1.0.0: DEFINITION OF PHYSICS:

1.0.0:  DEFINITION OF PHYSICS:
Physics is a science subject that deals with the study of matter, the energy that act on matter and the behavior of matter when energy act on them.

1.0.1:   BRANCHES F PHYSICS:
The study of physics in senior secondary is grouped into the following:
              1. Mechanics
              2. Electricity
              3. General Physics
              4. Optics
              5. Wave
              6. Magnetism
              7. Nuclear PhysicsThe seven branches of physics are thought / covered for the three years duration of study of physics in senior secondary school.

1.1.0:   MEASUREMENT:
Measurement is the determination of the amount or size or numerical value of a physical quantity using an instrument.

1.1.1:   POSITION:
Position is the location of a point in space.

1.1.2:   LOCATION OF A POINT:
Rectangular coordinate is used to locate the position of a point in space.

1.1.3:   RECTANGULAR COORDINATE:
The rectangular coordinate contains positive and negative x- axis and positive and negative y-axis. The axes are calibrated based on the size of quantity that are to be plotted on the graph.                          
The rectangular coordinate system is as shown in the figure below:

1.1.4. COORDINATE OF A POINT:

The coordinate of a point is always generally stated in the form A(x , y). Where A is the name of the point, x represents a value or number on x – axis and y also represents a value or number on y – axis.

EXPLANATION:

For example, if the coordinate of a point is given as D(2 , - 5).

D is the name of the point. That is point D.

2 is the number to be marked or plotted on the positive side of  x – axis while – 5 is the number to be marked or plotted on the negative side of y – axis.

1.1.5. USING RECTANGULAR COORDINATE TO LOCATE POINTS:
Rectangular coordinate system is used to locate the position of a point as illustrated in the worked example below:

Worked Examples:

Use rectangular coordinate and locate the position of the points whose coordinate are given below:

I) A(-5 , 3).    II)   C( 1 , 4 ).        III).   D( 2, 1 ).   IV)  F(2.5, -3).  V).  G(-3, -2)

Note: the dots represent the location of the points whose coordinates are given in the questions above.

EXCERCISE:

Use rectangular coordinate and locate the points whose coordinate are

A( 3, 4).        D( -2, 4).        P( -3, -4).        Z( 2, -5).          L( 0, 3)

2.0.0.  DISTANCE:

Distance is the space or gap or length between two points. The S.I unit of distance is meter and symbol of meter is .

2..1.0  DISPLACEMENT:

Displacement is a distance travelled in a given or specified direction. Its unit is meter. Displacements is a vector quantity because it has magnitude and direction.

2.1.1.  CALCULATION IN DISPLACEMENT:

Since displacement is a distance moved in a given direction, it implies that in the calculation of displacement, the distance and the angular displacement or bearing of the line from the start point must be calculated. This we will explain in the work example below.

Example 1.

A man walked 30km due East. He then moved 40km due North. What is the displacement of the man.                                                                                   

Solution:

Step I.  

Draw the cardinal point that indicates North, South, West and East.

Step II. 

Use the cardinal points, draw a vector triangle such that the two sides represent the motion while the hypotenuse represents displacement distance. That angle represents the bearing.

Step III.  

Use Pythagoras theory to find the hypotenuse of the triangle,

Which is the displacement and calculate the angular bearing from the start point.

Step IV.  

Use sohcatoa to calculate the bearing ( angular displacement) of the man from his starting point, thus:

From the question:

H = displacement distance, A = 30 km due east,  O = 40 km due north.

From the cardinal point, you draw a vector triangle as shown below.  The two sides of the triangle represent the 30 km distance due east and 40 km distance due north respectively, while the hypotenuse of the triangle represents the displacement distance from the start point.

H² = A² + O²
h² = 30² + 40²,    h² = 30 x 30 + 40 x 40,   h ²= 900 + 1600
h²  = 2500,    h = √2500.  h = 50km
Therefore distance = 50km

Now, let us find the bearing or angular displacement of the distance thus:
Tan P = opposite ÷ adjacent,  Tan P = 40/30,  
Tan P = 1.3333.   P = Tan–¹ 1.3333
Therefore, P = 53.12° degrees.

Therefore,
Displacement = a distance of 50km in a bearing of 36.88°
Or
displacement = 50km N36.88E or 50km E53.12°N

Explanation :
N36.88°E means you start from the North pole and measure 36.88° to the East pole while E53.12°N means you start from the East pole and measure 53.12° to the North pole.

3.0.0.  INSTRUMENTS FOR MEASURING DISTANCE:
The instruments for measuring distance are :
I) Meter rule.    II)  Callipers   III) Vernier calliper.   IV)    Micrometre screw gauge.   Etc.

3.1.0.  METER RULE:
Meter rule is used to measure distance on paper. It is calibrated in millimeters, centimeters, . some meter rules are calibrated in inch or feet. The accuracy of meter rule is 0.1 cm or 1.0 mm.

3.1.1.  IMAGE OF METER RULE:

3.1.2.  HOW TO MEASURE DISTANCE WITH METER RULE:
Distance is measured starting from the zero reference mark of the meter rule to any required length.  The measured distance or length is read and recorded.

3.2.0.  CALLIPERS:
Callipers are not used to measure distance or length directly as the case is in meter rule. The claws of the callipers are adjusted to touch the edges of the object that is being measured. The span of the jaws of the callipers is then measured against a calibrated scale or meter rule to determine or estimate the size of the measured distance or length.

3.2.1.   IMAGES OF CALLIPERS:

3.3.0.  VERNIER CALLIPER:
Vernier calliper is used to measure small distance.

3.3.1.  DIAGRAM OF VERNIER CALLIPER:

A Vernier calliper has two scale : the main scale and the Vernier scale.

3.3.2.  MAIN SCALE:
The calibrations or scales of the main scale of Vernier callipers vary depending on the range of measurement that the Vernier callipers are meant to measure. I will not center on the main scale.

3.3.3.  VERNIER SCALE:
The vernier scale of a vernier calliper is obtained by dividing a 9mm length into 10 equal divisions or intervals such that each divisions is 0.9mm or 0.09cm.

3.3.4.  ACCURACY OF VERNIER CALLIPER :
The accuracy of a vernier caliper is 0.1mm or 0.01cm. This means the that the smallest distance that a vernier calliper can measure is 0.1mm or 0.01cm

3.3.5.  ACCURACY OF AN INSTRUMENTS :
Accuracy of an instrument is the smallest distance (or value of a particular quantity) that the instrument can measure correctly.

3.3.6.  STEPS TO FOLLOW WHEN READING A VERNIER CALLIPER:
Step I.
Read and record the reading of the main scale
Step II.
Read and record the reading of the vernier scale at the point where the main scale and the vernier scale coincide.
Step III.  
Multiply the reading of the vernier scale by the accuracy of the vernier scale which is 0.01cm.
Step IV.  
Add the result of step III to the main scale reading of step I to get the final answer which is the reading of the vernier calliper.

Worked Example 1:
What is the reading of the vernier calliper as shown in the figures below?
Figures of vernier calipers.

Solution:
Reading of main scale.           =     10.0   cm
Reading of vernier scale        =     2                    
Reading of vernier scale x accuracy of vernier calliper  =                               2 x 0.01 cm  =    0.02   cm
Add the result of step III and step I to get the reading.       =    10.0 cm + 0.02 cm
Therefore,

Rcm ding of vernier calliper is  10.02cm              

 Worked Example 2:
Estimate the reading of the vernier calliper as shown in the figure below:

Solution:
Reading of main scale.           =     4.3   cm
Reading of vernier scale        =     5                    
Reading of vernier scale x accuracy of vernier calliper  =                               5 x 0.01 cm  =    0Re   cm
Add the result of step III and step I to get the reading.                                                                                                               =    10.0 cm + 0.02 cm
Therefore,     Reading of vernier calliper is  10.02 cm.    

EXERCISES:
what are the readings of the vernier callipers as shown in the figures below?.
1.                                                           2.  

 . 

3.                                              4.