PLAYWORK †WORKING WITH CHILDREN - 1935 Words

Assignment 310 – Understand Children and Young People’s Self-Directed Play Task A – Briefing Notes 1. Explain the characteristics of freely chosen, self directed play Freely chosen play is where children are able to choose what they want to do Personally directed play is where children will choose how they want to do what they have chosen to play. Intrinsically motivated is why children choose why they choose certain types of play. Goalless means that children will play for no particular reason in terms of where goals or rewards are concerned. 2 . Explain the importance of observing and analysing children’s and young people’s play. Through observing a child, it helps you to understand and be aware of the child’s interests†¦show more content†¦7- Explain the concept of acceptable and unacceptable risk in the context of different play types. The concept of acceptable and unacceptable risk are, a child learning to jump higher and higher from a tree, this is an acceptable risk as the child is learning what their limitations and boundaries are. Whereas an unacceptable risk would be two children sword fighting as normally this gets out of hand and can turn serious. 8. Evaluate different approaches to managing risk during children and young peoples play. An approach to managing risk could be to talk to those children involved with e.g. play fighting and if they got too carried away I would stop them from playing together. Another approach would be to provide crash mats if children are playing on an indoor climbing frame, I would place them underneath to steady their fall so they don’t injure themselves too badly. 9. Explain the value of enabling children and young people to manage risk for themselves. The value of enabling children and young people to manage risk for themselves is to get the child to push their own limits i.e. if they were playing on a climbing frame, they would know how high to climb or jump from the climbing frame. 10. Explain how play work organisations seek to balance the health, safety and security of the play environment with children and young people’s need for stimulation, risk and challenge. A play work provision

Physics Study Notes Hsc Free Essays

string(121) " the velocity is zero and the potential energy is zero because this is where we selected the zero of potential energy\)\." Physics Summary Contents Core Topic One: Space 1. 2. 3. We will write a custom essay sample on Physics Study Notes Hsc or any similar topic only for you Order Now 4. Gravity Space Launch and Return Future Space Travel Special Relativity 2 3 13 14 Page Core Topic Two: Motors and Generators 1. 2. 3. 4. 5. The Motor Effect Electromagnetic Induction Electric Generators Transformers Electric Motors 19 24 27 29 31 Core Topic Three: From Ideas to Implementation 1. 2. 3. 4. Cathode Rays Quantum Theory Solid State Devices Superconductivity 32 37 43 48 Option Topic: Quanta to Quarks 1. . 3. 4. 5. 6. Models of the Atom Quantum Physics The Electron Microscope Applications of Radioactivity Nuclear Applications The Structure of Matter 53 57 59 61 66 67 William Kim HSC Physics Summary | page 1 Core Topic One: Space 1. The Earth has a gravitational field that exerts a force on objects both on it and around it  § Define weight as the force on an object due to a gravitational field The weight of an object is the force of gravity acting on it. r W = mg Where W is the weight in newtons (N), m is the mass in kilograms (kg) and g can be either: 1. The accelerati on due to gravity (= 9. 8 m/s/s at the Earth’s surface); or 2. The gravitational field strength (= 9. 8 N/kg at the Earth’s surface).  § Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field. As we lift an object from the ground to a height above the ground we do work on it. This work is stored in the object as gravitational potential energy. For an object of mass m at a height h above the Earth’s surface the gravitational potential energy E is given by: E p = mgh However this equation is valid only when the object is near the Earth’s surface. The gravitational potential energy is a measure of the work done in moving an object from infinity to a point in the field. The general expression for the gravitational potential energy of an object of mass m at a distance r from the centre of the Earth (or other planet) is given by: E p = ? G mM E r Newton’s Law of Universal Gravitation m F = G 12 2 r where G is the universal gravitational constant. The Gravitational Field Surrounding any object with mass is a gravitational field. g= Gm r2 Where M is the mass of the Earth (or other planet). Change in Gravitational Potential Energy The change in potential energy of a mass m1 as it moves from infinity to a distance r from a source of a gravitational field (due to a mass m2) is given by: mm ? E p = G 1 2 r Change in Gravitational Potential Energy Near the Earth (when radius increases from A to B) ?1 1? ?E p = GmM E ? ? ? ?r ? ? A rB ? William Kim HSC Physics Summary | page 2 . Many factors have to be taken into account to achieve a successful rocket launch, maintain a stable orbit and return to Earth  § Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components Any moving object that moves only under the force of gravity is a projectile. The horizontal motion of a projectile is independent to the vertical motion. The reason for this result is that gravity is the only force acting on the objects and this always acts towards the centre of the Earth. Projectile motion can be analysed by realising that: 1. The horizontal motion is constant velocity. 2. The vertical motion of constant acceleration (with acceleration of g). Equations of Uniformly Accelerated Motion r r r v = u + at r r 1r s = ut + at 2 2 2 2 v = u + 2 as The Path of a Projectile The velocity at any point of the path of a projectile is simply the vector sum of the horizontal and vertical velocity components at that point. ?y = k (? x ) 2 ? ag ? k =? 2 ? ? 2u ? ? x? The horizontal component is constant. The vertical component changes at g, the acceleration due to gravity. Trajectories The path followed by a projectile – its trajectory – is a parabola (or linear) (1) Horizontal motion: ? x = u x t 1 (2) Vertical motion: ? y = a g t 2 2 From (1): ? x t= ux Combining (2) (3): 1 ? ?x ? 1 ag ? y = a g ? ? = (? x )2 2 ? u ? 2 ? x? 2 ux 2 ux ? y ? x William Kim HSC Physics Summary | page 3  § Describe Galileo’s analysis of projectile motion Galileo was responsible for deducing the parabolic shape of the trajectory of a projectile. Galileo’s analysis of projectile motion led him to consider reference frames. These are what all measurements are compared to. The concept of Galilean relativity refers that the laws of mechanics are the same in a frame of reference that is at rest or one that moves with constant velocity.  § Explain the concept of escape velocity in terms of the: o gravitational constant o mass and radius of the planet If an object is projected upward with a large enough velocity it can escape the gravitational pull of the Earth (or other planet) and go into space. The necessary velocity to leave the Earth (or other planet) is called the escape velocity. Escape velocity depends on the gravitational constant, the mass and radius of the planet. Suppose an object of mass m is projected vertically upward from the Earth’s surface (mass of M and radius R) with an initial velocity u. The initial mechanical energy, that is, kinetic and potential energy is given by: E k i + E pi = 1 M m mu 2 ? G E 2 RE Let us assume that the initial speed is just enough so that the object reaches infinity with zero velocity. The value of the initial velocity for which this occurs is the escape velocity ve . When the object is at infinity the mechanical energy is zero (the kinetic energy is zero since the velocity is zero and the potential energy is zero because this is where we selected the zero of potential energy). You read "Physics Study Notes Hsc" in category "Essay examples" Hence 1 M m mve2 ? G E = 0 which leads to: 2 RE ve = 2GM E RE William Kim HSC Physics Summary | page 4  § Discuss Newton’s analysis of escape velocity Circular Motion The motion of an object in a circular path with constant speed is called uniform circular motion. Although the speed remains the same in uniform circular motion, it follows that an object travelling in a circular path must be accelerating, since the velocity (that is, the speed in a given direction) is continually changing. r r v1 ? v2 v1 = v 2 r v2 r The change in velocity is given by: v2 r r r r v1 ? v = v 2 ? v1 r ? v1 r r ? v r and since: a = ? v r ? t ? v it follows that the object is accelerating. Isaac Newton proposed the idea of artificial satellites of the Earth. He considered how a projectile could be launched horizontally from the top of a high mountain so that it would not fall to Earth. As the launch velocity was increased, the distance that the object would travel before hitting the Earth would increase until such a time that the velocity would be sufficient to put the object into orbit around the Earth. (A higher velocity would lead to the object escaping from the Earth. ) Centripetal Acceleration As can be seen, when the change in velocity is placed in the average position between v1 and v2, it is directed towards the centre of the circle. When an object is moving with uniform circular motion, the acceleration (the centripetal acceleration) is directed towards the centre of the circle. For an object moving in a circle of radius r with an orbital velocity of v, the centripetal acceleration a is given by: v2 ac = r Earth Orbits A satellite can be put into Earth orbit by lifting it to a sufficient height and then giving it the required horizontal velocity so that it does not fall back to Earth. For the satellite to circle the Earth, the centripetal force required is provided by the gravitational attraction between the satellite and the Earth. Hence the centripetal acceleration is given by: v2 g= R William Kim HSC Physics Summary | page 5  § Use the term ‘g forces’ to explain the forces acting on an astronaut during launch g-forces on Astronauts Humans can withstand 4g without undue concern. Accelerations up to ~10g are tolerable for short times when the acceleration is directed parallel to a line drawn between the person’s front and back. The human body is relatively unaffected by high speeds. Changes in speed, however, that is, accelerations, can and do affect the human body creating ‘acceleration stress’. g-forces Acceleration forces – g-forces – are measured in units of gravitational acceleration g. For example, a force of 5g is equivalent to acceleration five times the acceleration due to gravity. If the accelerations are along the body’s long axis then two distinct effects are possible: 1. If the acceleration is in the direction of the person’s head they may experience a ‘black out’ as the blood rushes to their feet; or 2. If the acceleration is towards their feet, they may experience a ‘red out’ where the blood rushes to their head and retina.  § Compare the forces acting on an astronaut during launch with what happens during a roller coaster ride As you ‘fall’ from a height, you experience negative g-forces (you feel lighter). When you ‘pull out’ of a dip after a hill or follow an ‘inside loop’, you experience positive g-forces (you feel heavier). The positive g-forces are like those astronauts experience at lift-off. Consider a rider in a car at the bottom of an inside loop. The rider has two forces acting on them: 1. Their normal weight (mg) acting down; and 2. The ‘normal reaction force’ (N) acting up. This is the push of the seat upwards on their bottom. Assume that the loop is part of a circle of radius R. A centripetal force is required for the rider to travel in a circle. This is the difference between the normal force and the weight force, that is: mv 2 mv 2 N ? mg = : N = mg + R R The g-forces are found from the ‘normal force’ divided by the weight. That is: N = mg mg + mv 2 2 R = 1+ v mg gR N mg g’s felt by rider = William Kim HSC Physics Summary | page 6  § Discuss the impact of the Earth’s orbital motion and its rotational motion on the launch of a rocket A moving platform offers a boost to the velocity of a projectile launched from it, if launched in the direction of motion of the platform. This principle is used in the launch of a rocket by considering that the Earth revolves around the Sun at 107,000km/h relative to the Sun and rotates once on its axis per day so that a point on the Equator has a rotational velocity of approximately 1,700km/h relative to the Sun. Hence, the Earth is itself a moving platform with two different motions which can be exploited in a rocket launch to gain a boost in velocity. Earth Orbit A rocket heading into orbit is launched to the east to receive a velocity boost from the Earth’s rotational motion. An Interplanetary Trip The flight of a rocket heading into space is timed so that it can head out in the direction of the Earth’s motion and thereby receive an extra boost.  § Analyse the changing acceleration of a rocket during launch in terms of the: Law of Conservation of Momentum Forces experienced by astronauts Law of Conservation of Momentum Rocket engines generate thrust by burning fuel and expelling the resulting gases. Conservation of momentum means that as the gases move one way, the rocket moves the other. (Momentum before the burning is zero; hence the momentum after is also zero. The gases carry momentum in one direction down, and so the rocket carries an equal momentum in the opposite direction up. ) As fuel is consumed and the gases expelled, the mass of the system decreases. Since acceleration is proportional to the thrust and inversely proportional to the mass, as the mass decreases, the acceleration increases. Hence the forces on the astronauts increase. Forces Experienced by Astronauts g forces varied during the launch of Saturn V, a large three-stage rocket used to launch the Apollo spacecraft. This is attributed to the sequential shutdown of the multiple rocket engines of each stage – a technique designed specifically to avoid extreme g forces. William Kim HSC Physics Summary | page 7  § Analyse the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth Motion Whirling rock on a string Electron orbiting atomic nucleus Car cornering Moon revolving around Earth Satellite revolving around Earth Fc Provided By†¦ The string Electron-nucleus electrical attraction Friction between tyres and road Moon-Earth gravitational attraction Satellite-Earth gravitational attraction  § Compare qualitatively and quantitatively low Earth and geostationary orbits Low Earth Orbit A low Earth orbit is generally an orbit higher than approximately 250 km, in order to avoid atmospheric drag, and lower than approximately 1000 km, which is the altitude at which the Van Allen radiation belts start to appear. The space shuttle utilises a low Earth orbit somewhere between 250 km and 400 km depending upon the mission. At 250 km, an orbiting spacecraft has a velocity of 27,900km/h and takes just 90 minutes to complete an orbit of the Earth. Geostationary Orbit A geostationary orbit is at an altitude at which the period of the orbit precisely matches that of the Earth. If over the Equator, such an orbit would allow a satellite to remain ‘parked’ over a fixed point on the surface of the Earth throughout the day and night. From the Earth such a satellite appears to be stationary in the sky, always located in the same direction regardless of the time of day. This is particularly useful for communications satellites because a receiving dish need only point to a fixed spot In the sky in order to remain in contact with the satellite. The altitude of such an orbit is approximately 38,800 km. If a satellite at this height is not positioned over the Equator but at some other latitude, it will not remain fixed at one point in the sky. Instead, from the Earth the satellite will appear to trace out a ‘figure of eight’ path each 24 hours. It still has a period equal to the Earth’s, however, and so this orbit is referred to as geosynchronous. William Kim HSC Physics Summary | page 8  § Discuss the important of Newton’s Law of Universal Gravitation in understanding and calculating the motion of satellites Using Newton’s Law of Universal Gravitation combined with the expression for centripetal force, we can see that the orbital velocity required for a particular orbit depends nly on the mass of the Earth, the radius of the Earth and the altitude of the orbit (distance from the surface of the Earth). Given that the mass and radius of the Earth have fixed values, this means that altitude is the only variable that determines the specific velocity required. In addition, the greater the radius of the orbit, the lower the orbital velocity required. Once a launched rocket has achieved a suff icient altitude above the surface of the Earth, it can be accelerated into an orbit. It must attain a specific speed that is dependent only upon the mass and radius of the Earth and the altitude above it. If that speed is not reached, the spacecraft will spiral back in until it re-enters the atmosphere; if the speed is exceeded, it will spiral out. This can be considered by appreciating that the simplest orbital motion is a uniform speed along a circular path around the Earth. Uniform circular motion, as already mentioned, is a circular motion with a uniform orbital velocity. According to Newton’s First Law of Motion, a spacecraft in orbit around the Earth, or any object in circular motion, requires some force to keep it there, otherwise it would fly off at a tangent to the circle. This force is directed back towards the centre of the circle. In the case of spacecraft, it is the gravitational attraction between the Earth and the spacecraft that acts to maintain the circular motion that is the orbit. The force required to maintain circular motion, known as centripetal force, can be determined using the following equation: mv 2 FC = r The application of Newton’s Law of Universal Gravitation to the orbital motion of a satellite will produce an expression for the critical orbital velocity mentioned earlier. Recall that this law states that the gravitational attraction between a satellite and the Earth would be given by the following expression: m m FG = G E 2 S r This gravitational force of attraction also serves as the centripetal force for the circular orbital motion, hence: FG = FC Therefore, we can equate the formula for FG with that for FC: m E mS m S v 2 G = r r2 ? v = GmE r where v = orbital velocity (ms-1) where r = rE + altitude (m) William Kim HSC Physics Summary | page 9 Kepler’s Third Law: The Law of Periods Further, we can use the expression for orbital velocity to prove Kepler’s Third Law – the Law of Periods. The period or the time taken to complete one full orbit can be found by dividing the length of the orbit (the circumference of the circle) by the orbital velocity, v. 2? r T= v Changing the subject of this expression to v and then substituting into the formula for v given above: 2? r Gm E = T r 3 Gm E r ? 2 = T 4? 2 This means that for any satellite of the Earth at any altitude, the ratio r3:T2 always equals the same fixed value. William Kim HSC Physics Summary | page 10  § Describe how a slingshot effect is provided by planets for space probes Many of today’s space probes to distant planets such as Jupiter use a gravitational ‘slingshot’ effect (also known as a gravity-assist trajectory) that brings the probe close to other planets to increase the probe’s velocity. In 1974, Mariner 10 was directed past Venus on its way to Mercury. The Pioneer and Voyager probes also used this method. Consider a trip to Jupiter such as the Galileo probe that involved a single fly-by of Venus and two of the Earth. As the probe approaches Venus, it is accelerated by Venus’ gravitational attraction, causing it to speed up relative to Venus. By Newton’s Third Law, Venus will also experience a force slowing it down. It’s mass, however, is so much greater than that of the probe that the velocity decrease is imperceptible. ) As the probe passes Venus, its speed is reduced (relative to Venus). Relative to the Sun, however, its speed has increased. The probe picks up angular momentum from the planet (w hich loses an equal amount of an angular momentum). Gravity allows the ‘coupling’ between the probe and planet to facilitate the transfer. For this reason, gravity-assist trajectories should more correctly be called angular momentum-assist trajectories. f = vi + 2Vi Planet vi Spacecraft Vi Vf  § Account for the orbital decay of satellites in low Earth orbit All satellites in low Earth orbit are subject to some degree of atmospheric drag that will eventually decay their orbit and limit their lifetimes. As a satellite slows, it loses altitude and begins a slow spiral downwards. As it descends, it encounters higher density air and higher drag, speeding up the process. By the time the satellite is below an altitude of 200 km it has only a few hours left before colliding with the Earth. The re-entry process generates much heat and most satellites burn up (vaporise) before impacting. William Kim HSC Physics Summary | page 11  § Discuss issues associated with safe re-entry into the Earth’s atmosphere and landing on the Earth’s surface There are significant technical difficulties involved in safe re-entry, the most important being: 1. The heat generated as the spacecraft contacts the Earth’s atmosphere; and 2. Keeping the retarding-forces (g-forces) within safe limits for humans. Heating Effects The Earth’s atmosphere provides aerodynamic drag on the spacecraft and as a result high temperatures are generated by friction with air molecules. Identify that there is an optimum angle for re-entry into the Earth’s atmosphere and the consequences of failing to achieve this angle. g-Forces The angle of re-entry is critical: too shallow and the spacecraft will bounce off the atmosphere back into space; too steep and the g-forces will be too great for the crew to survive (and the temperatures generated with the atmosphere will be too high even for the refracting materials used). The ‘allowed’ angle of re-entry is –6. 2 °  ± 1 ° relative to the Earth’s horizon. William Kim HSC Physics Summary | page 12 3. Future space travel and exploration will entail a combination of new technologies based on current and emerging knowledge  § Discuss the limitation of current maximum velocities being too slow for extended space travel to be viable Scientists have not yet been able to produce speeds of spacecraft more than a few tens of thousands of kilometres per hour. When travelling to distant planetary objects, the engines of spacecraft are not on as spacecraft rely on inertia to move along. To increase the speed significantly would require the engines to be operating, which would require more fuel. More fuel would require more thrust putting the spacecraft into orbit, which would require more fuel and so on. To increase the speed of spacecraft to values that would make interplanetary travel feasible requires a whole new technology (one not based on the emission of gases produced by combustion). Clearly, while current maximum velocities are just adequate for interplanetary travel, they are entirely inadequate for interstellar travel.  § Describe difficulties associated with effective and reliable communications between satellites and earth caused by: – distance – van Allen radiation belts – sunspot activity Distance Microwaves and radio waves, like all EM waves, travel through space at the speed of light. This is the fastest speed possible in our universe and therefore places a limit on the speed and response time of space communications. The immense distance involved in space communications creates a distance-related time lag. Also, as EM radiation obeys an inverse square law, there is a loss of signal strength as distance increases. This is referred to as space loss. Van Allen radiation belts There are two belts of energetic charged particles, mainly electrons and protons, lying at right angles to the equator of the Earth. Some of the solar wind particles become trapped in the Van Allen radiation belts. Intense solar activity can disrupt the Van Allen Belts. This in turn is associated with auroras and magnetic storms. The charged particles drifting around the Earth in the outer belt corresponds to an electric current and hence has an associated magnetic field. Once or twice a month this current increases and as a result its magnetic field increases. This can lead to interference of short wave radio communication, errors in communication satellites and even failure of electrical transmission lines. Sunspot activity Sunspots are associated with the solar wind (consisting of a stream of charged particles). The solar wind affects the Earth’s magnetic field and this in turn affects radio communication. William Kim HSC Physics Summary | page 13 4. Current and emerging understanding about time and space has been dependent upon earlier models of the transmission of light  § Outline the features of the ether model for the transmission of light It was believed that light waves require a medium to propagate. Although nobody could find such a medium, belief in its existence was so strong that it was given a name – the ether. The ether: – Filled all of space, had low density and was perfectly transparent – Permeated all matter and yet was completely permeable to material objects – Had great elasticity to support and propagate the light waves  § Describe and evaluate the Michelson-Morley attempt to measure the relative velocity of the Earth through the ether The Ether Wind Because the Earth was moving around the Sun, it was reasoned that an ether wind should be blowing past the Earth. However, if a wind blows, the speed of sound relative to the stationary observer would vary. Thus it was believed that the speed of light should vary due to the presence of the â€Å"ether wind†. It was in an attempt to detect this difference that Michelson and Morley did their famous experiment. The Michelson-Morley Experiment Light sent from S is split into two perpendicular beams by the half-silvered mirror at A. These two beams are then reflected back by the mirrors M1 and M2 and are recombined in the observer’s eye. An interference pattern results from these two beams. The beam AM1 travelled across the ether, whilst AM2 travelled with and against the ether. The times to do this can be shown to be different and so introduce a phase difference between the beams. When the entire apparatus was rotated through 90 °, a change in the interference pattern was expected. None was observed. The result of the Michelson-Morley experiment was that no motion of the Earth relative to the ether was detectable. M1 A S M2 Ether Wind  § Discuss the role of critical experiments in science, such as Michelson-Morley’s, in making determinations about competing theories From a hypothesis, predictions are made of what should happen if a particular experiment is performed. If the results are not in agreement with the prediction, the hypothesis is incorrect. As we have seen, the fact that a null result was found from this experiment showed the ether hypothesis to be invalid. This opened up a completely revolutionary view of space and time with the work of Einstein. William Kim HSC Physics Summary | page 14  § Outline the nature of inertial frames of reference Frames of Reference Frames of reference are objects or coordinate systems with respect to which we take measurements. Position In maths, the Cartesian coordinate system is used and position is referred to the axes x, y and z. In experiments in class, the laboratory is the frame of reference. S S’ r u r P v Velocity An object P travels with velocity v with respect to a reference frame S. Another frame S’ moves with velocity u relative to S. The velocity of P relative to S’ is v’ = v – u. Velocity thus depends upon the reference frame. Inertial Frames of Reference An inertial frame of reference is one that is moving with constant velocity or is at rest (the two conditions being indistinguishable). In such reference frames, Newton’s Law of Inertia holds. A non-inertial frame of reference is one that is accelerating.  § Discuss the principle of relativity Three hundred years before Einstein, Galileo posed a simple idea, now called the principle of relativity, which states that all steady motion is relative and cannot be detected without reference to an outside point. This idea can be found built into Newton’s First Law of Motion as well. Two points to be reinforced:  § The principle of relativity applies only for non-accelerated steady motion  § This principle states that within an inertial frame of reference you cannot perform any mechanical experiment or observation that would reveal to you whether you were moving with uniform velocity or standing still. William Kim HSC Physics Summary | page 15  § Identify the significance of Einstein’s assumption of the constancy of the speed of light In 1905, Albert Einstein proposed that the speed of light is constant and is independent of the speed of the source or the observer. This premise explained the ‘negative’ result of the MichelsonMorley experiment and showed that the ether concept was not needed. As a consequence of this ‘law of light’ it can be shown that there is no such thing as an absolute frame of reference. All inertial reference frames are equivalent. That is, all motion is relative. The laws of physics are the same in all frames of reference; that is, the principle of relativity always holds.  § Recognise that if c is constant then space and time become relative In Newtonian physics, distance and velocity can be relative terms, but time is an absolute and fundamental quantity. Einstein radically altered the assumptions of Newtonian physics so that now the speed of light is absolute, and space and time are both relative quantities that depend upon the motion of the observer. (Our reality is what we measure it to be. Reality and observation cannot be separated. Remember this as we proceed).  § Discuss the concept that length standards are defined in terms of time with reference to the original meter †¦In other words, the measured length of an object and the time taken by an event depend entirely upon the velocity of the observer. (This is why our current standard of length is defined in terms of time – the metre is the distance travelled by light in a vacuum in the fraction 1/299792458 of a second).  § Identify the usefulness of discussing space/time, rather than simple space †¦Further to this, since neither space nor time is absolute, the theory of relativity has replaced them with the concept of a space-time continuum. Space and time, not just space, are relative quantities).  § Account for the need, when considering space/time, to define events using four dimensions †¦Any event then has four dimensions (three space coordinates plus a time coordinate) that fully define its position within its frame of reference. William Kim HSC Physics Summary | page 16  § Explain qualitatively and quantitatively the consequence of special relativity in relation to: The relativity of simultaneity The equivalence between mass and energy Length contraction Time dilation The Relativity of Simultaneity (simultaneity and the velocity of light) Observers in relative motion will disagree on the simultaneity of events separated in space. The Equivalence Between Mass and Energy The mass of a ‘moving’ object is greater than when it is ‘stationary’ – it experiences mass dilation (covered later). Since c is the maximum speed in the universe it follows that a steady force applied to an object cannot continue to accelerate. It follows that the inertia, that is the resistance to acceleration, must increase. But inertia is a measure of mass and so the mass has increased. It is this increase in mass that prevents any object from exceeding the speed of light, because as it accelerates to higher velocities its mass increases, which means that further accelerations will require even greater force. This is further complicated by time dilation because, as speeds increase to near light speed, any applied force has less and less time in which to act. The combined effect is that as mass becomes infinite and time dilates, an infinite force would be required to achieve any acceleration at all. Sufficient force can never be supplied to accelerate beyond the speed of light. If force is applied to an object, then work is done on it – energy is given to the object. This energy would take the form of increased kinetic energy as the object speeds up. But at near light speed the object does not speed up. The applied force is giving energy to the object and the object does not acquire the kinetic energy we would expect. Instead, it acquires extra mass. Einstein made an inference here and stated that the mass (or inertia) of the object contained the extra energy. Relativity results in a new definition of energy as follows: E = E k + mc 2 where E = total energy, Ek = kinetic energy, m = mass, c = speed of light When an object is stationary, it has no kinetic energy, but still has some energy due to its mass. This is called its mass energy or rest energy and is given by: 8 -1 E = mc 2 where E = rest energy (J), m = mass (kg), c = speed of light (3 x 10 m s ) William Kim HSC Physics Summary | page 17 Implications of Special Relativity: To measure speed we need to measure distance and time. If c remains constant, then it follows that distance (length) and time must change. Space and time are relative concepts. Length Contraction (the Lorentz-FitzGerald Contraction) The length of a ‘moving’ rod appears to contract in the direction of motion relative to a ‘stationary’ observer. l = l0 1 ? v2 c2 where l is the moving length, l0 is the ‘rest’ length (that is, the length as measured by an observer at rest with respect to the rod) and v is the speed of the rod. Time Dilation Time in a ‘moving’ frame appears to go slower relative to a ‘stationary’ observer t= t0 1? v c2 2 where t is the observed time for a ‘stationary’ observer and t0 is the time for an observer travelling in the frame. 0 is called the proper time (this is the time measured by an observer present at the same location as the events that indicate the start and end of an event). Mass Dilation The mass of a ‘moving’ object is greater than when it is ‘stationary’. m= m0 1? v2 c2 where m is the m ass for a ‘moving’ object and m0 is the mass for that object when it is ‘stationary. ’  § Discuss the implications of time dilation and length contraction for space travel The relativity of time allows for space travel into the future but not into the past. When travelling at relativistic speeds (0. 1c or faster), relativity influences the time that passes on the spacecraft. Astronauts on a relativistic interstellar journey would find their trip has taken fewer years than observed on Earth. William Kim HSC Physics Summary | page 18 Core Topic Two: Motors and Generators 1. Motors use the effect of forces on current-carrying conductors in magnetic fields  § Identify that moving charged particles in a magnetic field experience a force Charged particles moving in an external magnetic field will experience a force. If the moving charged particles are flowing through, and confined within, a conductor that is in an external magnetic field, the conductor will also experience a force. This effect is known as the motor effect. F = qvB Use left hand â€Å"FBI gun† An example: Van Allen Radiation Belts The Earth’s magnetic field captures charged particles from the solar wind (low energy) and cosmic rays (high energy). The charges are force to spiral along the field lines accumulating into two doughnut-shaped belts of â€Å"radiation† called the upper and lower Van Allen radiation belts. William Kim HSC Physics Summary | page 19  § Discuss the effect, on the magnitude of the force on a current-carrying conductor, of variations in: The strength of the magnetic field in which it is located The magnitude of the current in the conductor The length of the conductor in the external magnetic field The angle between the direction of the external magnetic field and the direction of the length of the conductor †¦The force is proportional to the magnetic field strength, B †¦The force is proportional to the current, I †¦The force is proportional to the length, L The force is at a maximum when the conductor is at right angles to the field, and is zero when the conductor is parallel to the field. The magnitude of the force is proportional to the component of the field that is at right angles to the conductor. F = BIl sin ? William Kim HSC Physics Summary | page 20  § Describe qualitatively and quantitatively the force on long parallel current-carrying conductors : Ampere’s Law Two parallel wires, each carrying a current, will exert a force on the other. This happens because each current produces a magnetic field (as in Oersted’s experiment). Therefore each wire finds itself carrying a current across the magnetic field produced by the other wire and hence experiences a force. Determining the magnitude of the force between two parallel conductors The magnetic field strength at a distance, d, from a long straight conductor carrying a current, I, can be found using the formula: kI B= d -7 -2 where k = 2. 0 x 10 N A The magnitude of the force experienced by a length, l, of a conductor due to to an external magnetic field is: F = I 2 lB or ? kI ? F = I 2l ? 1 ? ? d ? rearranged F II =k 1 2 l d F II =k 1 2 l d (Ampere’s law) I1 I2 If currents are in the same direction, then the conductors will attract. If currents are in opposite directions, then the conductors will repel. William Kim HSC Physics Summary | page 21  § Define torque as the turning moment of a force using: Torque is turning force. Its’ units are Newton-metres (Nm). ? = Fd where ? = torque, in Nm F = force, in N D = distance from rotational axis, in m F = BIl ? =BIld Rotational axis d ? = Fd  § Identify the forces experienced by a currentcarrying loop in a magnetic field and describe the net result of the forces b max ? = nBIA zero ? b Current Loops N I F = BIl ? =BIld ? = ? 1 + ? 2 = BIld + BIld = 2 BIld = BI (l ? 2d ) = BIA S ? = nBIA cos ? ? b (For each turn of the loop) Generally, ? =nBIA cos ? William Kim HSC Physics Summary | page 22  § Account for the motor effect due to the force acting on a current-carrying conductor in a magnetic field The motor effect Recall that charged particles moving in an external magnetic field will experience a force. If the moving charged particles are flowing through, and confined within, a conductor that is in an external magnetic field, the conductor will also experience a force. An electric motor is a device that transforms electrical potential energy into rotational kinetic energy.  § Describe the main features of a DC electric motor  § Discuss the importance of the invention of the commutator for developing electric motors  § Describe the role of the metal split ring and the brushes in the operation of the commutator Anatomy of a DC motor – Permanent magnets: provide an external magnetic field in which the coil rotates. As the magnets are fixed, they are known as the stator. – Rotating coil: carries a direct current that interacts with the magnetic field, producing torque. Armature: is made of ferromagnetic material and allows the coil to rotate freely on an axle. The armature and coil together are known as the rotor. The armature protrudes from the motor casing, enabling the movement of the coil to be used to do work. – Commutators: reverse the current of the coil every half turn to maintain consistent direction and torque. It i s a mechanical switch that automatically changes the direction of the current flowing through the coil when the torque falls to zero. – Brushes: maintain electrical contact of coils with the rest of the circuit. The development of DC motors outstripped that of AC motors and generators for two reasons: – Voltaic batteries could supply power – They could use powerful electromagnets that were far stronger than permanent magnets The development of the commutator was important because it led to the development of modern electric motors and generators. It enabled motors to provide steady circular motion of a drive shaft.  § Describe how the required magnetic fields can be produced either by currentcarrying coils or permanent magnets The magnetic field of a DC motor can be provided either by permanent magnets or by electromagnets. William Kim HSC Physics Summary | page 23 2. The relative motion between a conductor and magnetic field is used to generate an electrical voltage  § Outline Michael Faraday’s discovery of the generation of an electric current by a moving magnet Faraday had found that 3 things are necessary to generate (or â€Å"induce†) an EMF (voltage supply): – A magnetic field (from some magnets or electromagnet) – A conductor (eg. wire or coil of wire) – Relative motion / change between the field and the conductor If the conductor formed a closed loop then an induced current would also flow. ire If this wire is dropped so that it cuts flux lines, then a voltage appears between the ends because electrons are forced to the right. They eventually stop moving because they create an electric field pushing them back. As long as the magnet is moving, an emf and current is induced. Faraday’s Law ? =? n ? where ? = induced EMF, in V n = number of turns on coil ? = change in ? = magnetic flux, in Wb = BA B = magnetic flux density (field strength), in T The induced voltage can be increased by: Increasing n: more turns on the coil Increasing B: use strong magnets Increasing A: use a bigger coil Decreasing t: go faster! B A = area of coil in m 2 ? = time taken for to occur William Kim HSC Physics Summary | page 24  § Define magnetic field strength B as magnetic flux density Magnetic flux density is the magnetic flux per unit area and is a measure of the magnetic field strength.  § Explain the concept of magnetic flux in terms of magnetic flux density and surface area ? = BA sin ? where B = magnetic flux density, in T A = area, in m 2 ? = magnetic flux, in Wb  § Explain generated potential difference as the rate of change of magnetic flux through a circuit The induced emf is proportional to the rate of change of flux through the circuit. See Faraday’s Law (above).  § Account for Lenz’s Law in terms of conservation of energy and relate it to the production of back emf in motors Lenz’s Law This is a supplementary law to Faraday’s Law. It says that any induced emf or current will have a direction that opposes the change that caused it. This is really just a restatement of the law of conservation of energy because the induced electrical energy has come from the thing that causes the original motion. Eg. In a hydroelectric power station, the kinetic energy of flowing water is converted into electrical energy. N William Kim HSC Physics Summary | page 25  § Explain that, in electric motors, back emf opposes the supply emf Back emf Back emf is generated in any coil that experiences changing B fields, even though it is producing them. Note that back emf is frequency dependent – the higher the frequency of the changing field, the greater the back emf produced. Back emf is also produced in the rotating coil of a motor: – When the motor is spinning at its operating speed, back emf will have its max value, but†¦ – When the motor is just turned on it isn’t spinning yet so there is no back emf. This can lead to excessive current so the motor may be protected by using a â€Å"starting resistance† that limits current. When up to speed the resistor is taken out of the circuit. The coil becomes an electromagnet and generates an alternating B field BUT it also experiences the changing B field and generates its own emf that opposes the applied emf.  § Apply Lenzà ¢â‚¬â„¢s Law to the production of eddy currents Eddy Currents – are induced currents (usually unwanted or unintended) in two-dimensional conductors (eg. sheet metal) or three-dimensional conductors (eg. a block of steel). Sometimes it is necessary to design against them. Eg. the core of a motor is made of soft iron, and is made of thin layers (laminated) to prevent eddy currents. Some devices rely on eddy currents to work: Electromagnetic braking – a moving conductor near magnets will slow down because the eddy currents oppose its motion. Electromagnetic switching – security ‘gates’ that are really coils with AC generate a high frequency B field. Metal in this field develops eddy currents that work against the field, slowing it down. A detector circuit picks up on this and sets off an alarm. Induction Cooktops- are an application of Faraday’s Law. Instead of a heating element, this cooktop contains a set of coils with alternating current passing through them. This produces a changing B field above the cooktop. A metal saucepan placed on the cooktop is a conductor in the changing B field and therefore an electric current is induced in the base of the pan. The current heats the pan, and this heat cooks the food. Induction cooktops are approximately twice as efficient as a gas cooktop, but are expensive to purchase. William Kim HSC Physics Summary | page 26 ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? Movement of metal Eddy current loop 3. Generators are used to provide large scale power production in isolated areas and as a backup in emergency situations  § Identify the main components of a generator An electric generator (dynamo) is a device that includes all of the elements necessary to transform mechanical kinetic energy to electricity according to Faraday’s Law: – A magnetic field (provided by a set of permanent magnets); – A conductor (a coil mounted on an axle, so it can spin); – Relative motion (the coil is made to spin by some other form of energy). Compare the structure and function of a generator to an electric motor In fact, most generators are constructed just like a motor, however the flow of energy through them is different. Motor: electrical energy a kinetic energy Generator: kinetic energy a electrical energy  § Describe the operation of an AC and a DC generator EMF is generated in the coil and a circuit is completed to the outside world through ring connectors, just like motors. If standard slip rings are used then a dynamo naturally produces alternating current AC. I t E t Doubling the frequency of rotation doubles the maximum induced emf If a split ring commutator is used instead, then the direction of the current flowing from the coil is reversed every ? c ycle. This produces a pulsing type of direct current DC. + I t William Kim HSC Physics Summary | page 27  § Discuss the energy losses that occur as energy is fed through transmission lines from the generator to the consumer Analyse the effects of the development of AC and DC generators on society and the environment Even good electrical conductors like copper used to supply electricity, sometimes through considerable cable lengths to towns and cities, generate substantial resistances. It follows that to minimise energy loss in the wires, the current needs to be kept low (heating losses vary as the square of the current). This is achieved by transmitting the energy at high voltages.  § Impact on society – Impact on environment – Positive Improved lifestyle Street lighting Electric trains Communication Computerisation of many systems eg. anking, stock market Industrial development a more jobs Lots of electric trains have reduced pollution from steam trains and made public transport more available Electricity has replaced older, more-polluting technologies eg. electrical heating instead of coal burning in fireplaces – Negative Possible risk of cancer living near distribution cables Longer working hours Reliance on electricity leaves us vulnerable to systems loss due to electr ical failure – – – Burning coal in power stations produces smoke and CO2 (a greenhouse gas) Nuclear power stations produce radioactive water have a risk of nuclear accident Hydroelectric schemes redirect water away from river habitats Mining impacts negatively on environment Visual pollution of cables  § Assess evidence about the physiological effects on humans living near high voltage power lines 1979 study found children living near high voltage power lines appeared to develop a particular form of cancer. 997 study showed no evidence of an increase risk of childhood cancer at residential magnetic field levels. 1998 panel stated that EM fields should be considered â€Å"possible human carcinogens† and that there is â€Å"no conclusive and consistent evidence that EM fields cause any human disease. † William Kim HSC Physics Summary | page 28 4. Transformers allow generated voltage to be either increased or decreased before it is used  § Explain the purpose and principles of transform ers in electrical circuits A transformer is a device that alters the voltage and current of an electricity supply. The AC voltage source produces an alternating current in the primary coil. This produces an alternating B field that threads through the secondary coil. The secondary coil now has: – Conductor – B field – Change and therefore generates its own voltage. If there is a closed loop then an alternating current will flow as well.  § Compare step-up and stepdown transformers Step-up transformers: increase voltage and decrease current Step-down transformers: decrease voltage and increase current  § Determine the relationship between the ratio of the number of turns in the primary and secondary coils and the ratio of primary to secondary voltage Vp = primary voltage (voltage in) Ip = primary current np = number of turns on primary coil Vs = secondary voltage (voltage out) Is = secondary current ns = number of turns on secondary coil Vp Vs = np ns If 100% efficient (this needs perfect â€Å"flux linkage†, usually using an iron core) then: Power in primary = Power in secondary ? V p I p = Vs I s ? ? Vp Vs Vp Vs = = Is Ip np ns = Is Ip William Kim HSC Physics Summary | page 29  § Explain why voltage transformations are related to the conservation of energy The Principle of Conservation of Energy states that energy cannot be created or destroyed but that it can be transformed from one form to another. This means that if a step-up transformer gives a greater voltage at the output, its current must be decreased: i. e. power in = power out.  § Explain the role of transformers in electricity sub-stations NSW power stations produce electricity with a voltage of about 23,000 V and a current of about 30,000 A. Unfortunately, this amount is too high to be sent through a cable. This is because it heats the cable causing energy loss. This is called joule heating and happens because: P = I2R So to reduce joule heating, the current must be reduced as much as possible with a step-up transformer. Additional transformers between the power station and consumer (in sub-stations) gradually stepdown the voltage, to 240 V by the time it gets to household users. This is because at high voltages, electricity can conduct through air, making it dangerous for use in the home. Discuss why some electrical appliances in the home that are connected to the mains domestic power supply use a transformer Most electronic circuits are designed to operate at low DC voltages of between 3 V and 12 V. Therefore, household appliances that have electronic circuits in them will have either a plug-in transformer or an inbuilt transformer to step down the domestic 240 V supply. These transformers also have a rectifier circuit built into them that converts AC to DC. TVs also contain a step-up transformer for producing the high voltages needed for the CRT.  § Analyse the impact of the development of transformers on society The development of the generator and transformer has allowed for the setting up of national power grids in almost every country, making that most convenient and flexible form of energy, electricity, accessible from many miles away. The transformer’s role is to step voltage up and down to make efficient transportation and distribution possible. William Kim HSC Physics Summary | page 30 5. Motors are used in industries and the home usually to convert electrical energy into more useful forms of energy  § Describe the main features of an AC electric motor AC induction motor: – The rotor – end rings short circuit non-ferrous rotor bars, that is sealed i. e. no external connections at all (usually a â€Å"squirrel cage†). Encased in a laminated iron armature. – The stator – surrounding electromagnet. – Connection to stator – the surrounding electromagnet receives the AC. In an AC induction motor, the principle of operation is: 1. AC to surrounding electromagnet, which†¦ 2. Produces an oscillating (rotating) B field, which†¦ 3. Induces a current in the rotor, which†¦ 4. Turns the rotor into an electromagnet that†¦ 5. Tries to oppose the field being generated by the stator. 6. The stator and the rotor push against each other (using their B fields), which†¦ 7. Causes the rotor to turn! Brilliant! AC (synchronous) motor: – A rotating coil – Surrounding magnets – Connection to coil via slip rings (commutator for DC motor)  § Explain that AC motors usually produce low power and relate this to their use in power tools Power is the rate of work. Work is done when energy is transformed from one type to another. Induction motors are considered to produce low power because the amount of mechanical work they achieve is low compared with the electrical energy consumed. The ‘lost power’ of induction motors is consumed in magnetising the working parts of the motor and in creating induction currents in the rotor. AC induction motors are considered to be unsuitable for use in heavy industry because their low power rating would make them too expensive to run when performing a specific task. However, they are used extensively in power tools and electric domestic appliances where the loss of power is not economically significant. Explain the advantages of induction motors Advantages of AC induction motors: 1. Simplicity of design; 2. High efficiency (hence low maintenance – there are no brushes or commutators to wear out); 3. Relatively low cost William Kim HSC Physics Summary | page 31 Core Topic Three: From Ideas to Implementation 1. Increased understandings of cathode rays led t o the development of television  § Explain that cathode ray tubes allowed the manipulation of a stream of charged particles Discharge Tubes – Investigation of vacuum tubes could not occur until good vacuum pumps had been invented. A vacuum tube is a glass tube fitted with an electrode at either end, and almost all of the air sucked out. – The positive electrode is the â€Å"anode†; The negative electrode is the â€Å"cathode†. When a high voltage is connected between the electrodes, an invisible ray travels from the cathode to the anode. They were called â€Å"cathode rays†. Cathode rays cause glass to glow green. – A discharge tube is a cathode ray tube with a vacuum pump fitted, so that the air pressure inside the tube can be varied. At different air pressures, different bright effects appear in the tubes e. . bands, striations and dark spaces. These are caused by cathode rays striking atoms in the air inside the tube. The atoms become excited then release photons of visible light – A beam of electrons travels from the cathode to the anode and can be deflected by electrical and/or magnetic fields. Anode Glass glows here Cathode  § Explain why the apparent inconsist ent behaviour of cathode rays caused debate as to whether they were charged particles or electromagnetic waves In 1892 Hertz demonstrated that cathode rays could penetrate thin metal foils. This he believed supported a wave nature. In 1895 Jean-Baptise Perrin showed that cathode rays deposited negative charges on impact with an object, suggesting a particle nature. There was controversy over the nature of cathode rays – waves or particles. William Kim HSC Physics Summary | page 32  § Identify that charged plates produce an electric field If metal plates are separated by a distance and are attached to a power source, an electric field will be produced between them. E = V/d  § Describe quantitatively the force acting on a charge moving through a magnetic field Recall that the force (F) acting on a charge (q) moving with a velocity (v) at an angle to a magnetic field (B), is given by: FB = qvB Where FB = magnetic force (N) q = charge (C) v = velocity of charge (ms-1) B = magnetic field strength (T)  § Discuss qualitatively the electric field strength due to a point charge, positive and negative charges and oppositely charged parallel plates. Describe quantitatively the electric field due to oppositely charged parallel plates ++++++++ If a positive charge is placed near another positive charge, it will experience a force of repulsion. A positive charge placed in a field will experience a force in the direction of the arrow. A negative charge placed in a field will experience a force opposite to the direction of the arrow.  § FE = qE Where FE = electric force (N) q = charge (C) E = electric field strength (NC-1) ———– William Kim HSC Physics Summary | page 33  § Outline Thomson’s experiment to measure the charge/mass ratio of an electron Cathode ray particles ? v? B ? ? ? ? ? ? ? FM ? ? FE ? ? ? v? E ? ? ? J. J. Thomson’s Experiment – By fitting plates to his CRT, he could subject the cathode rays to an electric field. The rays deflected, proving that they were charged particles, not electromagnetic waves. – He noticed that the rays deflected toward the positive plate, proving that they were negatively charged particles. – By crossing electric and magnetic fields, Thomson was able to deduce the velocity of the cathode rays. By turning off the E field, the particles followed a circular arc caused by the B field. The magnetic force was acting like a centripetal force. mv qvB = r q v ? = m Br – 2 ? FM = FE qvB = qE E ? v = B Thomson adjusted the strength of the fields so that the particles were not deflected. By carefully measuring the strength of the fields, Thomson could calculate v. Thomson had already measured B and worked out v. By measuring the radius of curvature r, he could then calculate q/m, i. e. the charge/mass ratio of an electron. q/m for these particles was 1800 times greater than for a hydrogen ion, the simplest known atomic ion. Thomson quickly compared the charges and found them to be about the same (though opposite in sign) Therefore mass for cathode ray particles was 1800 times smaller than hydrogen Therefore cathode ray particles were subatomic particles! This was the first discovery of subatomic particles They were later called electrons. William Kim HSC Physics Summary | page 34  § Outline the role in a cathode ray tube of: Electrodes in the electrode gun The electric field The fluorescent screen The Cathode Ray Tube Each CRT has a vacuum tube/chamber, a cathode, an anode, and a target. Electrodes in the electron gun The electron gun produces a narrow beam of electrons. It consists of a filament, a cathode and two open-cylinder anodes. The anodes help to accelerate and focus the electrons. A ring shaped electrode – the grid – between the cathode and anodes controls the brightness of the spot by controlling the number of electrons emitted by the gun. By making the grid negative with respect to the cathode the number of electrons, and hence the brightness is reduced. The electric field Acts as a deflection system. It consists of two sets of parallel plates connected to a parallel plates connected to a potential difference. This produces an electric field between the plates. The Y-plates control the vertical deflection and the X-plates the horizontal deflection. The fluorescent screen The inside glass of the end of the tube is coated with a fluorescent material for example, zinc sulphide. When an electron beam hits the screen, the coating fluoresces and a spot of light is seen on the screen. The screen acts as a detector of cathode rays. Electrons Cathode Anode To plates and screen William Kim HSC Physics Summary | page 35  § Outline applications of cathode rays in oscilloscopes, electron microscopes and television sets The Cathode Ray Oscilloscope (CRO) Is an electronics diagnostics device because it can show a graph of how voltages vary over time. Deflection of the electron beam is achieved by two sets of plates. Horizontal plates cause vertical deflection while vertical plates cause horizontal deflection. TV Tube An electron gun again produces the electron beam. Coils are used instead of plates, however. Electric current through the coils produce magnetic fields that can deflect the beams quickly from side to side, and more slowly from bottom to top. In this way the beam scans the entire screen. By varying the intensity of the beam, a picture is built up. The picture is refreshed 50 times / second, which is too fast to be noticed by the human eye. The Electron Microscope Uses electrons instead of light. Their wavelength is 100,000 times smaller than visible light, therefore their resolving power is 100,000 times greater. – A â€Å"sample† is placed inside the chamber (which is really the CRT) – The air is then sucked out – An electron gun produces the electron beam – Coils produce magnetic fields to focus the beam (â€Å"magnetic lenses†) – The beam scans over the surface of the sample – Detectors pick up the reflected and scattered electron beam, and from this information a 3 dimensional image is constructed  § Discuss the impact of increased understandings of cathode rays and the development of the oscilloscope on experimental physics The introduction of electronic control systems into all forms of science and industry has seen the cathode ray oscilloscope (CRO) become one of the most widely utilised test instruments. Because of its ability to make ‘voltages’ visible, the cathode ray oscilloscope is a powerful diagnostic and development tool. William Kim HSC Physics Summary | page 36 2. The reconceptualisation of the model of light led to an understanding of the photoelectric effect and black body radiation  § Explain qualitatively Hertz’s experiments in measuring the speed of radio waves and how they relate to light waves Recall: Maxwell’s theory of electromagnetic waves In 1864 Maxwell, through a set of four brilliant equations, predicted a range of invisible waves made up of an electric and magnetic wave that regenerate each other. The speed of these waves was calculated to be 3 x 108 ms-1 and probably included light. E v B Heinrich Hertz’s Experiment: (proving Maxwell’s theory) Performed in 1886, Hertz built equipment to generate and transmit EM waves with ? ? 1m. He also had a separate receiver (a loop of wire) located about 20m away. Spark gaps were included to show when high voltage AC was present in the transmitter or receiver. The receiver spark only appeared when the transmitter spark was present. Hertz hypothesised that the sparks set up changing electric and magnetic fields that propagated as an electromagnetic ave, as postulated by Maxwell. He showed that these were waves being transmitted because he could reflect, refract and polarise them. By measuring the frequency, he calculated v (v = f ? ) and it came out as 3 x 108 ms-1. These properties proved Maxwell’s theory and as they are also exhibited by light, Hertz was able to provide experimental evidence that light is a form of transverse electromagnetic wave. to induction coil 1mm gap transmitter charged plates re ceiver William Kim HSC Physics Summary | page 37  § Describe Hertz’s observation of the effect of a radio wave on a receiver and the photoelectric effect he produced but failed to investigate Outline applications of the production of electromagnetic waves by oscillating electric charges in radio antennae Hertz observed that the transmitter spark was producing s How to cite Physics Study Notes Hsc, Essay examples

Statement on Monetary Policy

Questions: Question 1a.) Which monetary policy is more effective in moderating the business cycle, tight or easy? Give reasons for your answers.b.) What is the current monetary policy stance of the RBA? What factors do the RBA take into consideration, before a decision is made as to whether to implement a tight or easy monetary policy?c.) Using AD-AS model, explain how interest rates affect the key macroeconomic variables. Question 2a.) Comment on the recent factors that are affecting the value of the Australian dollar. Use diagrams to illustrate your answer.b.) Who gains and who loses when the Australian dollar depreciates? Justify your answer.c.) In your opinion, is a depreciating $A good or bad for the Australian economy? Justify your answer. Answer: 1. a) In order to moderate the business cycle, a tight monetary policy would be more effective in moderating the business cycles since it would prevent an overheating of the economy which could give rise to consistent high inflation and thus could ultimately pave way to recessionary conditions. Expansionary monetary policy even though helps in rapid growth of the economy and businesses cause money supply to increase which fuels inflation and erodes purchasing power in the long term. Tight monetary policy prevents this and hence leads to lower fluctuations in the business cycle albeit at the cost of the moderation of economic growth (Mankiw, 2012).b) RBA currently is maintaining an easy monetary policy which is apparent from the cut of 25 basis points in the cash rate in February 2015 despite very low interest rates. The primary factors that RBA takes into consideration which deciding on the monetary policy are the GDP growth, related macroeconomic indicators, inflation, employment tr ends, global investment and growth prospects especially China (because the Australian economy is significantly dependent on mining) and US, housing prices, commodity prices particularly coal, oil , iron ore, aluminium and other minerals and also try to predict the likely movements in the above variables going forward in the near to medium term (RBA, 2015).c) Interest rates have significant impact on the inflation level and real output level. If the interest rates decline, there is an increase in the aggregate demand which leads to inflation in the short term and thus increase in price level. Further due to easy availability of credit, investment cycle is enhanced which leads to an enhancement in the real output as well in the medium to long term. However an increase in the interest rates leads to decrease in aggregate demand which leads to deflation in the short term and adversely impacts the investment cycle and leads to decline in real output in the medium to long term (Dombusch, Fischer Startz, 2012).2. a) One of the major factors which have caused the depreciation of the Australian Dollar in the recent times is the decline in the price of commodities such as iron ore, coal, copper and uranium primarily due to a slowdown in demand in China which is the worlds largest consumer of many commodities and accounts for sizable exports from Australia. This has led to a current account deficit and thus increased demand of USD for covering the deficit. Additionally recovery of the US economy in the last year has additionally fuelled the depreciation of the AUD versus the USD. However the depreciation of AUD versus EUR and YEN has been lesser since the economies of Eurozone nations and Japan are underperforming (RBA, 2015). This depreciation of AUD is captured in the graph shown below. b) When the Australian Dollar depreciates, it augers well for the exporters belonging to the mining and dairy industry as it increases the profitability and competitiveness of these businesses. However the various import based businesses would be hurt since imports would tend to get more expensive and thus the consumers may be hurt. Additionally inflow of foreign capital may be stalled but outflow of capital may increase since usually depreciation is associated with an underperforming economy (Mankiw, 2012). c) I believe that a depreciating dollar is good for the Australian economy since it augers well for the various exporters particularly those belonging to the mining sector, dairy which forms a significant chunk of the Australian economy and exports and thus provide a boost to the economy which would also lead to a trade surplus and thus enhance the availability of financial resources for the government to deploy for fostering national growth (Dombusch, Fischer Startz, 2012). References Dombusch, R, Fischer, S Startz, R 2012.Macroeconomics, 10th edn, McGraw Hill Publications, New YorkMankiw, G 2012. Principles of Macroeconomics, 6th edn, Worth Publishers, London RBA 2015. Statement on Monetary Policy, Reserve Bank of Australia, Available from: https://www.rba.gov.au/publications/smp/2015/feb/pdf/0215.pdf (Accessed on February 11, 2015)