Rules to perform mathematical operations
Uncertainty and errors
Rules to perform mathematical operations
Uncertainty and errors
What is an ammeter?
What is the voltmeter?
A capacitor is simply an device that is composed of two metal conductors that are separated by an insulator. The insulator is called a dielectric. Fig 1 below shows a simplified version of a capacitor.
In fig 2 below two metal foils are separated by a paper dielectric. The three are them are then rolled as shown.
The fig 3 below shows a capacitor as sold commercially. It is a capsule that contains the rolled up aluminium foils.
The two plates are then connected to a power supply. The positive plate is connected to the positive terminal while the other negative plate is connected to the negative terminal as shown in fig 4 below.
After the capacitor is connected to the power supply, the capacitor will be charged by the power supply. That is charges will accumulate on the two plates. Positive charges will accumulate on the positive plate of the capacitor while negative charges will accumulate on the negative plate of the capacitor. The charges on the two plates will create an electric field between the two plates where the energy will be stored in the form of electric potential energy.
The fig 6 below shows a capacitor whose terminals are connected to a light bulb.
When this happens the energy stored in the capacitor will be released i.e is “discharged” into the bulb.
Hence a capacitor can be simply said to be an electronic device that stores charges. It is used in rectification circuits that changes alternating current to direct current, in tuning circuits and in filter circuits to filter out direct current.
In many questions in Physics or often in a practical students are often required to draw a line of best fit.
So what is the line of best fit. The line of best fit is a line that is drawn through a series of points in a graph in order to determine the trend in the points. The line can then be used to determine the gradient and the y-intercept of the line if it is a straight line.
The following are tips that you can use in order to obtain the best line of best fit from a few points. You should read all of them and choose which tips is appropriate.
Sometimes you will obtains points that are as shown below in fig 1. Although I must say that such occurrence is quite rare. In fig 1 below you can see that the points forms an almost perfect straight line. In such a case it would be enough to choose two points far away from each other and then draw a line from one to the the other. Make sure that the line extends on both side to cover all the points.
Most of the time you will be faced with points as in fig 2 below. As such you will have to draw a line such that the points are on both sides of the lines as such. A you see from one side of the line to the other the points alternates up and down. Also make sure that two points that are closest to each other and are on different sides of the line are the same distance from the line.
In case you are unable to make the points alternate on both sides of the line from one end of the line to the other as in fig 3 below. It is acceptable to have them two by two on the same side of the line.
It may happen as in fig 4 below that one point is so far from the others that it would make no sense to include it when drawing the line of best fit. This point is called an outlier and it is often the result of an error in calculation, an error in taking readings such as parallax error or zero error. It is advisable to check the data again.
Although I had said that it is possible to have the points two by two on both side of the line. It is not possible to have them as in fig 5 below.
If you are in the situation that I described as in fig 5 above then you can draw the line as in the fig 6 below.
So if you have any difficulties drawing the line then review the five different tips above you will surely find a tip for you. Otherwise send me a picture of the points and we will see it together.
We have seen in a previous post about the gradient and how it is calculated. We have also learned about the y-intercept. We are now going to combined the two concept to learn about the equation of a straight line.
All straight line can be represented by a equation that will be in the form of
y = mx + c
We are now going to use a straight line in order to see how the equation above can be determine and how the x-coordinate or the y-coordinate could be determined if one of the two and the equation is available.
Fig 1 above shows a straight line graph.
The gradient of the graph can be determined using the method described here.
Two coordinates that can be used are (1,6) and (5,18)
Thus Gradient = (18 –6)/(5-1)
And using the methods described here we can deduce that the y-intercept is 3.
Now the gradient = 3
and y-intercept =3
If we say that gradient = m then m= 3
and if we say that y-intercept = c then c=3
replacing the values for m and c in y= mx +c
the equation of the line becomes
y = 3x + 3
All straight lines can be represented with an equation and the way to obtain the equation is by using the method above.
We are going to use the equation to determine
(i) the value of the y-coordinate when the x-coordinate is 3
(ii) the value of the x coordinate when the y-coordinate is 15
(i) Equation of the line is y = 3x + 3 and x=3
replacing x=3 in the equation we get
y =3(3) +3
=12 check on the graph!!!!!!
(ii) y =15
Replacing into the equation we get
15 = 3x +3
15 – 3 = 3x
12 = 3x
x = 4 Check on the graph!!!!!!
We have seen in this post about the concept of pressure that a solid exerts on a surface.
Today we are going to look at the pressure that a liquid exerts on the bottom of a container in which it is present.
Fig 1 below shows a container containing an unknown orange liquid. The depth of liquid in the container is h. The density of the liquid is denoted by the letter ρ.
Fig 2 below represent the same container but this time in 3 dimensions. The container is in the form of a cuboid. The liquid will thus be in the form of a cuboid with cross-sectional area and height h.
Now how do you obtain the pressure that the liquid is exerting on the bottom of the container?
The equation for pressure is
Pressure = Force Applied /Surface area
= Force applied by liquid/surface area over which pressure is applied
= Weight of liquid /Cross-sectional area A
= mg /A
= (Volume of liquid *density*g)/A
= (Ah *ρ)*g/A
Where g is the acceleration due to gravity.
A bottle of water of density 1000 kg/m3 has water of depth 0.10 m in it. If the acceleration due to gravity,g, is 9.81 m/s2 , calculate the pressure due to the water on the bottom of the container.
Pressure due to water =hρg
=981 Pascal (Pa)
We have seen in a previous post the gradient of a straight line graph. Today we are going so have a look at the y-intercept.
Now as you know a graph is composed of two axes. The y-axis and the x-axis as shown in fig 1.
Now in order to find the y-intercept you will need a straight line that passes through the y-axis, i.e it intercept the y-axis, as seen in fig 2 below.
As you can see in fig 2 above all three lines crosses the y-axis. As a result the three lines would have a y-intercept.
How to obtain the y-intercept of a straight line?
1. If the x-axis starts at 0
As you can see in fig 3 below the point at which the line crosses the y-axis is at the y=2 coordinate. Hence the y-intercept is 2.
We can thus define the y-intercept as being the value of the y-coordinate when the x-coordinate is 0.
2. If the x-axis does not start at 0
As you can see in fig 4 below the x-axis does not start at 0. Hence as you may have guess wrongly the y-intercept is not 2 since according to the definition the y-intercept is the y-coordinate when the x-coordinate is 0.
So how do you obtain the y coordinate.
You will first have to calculate the gradient of the line using the method described in this post.
The gradient in this case is 1.
You will use the equation y = mx + c and a coordinate on the line in this case (3,2).
y = 2
gradient = m =1
Hence the only variable left is c, the y-intercept.
3 = 1*2 + c
c = 3 – 2 =1
The y-intercept of the line is thus 1.
You should thus be very careful to check that the x-axis starts with 0 or does not start with 0 so as to choose which of the two methods to use.
Today we are going to see the second equation of motion.
So how do we derive the second equation of motion?
Let us take an object that starts from an initial velocity u and then accelerates during a time t until it reaches a final velocity v.
Now during the time the object was accelerating the object has moved a distance s. In the second equation of motion we will try to derive the equation to calculate this distance s.
Fig 2 below shows what is happening to the object.
Now you would remember that the area under a velocity-time graph is the distance travelled by the object. Hence we are going to calculate the area under the graph as indicated by the shaded area below.
Now if s the distance travelled is the area under a velocity-time graph then we can say that
Area under graph = 1/2( v + u)*t
hence s = 1/2( v + u)*t
Now if you can remember the first equation of motion is
v = u +at
Hence if we replace v = u +at into s = 1/2( v + u)*t we get
s = 1/2( v + u)*t
s = 1/2( [u +at] + u)*t
s = 1/2( u + at + u)*t
s = 1/2(2 u + at )*t
s =1/2(2ut +at2)
S = ut + 1/2at2
Hence as you can see above the second equation of motion is
S = ut + 1/2at2
We are now going to look at two examples where the second equation of motion can be used.
A bus starts from rest and accelerates at a rate of 2.5 m/s2 for a time of 50 s. Determine the distance travelled by the bus during the acceleration phase.
Now let us identify the different variables involved.
u the initial speed = 0 m/s
t time taken = 50 s
a acceleration = 2.5 m/s2
s distance travelled = ???
If we want calculate s the distance travelled we will have to use the second equation of motion
S = ut + 1/2at2
Substituting the different values we get
S = ut + 1/2at2
= o*50 + 0.5* 2.5*502
= 0 + 0.5*2500
An aero plane travelling at a speed of 300 m/s lands on a track 2000 m long and decelerates for a time of 25 s until it comes to rest. Calculate the deceleration needed to stop the plane.
We must now identify the different variables involved as in the first example.
initial speed u = 300 m/s
time taken t = 25 s
Distance travelled s = 2000 m
acceleration a = ????
If we want calculate the acceleration a we will have to use the second equation of motion
S = ut + 1/2at2
and make a the subject of formula.
s – ut = 1/2at2
a =2(s - ut)/t2
= 2(2000 – 300*25)/252
= –8.8 m/s2
As you can see using the second equation of motion is not that difficult. Should you enter into any difficulties just post the question into the comment section.
An ammeter is a device that is used to measure the electric current. Now the ammeter can either be digital or analogue. It can be a single range or dual range.
The digital ammeter is one on which there is a screen and you can read the value of the electric current. An example of a digital ammeter is in fig 1 below. It the easiest to use as you just to connect it and read it.
The analogue ammeter is the one that was used in the old days. It has a pointer in it that will indicate the current in the circuit. It is more difficult to use and the chance of committing a parallax error is greater. However it does not require maintenance like the digital ones that require a dry cell in it to work. The dry cell require frequent replacing. Fig 2 below is an example of an analogue ammeter.
A single range ammeter is one that ca measure electric current within a particular range. For example it can measure from 0 to 5 A. However there are ammeters that can measure current on two different range. For example it can measure current on the 0 to 1 A range. However if the current being measured exceed the 1 A vale it can be switched to a higher range like the 0 to 5 A. An examples of a single range ammeter is in fig 3 and that of a double range is in fig 4.
Fig 3 A single range ammeter
Fig 4 A dual range ammeter.
How to use a dual range ammeter?
Most ammeter as you can see above have only two terminals and you just plug the wires in. However a dual range ammeter has three terminals. So how would you plug in the wires in that case? I would take the case of an ammeter that have two ranges 0 – 1 A and 0 – 5 A. If you want to use the 0 – 1 A range then you would connect the wires as in the fig 5 below.
Fig 5 Connecting an ammeter on the first range.
However if you want to connect the Ammeter on the second range then you would have to connect the wires as shown in fig 6 below.
Fig 6 Connecting an ammeter on the second range
The magnet is a very important object that have many applications in physics. The most important ones being the compass, motors and generators.
So what is a magnet?
It is simply a material that has a magnetic field around it. In physics a field is a region around an object that enable that object to exert a force at a distance. As a result due to its magnetic field, the magnet is able to exert a magnetic force on metallic objects as shown in fig 1 below, on other magnets or on electromagnets.
There are two main characteristics that a that a magnet must have.
1. Firstly the magnet must be made up of a magnetic material. A magnetic material must be a metal and it will be attracted to a magnet. However some metals are not attracted to a magnet and as a result are not magnetic materials. Aluminium, copper or gold objects for example cannot be turned into magnet and they are not attracted to a magnet. Magnetic material can be made into a magnet using an appropriate methods that we are going to see in a future post.
2. The magnet must have at least two poles. Fig 2 below shows a magnet with two poles. For the time being there has not been a magnet with one pole.
Fig 2 A magnet
As you can see in fig 2 the two sides of the two-pole magnet is labeled “N” and “S”. So how do you determine which side is the “N” and which side is the “S”.
In order to do that we suspend the magnet such that it is able to move freely as shown in fig 3 below.
A magnet that is suspended like that will always come to rest along a north-south axis. One part of the magnet will always point towards the north. Hence this side of the magnet is called the north-seeking pole or simply the north pole (N) . The other side of the magnet will always point towards the south. Hence this side of the magnet is called the south-seeking pole or simply the south pole(S).
Hence if any material has the two characteristics then it will be be a magnet.
As we have seen it a previous post a power supply has an electromotive force (e.m.f). The e.m.f. is the energy that the power supply will provide to 1 C of charge to make the 1 C of charge move around the circuit.
Now in a circuit there may be circuit components. Fig 1 below shows a circuit where there is a power supply and a bulb.
Now the 1 C is moving from the positive terminal and is moving around the circuit. While passing through the bulb it will move form the point A to the point B. Now the 1 C of charge has electrical energy. When it will pass through the bulb it will lose the electrical energy which will be transformed into another form.We say that the energy is dissipated.
Hence we can define the potential difference between point two points is the energy dissipated when 1 C of charge move from one of the point to the other.
Hence if the potential difference across the bulb is 3 V it means that when 1 C of charge flows from point A to point B 3 J of energy is dissipated in the bulb.
Hence if the potential difference across a circuit element is V and Q charge flows in the circuit element then the energy dissipated W can be calculated using the equation
W = VQ
Thus if we consider that an electric current is a flow of positive charges according to the conventional definition of current, for the positive charges to move from the positive terminal to the negative terminal of the power supply they need energy. It is this energy that is supplied by the power supply to the positive charges. The positive charges can thus use this energy to move around the circuit. It is this energy that the positive charges possesses that is called electrical energy.
In fig 2 you can see 1 c of charge leaving the positive terminal of the power supply. Now for this 1 C of charge the power supply will provide a certain amount of energy.
Suppose that for every one coulomb of charge that leaves the power supply the power supply provides 10 J of energy. Then the electromotive force (e.m.f.) of the power supply will be 10 Volt (V).
Hence if the e.m.f. of a dry cell is 1.5 V it means that when the dry cell is in a circuit for every 1 C of charge then the power supply will provide 1.5 J of energy.
Hence we can define the electromotive force as the energy that the power supply provide to a unit charge to move it around the circuit.
Hence if the a charge Q is moved around a circuit and the power supply supplied W amount of energy, the the e.m.f. E can be calculated as shown:
E = W/Q
If 4 C of charge needs 10 J of energy is provided by the power supply. Calculate the e.m.f. of the power supply?
E = W/Q
= 10 /4
We have seen in a previous post that when you have a complete circuit an electric current will flow. We have also see that an electric current is due to a flow of electrons that moves from the negative terminal to the positive terminal of the power supply.
Fig 1 below shows how the electron flows from the negative terminal to the positive terminal of the power supply.
However in a circuit the direction of the electric current is not defined as moving from the negative terminal to the positive terminal just as the flow of electrons. It is defined as moving from the positive terminal to the negative terminal of the power supply as shown in fig 2 below.
So why is it that despite knowing that the electric current is due to a flow of electrons that moves from the negative terminal to the positive terminal as shown in fig 1we set the direction of the electric current as being from the positive terminal to the negative terminal as shown in fig 2.
The answer lies in convention. A convention is like an assumption. When electricity was first discovered it was assumed that electricity is a flow of positive charges that the positive terminal to the negative terminal. Hence in all physics books at the time the electric current was taken as moving from the positive terminal to the negative terminal.
However it was discovered later on that in a circuit only electrons move. they move from the negative terminal to the positive terminal of the power supply. However the concept of conventional current was so ingrained that it could not be altered.
Hence as from now we will say that electric current is a flow of positive charges even though we know that it is in fact due to the flow of electrons.
Fig 1 below show a complete circuit. As you can see from the circuit electrons will flow through the bulb on its way to the positive terminal of the power supply.
Now depending on the circuit a certain number of electrons will flow through the bulb every second.
Hence if we know the number of electrons that is flowing through the bulb every second, then it means that we cal calculate the amount of charge that flow through the bulb in one second.
If the number of electrons that flows through the bulb in 10 s is 3.0x1022 and the charge of one electron is 1.6 x 10-19 C,
Calculate (i) the amount of charge that flows through the bulb in 10 s.
(ii) the amount of charge that flows through the bulb in 1 s.
(i) The amount of charge flowing though the bulb in 10 s is
Q = 3.0 x 1022 *1.6x10-19
= 4800 C
(ii) The amount of charge flowing through the bulb in 1s is
Q = 4800/10 =480 C
Now the quantity electric current is defined as the rate of flow of electric charge.
The unit of electric current is the Ampere (A) .
Thus if we are able to determine the rate of flow of electric charge or the amount of charge that flows in a circuit element like the bulb every second it means that we have determined the electric current.
If the amount of charge flowing through the bulb every second is 480 C then
The electric current = 480 C /s or 480 A.
Thus we can conclude that the electric current is merely an indication of the amount of charge flowing per second in a circuit element. The more charge flowing per second the greater the electric current flowing through the circuit element.
As you know in order to have electricity you will need a complete. So in order to have a complete circuit there are several features that must be present.
The power supply
A Power supply may comes in different forms such as:
1. The power pack that you have in your school’s physics lab. An example if in fig 1 below.
2. A dry cell such as the one in your portable radio. An example is shown in fig 2 below.
3. A battery such as the one that you have in your car.
The purpose of the power supply is to give energy to the electrons such that they can move around the whole circuit and come back to the power supply.
The symbol of the power supply is shown in fig 3.
As you know all power supplies have a positive terminal usually red in colour while the negative terminal is black in colour. Hence in the symbol the long vertical bar is the positive terminal while the short vertical bar is the negative terminal.
A circuit must also contain a switch. The purpose of the switch is to allow to to stop the flow of current in the circuit by opening it or to allow the flow of current by closing it. An example of a switch is shown in fig 4.
The symbol of the switch is as follows:
A closed switch –current flowing
An open switch- no current flowing
A load is a circuit component that will use the energy supplied by the power supply and will thus convert the electrical energy. Fig 5 below show a bulb that convert electrical energy to light energy.
Each load has a symbol. The symbol for the bulb is as shown below:
The complete circuit
Now let us combine all this together. In order to have a complete circuit we need the power supply the switch and the bulb to be connected together with wires as shown in fig 6.
Let us try to understand what happens when the switch is closed and a current is flowing on the circuit.
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As you know an electric current is due to a flow of electrons. When the switch is closed electrons can leave the negative terminal of the power supply as shown in fig 7 and move in a clockwise direction. The electrons will move through the switch, through the bulb and finally will reach the positive terminal of the power supply. The movement of the electrons are represented on fig 7 using arrows.The electron will lose energy to the bulb and this energy will be lost in the form of light. Should the switch be open then the electron will not be able to move from the negative terminal to the positive terminal.
If the circuit is represented using symbols then the circuit will be as shown below in fig 8.
The gravitational force is an attractive force that a mass exerts on another mass when the former is placed in the latter’s gravitational field.
In fig 1 above we have two masses m1 and m2 that are situated at a distance r apart. As you know these two masses will create gravitational fields around themselves such that mass M1 will exert a gravitational force F2 on mass M2 while the mass M2 will exert a gravitational force F1 on mass M1. That is each mass will exert a gravitational force on the other mass.
And from the third law of motion we know that “For every action there is an equal and opposite reaction.
Hence the gravitational force F2 will be equal to the gravitational force F1 .
F1 = F2
If the two forces are equal then they have the same magnitude and as a result
|F1 | = |F2 | = F
We can thus replace F1 and F2 by F as shown in fig 2 below.
Law of gravitation
In order to calculate the magnitude of the force that each mass will exert on the other the law of gravitation must be used.
It merely states that the gravitational force that the two objects will exert on each other is directly proportional to the product of the two masses and inversely proportional to the square of their distance of separation.
Simply written in an equation it will be as shown below:
Removing the proportionality sign you will obtain the equation below:
Where G is the universal gravitational constant
G = 6.67 x 10-11N m2 kg-2
The moon and the earth are separated by a distance of 3.8x 108 m. The mass of the moon is 6.4 x 1022 kg while that of the earth is 6.0x1024 kg. Calculate the gravitational force between the moon.