_{}^{}olanrewaju(m) Says (04:52pm On 3 Mar) (1364 Views)

Hello Aspirants in the house, This thread is created for all physicist students. We will all brainstorm on Various Aspect of physcis, I will Take you some tutorials. I am also an Aspirants oooo, I stand to be corrected. #Physics
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_{}^{}olanrewaju(m) Says (04:56pm On 3 Mar) (1364 Views)

SYLLABUS
TOPICS/CONTENTS/
NOTES OBJECTIVES
1. MEASUREMENTS AND
UNITS
(a) Length, area and
volume: Metre rule,
Venier calipers
Micrometer
Screwguage,
measuring cylinder
(b) Mass
(i) unit of mass
(ii) use of simple
beam balance
(iii) concept of beam
balance
(c) Time
(i) unit of time
(ii) timemeasuring
devices
(d) Fundamental
physical quantities
(e) Derived physical
quantities and their
units
(i) Combinations of
fundamental
quantities and
determination of their
units
(f) Dimensions
(i) definition of
dimensions
(ii) simple examples
(g) Limitations of
experimental
measurements
(i) accuracy of
measuring
instruments
(ii) simple estimation
of errors.
(iii) significant figures.
(iv) standard form.
(h) Measurement,
position, distance and
displacement
(i) concept of
displacement
(ii) distinction
between distance and
displacement
(iii) concept of
position and
coordinates
(iv) frame of reference
Candidates should be
able to:
i. identify the units of
length, area and
volume;
ii. use different
measuring
instruments;
iii. determine the
lengths, surface areas
and volume of regular
and irregular bodies;
iv. identify the unit of
mass;
v. use simple beam
balance, e.g Buchart's
balance and chemical
balance;
vi. identify the unit of
time;
vii. use different time
measuring
devices;
viii. relate the
fundamental physical
quantities to their
units;
ix. deduce the units of
derived physical
quantities;
x. determine the
dimensions of
physical quantities;
xi. use the dimensions
to determine the units
of physical quantities;
xii. test the
homogeneity of an
equation;
xiii. determine the
accuracy of
measuring
instruments;
xiv. estimate simple
errors;
xv. express
measurements in
standard form.
Candidates should be
able to:
i. use strings, meter
ruler and engineering
calipers, vernier
calipers and
micrometer, screw
guage
ii. note the degree of
accuracy
iii. identify distance
travel in a specified
direction
iv. use compass and
protractor to locate
points/directions
v. use Cartesians
systems to locate
positions in xy plane
vi. plot graph and
draw inference from
the graph.
2. Scalars and Vectors
(i) definition of scalar
and vector quantities
(ii) examples of scalar
and vector quantities
(iii) relative velocity
(iv) resolution of
vectors into two
perpendicular
directions including
graphical methods of
solution.
Candidates should be
able to:
i. distinguish between
scalar and vector
quantities;
ii. give examples of
scalar and vector
quantities;
iii. determine the
resultant of two or
more vectors;
iv. determine relative
velocity;
v. resolve vectors into
two perpendicular
components;
vi. use graphical
methods to solve
vector problems;
3. Motion
(a) Types of motion:
translational,
oscillatory, rotational,
spin and random
(b) Relative motion
(c) causes of motion
(d) Types of force
(i) contact
(ii) force field
(e) linear motion
(i) speed, velocity and
acceleration
(ii) equations of
uniformly accelerated
motion
(iii) motion under
gravity
(iv) distancetime
graph and velocity
time graph
(v) instantaneous
velocity and
acceleration.
(f) Projectiles:
(i) calculation of
range, maximum
height and time of
flight from the
ground and a height
(ii) applications of
projectile motion
(g) Newton's laws of
motion:
(i) inertia, mass and
force
(ii) relationship
between mass and
acceleration
(iii) impulse and
momentum
(iv) force  time graph
(v) conservation of
linear momentum
(Coefficient of
restitution not
necessary)
(h) Motion in a circle:
(i) angular velocity
and angular
acceleration
(ii) centripetal and
centrifugal forces.
(iii) applications
(i) Simple Harmonic
Motion (S.H.M):
(i) definition and
explanation of simple
harmonic motion
(ii) examples of
systems that execute
S.H.M
(iii) period, frequency
and amplitude of
S.H.M
(iv) velocity and
acceleration of S.H.M
(v) simple treatment
of energy change in
S.H.M
(vi) force vibration
and resonance
(simple treatment)
(iii) conservative and
nonconservative
fields
(iv) acceleration due
to gravity
(v) variation of g on
the earth's surface
(iv) distinction
between mass and
weight
(v) escape velocity
(vi) parking orbit and
weightlessness
Candidates should be
able to :
i. identify different
types of motion ;
ii. solve numerical
problem on collinear
motion;
iii. identify force as
cause of motion;
iv. identify push and
pull as form of force
v. identify electric and
magnetic attractions,
gravitational pull as
forms of field forces;
vi. differentiate
between speed,
velocity and
acceleration;
vii.deduce equations
of uniformly
accelerated motion;
viii. solve problems of
motion under gravity;
ix. interpret distance
time graph and
velocitytime graph;
x. compute
instantaneous velocity
and acceleration
xi. establish
expressions for the
range, maximum
height and time of
flight of projectiles;
xii. solve problems
involving projectile
motion;
xiii. solve numerical
problems involving
impulse and
momentum;
xiv. interpretation of
area under force 
time graph
xv. interpret Newton's
laws of motion;
xvi. compare inertia,
mass and force;
xvii. deduce the
relationship between
mass and
acceleration;
xviii. interpret the law
of conservation of
linear momentum and
application
xix. establish
expression for
angular velocity,
angular acceleration
and centripetal force;
xx. solve numerical
problems involving
motion in a circle;
xxi. establish the
relationship between
period and frequency;
xxii. analyse the
energy changes
occurring during
S.H.M
xxiii. identify different
types of forced
vibration
xxiv. enumerate
applications of
resonance.
Candidates should be
able to:
i. identify the
expression for
gravitational force
between two bodies;
ii. apply Newton's law
of universal
gravitation;
iii. give examples of
conservative and non
conservative fields;
iv. deduce the
expression for
gravitational field
potentials;
v. identify the causes
of variation of g on
the earth's surface;
vi. differentiate
between mass and
weight;
vii. determine escape
velocity
5. Equilibrium of
Forces
(a) equilibrium of
particles:
(i) equilibrium of
coplanar forces
(ii) triangles and
polygon of forces
(iii) Lami's theorem
(b) principles of
moments
(i) moment of a force
(ii) simple treatment
and moment of a
couple (torgue)
(iii) applications
(c) conditions for
equilibrium of rigid
bodies under the
action of parallel and
nonparallel forces
(i) resolution and
composition of forces
in two perpendicular
directions,
(ii) resultant and
equilibrant
(d) centre of gravity
and stability
(i) stable, unstable
and neutral equilibra
Candidates should be
able to:
i. apply the conditions
for the equilibrium of
coplanar forces to
solve problems;
ii. use triangle and
polygon laws of
forces to
solve equilibrium
problems;
iii. use Lami's theorem
to solve problems;
iv. analyse the
principle of moment
of a
force;
v. determine moment
of a force and couple;
vi. describe some
applications of
moment of a force
and couple;
vii. apply the
conditions for the
equilibrium
of rigid bodies to
solve problems;
viii. resolve forces into
two perpendicular
directions;
ix. determine the
resultant and
equilibrant
of forces;
x. differentiate
between stable,
unstablend neutral
equilibra.
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_{}^{}olanrewaju(m) Says (05:06pm On 3 Mar) (1364 Views)

Wokdone, Power, and Energy.
Work.
In ordinary meaning, work refers to any kind of physical or mental activity. But in physics, work is always associated with movement. Thus, work is said to be done if a force applied makes the object to move a distance in the direction of the applied force.
Work is defined as the product of force and perpendicular distance in the direction of the applied force. It is denoted by w, measured in Joule and a scaler quantity.
Mathematically;
work = force X distance
w = f X s
since, force = mass X acceleration
work = m X a X s.
Example: Find the workdone when a force of 5N is exerted on a body through a distance of 10m.(Take g=10m/s2)
Solution.
In Physics you have to write your Data(I mean what you have been given in the question)
F=5N
s=5m
W=F*D
W=5*10
=50J (Don't forget your units)
So the workdone is 50J.
Example; A crane lifts a load of 2kg very slowly through a vertical distance of 100cm. Calculate the workdone against gravity if g = 10m/s2.
Solution;
load=2kg
s=100cm=1m
a=g=10m/s2.
Workdone=F X S
w=mas
w=2 X 10 X 1
w=20J.
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_{}^{}olanrewaju(m) Says (10:28am On 3 Mar) (1364 Views)

Morning all, Before we proceed in today's class, We need to know some conversions.
1. To convert from mm to cm, "we" divide by 10.
2. To convert from cm to m, "we" divide by 100.
3. Therefore; to convert from mm to m, "we" divide by 1000.
4. To convert from mm2 to cm2, "we" divide by 10 and another 10, why? It is beacuse of the square.
5. To convert from cm2 to m2, "we" divide by 100 and another 100, why? Same as above.
6. To convert from mm2 to m2, "we" divide by 1000 and another 1000, why? Same as above.
7. The standard unit of measurement of length is m not mm.
Bonus; To convert mm4 to cm4, "we" divide by 10, and 10, and 10 and 10. Why? The power is 4.
8. To convert form g to kg, we "divide" by 1000
I will updates it as we move on.
Before we move on Let us know about Scalar and Vector quantities.
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_{}^{}olanrewaju(m) Says (10:38am On 3 Mar) (1364 Views)

Scalars and Vectors
What are Scalar Quantities.
A scalar quantity is a quantity that has magnitude but no Direction. Scalar quantity are known for numerical value.
Example of Scalar quantity are Speed, Distance etc
What are Vector quantities.
Vector quantites are quantities that has both magnitude and direction. Eg Velocity, Acceleration, Displacement etc.
Let us test our selves.
Example: Identify each of the following whether they are vector or scalar quantity.
a. 5m It is a scalar quantity because it has no direction
b. 30 m/sec, East  It is a vector quantity because it has direction which is "EAST"
c. 5 mi., North It is a vestor quantity because it has directio which is "NORTH"
Classworks
Identify the following quantities.
a Temperature
b 256 bytes
c Force
d Momentum
e Work
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_{}^{}youngscholar(m) Says (01:34pm On 3 Mar) (1364 Views)

Okay sir!
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_{}^{}youngscholar(m) Says (01:39pm On 3 Mar) (1364 Views)

olanrewaju:Scalars and Vectors
What are Scalar Quantities.
A scalar quantity is a quantity that has magnitude but no Direction. Scalar quantity are known for numerical value.
Example of Scalar quantity are Speed, Distance etc.
What are Vector quantities.
Vector quantites are quantities that has both magnitude and direction. Eg Velocity, Acceleration, Displacement etc.
Let us test our selves.
Example: Identify each of the following whether they are vector or scalar quantity.
a. 5m It is a scalar quantity because it has no direction
b. 30 m/sec, East  It is a vector quantity because it has direction which is "EAST"
c. 5 mi., North It is a vestor quantity because it has directio which is "NORTH"
Classworks
Identify the following quantities.
a Temperature
b 256 bytes
c Force
d Momentum
e Work
Solution
1.Temperature A scalar unit because it has no direction
2.256 byets A scalar unit because it has no direction
3.Force A vector unit becuase it has direction "A" accerleration
4.Momentum A vector unit because it has direction "p=mv"
e.Work A vector unit because it has direction.
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_{}^{}olanrewaju(m) Says (01:41pm On 3 Mar) (1364 Views)

Solution
1.Temperature A scalar unit because it has no direction
2.256 byets A scalar unit because it has no direction
3.Force A vector unit becuase it has direction "A" accerleration
4.Momentum A vector unit because it has direction "p=mv"
e.Work A vector unit because it has direction.
Nice one.. :D
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_{}^{}Heavens lawrence(m) Says (05:55pm On 3 Mar) (1364 Views)

olanrewaju:Solution 1.Temperature A scalar unit because it has no direction 2.256 byets A scalar unit because it has no direction 3.Force A vector unit becuase it has direction \"A\" accerleration 4.Momentum A vector unit because it has direction \"p=mv\" e.Work A vector unit because it has direction.
Nice one.. :D work is a scaler quantity
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_{}^{}olanrewaju(m) Says (06:29pm On 3 Mar) (1364 Views)

Heavens lawrence:Nice one.. :D work is a scaler quantity[/quote]
Yes Havens, work is a Scalar quantity, because it has no direction.
I just appreciated @ youngscholar( m) answers.
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_{}^{}youngscholar(m) Says (06:32pm On 3 Mar) (1364 Views)

Noted Sir!
Proceed.
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_{}^{}olanrewaju(m) Says (06:40pm On 3 Mar) (1364 Views)

Another to note is that, scalar quantities don't have Velocity, while vector quantities do.
Like the diagram below.
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_{}^{}olanrewaju(m) Says (12:16pm On 3 Mar) (1364 Views)

Morning Class, Today I will treat Circular Motion.
First of all what is motion?
Motion involves a change of position of a body with time. It also involves how things move and what makes them to move. It exists in different forms and occurs in solids, liquids, and gases.
Many scientists have studied motion and its properties because of its importance to life. The Italian, Galileo Galilei did the first systematic study of motion. Also, Sir Isaac Newton did more detailed work on the study of motion.
Motion is defined as the change of position of a body with time.
Types of Motion.
There are five types of motion, which are; translational rectilinear motion, random motion, oscillatory motion, rotational motion and relative motion.
1. Translational rectilinear motion; This type of motion occurs when an object moves in a fixed direction. When a body moves in a fixed direction, the body is said to be translated. Examples are; movement of a car from a point to another, human movement from a point to another:
2. Random motion; This sis a type of motion in which a body moves in a nonlinear manner and changes direction continuously. Examples are; molecular movement in gases, dashing of objects in air eg feather thrown into air.
3. Oscillatory motion; This is a to and fro movement about a fixed point. It arises due to a slight displacement of an object from the initial position to a new position and back to the initial position. Examples are; simple pendulum, vibration of plucked guitar strings.
4. Rotational motion; This is the movement of a body in a rotational manner about its axis. Examples are; rotation of electric fan blades, movement of car wheels.
5. Relative motion; Is the movement of object in relation to another body. A car travelling on a road side is said to be in motion with respect to the poles and the trees on the road side. Examples are; pulling on a spinal spring, bending a ruler on a beam.[/b]
Causes of Motion.
Force causes motion. When force is applied on any object, it causes change. Force is a vector quantity denoted by F and measured in Newton(N).
Types of Force.
There are two types of force, which are the contact force and force field.
1. Contact force; This exists when there is a touch, hold or a contact with the object in question. It is further divided into like force or push, unlike force or pull and fictional force or viscous.
2. Force field; This exists in or is confined/restricted to a given region or space, i.e., a region in space where a body experiences the effect of a force which occurs as a result of the influence of some physical agencies. It is further divide into; magnetic force, gravitational force and electric force.
a. Magnetic force; This is a force that attracts magnetic substance eg nails.
b. Gravitational force; This is a force that attracts or pulls objects irrespective of their masses towards the centre of the Earth's surface.
c. Electric force; This is a force that keeps current through a conductor of electricity.[/b]
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_{}^{}olanrewaju(m) Says (12:23pm On 3 Mar) (1364 Views)

We want to treat the Calculations under Circular Motion.
Any where I missed it, correct me please.
Circular Motion.
Circular motion is a motion in a circular path in which the speed of the body remains constant while its direction continuously changes eg the Earth moving round the sun, a racing car moving round a circular track.
The velocity of a body undergoing circular motion is given by;
v = wr
where v is linear velocity (m/s)
w is angular velocity (rad/sec)
r is radius (m).
The acceleration is given by;
a = vw, or
a = (wr)w = w ^{2}r, or
a = v(v/r) = v ^{2}/r.
Centripetal force.
This is the force required to keep object moving in a circular path.
Therefore Centripetal force is
f = mw ^{2}r.
where f=force
m=mass
w=angular velocity
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_{}^{}olanrewaju(m) Says (12:31pm On 3 Mar) (1364 Views)

Formulas to note when solving.
v=wr
a=v ^{2}/r
f=mw ^{2}/r
Exercise.
An object of mass 10kg moves in a circle at radius 2m at unformed speed of 36m/s. Calculate; (a) the angular velocity (b) the centripetal force.
Solution.
Number one thing to state is your data.
m=10kg
r=2m
v=36m/s
a=?
F=?
a. V=wr
w=36/2=18
b. F=mw ^{2}/r
=10*18 ^{2}/2
=1620N
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_{}^{}Heavens lawrence(m) Says (10:11pm On 3 Mar) (1364 Views)

great!.. Following...
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_{}^{}youngscholar(m) Says (12:44pm On 3 Mar) (1364 Views)

Heavens lawrence: great!.. Following...
Yes boss! ;D
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_{}^{}olanrewaju(m) Says (06:30pm On 3 Mar) (1364 Views)

Evening To all Following !.
Please try and invite your friends to this thread. There's a reward for that.
Today we will be treating Electrical Power and Energy
What is Electrical Energy
Electrical energy is the workdone when a quantity of charge moves between two points of potential differences measured in Joule.
Mathematically;
W = QV;
Recall that Q = It.
W = IVt(1).
Recall that V = IR.
W = I ^{2}Rt(2).
W = V ^{2}t/R (3).
Example; Calculate the electrical energy produced by a heater with a voltage supply of 280V, when a current of 15amps passed through it for 8minutes.
Solution.
Write your data out.
V=280v
I=15A
T=8 * 60=480s
W=IVT
=280*15*480
=2016000J
Electrical Power.
Electrical power is the rate at which the energy is used up. Power is measured in watts with wattmeter.
Mathematically,
Power = workdone/time
P = IVt/t.
P = IV (1).
P = I2R (2).
P = V2/R.
Example; Calculate the power dissipated by a heater of 280V and a resistance of 18 Ohms.
Solution;
P=V2/R
=280 ^{2}/18
=4335.55W
Classwork.
1.Calculate the power dissipated by a iron of 300V and a resistance of 7 Ohms.
2.Calculate the electrical energy produced by a iron with a voltage supply of 380V, when a current of 6amps passed through it for 15minutes.
Solve #Physics Students.
Waiting @Heavens_Lawrence @ youngscholar( m) @ brainiac @ Kamccy and others
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_{}^{}brainiac Says (11:22pm On 3 Mar) (1364 Views)

not bad....nice post
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_{}^{}Heavens lawrence(m) Says (07:28am On 3 Mar) (1364 Views)

P=IV...
WHERE I=V/R...
P=(V/R) * V...
P=(300/7) * 300....
P=42.86 * 300...
P=12,857.1WATT..
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_{}^{}Heavens lawrence(m) Says (07:36am On 3 Mar) (1364 Views)

2. Energy=IVt...
I=6...
t=15 * 60=900...
V=380...
Energy=6 * 900 * 380...
Energy=2,052kj
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_{}^{}olanrewaju(m) Says (11:00am On 3 Mar) (1364 Views)

Heavens lawrence:
2. Energy=IVt...
I=6...
t=15 * 60=900...
V=380...
Energy=6 * 900 * 380...
Energy=2,052kj
Nice one Heaven's, but the Workdone will be in Joules
Heavens lawrence:
P=IV...
WHERE I=V/R...
P=(V/R) * V...
P=(300/7) * 300....
P=42.86 * 300...
P=12,857.1WATT..
Keep it up...
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_{}^{}olanrewaju(m) Says (11:25am On 3 Mar) (1364 Views)

Morning Class, How was night ?, hope you all slept well ? and hope we are all preparing ahead of of Jamb Exam ?.
2017 Jamb registration begins today, in case you are not aware.
Let me go straight to the point, Today we are treating "Measurement of Electrical Power', we already explained what we meant by electrical Power, so let us solve some calculations under it.
Electrical power consumed by an electrical appliance is measured in watt. The unit of electrical energy is kilowatthour.
Note these formula's as you will use them.
P = IV.
W=IVT
Where P =Power(Watts)
I=Electricity(A)
V=Voltage(V)
T=Time(S)
The formula above is used when cost/money is not involved
The formula below is used when money is involved.
W=pt
where
P=(Kwh).
I=(A)
T(hr)
Note: To convert from watt to kilowatt you divide by 1000
So let us go the main business.
Example; A lamp is rated 8V, 16w. How many Joules does it consume in an hour?
Solution:
Data
V=8v
P=16W
T=1hour*60*60=3600s( Money is not involved)
W=IVT
Remeber P=IV
I=16/8
I=2A
W=2*8*3600
=57600Joules
Example; A household refrigerator is rated 240w. If electricity costs 5K per Kwh, what is the cost of operating it for 30days?
Solution:
We can see that the above question, money "5K" is involved.
Data
P=240w=240/1000=0.24kw
t=30days=24hour*30=720 hour
W=pt
W=0.24*720
=172.8kwh
Then if 5k = 1kwh
X= 172.8
X=172.8 *5
=864k
1) Find the cost of running five 50w lamps and three 100w lamps for 6hrs if electric energy costs #20 per unit.
P=50*5=250w=0.25kw
P=100*3=300w=0.3
P=0.25+0.3=0.55
T=6hr
W=pt
=0.55*6=3.3kwh
If 1kwh=#20
3.3=X
X=20*3.3=
Electrical Installation.
Electrical installation is the branch of physics that leads to electrical engineering. It deals with how two wires, one live and the other neutral can be connected in a building.
Live wires are usually insulated with red colour insulator, while neutral wires are insulated with black colour insulator
Classwork.
1. An electric lamp rated at 48w, 12v supply. Calculate the current flowing in the lamp and the resistance.
2. It takes 4minutes to boil a quantity of water electrically. How long will it take to boil same quantity of water using the same heating coil, but with the current doubled?
3. Calculate the amount of of heat generated in an external load of resistance 8 Ohm if an alternating current of peak value 5A is passed through it for 1min 40sec.
4. An electric heater takes 4A when operated on a 250V supply. What is the cost of the electricity consumed at 10k per kwh when the heater is used for 5hours?
5. A man has a 40w and two 60w bulbs in a room. How much will it cost him to keep them lit for 8hours if the cost of a unit is 20kobo?
@ youngscholar( m) @ brainiac @ Esix @ Eazi @Splendid baby @ raquel97 @HeavensLawrence
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_{}^{}femmy4j Says (01:46pm On 3 Mar) (1364 Views)

1. Since P = IV
p=48
V=12
I=?
solu.
I=48/12, =4A.
2. from reasoning since H=I²RT, for same heating effect and when current is doubled it T(2) becomes
I²(1)T(1)=I²(2)T(2), substituting T(1)=4 mins or 4×60s
I²(1)×4×60=(2I)²T(2)
T(2)=(I²(1)×4×60)/4×I²
T(2)=60s or 1mins
3.W=IVT
W=I²RT from (V=IR)
R=8ohm
I=5A
T=1mins 40s to 100s
substitute
W=(5)²×8×100
W=20000watt or 20kW
4.
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_{}^{}Youngscholar(m) Says (05:43pm On 3 Mar) (1364 Views)

kk
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_{}^{}olanrewaju(m) Says (01:14pm On 3 Mar) (1364 Views)

Nice one @ femmy4j
No 1.
P=48w
V=12v
I=?
I=P/v
=48/12
=4A
You were asked to find resistance also
Therefore V=IR
R=12/4=3ohms
No 2.
2minutes
No 3
Correct @ femmy4j
4.
I=4A
V=250v
t=5 hours
P=?
P=IV
=4*250=1000watt=1kw
W=pt
=1*5=5kwh
Then if 1kw=10k
5 = X
X=5*10=50k
5. A man has a 40w and two 60w bulbs in a room. How much will it cost him to keep them lit for 8hours if the cost of a unit is 20kobo?
P1=40w
P2=60*2=120w
P=40+120=160w=0.16kw
t=8 hours
W=pt
=0.16*8=1.28kwhr
Therefore If 1=20kobo
1.28=X
X=1.28*20
=25.6k
That is the solution class.
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_{}^{}olanrewaju(m) Says (06:01pm On 4 Apr) (1364 Views)

Evening Class...Long time...
I'm really sorry for not been active in this thread. Hope my apology is accepted ?.
That is why we need you to invite your friends to this thread..And if you can take us physics, feel free.
Today we will be treating Pressure..
PRESSURE
a) AtmosPheric Pressure
i) definition of atmosPheric Pressure
ii) units of Pressure
iii) measurement of Pressure
iv) simPle mercury barometer, aneroid barometer and manometer
v) variation of Pressure with height
vi) the use of barometer as an altmeter
b) Pressure in liquids
i) the relationshiP between Pressure, dePth and density (P = pgh)
ii) transmission of pressure in liquids ( pascal's principle)
iii) application.
Pressure.
Pressure is defined as the force acting at right angle, normal or perpendicular per unit surface area in contact with a substance.
Thus, the force applied on the surface area of a substance is called pressure. It is a scale quantity and exists in solid, liquid and gases.
Mathematically;
Pressure = Force/Area.
Example; The weight on the narrow heel of a girl's shoe is 250N and the surface area of the heel in contact with the floor is 50mm2. Determine the pressure exerted on the heel.
Solution;
F = 250N.c.
A = 50mm2 = 50 X 106m2.
Pressure = Force/ Area = 5 X 106N/m2.
Example; Calculate the pressure on the surface of a rectangular box of weight 100N if the base of this box has an area of 2m2.
Solution;
Weight (Force) = 100N.
Area = 2m2.
Pressure = Force/Area = 50N/m2.
Pressure in Liquid.
Consider a baker containing liquid at level h. The total force acting at the base of the beaker is called thrust. Thrust per unit area is called pressure of the liquid.
Mathematically;
Pressure = Force/Area.
Pressure = mg/A.
Recall that,
Density = mass/volume.
Therefore;
Pressure = density X volume X g /A.
Recall that;
Volume = Area X Height.
Therefore;
Pressure = density X height X acceleration due to gravity.
Example; A cylindrical jar of radius 7cm and height 25cm is filled with a liquid 0.80g/cm3. What is the pressure exerted at the bottom of the jar by the liquid?
Solution;
H = 25cm = 0.25m
r = 7cm
Density = 0.8g/cm3 = 8000kg/m3.
Pressure = density X height X g = 20,000N/m2.[/b]
Pascal's Law.
A scientist called Blaise Pascal propounded a law relating to pressure. According to him, Liquid transmits pressure equally in all direction. Pascal demonstrated this by putting water inside a rubber ball and ensured that the water was evenly distributed. He pressed the ball with equal pressure and water was seen out of the rubber ball at the same time. This shows that the increase in pressure or pressure applied equally on the same surface is equally distributed in all directions.
Atmospheric Pressure.
The surrounding environment is full of air blowing from one part to the other. The force due to the vertical column of air on a unit area of the surface gives the atmospheric pressure.
Fortin Barometer.
Fortin barometer is used for more accurate measurement of atmospheric pressure than the simple or Torricellian barometer.
It is made up of a tube containing mercury which is protected by enclosing it in a brass tube.
Aneroid Barometer.
Aneroid barometer has no liquid and is widely used in the home for showing weather changes. The essential part of the barometer is a flat cylindrical metal box or capsule, corrugated for strength, and sealed after having been partially exhausted of air. Increase in atmospheric pressure causes the box to cave in slightly, while a decrease allows it to expand.
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_{}^{}youngscholar(m) Says (12:51pm On 4 Apr) (1364 Views)

Nice one @ olanrewaju( m)..
Proceed Boss..
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_{}^{}olanrewaju(m) Says (02:27pm On 5 May) (1364 Views)

Afternoon Class.
It seems no one is following me..
If you want me to continue, can you please notify so I can know.
Thanks
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_{}^{}olanrewaju(m) Says (11:23am On 5 May) (1364 Views)

Hello
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