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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