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Physics Stage 6 Formulas

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Formula Sheet for Stage 6 Physics

Preliminary Course

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Energy = VIt

P=VI

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H.S.C. Course - Core

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H.S.C. Course - Options

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Constants

How to Use the Formulas for Stage 6 Physics

Preliminary Course

|Formula |Name |Comments |Typical Problem |Typical Answer |
|[pic] |Wave Equation |v= velocity (m/s) |Calculate the wavelength of a water wave |[pic] |
| |8.2.1 |f = frequency (hz) |travelling at 3 m/s whose frequency is 6 | |
| | |( = wavelength (m) |Hz. | |
|[pic] |Intensity Law |I = intensity (no units) |A globe is viewed from 2 metres at a |[pic] thus it appears 9 times less bright. |
| |8.2.3 |d = distance (m) |certain brightness. How bright does it | |
| | | |appear 6 metres away? | |
|[pic] |Snell’s Law |v1 = speed in first medium |A light ray travelling at 3 X 108 m/s in |[pic] |
| |8.2.4 |v2 = speed in second medium |air enters a pool of water at 45( to the |[pic] |
| | |sin i = angle in first medium |normal. If it slows down to 2.4 X 108 | |
| | |sin r = angle in second medium |m/s, what is its angle in the water? | |
| | |Angles are always measured to the normal | | |
|[pic] |Electric Field |E = Electric field (N/C, V/m) |What is the force exerted on a 2 X 10-6 C|[pic] |
| |8.3.2 |F = Force (N) |charge moving in an electric field of | |
| | |q = charge (C) |size 5 X 103 V/m? | |
|[pic] |Ohm’s Law |R = resistance (() |What is the current through a 6( resistor|[pic] |
| |8.3.2 |V = voltage (V) |when a voltage of 18 V is applied across | |
| | |I = current (A) |it? | |
|Energy = VIt |Electrical Energy |Energy (J) |How much energy is delivered to an |Energy = VIt = 240 X 2 X 3 X 60 (sec) |
| |8.3.4 |V = voltage (V) |electric kettle if a current of 2A is |= 86400 = 86.4 kJ |
| | |I = current (A) |used for 3 minutes plugged into the mains| |
| | |T = time (s) |(240V)? | |
|P=VI |Electrical Power |P = power (W) |What current is required by a 2400W |[pic] |
| |8.3.4 |V = voltage (V) |electric heater plugged into the mains | |
| | |I = current (A) |(240V)? | |
|[pic] |Average Velocity |vav = average velocity (m/s) |How much distance is covered by a car |[pic] |
| |8.4.1 |r = distance covered (m) |travelling for 4 mins at an average speed| |
| | |t = time (s) |of 16 m/s? | |
|[pic] |Average Acceleration |[pic] = average acceleration (ms-2) |A train accelerates from 12 m/s to 18 m/s|[pic] |
| |8.4.2 |(v = change in velocity (m/s) |in 12 s. What is the value of this | |
| | |(t = change in time (s) |acceleration? | |
|[pic] |Newton’s Second Law |[pic] = sum of all forces (N) | | |
| |8.4.2 |m = mass (kg) | | |
| | |[pic] = acceleration (ms-2) | | |
|[pic] |Kinetic Energy |Ek = kinetic energy (J) |A ball possesses 10 J of energy. What is|[pic] |
| |8.4.3 |m= mass (kg) |its mass if it is moving at 2 m/s? | |
| | |v = speed (m/s) | | |
|[pic] |Momentum |P = momentum (Ns, kgm/s) |What is the momentum of 1400 kg car |[pic] |
| |8.4.4 |m = mass (kg) |moving at 6 m/s? | |
| | |v = velocity (m/s) | | |
|[pic] |Impulse |Impulse = change in momentum (Ns, kgm/s) |A ball of mass 0.5 kg travelling at 3 m/s|[pic] |
| |8.4.4 | |hits wall and bounces back at the same |[pic] |
| | | |speed. If it is in contact with the wall| |
| | | |for 0.1 s, what is the force exerted by | |
| | | |the wall? | |
|[pic] |Universal Gravitation |F = force (N) |What is the gravitational force between |[pic] |
| |8.5.4 |G = universal gravitation constant (6.67 X 10-11 |the Earth (m=6X1024 kg) and 2 kg ball | |
| | |Nm2 kg-2) |given that the radius of the Earth is | |
| | |m1 = mass of body 1 (kg) |5.8X106 m | |
| | |m2 = mass of body 2 (kg) | | |
| | |d = separation between the two bodies (m) | | |
|[pic] |Kepler’s Third Law |r = radius of motion (m) |What is the period of a 1012 kg comet |[pic] |
| |8.5.4 |T = period of motion (s) |that orbits at 5 X 108 m? | |
| | |G = universal gravitation constant (6.67 X 10-11 | | |
| | |Nm2 kg-2) | | |
| | |M = mass of system (kg) | | |

How to Use the Formulas for Stage 6 Physics

HSC Course (Core)

|Formula |Name |Comments |Typical Problem |Typical Answer |
|[pic] |Gravitational Potential |Ep = Potential energy |What is the gravitational |[pic] |
| |Energy |G = universal gravitation constant (6.67 X 10-11 Nm2 |potential energy between two 100 | |
| |9.2.1 |kg-2) |kg masses through a distance of | |
| | |m1 = mass of body 1 (kg) |2000 m? | |
| | |m2 = mass of body 2 (kg) | | |
| | |r = separation between the two bodies from infinity to | | |
| | |r (m) | | |
|[pic] |Gravitational Force |F = force (N) |What is the weight of a 100 kg |[pic] |
| |9.2.1 |M = mass (kg) |person? | |
| | |g = gravitational constant at the surface of the Earth | | |
| | |9.81 ms-2 | | |
|[pic] |Newtons’ Laws of Motion |ux = initial speed in x direction (m/s) |What is the maximum height of a |Let up be positive. |
| |9.2.2 |vx = final speed in x direction (m/s) |projectile launched at 45( to the |At max. height, vy=0, thus |
| | |uy = initial speed in y direction (m/s) |horizontal at 50 m/s? |[pic] |
| | |vy = final speed in y direction (m/s) | |[pic] |
| | |a = constant acceleration (ms-2) | | |
| | |(x, (y = change in displacement (m) | | |
| | |t = time (s) | | |
|[pic] |Kepler’s third law |r = radius of motion (m) |What is the period of a 1012 kg |[pic] |
| |9.2.2 |T = period of motion (s) |comet that orbits at 5 X 108 m? | |
| | |G = universal gravitation constant (6.67 X 10-11 Nm2 | | |
| | |kg-2) | | |
| | |M = mass of system (kg) | | |
|[pic] |Relativistic Length |Lv = apparent length (m) |What is the apparent length of a |[pic] |
| |Contraction |Lo = “rest” length (m) |spaceship of rest length 150m | |
| |9.2.4 |v = relative velocity (m/s) |travelling at 0.9c? | |
| | |c = speed of light (3 X 108 m/s) | | |
|[pic] |Relativistic time dilation|tv = apparent time (s) |How much slower does an astronaut |[pic]times slower |
| |9.2.4 |to = “rest” time (s) |travelling at 0.9c appear to an | |
| | |v = relative velocity (m/s) |observer “at rest” | |
| | |c = speed of light (3 X 108 m/s) | | |
|[pic] |Magnetic force on a |F = force (N) |Calculate the force on 2m of wire |[pic] |
| |current-carrying wire of |B= Magnetic Field (T) |carrying a current of 4A in a | |
| |length l in a magnetic |I = current (A) |magnetic field of 0.1T. | |
| |field |l = length (m) | | |
| |9.3.1 | | | |
|[pic] |Force per unit length |F = force (N) |What is the force per unit length |[pic] |
| |9.3.1 |l = length (m) per unit |on two wires, both carrying 10A, | |
| | |I1, I2 = two currents |separated by a distance of 3m? | |
| | |parallel=repulsive, antiparallel=attractive | | |
| | |d = separation of the two currents (m) | | |
| | |k=magnetic constant (2 X 10-7 NC-1m-1) | | |
|[pic] |torque |( =torque (Nm) |What is the torque on a nut when a|[pic] |
| |9.3.1 |F =force (N) |0.6 m spanner has a force of 80 N | |
| | |d=distance (m) |applied on it? | |
|[pic] |torque on a coil immersed |( =torque (Nm) |What is the torque on a 0.20 m2 |[pic] |
| |in a magnetic field |n =number of turns of coil |coil of 200 turns immersed in a | |
| |9.3.1 |B=magnetic field (T) |magnetic field of 0.2 T carrying a| |
| | |I = current (A) |current of 3 A? | |
| | |A = area of coil immersed in magnetic field (m2) | | |
| | |cos (=angle between the coil and the magnetic field | | |
|[pic] |Transformer equation |Vp = primary voltage (V) |A transformer is required to step |[pic]turns. |
| |9.3.4 |Vs = secondary voltage (V) |down mains voltage (240V) to 12 V.| |
| | |Np = number of turns in the primary coil |If the primary coil has 960 turns,| |
| | |Ns = number of turns in the secondary coil |how many turns are required in the| |
| | | |secondary coil? | |
|[pic] |Magnetic force on a charge|F = force (N) |What is the force on an electron |[pic] |
| |in a magnetic field |q = charge (C) |travelling at 105 m/s in a | |
| |9.4.1 |v = velocity (m/s) |magnetic field of 3 T at an angle | |
| | |B = magnetic field (T) |of 30( to the field? | |
| | |Sin ( = angle between the velocity and the magnetic | | |
| | |field | | |
|[pic] |Electric Field |E = Electric Field (N/C, V/m) |What is the electric field between|[pic] |
| |9.4.1 |V = Voltage (V) |the prongs of a mains outlet | |
| | |d = distance (m) |(240V) if its separation is 18 mm?| |
|[pic] |Energy of a Photon |E = Energy (J) |What is the energy of a photon of |[pic] |
| |9.4.2 |h = Planck’s constant 6.626 X 10-34 Js |yellow light (f=5.1X1015 Hz) ? | |
| | |f = frequency (Hz) | | |
|[pic] |Wave Equation |c = speed of light 3 X 108 m/s |What is the frequency of yellow |[pic] |
| |9.4.2 |f = frequency (Hz) |light given that its wavelength is| |
| | |( = wavelength (m) |590 nm? | |

HSC Course (Options)

|Formula |Name |Comments |Typical Problem |Typical Answer |
|[pic] |Acoustic Impedance |Z = Acoustic impedance (kgm2s-1 =Rayls) |What is the acoustic impedance of vaseline |[pic] |
| |9.6.1 |[pic]= acoustic density (kgm) |given that its acoustic density is 0.003 kgm | |
| | |[pic]= speed of sound in medium (m/s) |and the speed of sound in vaseline is 650 m/s? | |
|[pic] |Reflection Intensity |Ii=initial intensity |What is the reflection intensity at the |[pic] |
| |9.6.1 |Io= output intensity |interface of air (acoustic impedance = 0.5 | |
| | |Z2 = acoustic impedance (medium 1) |kgm2s-1) and vaseline (Z=1.95 kgm2s-1)? | |
| | |Z1 = acoustic impedance (medium 2) | | |
|[pic] |Astronomical Distance |M = absolute magnitude |How far away is a star that appears m=4.5 on |[pic] |
| |9.7.4 |m = relative magnitude |Earth while its absolute magnitude is –3.4? |[pic] |
| | |d = distance in parsecs | | |
| | |[pic] | | |
|[pic] |Ratio of intensity |IA,B = intensity of objects A and B |How much dimmer is Sirius A (M=-4.3) compared |[pic] |
| |9.7.4 |MA,B = absolute magnitude of A & B |to Echelon (M=-2.1)? | |
| | |[pic] | | |
|[pic] |Kepler’s Third Law |r = radius of motion (m) |Sirius A and B orbit each other every 61 years.|T = 61 X 365.25 X 24 X 60 X 60 |
| |9.7.5 |T = period of motion (s) |What is the radius of this orbit if Sirius A |= 1925013600 s |
| | |G = universal gravitation constant (6.67 X 10-11 |has a mass of 1027 kg and Sirius B is 1029 kg? |[pic] |
| | |Nm2 kg-2) | |[pic] |
| | |M1+M2 = total mass of system (kg) | |r = 6.9 X 109 m |
|[pic] |Ryberg’s equation |( = wavelength (m) |What wavelength of light is produced in |[pic] |
| |9.8.1 |ni,f = quantum states (shells) |hydrogen by a transition between the first and |[pic] |
| | |R = Ryberg’s constant 1.10 X 107 m-1 |fourth orbitals (shells)? | |
|[pic] |Wavelength of a particle |( = wavelength (m) |What is the wavelength associated with an |[pic] |
| |9.8.2 |h = Planck’s constant 6.626 X 10-34 Js |elephant (m=1400 kg) moving at 4 m/s? | |
| | |m = mass of particle (kg) | | |
| | |v = speed of particle (m/s) | | |
|[pic] |Amplifier Gain |Vout = output voltage (V) |A BC547 transistor has a gain of 120. What is |[pic] |
| |9.9.6 |Vin = input voltage (V) |the output voltage if the input voltage is 60 | |
| | | |mV? | |
|[pic] |Open Gain Loop |Ao = Amplifier gain |A 741 op-amp has an output voltage of 12 V when|[pic] |
| |9.9.6 |Vo = output voltage (V) |its V+ is 2.3 V and its V- is 1.2 V. What is | |
| | |V+ = positive input voltage (V) |its gain in this configuration? | |
| | |V- = negative input voltage (V) | | |

-----------------------
Speed of light c 3 X 108 ms-1 in vacuum

Acceleration g 9.81 ms-2 due to gravity

Universal Gravitation G 6.67 X 10-11 Nm2 kg-2

Electric constant k 9 X 109 NC-1m-1

Magnetic constant k 2 X 10-7 Ns2C-2

Planck’s Constant h 6.626 X 10-34 Js

Ryberg’s Constant R 1.10 X 107 m-1

mass of proton mp 1.67 X 10-27 kg

mass of neutron mn 1.68 X 10-27 kg

mass of electron me 9.11 X 10-31 kg

charge on an electron e 1.602 X 10-19 C

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...Questions; 1. When looking at the spacing of the dots. I notice at the beginning stages the dots are nearly connected and close but as the time increase the dots begin to separate which creates more spacing between the dots. As the space increase between the dots,the displacement(metres) increases as well. This tells us that the object under accelerated motion is covering the grounds in each time segment equally. 2. I feel that we use the line of best fit in order to avoid the errors that can occur while we are graphing. When we add our data to the table, we may have made a few mistakes which can change our analysis drastically so when we use the line of best fit we can see if our data is going in the pattern trend that we have planned out. If the line of best fit is avoiding some points that means our plan could be wrong or that we have a data that’s different...

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Quantum Computer by Quantum Mathematics

...and Development Private Limited. HaritaBhargava haritabhargava@ymail.com Abstract-. India gives its greatest contribution toward Intellectual property in form of SANHKYA YOGA. We know very new name of INDIA by BHARAT IT gives us necessary clue. BHA = LIGHT + KNOWLEDGERATA = DEVOTED So BHARAT means DEVOTED to LIGHT & KNOWLEDGE. The model for which modern science search is already available in SANKHYA YOGA. Sankhya yoga like INTELLECTUALPROPERTY is unparallel in the world. Sankhya is a philosophical doctrine which counts the categories which constitute this world. Computer science gives it the name “DIGITAL DECADE”. Cosmology presents it in name of “UNCERTANITY PRINCIPLE” IT is being researched in the form of Quantum (wave & particle) in physics department. In future, it will be known as Medical compiler. [R4] Terms and condition : INNOVATION – RESPECT IT Under the WTO provision of TRIPS I. INTRODUCTION Article 8.1 - That agreement is to be put into effect by member countries through implementing laws. The provision of the agreement is addressed to member countries. It does not directly modify the legal sanction of private parties, who cannot claim rights based on the agreement until they are translated into national law. Article 9.2 - The concept of fair use seeks to balance granting authors sufficient incentives to create new works against the interest of the public and future authors to have access & to to use of the work Article 10 – The agreement introduces multilateral...

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