Exercise
Exercise
Note:
Exercise
Disadvantages of single stroke method
Exercise
With the aid of a diagram explain how you would magnetize a steel bar so as to obtain a south pole at marked end of the bar by
Which of the above method produced a stronger magnet? Give a reason.
Main scale reading: 10.0 cm
Vernier scale reading: 0.02 cm (Alignment of scale lines)
Vernier Callipers reading = Main scale reading + Vernier scale reading: = 10.0 cm + 0.02 cm = 10.02 cm
Example
This vernier callipers has a zero error of +0.13cm
Correction of the Positive Error
Example
This vernier callipers has a negative error of -0.02 cm
Correction of the Negative Error
Exercise
The figure below shows a vernier callipers
State the correct reading of the scale if the instrument has a zero error of –0.02cm.
Example
Sleeve scale reading: 2.5 mm
Thimble scale reading: 0.38 mm
Micrometer reading: 2.88 mm
It occurs if the zero mark of the thimble scale does not coincide with the horizontal(centre) line of the sleeve scale when the micrometer is closed without an object.
Note:
Exercise
Write down the number of significant figures in each of the following
Exercise
Express the following in cm giving the answers in standard form
Exercise
Find the volume of a cube whose side is 2.22cm. Express your answer correct to 3d.p
Submultiple/ multiple | prefix | Symbol for prefix |
10^{-1} | deci | d |
10^{-2} | centi | c |
10^{-3} | milli | m |
10^{-6} | micro | μ |
10^{-9} | nano | n |
10^{-12} | pico | p |
10^{-15} | femto | f |
10^{-18} | atto | a |
10^{1} | deca | Da |
10^{2} | hecto | H |
10^{3} | kilo | K |
10^{6} | mega | M |
10^{9} | giga | G |
10^{12} | tera | T |
10^{15} | peta | P |
10^{18} | exa | E |
Exercise
Revision Questions
Examples
Find the moment of the force about the pivot in the figure below
Moment of a force = force × perpendicular distance
∴ moment of force about pivot=20N×0.4m = 8Nm
Examples
Exercise
A half meter rule is suspended vertically from a pivot at the 0cm mark. It is maintained in the vertical position by four horizontal forces acting in the directions shown in the figure below.
The 10.0N force acts through the 15cm mark, 4.0N force through the 20cm mark and 5.0N force through the 40cm mark. Calculate F which acts through the 30 cm mark.
Exercise
A uniform metal rod of length 80cm and mass 3.2kg is supported horizontally by two vertical spring’s balances C and D balance C is also from one end while balance D is 30cm from the other end. Find the reading on each balance.
Example
Two vertical equal and opposite forces act on a meter rule at 20cm and 90cm marks respectively. If each of the forces has a magnitude of 4.0N, calculate their moment on the meter rule about the 40cm mark.
solution
Total moment = one of the force,F x perpendicular distance between the forces,d
=4.0N×(0.9-0.2)m
=4.0N ×0.7
=2.8Nm
Revision Exercise
Examples
Exercise
Exercise
Exercise
Experiment
Aim: To show that sound requires a material medium to travel. (I.e.sound does not travel in vacuum)
Apparatus
Procedure
Conclusion
The above observations show that the sound cannot travel through a vacuum. It needs a material medium for propagation.
Example
Exercise
A fishing boat uses ultra-sound of frequency 6.0x10^{4}Hz to detect fish directly below. Two echoes of the ultra sound are received, one after 0.09s coming from the shoal of fish and other after 0.12s coming from the sea bed. If the sea bed is 84m below the ultrasound transceiver, calculate:
Used:
Revision Exercise
Examples
Weight, W (N) |
0.5 | 1.0 | 1.5 | 2.0 | 2.5 | 3.0 |
Total length, L (x10^{-2}m) |
7.5 | 8.0 | 8.5 | 9.0 | 9.5 | 10.0 |
Exercise
Force of compression, F (N) | 0.0 | 5.0 | 10.0 | 15.0 | 17.5 | 22.5 | 25.0 | 30.0 | 35.0 | 40.0 | 45.0 | 50.0 |
Length of spring, L(cm) compression | 14.50 | 13.00 | 11.50 | 10.00 | 9.25 | 7.75 | 7.00 | 6.50 | 6.25 | 6.00 | 6.00 | 6.00 |
Exercise
Two springs of negligible weights and of constants k_{1}= 50Nm^{-1 }and k_{2}=100Nm^{-1} respectively are connected end to end and suspended from a fixed point. Determine
Revision Exercise
Mass(g) | 0 | 25 | 50 | 75 | 100 | 125 |
Reading(cm) | 10 | 11.5 | 12.5 | 13.5 | 14.4 | 16.0 |
Force(N) | ||||||
Extension(mm) |
Notes:
Exercise
The figure below shows a compass placed under a vertical wire XY.
A large current is passed from X to Y. Draw the position of the magnetic compass needle.
Exercise
Show the magnetic field pattern inside the loop below
Exercise
Number of turns, n, | 0 | 4 | 8 | 12 | 16 | 20 | 24 | 28 | 32 | 36 |
Weight, W, of pins x 10^{-3} (N) | 0 | 4 | 14 | 30 | 58 | 108 | 198 | 264 | 296 | 300 |
Exercise
Show the resultant magnetic field and direction of force of the conductor in each of the following.
Exercise
Revision exercise
Examples
Exercise
Example
The figure below shows a displacement-time graph of a wave travelling at 2500cms^{-1}
Determine for the wave:
a) Amplitude
Solution
A = maximum displacement from mean position
=3cm OR 0.03m in SIunits
b) Periodic time
Solution
T=(9-1) x 10s^{-3}s
=8x10s^{-3}
c) Frequency
Solution
d) Wave length
Solution
Revision Exercise
Notes:
Assumptions made in deriving the equation of the continuity
Examples
Exercise
Revision Exercise
Exercise
The figure below shows two parallel rays incident on a concave mirror. F is the focal point of the mirror.
Sketch on the same diagram the path of the rays after striking the mirror.
It can be shown through geometry that the radius of curvature is twice the focal length i.e.
r=2f
Example
A lady holds a large concave mirror of focal length 1m, 80cm from her face.
Exercise
Examples
Solution
Solution
Exercise
The object distance u, the focal length f, and the image distance v related by the mirror formula:
^{1}/_{f} = ^{1}/_{u} + ^{1}/_{v}
Examples
Solution
Exercise
Exercise
A concave mirror and an illuminated object are used to produce a sharp image of the object on a screen. The object distances and image distances are given below.
Object distance, u(cm) |
80.0 | 26.7 | 22.4 | 20.6 | 19.6 |
Image distance, v(cm) | 20.0 | 40.0 | 56.0 | 72.0 | 88 |
u + v (cm) | |||||
uv(cm^{2}) | |||||
Magnification, m |
Note: The defect of spherical mirrors in which marginal rays are not brought into focus at the principal focus resulting in blurred images is called spherical aberration.
Revision questions