatus shown below may be used to find how a fixed mass of gas, at constant volume varies with temperature.
The air is contained in a round bottom flask, which has a constant volume, and the pressure is measured using the pressure sensor attached to the flask by a short tube. It is important to have a short tube so that the temperature in the whole system is equal at any point. The round bottom flask is placed in a water bath which is used to vary the temperature of the water, and hence the air in the flask. Recent results have shown that the thermometer is best placed in the water bath as this gives more accurate results for the temperature of the air in the flask. Placing the thermometer in the flask usually results in a time delay in measuring the temperature of the air inside the flask. The flask should be fully immersed in the water, to ensure all the air in the flask is at the same temperature The temperature must be recorded as Absolute Temperature (in Kelvin) to find a relationship.
The pressure increases proportionally with the absolute temperature (i.e if you double the absolute temperature you will double the pressure provided the mass and volume remain constant). This can be expressed as:
If the absolute temperature of a gas increases, the speed of the molecules increases. The force and frequency of the impacts on the walls of the container increases, as this is the cause of pressure,then pressure increases.
How Temperature is related to Pressure for a constant mass and volume of gas.
Temperature ( o C) | 0 | 20 | 50 | 80 | 100 |
---|---|---|---|---|---|
Pressure (kPa) | 93 | 100 | 110 | 120 | 127 |
Use this to show a relationship! |
This is about the variation of volume with temperature at constant pressure. The apparatus shown is used to find how the volume of air varies with temperature provided the pressure remains constant.
The pressure remains constant, since it is equal to the atmospheric pressure plus the small additional pressure due to the weight of the mercury bead on to of the trapped air. This is because the capillary tube is open at one end. The volume of air is measured on the scale and the water bath is used to vary the temperature. The absolute temperature must be used to find a relationship between pressure and volume.
It is found that the volume of the gas increases proportionally with the absolute temperature provided the pressure and mass remain constant. This can be expressed as:
Kinetic theory of Charles’ law
The pressure of a gas is caused by the molecules hitting the walls of the container. If the absolute temperature of the gas increases the speed of the molecules increases. This would result in more forceful and frequent collisions on the walls. However, to maintain the pressure then there must be no increase in the frequency and magnitude of the collisions, so the volume must increase.
How Temperature is related to volume for a constant mass and pressure of gas.
Temperature ( o C) | 0 | 20 | 40 | 60 | 80 | 100 |
---|---|---|---|---|---|---|
Length of air Column(cm) | 20 | 21.5 | 22.9 | 24.4 | 25.9 | 27.3 |
Proportional to volume | ||||||
Use these figures to show a relationship! |
The three separate gas laws can be summarised by one equation, known as the General Gas Equation:
This is often written as:
Where p1, V1 and T1 refer to one set of conditions of pressure volume and temperature, and p2, V2 and T2 to another set of conditions for the same mass of the same gas.
An individual gas law can be found from this equation by covering up the variable which is kept constant (or cancelling out the variable as it remains constant).
Complete the three graphs above and for two of them try to work out the equation for the straight line, i.e. what is y = mx + c
The unit of pressure is the Pascal. I Pa is 1 Nm-2. You must remember you will not measure zero pressure as we have an atmosphere. I atmosphere is the pressure exerted due to our atmosphere and is approximately equal to 1 x105 Pa. This is equivalent to a weight of 105 N acting on a square of area 1m2. At ground level this is approximately the mass of 104 kg on a square metre which equates to about 10 Fiat 500 in 1m2.
Many teachers think that voltage is too difficult a concept for S1 students to understand. By the time students get to AH we expect them to be fully knowledgeable about voltage, but we don’t clearly explain it to them as we go along. I am as guilty as the next person of doing this so… . My new mission is to teach voltage as best and as fully as I can to S1 and build on the concept each year so that by AH they will feel confident about this work.
Having met Gill Arbuthnott at the Edinburgh International Book Festival (see post in Blog) I was really impressed with the way she tries to explain difficult concepts early on. She has given me permission to reproduce her page 16 on The Volt here.
The Volt
This was named after Alessandro Volta It is a unit of measurement in electricity. It tells us how much energy an electric charge has. You sometimes hear people saying things like, “The number of volts running through the circuit is…”. This doesn’t actually make sense! It’s like saying, “The height running through the mountain is 1000 metres.” Heights don’t run, and neither do volts. There is no Usain Volt!
What is a volt?
So what is a volt? Imagine you are in a building with stairs and a lift. You carry a tennis ball up one floor in the lift, and let it roll back to ground level down the stairs. A battery is like the lift – it’s a way of giving energy to something. In the building this is the ball – in electrical terms it’s an electron.
The ball rolling down the stairs is losing energy. In our circuit the equivalent is the electrons losing their energy to power a bulb. The voltage is equivalent to the height you take the ball up in the lift – more height is equivalent to greater voltage. And the distance the ball goes up in the lift must be the same as the distance it comes down by the stairs.
There are plenty of pictures in the book, but I didn’t think it was as easy to reproduce them. The book is full of more really interesting stuff, and even material about coins that Mr Chemistry opposite Mrs Physics didn’t know about (but then he’s far too young!)
http://www.bloomsbury.com/uk/a-beginners-guide-to-electricity-and-magnetism-9781472915740/
Definition: Potential difference is the amount of work done to move an electric charge from one point to another.
or
Definition: The definition of voltage is the electromotive force or the electrical potential difference between two points in a circuit expressed in volts.
Voltage is a scalar quantity. The SI unit of voltage is the volt, such that 1 volt = 1 joule/coulomb.
The easiest way to understand voltage is to use a water analogy. Using a hose as an example, think of voltage as the amount of pressure forcing water through a garden hose. The higher the pressure in the pipe the more water is forced through the pipe each second. The greater the voltage, the greater the flow of electrical current (that is, the quantity of charge carriers that pass a fixed point per unit of time Q=It) through a conducting or semiconducting medium for a given resistance to the flow.
One volt will drive one coulomb (6.24 × 10 18 ) charge carriers, electrons, through a resistance of one ohm in one second.
Voltage can be direct or alternating. A direct voltage maintains the same polarity at all times. So charges always flow in one direction. In an alternating voltage, the polarity reverses direction periodically. The number of complete cycles per second is the frequency, which is measured in hertz (one cycle per second). An example of direct voltage is the potential difference between the terminals of a cell. Alternating voltage exists between the mains positive and negative.
Changes have been introduced this session and you can read details by following the link below, although it is very much “teacher speak”
http://www.sqa.org.uk/sqa/files_ccc/PhysicsN5GuidanceSession1617.pdf
We will be continuing to use the old system of marking for the unit assessments as there is a possibility that these tests might be abandoned in future years, there is no point re-writing them.
Part of the National 5 course requires you to produce an experimental write up. It is good scientific practice. A threshold was introduced in the session 2016-2017, you are now only required to pass 5 out of 6 assessment standards to pass the overall outcome 1.
Below is the mark sheet to ensure that you’ve covered all the requirements, in both pdf and word format.
There are also two documents to help with your write up.
N5 Outcome 1 word N5 Outcome 1 pdf
Outcome 1 – Guide Nat 5 Thanks to Banff Academy for this document
WRITING UP O1 word WRITING UP O1 pdf
Here it is in simple form
Your best work. Rulers, sharp pencils etc.
Correct use of terminology and units at all times.
(No voltage in/ through) (No current across)
NO WAFFLE!
Title Short and relevant with date.
Aim What are you trying to find out/prove?
To find out how “something” affects “something else”.
Method Instructions on how to complete the experiment; make it reliable and make it a fair test:
Set up the following apparatus (draw a good labelled diagram).
The “something” was set at a “value” and increased by an “amount” using the “piece of equipment”. The “something else” was noted at each value using the “other piece of equipment”. Other variables were kept constant by…….
Results Display the findings.
A neat table with headings and units.
An appropriate graph of “something” on the x-axis and “something else” on the y-axis.
Conclusion What did you find out?
As the “something” is increased / decreased, the “something else” increased / decreased / stayed the same. Also include “directly/inversely proportional” if appropriate.
Evaluation Are there any improvements that could be made to your experiment to reduce uncertainties?
I was rather shocked to find that my new S1 class are really lovely but had never heard of the term average. Maybe I ought to have used the term mean, but I don’t think that would make much difference. So out came the calculators and I explained that mean average was all the numbers shared out evenly.
The mean is the average of the numbers.
It is easy to calculate:
add up all the numbers, then divide by how many numbers there are.
I asked student to use calculators to find the mean average of:
600, 100, 900, 450, 50
The mean average should be 420, the mean must be bigger than the smallest number and smaller than the largest number.
Some students got 2060. These are the ones that didn’t push the equals button between adding them up and dividing by 5 (as there are 5 numbers). So the calculator did the sum
600+100+900+450+ (50÷5)=600+100+900+450+ 10=2060
So using your calculator either do
600+100+900+450+ 50= ans÷5 =420
or
(600+100+900+450+ 50)÷5=420
Now this is OK with nice round numbers but we were using time, so students needed to fix their calculators to prevent calculator diarrhoea! (this is a Mrs Physics term and not a recognised scientific term)
We can FIX calculator diarrhoea (the tendency to write down everything that comes up on the display) using the FIX button on your calculator.
Here goes (I’m using my Casio fx85 or fx83)
Step one: Press the SHIFT and then MODE button on this line under the screen. This will bring up the menu.
Step 2: Make sure that the calculator is in line mode, so press 2
Step 3: Repeat step one but this time press the 6 on your calculator which brings up the FIX button. It will then ask you to record how many decimal places you need. This will depend on your values and measurements, but a suggest 2 is a safe bet.
Step 4: Select the number of decimal places you want to use.
Step 5: You’re ready to go.
Step 6: If you want to return to normal. Press the SHIFT followed by the MODE button and then 8 for return to normal. You will have 2 options, 1 (maths mode, good if you want fractions), 2 (line mode, good if you don’t want the fractions)
Try it and let me know how you get on!
Whilst working with the Police Crash investigator team I had a meeting with Pete Monteith, now known as calculator Pete, whom I decided must be of a similar age to me, as he had the same calculator I’d had at school. The difference was, he was still using his, whilst I was enjoying the delights of an updated calculator.
These days calculators are more technical than the computers I was brought up with. However, I think fewer people know how to use them, which isn’t surprising given the poor quality of the instruction manual. I am on a one woman campaign to get my students to fully utilise this great resource. When I started teaching Physics it used to take two weeks to teach resistance in parallel, now the students are happy in under a lesson- the reason? Their calculators “do as it says on the tin”!
Here are a few things to check out and try. (I am using a Casio, and I know the brighter amongst you would much prefer a Sharp, but I’ve never got on with them). You can draw your own conclusions!
Let’s check out using the calculator how to find total resistance in parallel. The equation is
1/Rt = 1/R1 + 1/R2 + 1/R3… etc.
How can your calculator to do this easily?
Let’s try adding a 7 ohm resistor in parallel with a 28 ohm resistor.
Yes that really did take nearly two weeks to teach before the age of super calculators.
Remember, your calculator can be a great asset to you during your exam and your career but only if you know how to use it.
Here is another of my favourite buttons, the degrees, minutes and second button.
My final fun button that is now a favourite of my classes, although they are right to be a little nervous, if they don’t know what they are doing. These tips cut the time to find resultants or components of vectors and their angles. The teaching of SOHCAHTOA (which incidentally I have to spell out using Six Old Horses, Clumsy And Heavy, Trod On Albert because I can’t spell SOHCAHTOA) appears to have gone by the wayside in Maths. But once you know your Pythagoras and SOHCAHTOAs from other Greek Philosophers then this button can save loads of time, but please don’t use it unless you’re sure you know what you’re doing.
I am sure many of you know that a right angled triangle with sides 3cm and 4cm will have a hypotenuse (or large side) of 5cm, but what will the angle be between the X axis and hypotenuse?
For this we want to use the rectangular to polar coordinate buttons.
We want to find the hypotenuse of a 3,4 triangle and the angle it subtends. so therefore we need the Pol button.
Practice using these buttons and when you need some more handy calculator hints let me know!
To Calculator Pete, if you ever want a race, I’ll time you with your factorial button anytime! When I was at school my Casio beat my friends Texas calculator by about 45 seconds every time I pressed 69 and then the factorial button (!). This then multiplied 69 by 68,by 67 etc. all the way down to 1. Funny, my new super calculator can still only manage factorials up to 69 before running out of digits, but the speed is remarkable. My question to myself, is why did me and my friend Deb Faulder, ever race regularly- as if we expected one day her calculator to beat mine.
Here are some documents to help you through the maths that you’ll need for your Physics. Don’t panic! You can easily learn the maths, but you’ll need to practice it regularly.
Need a piece of graph paper? Download it here multiwidth graph grey
The Relationship Sheet and Data Sheet. Why not make some flashcards using the information and leave them as a pile by the biscuit tin. Before tucking in set yourself a target of how many to get right!
Physics N5 Relationship Sheet The relationship sheet you’ll have during your SQA exams. Make sure you’ve got an annotated copy.
Physics N5 Data Sheet The data sheet will be on page two of your section 1. It is important that you get used to using it.
basic maths with answer. Here is a document for you to test out whether you can know your prefixes, scientific notation and significant figures.
Prefix | Symbol | Multiple | Multiple in full |
Tera | T | × 1012 | × 1 000 000 000 000 |
Giga | G | × 109 | × 1 000 000 000 |
Mega | M | × 106 | × 1 000 000 |
kilo | k | × 103 | × 1 000 |
centi | c | × 10-2 | ÷ 100 |
milli | m | × 10-3 | ÷ 1 000 |
micro | m | × 10-6 | ÷ 1 000 000 |
nano | n | × 10-9 | ÷ 1 000 000 000 |
pico | p | ×10-12 | ÷ 1 000 000 000 000 |
Above is a table of prefixes, which you will commonly find in Physics.
NB THE STANDARD UNIT FOR MASS IS THE KILOGRAM. Do not try changing it to grammes!
Watch out for ms which is not metres per second but milli seconds
Always set out maths problems using the structure given below. It may seem to take longer but it will save time in the long run as it makes the question clearer.
Here is the same information but in a little more detail for those whom require it.
A little out of date, maybe he’ll update it for the three mark question
The speed of light in air is 300 000 000 ms-1 (fast) We will use this number loads of times over the next year. It is a big number and must be entered carefully into your calculators.
300 000 000 means 3 ×108 or 3 ×10×10×10×10×10×10×10×10 THIS IS NOT THE SAME AS 38 WHICH EQUALS 6561
There are various ways of putting this number into your calculator.
1. Obviously you can do 300 000 000
2. you can use the xy or yx Here you would do 3×10 yx 8. This should give you the correct answer.
3. The EASIEST WAY IS USING THE exp / ee/ 10x button. Here you go 3exp8 or 3ee8 or 3×108 PLEASE NOTE. The exp / ee/ 10x button means 1×10x . DO NOT ADD TOO MANY 10S ON HERE!
Q: What happened to the plant in math class?
A: It grew square roots.
Q: How do you make seven an even number?
A: Take the s out!
Q: Why should the number 288 never be mentioned?
A: It’s two gross.
Q: Why is a math book always unhappy?
A: Because it always has lots of problems.
Q: Why did I divide sin by tan?
A: Just cos.
What Type of Learner are You?
Before you start a revision plan check out what kind of learner you are. This will then direct you to the best way for YOU to revise. Find your learning style
Once you’ve found your learning style take a look at the advice in the documents below. How to work to your learning style . learning style
study skills , study skills here is a powerpoint and pdf document of the power point to give some tips on revision.
Revision Documents
Nat 5 – Physical Quantities – Quantities and equations used at Nat 5
N5 WORKSHOP booklet 2017– Some hints to revision and relaxation ready prepared for your exam on 17th May 2017
From the lessons see the notes on voltage divider circuits
voltage divider Q voltage divider Q pdf
potential-dividers potential-dividers pdf
Uploading from school is 100 × quicker than home or the mobile!
N5 Energy and Electricity National 5 Physics E&E Revision (INT2)