Download year_11_revision_mechanics_electricity_notes

A MOTION AND FORCES
1 Speed is the rate of change of distance
speed = distance / time
distance = speed x time
m/s
or
ms-1
time = distance/speed
2 velocity is the rate of change of displacement
velocity = displacement / time
displacement = velocity x time
time = displacement / velocity
m/s
or
ms-1
2 acceleration is rate of change of velocity
Acceleration = change in velocity / time
a = v-u/t
v=u+at
u=v-at
m/s2 or ms-2
EXAMPLE 1
car speeds from 10m/s to 25m/s in 10s. its acceleration is:
change in velocity: 25m/s – 10m/s
time: 10s
a = v-u/t
a = 25-10/10
a=15/10
a=1.5m/s2
EXAMPLE 2
A ball falls from a height in 0.63s. what is its final velocity (v) is acceleration due to gravity is 9.8m/s2 :
v=u+at
v=0+9.8 x 0.63 v=6.174m/s (6.2m/s in 2 sig fig)
3 distance time graph
Distance-time graph features
Steep line
Flat line
higher speed
stopped at distance x metres
in time 0-15 minutes (A to B):
6km/15min = 0.4km/minute
OR
change minutes to seconds 15 min is 900s and 6km is 6000m
6000m/900s = 6.7m/s
Constant speed because the line is straight
Line A to B is steep so higher speed
In time 15min to 35min (B to C):
Stopped at distance 6km for 20minutes
in time 40-45 minutes (C to D):
moves from 6km to 8km, from 35 to 45 minutes
this is 8-6= 2km = 2000m
45-35=
10min = 600s
Speed is 2000m/600s = 3.33m/s
Line C to D is not steep compared to between A and B, so it is a lower speed
AREA UNDER LINE IS DISTANCE
4 velocity time graph
6 m/s
4 m/s
Find area of under the line using
triangles and rectangles
(½.3.4=6m) + ( ½.2.2=2m) +
(½.3.6=9m) + (5.4=20m) = 37m
3s
6s
8s
Average velocity = 37m/11s = 3.36m/s
The area under speed time graph is total distance travelled.
Steepness (gradient) of line shows acceleration
5 Force
FORCE CAN BE A PUSH PULL TWIST
THRUST (DRIVING FORCE OF ENGINE)
DRAG (AIR RESISTANCE)
FRICTION (force opposing sliding motion of tires)
WEIGHT (GRAVITY PULLS ON MASS OF CAR)
REACTION (ROAD PUSHES UP)
DEFINTION: A FORCE IS AN ACTION THAT ACCELERATES A MASS, CHANGE ITS DIRECTION OR SHAPE
IT CAN BE A PUSH PULL TWIST
TYPES OF FORCES: GRAVITATION, MAGNETIC, ELECTROSTATIC, NUCLEAR FORCE
Weight of gravity on a mass is weight
Weight = mass x acceleration due to gravity
Weight is measured in NEWTON
Gravity on a mass (WEIGHT) makes object move slowly at first, then faster and faster (accelerate)
As object fall through air AIR RESISTANCE increases
OBJECT IN FREE FALL REACH TERMINAL VELOCITY WHEN FORCES ARE BALANCED (DRAG = WEIGHT)
OBJECTS IN FREE FALL ACCELERATE IF DRAG IS LESS THAN WEIGHT
OBJECTS IN FREE FALL DECELERATE IF DRAG IS GREATER THAN WEIGHT
EXAMPLE 1
An object has mass of 3kg. what is its weight? (assume g=10m/s2)
W=mg
W=3kg x 10m/s2
W=30N
EXAMPLE 2
What is the mass of object if its weight is 600N (assume g= 9.8 m/s2)?
W=mg
m=W/g
m=600N/9.8
m=61.2kg
EXAMPLE 3
What is the unbalanced force on the box mass 12kg, and which way is it directed?
The unbalanced force is 60-20 = 40N towards east
The acceleration is found using F=ma
m=F/a
m=40N/12kg
m=3.3ms-2
PARACHUTIST FALLING IN FREE FALL
EXPLAIN WHAT IS HAPPENING HERE AT EACH STAGE
THE ANSWER IS ON THE GRAPH AFTER THE ELECTRICITY SECTION
ELECTRICITY AND SAFETY
Electric shock can be fatal and the heat from exposed wires with current can cause fires.
Fuse symbol (fuses/breaks circuit, no more current flows)
Plug:
blue neutral wire (left pin)
brown live current providing wire connected to fuse (right pin)
yellow and green wire EARTH wire to take current away in case of fault
Fuse breaks a circuit when a current greater than fuse
rating flows through it. This prevents fatal shock if
appliance is LIVE due to a fault.
Hair driers, electric shavers, blenders, juicers, drills and
other appliances used near water are double insulated.
They have either plastic casing or the design does not
allow live wires to touch the outside casing.
OHM’S LAW
Ohms law: current is directly proportional to PD when temperature is constant
Straight line through origin
OHM’S LAW QUESTION
The current in a component with 20Ω resistance if the
voltage 230V?
What is a suitable resistor to use: 2A, 5A or 13A?
E energy in joules J
Use SI units
P Power in Watts W
1kJ = 1000J
t time in seconds s
1MJ = 106 J
1kW = 1000W
Q charge in coulombs C
1 hour = 3600s
I current in time in amps A
t time in seconds s
QUESTIONS
WHAT IS THE ENERGY USED BY AN ELECTRIC IRON (POWER 3000W) USED FOR 60 SECONDS?
WHAT IS THE ENERGY USED BY AN ELECTRIC IRON (POWER 2.5KW) USED FOR 1.5 MINUTE?
AN ELECTRIC IRON HAS 11.5A CURRENT FLOWING THROUGH IT FOR 60 SECONDS. CALCULATE THE
CHARGE THAT FLOW THROUGH THE IRON.
Connecting voltmeter and ammeter
Ammeter is placed in series with
components
Voltmeter is placed in parallel
(across the component) to
component where P.D. is to be
measured.
CURRENT IN A CIRCUIT 1
The two resistors have less current flowing through
The single resistor has higher current (0.3A) as it is shown in the diagram. This means the current in A1
will be 0.4A - 0.3A = 0.1A
CURRENT IN A CIRCUIT 2
The single bulb
is brighter, it
has same PD
but higher
current.
The two bulbs in
parallel are equally
bright because
both bulbs are
connected directly
to the cell
Predict which bulb/s are brighter.
Hint: find out which bulb/s have less
current flowing through them?
In the first 30s the parachutist is ACCELERATING (the velocity is increasing)
In 30s – 40s parachutist is in uniform or TERMINAL VELOCITY (constant velocity)
In 40s-50s parachutist is DECELERATING from 50m/s to 10m/s (velocity is decreasing)
After 50s, parachutist has reached a lower TERMINAL VELOCITY